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t up these hypotheses is shown below: H0 : µ = 10 hours HA : µ = 10 hours A two-sided test allows us to consider the possibility that the data show us something that we would ﬁnd 8 Introduction to linear regression 8.1 (a) The residual plot will show randomly distributed residuals around 0. The variance is also approximately constant. (b) The residuals will show a fan shape, with higher variability for smaller x. There will also be many points on the right above the line. There is trouble with the model being ﬁt here. 8.3 (a) Strong relationship, but a straight line would not ﬁt the data. (b) Strong relationship, and a linear ﬁt would be reasonable. (c) Weak relationship, and trying a linear ﬁt would be reasonable. (d) Moderate relationship, but a straight line would not ﬁt the data. (e) Strong relationship, and a linear ﬁt would be reasonable. (f) Weak relationship, and trying a linear ﬁt would be reasonable. 8.5 (a) Exam 2 since there is less of a scatter in the plot of ﬁnal exam grade versus exam 2. Notice that the relationship between Exam 1 and the Final Exam appears to be slightly nonlinear. (b) Exam 2 and the ﬁnal are relatively close to each other chronologically, or Exam 2 may be cumulative so has greater similarities in material to the ﬁnal exam. Answers may vary. surprising. 7.41 (a) These data are paired. For example, the Friday the 13th in say, September 1991, would probably be more similar to the Friday the 6th in September 1991 than to Friday the 6th in another month or year. (b) Let µdiﬀ = µsixth − µthirteenth. H0 : µdiﬀ = 0. HA : µdiﬀ = 0. (c) Independence: The months selected are not random. However, if we think these dates are roughly equivalent to a simple random sample of all such Friday 6th/13th date pairs, then independence is reasonable. To proceed, we must make this strong assumption, though we should note this assumption in any reported results. Normality: With fewer than 10 observations, we would need to see clear outliers to be concerned. There is a borderline outlier on the right of the histogram of the diﬀerences, so we would want to report this in formal analysis results. (d) T = 4.93 for df = 10 − 1 = 9 → p-value = 0.001. (e) Since p-value < 0.05, reject H0. The data provide strong evidence that the average number of cars at the intersection is higher on Friday the 6th than on Friday the 13th. (We should exercise caution about generalizing the interpetation to all intersections or roads.) (f) If the average number of cars passing the intersection actually was the same on Friday the 6th and 13th, then the probability that we would observe a test statistic so far from zero is less than 0.01. (g) We might have made a Type 1 Error, i.e. incorrectly rejected the null hypothesis. 8.7 (a) r = −0.7 → (4). (c) r = 0.06 → (1). (d) r = 0.92 → (2). (b) r = 0.45 → (3). 8.9 (a) The relationship is positive, weak, and possibly linear. However, there do appear to be some anomalous observations along the left where several students have the same height that is notably far from the cloud of the other points. Additionally, there are many students who appear not to have driven a car, and they are represented by a set of points along the bottom of the scatterplot. (b) There is no obvious explanation why simply being tall should lead a person to drive faster. However, one confounding factor is gender. Males tend to be taller than females on average, and personal experiences (anecdotal) may suggest they drive faster. If we were to follow-up on this suspicion, we would ﬁnd that sociological studies conﬁrm this suspicion. (c) Males are taller on average and they drive faster. The gender variable is indeed an important confounding variable. 501 (¯x, ¯y): ¯y = a+b× ¯x. Plug in ¯x, ¯y, and b, and solve for a: 51. Solution: travel time = 51 + 0.726 × distance. (b) b: For each additional mile in distance, the model predicts an additional 0.726 minutes in travel a: When the distance traveled is 0 miles, time. the travel time is expected to be 51 minutes. It does not make sense to have a travel distance of 0 miles in this context. Here, the y-intercept serves only to adjust the height of the line and is mean(c) R2 = 0.6362 = 0.40. About ingless by itself. 40% of the variability in travel time is accounted for by the model, i.e. explained by the distance traveled. (d) travel time = 51 + 0.726 × distance = 51 + 0.726 × 103 ≈ 126 minutes. (Note: we should be cautious in our predictions with this model since we have not yet evaluated whether it is a well-ﬁt model.) (e) ei = yi − ˆyi = 168 − 126 = 42 minutes. A positive residual means that the model underestimates the travel time. (f) No, this calculation would require extrapolation. 8.25 (a) murder = −29.901 + 2.559 × poverty%. (b) Expected murder rate in metropolitan areas with no poverty is -29. 901 per million. This is obviously not a meaningful value, it just serves to adjust the height of the regression line. (c) For each additional percentage increase in poverty, we expect murders per million to be higher on average by 2.559. (d) Poverty level explains 70.52% of the variability in 0.7052 = murder rates in metropolitan areas. (e) 0.8398. √ 8.27 (a) There is an outlier in the bottom right. Since it is far from the center of the data, it is a point with high leverage. It is also an inﬂuential point since, without that observation, the regression line would have a very diﬀerent slope. (b) There is an outlier in the bottom right. Since it is far from the center of the data, it is a point with high leverage. However, it does not appear to be aﬀecting the line much, so it is not an inﬂuential point. (c) The observation is in the center of the data (in the x-axis direction), so this point does not have high leverage. This means the point won’t have much effect on the slope of the line and so is not an inﬂuential point. 8.11 (a) There is a somewhat weak, positive, possibly linear relationship between the distance traveled and travel time. There is clustering near the lower left corner that we should take special note of. (b) Changing the units will not change the form, direction or strength of the relationship between the two variables. If longer distances measured in miles are associated with longer travel time measured in minutes, longer distances measured in kilometers will be associated with longer travel time measured in hours. (c) Changing units doesn’t aﬀect correlation: r = 0.636. 8.13 (a) There is a moderate, positive, and linear relationship between shoulder girth and height. (b) Changing the units, even if just for one of the variables, will not change the form, direction or strength of the relationship between the two variables. 8.15 In each part, we can write the woman’s age as a linear function of the spouse’s age. (a) ageW = ageS + 3. (b) ageW = ageS − 2. (c) ageW = 2 × ageS. Since the slopes are positive and these are perfect linear relationships, the correlation will be exactly 1 in all three parts. An alternative way to gain insight into this solution is to create a mock data set, e.g. 5 women aged 26, 27, 28, 29, and 30, then ﬁnd the spouses ages for each women in each part and create a scatterplot. 8.17 Correlation: no units. kg/cm. Intercept: kg. Slope: 8.19 Over-estimate. Since the residual is calculated as observed − predicted, a negative residual means that the predicted value is higher than the observed value. 8.21 (a) There is a positive, very strong, linear association between the number of tourists and spending. (b) Explanatory: number of tourists (in thousands). Response: spending (in millions of US dollars). (c) We can predict spending for a given number of tourists using a regression line. This may be useful information for determining how much the country may want to spend in advertising abroad, or to forecast expected revenues from tourism. (d) Even though the relationship appears linear in the scatterplot, the residual plot actually shows a nonlinear relationship. This is not a contradiction: residual plots can show divergences from linearity that can be diﬃcult to see in a scatterplot. A simple linear model is inadequate for modeling these data. It is also important to consider that these data are observed sequentially, which means there may be a hidden structure not evident in the current plots but that is important to consider. 8.23 (a) First calculate the slope: b = r × sy/sx = 0.636 × 113/99 = 0.726. Next, make use of the fact that the regression line passes through the point 502 APPENDIX A. EXERCISE SOLUTIONS 8.29 (a) There is a negative, moderate-to-strong, somewhat linear relationship between percent of families who own their home and the percent of the population living in urban areas in 2010. There is one outlier: a state where 100% of the population is urban. The variability in the percent of homeownership also increases as we move from left to right in the plot. (b) The outlier is located in the bottom right corner, horizontally far from the center of the other points, so it is a point with high leverage. It is an inﬂuential point since excluding this point from the analysis would greatly aﬀect the slope of the regression line. 8.31 (a) The relationship is positive, linear, and moderate. Due to the clear non-constant variance in the residuals, a linear model is not appropriate for modeling the relationship between hours worked and income. (b) Neither are a particularly: For the logged model, the scatterplot and residual plot show more constant variance in the residuals. However, the scatterplot with the logged model looks to have a bit of curvature. (c) For each hour increase hours works we would expect the income to increase on average by a factor of e0.058 ≈ 1.06, i.e. by 6%. 8.33 (a) The relationship is positive, moderate-tostrong, and linear. There are a few outliers but no points that appear to be inﬂuential. (b) weight = −105.0113 + 1.0176 × height. Slope: For each additional centimeter in height, the model predict
s the average weight to be 1.0176 additional kilograms (about 2.2 pounds). Intercept: People who are 0 centimeters tall are expected to weigh - 105.0113 kilograms. This is obviously not possible. Here, the y- intercept serves only to adjust the height of the line and is meaningless by itself. (c) H0: The true slope coeﬃcient of height is zero (β = 0). HA: The true slope coeﬃcient of height is diﬀerent than zero (β = 0). The p-value for the two-sided alternative hypothesis (β = 0) is incredibly small, so we reject H0. The data provide convincing evidence that height and weight are positively correlated. The true slope parameter is indeed greater than 0. (d) R2 = 0.722 = 0.52. Approximately 52% of the variability in weight can be explained by the height of individuals. 8.35 (a) H0: β = 0. HA: β = 0. The p-value, as reported in the table, is incredibly small and is smaller than 0.05, so we reject H0. The data provide convincing evidence that women’s and spouses’ heights are positively correlated. (b) heightS = 43.5755 + 0.2863 × heightW . (c) Slope: For each additional inch in woman’s height, the spouse’s height is expected to be an additional 0.2863 inches, on average. Intercept: Women who are 0 inches tall are predicted to have spouses who are 43.5755 inches tall. The intercept here is meaningless, and it serves only to adjust the height of the line. (d) The slope is positive, so r must also be positive. r = (e) 63.2612. Since R2 is low, the prediction based on this regression model is not very reliable. (f) No, we should avoid extrapolating. 0.09 = 0.30. √ 8.37 (a) H0 : β = 0; HA : β = 0 (b) The pvalue for this test is approximately 0, therefore we reject H0. The data provide convincing evidence that poverty percentage is a signiﬁcant predictor of murder rate. 18 = 2.10; 2.559±2.10×0.390 = (1.74, 3.378); For each percentage point poverty is higher, murder rate is expected to be higher on average by 1.74 to 3.378 per million. (d) Yes, we rejected H0 and the conﬁdence interval does not include 0. (c) n = 20, df = 18, T ∗ 8.39 (a) True. (b) False, correlation is a measure of the linear association between any two numerical variables. 8.41 There is an upwards trend. However, the variability is higher for higher calorie counts, and it looks like there might be two clusters of observations above and below the line on the right, so we should be cautious about ﬁtting a linear model to these data. 8.43 (a) r = −0.72 → (2) (b) r = 0.07 → (4) (c) r = 0.86 → (1) (d) r = 0.99 → (3) 8.45 (a) There is a weak-to-moderate, positive, linear association between height and volume. There also appears to be some non-constant variance since the volume of trees is more variable for taller trees. (b) There is a very strong, positive association between diameter and volume. The relationship may include slight curvature. (c) Since the relationship is stronger between volume and diameter, using diameter would be preferred. However, as mentioned in part (b), the relationship between volume and diameter may not be, and so we may beneﬁt from a model that properly accounts for nonlinearity. 503 Appendix B Data sets within the text Each data set within the text is described in this appendix. For those data sets that are in multiple sections in a chapter, only the ﬁrst section is listed in that chapter. If a data set is not listed here, e.g. Section 3.2.10 lists imagined probabilities for whether a parking garage will ﬁll up and whether there is a sporting event that same evening for an unnamed college, it may not be listed in this data appendix. When a raw data set is available vs just a description, there is a corresponding page for the data set at openintro.org/data. That webpage also includes many more data sets than are covered in this textbook, and each data set on the website includes a description, it’s source, a detailed overview of each data set’s variables, and download options. Chapter 1: Data collection 1.1 stent30, stent365 → The stent data is split across two data sets, one for the 0-30 day and one for the 0-365 day results. Chimowitz MI, Lynn MJ, Derdeyn CP, et al. 2011. Stenting versus Aggressive Medical Therapy for Intracranial Arterial Stenosis. New England Journal of Medicine 365:993-1003. www.nejm.org/doi/full/10.1056/NEJMoa1105335. NY Times article: www.nytimes.com/2011/09/08/health/research/08stent.html. 1.2 loan50, loans full schema → This data comes from Lending Club (lendingclub.com), which provides a large set of data on the people who received loans through their platform. The data used in the textbook comes from a sample of the loans made in Q1 (Jan, Feb, March) 2018. 1.2 county, county complete → These data come from several government sources. For those variables included in the county data set, only the most recent data is reported, as of what was available in late 2018. Data prior to 2011 is all from census.gov, where the speciﬁc Quick Facts page providing the data is no longer available. The more recent data comes from USDA (ers.usda.gov), Bureau of Labor Statistics (bls.gov/lau), SAIPE (census.gov/did/www/saipe), and American Community Survey (census.gov/programs-surveys/acs). 1.4 The study in mind regarding chocolate and heart attack patients: Janszky et al. 2009. Chocolate consumption and mortality following a ﬁrst acute myocardial infarction: the Stockholm Heart Epidemiology Program. Journal of Internal Medicine 266:3, p248-257. 1.4 The Nurses’ Health Study was mentioned. For more information on this data set, see www.channing.harvard.edu/nhs 1.5 The study we had in mind during the introduction of Section 1.5.1 was Anturane Reinfarction Trial Research Group. 1980. Sulﬁnpyrazone in the prevention of sudden death after myocardial infarction. New England Journal of Medicine 302(5):250-256. 504 APPENDIX B. DATA SETS WITHIN THE TEXT Chapter 2: Summarizing data 2.1 county → This data set is described in the data for Chapter 1. 2.1 email50, email → These data represent emails sent to David Diez. Each data set includes 21 variables. The email50 data set is a random sample of 50 emails from email. 2.2 loan50, county → These data sets are described in the data for Chapter 1. email50, email → These data sets are described in the data for Section 2.1. 2.2 2019 mean and median income. → https://data.census.gov/cedsci/table?hidePreview=true&tid=ACSST1Y2019.S1901 2.2 possum → The brushtail possum statistics are based on a sample of possums from Australia and New Guinea. The original source of this data is as follows: Lindenmayer DB, et al. 1995. Morphological variation among columns of the mountain brushtail possum, Trichosurus caninus Ogilby (Phalangeridae: Marsupiala). Australian Journal of Zoology 43: 449-458. 2.3 SAT and ACT score distributions → The SAT score data comes from the 2018 distribution, which is provided at reports.collegeboard.org/pdf/2018-total-group-sat-suite-assessments-annual-report.pdf The ACT score data is available at act.org/content/dam/act/unsecured/documents/cccr2018/P 99 999999 N S N00 ACT-GCPR National.pdf We also acknowledge that the actual ACT score distribution is not nearly normal. However, since the topic is very accessible, we decided to keep the context and examples. 2.3 nba players 19 → Summary information from the NBA players for the 2018-2019 season. Data were retrieved from www.nba.com/players. 2.4 loans full schema → This data set is described in the data for Chapter 1. 2.5 malaria → Lyke et al. 2017. PfSPZ vaccine induces strain-transcending T cells and durable protection against heterologous controlled human malaria infection. PNAS 114(10):2711-2716. www.pnas.org/content/114/10/2711 Chapter 3: Probability and probability distributions 3.1 email → This data set is described in the data for Chapter 2. 3.1 playing cards → A table describing the 52 cards in a standard deck. 3.2 Machine learning on fashion. → This is a simulated data set, not based on any speciﬁc machine learning classiﬁer. 3.2 smallpox → Fenner F. 1988. Smallpox and Its Eradication (History of International Public Health, No. 6). Geneva: World Health Organization. ISBN 92-4-156110-6. 3.2 family college → A simulated data set based on real population summaries at nces.ed.gov/pubs2001/2001126.pdf. 3.2 Mammogram screening, probabilities. → The probabilities reported were obtained using studies reported at www.breastcancer.org and www.ncbi.nlm.nih.gov/pmc/articles/PMC1173421. 3.4 stocks 18 → Monthly returns for Caterpillar, Exxon Mobil Corp, and Google for November 2015 to October 2018. 3.5 Blood type prevalence. → The fraction of people with O+ blood is about 38% according to https://www.redcrossblood.org/donate-blood/blood-types/o-blood-type.html We used 35% for simplicity in the examples. 505 Chapter 4: Sampling distributions 4.1 Blood type prevalence. → This data set is described in the data for Chapter 3. 4.2 run17, run17samp → These data set represent the full population and a sample of the runners and their run times in the 2017 Cherry Blossom Run in Washington, DC. For more details, see www.cherryblossom.org. 4.2 poker → The full data set includes poker winnings (and losses) for 50 days by a professional poker player, which represents their ﬁrst 50 days trying to play for a living. Anonymity has been requested by the player. Chapter 5: Foundations for inference 5.1 email → This data set is described in the data for Chapter 2. 5.1 pew energy 2018 → The actual data has more observations than were referenced in this chapter. That is, we used a subsample since it helped smooth some of the examples to have a bit more variability. The pew energy 2018 data set represents the full data set for each of the diﬀerent energy source questions, which covers solar, wind, oﬀshore drilling, hydrolic fracturing, and nuclear energy. The statistics used to construct the data are from the following page: www.pewinternet.org/2018/05/14/majorities-see-government-eﬀorts-to-protect-the-environmentas-insuﬃcient/ 5.2 pew energy 2018 → Se
e the details for this data set above in the Section 5.1 data section. 5.2 ebola survey → In New York City on October 23rd, 2014, a doctor who had recently been treating Ebola patients in Guinea went to the hospital with a slight fever and was subsequently diagnosed with Ebola. Soon thereafter, an NBC 4 New York/The Wall Street Journal/Marist Poll found that 82% of New Yorkers favored a “mandatory 21-day quarantine for anyone who has come in contact with an Ebola patient”. This poll included responses of 1,042 New York adults between Oct 26th and 28th, 2014. Poll ID NY141026 on maristpoll.marist.edu. 5.3 transplant → This is a made up data set about the health outcomes for a hypothetical medical consultant. Note that the data set on the website has 62 patients, not 142 patients, so there will a diﬀerence for what is covered in this book vs the data set on the website. 5.3 Alaska residents under 5 years old. → The 2010 statistic comes from the US census: https://data.census.gov. 506 APPENDIX B. DATA SETS WITHIN THE TEXT Chapter 6: Inference for categorical data 6.1 Supreme Court → The Gallup organization began measuring the public’s view of the Supreme Court’s job performance in 2000, and has measured it every year since then with the question: “Do you approve or disapprove of the way the Supreme Court is handling its job?”. In 2018, the Gallup poll randomly sampled 1,033 adults in the U.S. and found that 53% of them approved. https://news.gallup.com/poll/237269/supreme-court-approval-highest-2009.aspx 6.1 Life on other planets → A February 2018 Marist Poll reported: “Many Americans (68%) think there is intelligent life on other planets”. This is up from 52% in 2005. The results were based on a random sample of 1,033 adults in the U.S. http://maristpoll.marist.edu/212-are-americans-poised-for-an-alien-invasion 6.1 Congressional approval rating. → This survey data is from news.gallup.com/poll/237176/snapshot-congressional-job-approval-july.aspx 6.1 Tire inspection. → This is a hypothetical scenario not based on real data. 6.1 Toohey poll. → This is a hypothetical scenario not based on a real person or real data. 6.1 Support for nuclear energy. → The results are from the following Gallup poll: www.gallup.com/poll/182180/support-nuclear-energy.aspx 6.2 cpr → B¨ottiger et al. Eﬃcacy and safety of thrombolytic therapy after initially unsuccessful cardiopulmonary resuscitation: a prospective clinical trial. The Lancet, 2001. 6.2 gear company → This is a hypothetical scenario not based on real data. 6.2 healthcare law survey → Pew research survey on the Aﬀordable Care Act (aka Obamacare) that ran the survey question with two variants. http://www.people-press.org/2012/03/26/public-remains-split-on-health-care-bill-opposedto-mandate/ 6.2 fish oil 18 → Manson JE, et al. 2018. Marine n-3 Fatty Acids and Prevention of Cardio- vascular Disease and Cancer. NEJMoa1811403. 6.3 jury → Simulated data set of registered voter proportions and representation on juries from a population. 6.3 M&Ms → Rick Wicklin collected a sample of 712 candies, or about 1.5 pounds, and counted how many there were of each color. https://qz.com/918008/the-color-distribution-of-mms-as-determined-by-a-phd-in-statistics 6.4 gsearch → Simulated (fake) data set for Google search experiment. 6.4 ask → Experiment results from asking about iPods, where the original source is: Minson JA, Ruedy NE, Schweitzer ME. There is such a thing as a stupid question: Question disclosure in strategic communication. opim.wharton.upenn.edu/DPlab/papers/workingPapers/ Minson working Ask%20(the%20Right%20Way)%20and%20You%20Shall%20Receive.pdf 6.4 Obama and Congressional approval by political aﬃliation. → This survey was completed by Pew Research and the full results may be found at: http://www.people-press.org/2012/03/14/romney-leads-gop-contest-trails-in-matchup-with-obama 6.4 Attitudes on climate change → A Pew Research poll published in May of 2021 looks at how Americans’ attitudes about climate change diﬀer by generation, party and other factors. https://www.pewresearch.org/fact-tank/2021/05/26/key-ﬁndings-how-americans-attitudes-about-climate-change-diﬀer- by-generation-party-and-other-factors/ 507 Chapter 7: Inference for numerical data 7.1 Risso’s dolphins → Endo T and Haraguchi K. 2009. High mercury levels in hair samples from residents of Taiji, a Japanese whaling town. Marine Pollution Bulletin 60(5):743-747. Taiji was featured in the movie The Cove, and it is a signiﬁcant source of dolphin and whale meat in Japan. Thousands of dolphins pass through the Taiji area annually, and we will assume these 19 dolphins represent a simple random sample from those dolphins. 7.1 Croaker white ﬁsh → www.fda.gov/food/foodborneillnesscontaminants/metals/ucm115644.htm 7.1 run17samp → This data set is described in the data for Chapter 4. 7.2 textbooks, ucla textbooks f18 → Data were collected by OpenIntro staﬀ in 2010 and again in 2018. For the 2018 sample, we sampled 201 UCLA courses. Of those, 68 required books could be found on Amazon. The websites where information was retrieved: sa.ucla.edu/ro/public/soc, ucla.verbacompare.com, and amazon.com. 7.2 sat improve → This is a hypothetical (fake) data set for SAT improvement from an SAT preparation company. 7.3 Jennifer-John study. → Bertrand M, Mullainathan S. 2004. Science faculty’s subtle gender biases favor male students. PNAS October 9, 2012 109 (41) 16474-16479. https://www.pnas.org/content/109/41/16474 7.3 resume → Study for racial bias in hiring, where the study’s data is available in the resume data set. This data set is explored in great detail in the logistic regression section of the OpenIntro Statistics textbook (free PDF). The original source for this data is: Bertrand M, Mullainathan S. 2004. Are Emily and Greg More Employable than Lakisha and Jamal? A Field Experiment on Labor Market Discrimination. The American Economic Review 94:4 (991-1013). www.nber.org/papers/w9873 7.3 Exams variants. → This is a simulated (fake) data set for exam performance of students for two diﬀerent exam variations. 7.3 ncbirths → A random sample of 1000 NC births. A sample of that random sample was used for the example in the section. 7.3 stem cells → Menard C, et al. 2005. Transplantation of cardiac-committed mouse embryonic stem cells to infarcted sheep myocardium: a preclinical study. The Lancet: 366:9490, p10051012. https://www.thelancet.com/journals/lancet/article/PIIS0140-6736(05)67380-1/fulltext 508 APPENDIX B. DATA SETS WITHIN THE TEXT Chapter 8: Introduction to linear regression 8.1 simulated scatter → Fake data used for the ﬁrst three plots. The perfect linear plot uses group 4 data, where group variable in the data set (Figure 8.1). The group of 3 imperfect linear plots use groups 1-3 (Figure 8.2). The sinusoidal curve uses group 5 data (Figure 8.3). The group of 3 scatterplots with residual plots use groups 6-8 (Figure 8.8). The correlation plots uses groups 9-19 data (Figures 8.9 and 8.10). 8.1 possum → This data is described in the data for Chapter 2. 8.1 simulated scatter → The plots for things that can go wrong uses groups 20-23 (Figure 8.26). 8.2 elmhurst → These data were sampled from a table of data for all freshman from the 2011 class at Elmhurst College that accompanied an article titled What Students Really Pay to Go to College published online by The Chronicle of Higher Education: chronicle.com/article/WhatStudents-Really-Pay-to-Go/131435. 8.2 textbooks, ucla textbooks f18 → This data is described in the data for Chapter 7. 8.2 loan50 → This data set is described in the data for Chapter 1. 8.2 mariokart → Auction data from Ebay (ebay.com) for the game Mario Kart for the Nintendo Wii. This data set was collected in early October, 2009. 8.2 simulated scatter → The plots for types of outliers uses groups 24-29 (Figure 8.19). 8.3 county, county complete → These data sets are described in the data for Chapter 1. 8.4 midterms house → Data was retrieved from Wikipedia. 509 Appendix C Distribution tables C.1 Random Number Table Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1-5 44394 61578 18529 81238 11173 96737 63514 35087 00148 28999 37911 33624 93282 57429 65029 14779 52072 76282 46561 70623 03605 46147 09547 12899 21223 35770 04243 56989 53233 20232 6-10 76100 75037 73285 18321 58878 95194 55066 57036 73933 76232 50834 82379 63059 71933 24328 23173 12187 91849 33015 36097 08541 07603 77804 05005 38353 35697 65817 05587 48698 30909 11-15 85973 54792 95291 71085 25516 14419 65162 10001 49369 32637 10927 03625 10830 80329 06826 97183 35360 17138 04577 48780 17546 92057 95099 86667 56970 32281 81819 79995 59304 77126 Column 16-20 26853 74216 49606 08284 15058 22202 96016 39424 32403 95697 74075 58336 89432 56521 61448 59835 82925 59554 02178 06921 85790 87609 22158 72331 48965 53514 64381 36598 63566 50041 21-25 07080 31952 67174 39318 48639 92867 91723 50536 53850 63679 26558 27390 26917 97594 54760 69580 44923 35476 32915 60683 48413 52670 53279 09114 58371 10854 83509 02316 25352 96500 26-30 91603 31235 95905 31434 52723 73525 21160 77380 16291 54506 42311 00586 31555 92651 09351 94653 44532 67007 35912 22461 69382 96255 23161 28187 02697 16778 44316 81627 03322 24033 31-35 00476 31258 33679 26173 95864 94382 24285 45042 93619 11299 36483 06344 51793 14819 73930 55095 18251 02484 48974 36175 89785 96660 72675 97404 61417 56447 56316 50104 29938 77422 36-40 19681 57886 75811 07440 89673 29927 33264 48180 27557 94294 71820 89625 18718 86546 99564 80666 96991 10122 92985 61281 80206 83167 92804 26750 54746 46965 47742 47720 82306 20150 510 APPENDIX C. DISTRIBUTION TABLES C.2 Normal Probability Table A normal probability table may be used to ﬁnd percentiles of a normal distribution using a Z-score, or vice-versa. Such a table lists Z-scores and the corresponding percentiles. An abbreviated probability table is provided in Figure C.1 that we’ll use for the examples in this appendix. A full
table may be found on page 512. 0.00 0.01 0.02 Second decimal place of Z 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413 0.8643 ... 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 ... 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 ... 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485 0.8708 ... 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508 0.8729 ... 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531 0.8749 ... 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554 0.8770 ... 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0.8790 ... 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 ... 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 ... Z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 ... Figure C.1: A section of the normal probability table. The percentile for a normal random variable with Z = 1.00 has been highlighted, and the percentile closest to 0.8000 has also been highlighted. When using a normal probability table to ﬁnd a percentile for Z (rounded to two decimals), identify the proper row in the normal probability table up through the ﬁrst decimal, and then determine the column representing the second decimal value. The intersection of this row and column is the percentile of the observation. For instance, the percentile of Z = 0.45 is shown in row 0.4 and column 0.05 in Figure C.1: 0.6736, or the 67.36th percentile. Figure C.2: The area to the left of Z represents the percentile of the observation. negative Zpositive Z EXAMPLE C.1 SAT scores follow a normal distribution, N (1100, 200). Ann earned a score of 1300 on her SAT with a corresponding Z-score of Z = 1. She would like to know what percentile she falls in among all SAT test-takers. Ann’s percentile is the percentage of people who earned a lower SAT score than her. We shade the area representing those individuals in the following graph: 511 The total area under the normal curve is always equal to 1, and the proportion of people who scored below Ann on the SAT is equal to the area shaded in the graph. We ﬁnd this area by looking in row 1.0 and column 0.00 in the normal probability table: 0.8413. In other words, Ann is in the 84th percentile of SAT takers. EXAMPLE C.2 How do we ﬁnd an upper tail area? The normal probability table always gives the area to the left. This means that if we want the area to the right, we ﬁrst ﬁnd the lower tail and then subtract it from 1. For instance, 84.13% of SAT takers scored below Ann, which means 15.87% of test takers scored higher than Ann. We can also ﬁnd the Z-score associated with a percentile. For example, to identify Z for the 80th percentile, we look for the value closest to 0.8000 in the middle portion of the table: 0.7995. We determine the Z-score for the 80th percentile by combining the row and column Z values: 0.84. EXAMPLE C.3 Find the SAT score for the 80th percentile. We look for the are to the value in the table closest to 0.8000. The closest value is 0.7995, which corresponds to Z = 0.84, where 0.8 comes from the row value and 0.04 comes from the column value. Next, we set up the equation for the Z-score and the unknown value x as follows, and then we solve for x: Z = 0.84 = x − 1100 200 → x = 1268 The College Board scales scores to increments of 10, so the 80th percentile is 1270. (Reporting 1268 would have been perfectly okay for our purposes.) For additional details about working with the normal distribution and the normal probability table, see Section 2.3, which starts on page 101. −3−2−10123 512 APPENDIX C. DISTRIBUTION TABLES 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 Z Second decimal place of Z 0.0002 0.0003 0.0005 0.0007 0.0010 0.0014 0.0019 0.0026 0.0036 0.0048 0.0064 0.0084 0.0110 0.0143 0.0183 0.0233 0.0294 0.0367 0.0455 0.0559 0.0681 0.0823 0.0985 0.1170 0.1379 0.0003 0.0004 0.0005 0.0007 0.0010 0.0014 0.0020 0.0027 0.0037 0.0049 0.0066 0.0087 0.0113 0.0146 0.0188 0.0239 0.0301 0.0375 0.0465 0.0571 0.0694 0.0838 0.1003 0.1190 0.1401 0.0003 0.0004 0.0005 0.0008 0.0011 0.0015 0.0021 0.0028 0.0038 0.0051 0.0068 0.0089 0.0116 0.0150 0.0192 0.0244 0.0307 0.0384 0.0475 0.0582 0.0708 0.0853 0.1020 0.1210 0.1423 0.0003 0.0004 0.0006 0.0008 0.0011 0.0015 0.0021 0.0029 0.0039 0.0052 0.0069 0.0091 0.0119 0.0154 0.0197 0.0250 0.0314 0.0392 0.0485 0.0594 0.0721 0.0869 0.1038 0.1230 0.1446 0.0003 0.0004 0.0006 0.0008 0.0011 0.0016 0.0022 0.0030 0.0040 0.0054 0.0071 0.0094 0.0122 0.0158 0.0202 0.0256 0.0322 0.0401 0.0495 0.0606 0.0735 0.0885 0.1056 0.1251 0.1469 0.0003 0.0004 0.0006 0.0008 0.0012 0.0016 0.0023 0.0031 0.0041 0.0055 0.0073 0.0096 0.0125 0.0162 0.0207 0.0262 0.0329 0.0409 0.0505 0.0618 0.0749 0.0901 0.1075 0.1271 0.1492 0.0003 0.0004 0.0006 0.0009 0.0012 0.0017 0.0023 0.0032 0.0043 0.0057 0.0075 0.0099 0.0129 0.0166 0.0212 0.0268 0.0336 0.0418 0.0516 0.0630 0.0764 0.0918 0.1093 0.1292 0.1515 0.0003 0.0005 0.0006 0.0009 0.0013 0.0018 0.0024 0.0033 0.0044 0.0059 0.0078 0.0102 0.0132 0.0170 0.0217 0.0274 0.0344 0.0427 0.0526 0.0643 0.0778 0.0934 0.1112 0.1314 0.1539 0.1611 0.1788 0.1867 0.2061 0.2148 0.2358 0.2451 0.2676 0.2776 0.3015 0.3121 0.3372 0.3483 0.3745 0.3859 0.4129 0.4247 0.4522 0.4920 0.4641 ∗For Z ≤ −3.50, the probability is less than or equal to 0.0002. 0.1635 0.1894 0.2177 0.2483 0.2810 0.3156 0.3520 0.3897 0.4286 0.4681 0.1762 0.2033 0.2327 0.2643 0.2981 0.3336 0.3707 0.4090 0.4483 0.4880 0.1736 0.2005 0.2296 0.2611 0.2946 0.3300 0.3669 0.4052 0.4443 0.4840 0.1711 0.1977 0.2266 0.2578 0.2912 0.3264 0.3632 0.4013 0.4404 0.4801 0.1660 0.1922 0.2206 0.2514 0.2843 0.3192 0.3557 0.3936 0.4325 0.4721 0.1685 0.1949 0.2236 0.2546 0.2877 0.3228 0.3594 0.3974 0.4364 0.4761 0.0003 0.0005 0.0007 0.0009 0.0013 0.0018 0.0025 0.0034 0.0045 0.0060 0.0080 0.0104 0.0136 0.0174 0.0222 0.0281 0.0351 0.0436 0.0537 0.0655 0.0793 0.0951 0.1131 0.1335 0.1562 0.1814 0.2090 0.2389 0.2709 0.3050 0.3409 0.3783 0.4168 0.4562 0.4960 0.0003 −3.4 0.0005 −3.3 0.0007 −3.2 0.0010 −3.1 0.0013 −3.0 0.0019 −2.9 0.0026 −2.8 0.0035 −2.7 0.0047 −2.6 0.0062 −2.5 0.0082 −2.4 0.0107 −2.3 0.0139 −2.2 0.0179 −2.1 0.0228 −2.0 0.0287 −1.9 0.0359 −1.8 0.0446 −1.7 0.0548 −1.6 0.0668 −1.5 0.0808 −1.4 0.0968 −1.3 0.1151 −1.2 0.1357 −1.1 0.1587 −1.0 0.1841 −0.9 0.2119 −0.8 0.2420 −0.7 0.2743 −0.6 0.3085 −0.5 0.3446 −0.4 0.3821 −0.3 0.4207 −0.2 0.4602 −0.1 0.5000 −0.0 Negative Z 513 Second decimal place of Z 0.07 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.04 0.03 0.06 0.05 0.00 0.01 0.02 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8485 0.8708 0.8907 0.9082 0.9236 0.9370 0.9484 0.9582 0.9664 0.9732 0.8461 0.8686 0.8888 0.9066 0.9222 0.9357 0.9474 0.9573 0.9656 0.9726 0.8413 0.8643 0.8849 0.9032 0.9192 0.9332 0.9452 0.9554 0.9641 0.9713 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.9719 Z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 0.9988 3.1 0.9991 3.2 0.9994 3.3 0.9996 3.4 0.9997 ∗For Z ≥ 3.50, the probability is greater than or equal to 0.9998. 0.8508 0.8729 0.8925 0.9099 0.9251 0.9382 0.9495 0.9591 0.9671 0.9738 0.8531 0.8749 0.8944 0.9115 0.9265 0.9394 0.9505 0.9599 0.9678 0.9744 0.8554 0.8770 0.8962 0.9131 0.9279 0.9406 0.9515 0.9608 0.9686 0.9750 0.9793 0.9838 0.9875 0.9904 0.9927 0.9945 0.9959 0.9969 0.9977 0.9984 0.9778 0.9826 0.9864 0.9896 0.9920 0.9940 0.9955 0.9966 0.9975 0.9982 0.9803 0.9846 0.9881 0.9909 0.9931 0.9948 0.9961 0.9971 0.9979 0.9985 0.9772 0.9821 0.9861 0.9893 0.9918 0.9938 0.9953 0.9965 0.9974 0.9981 0.9798 0.9842 0.9878 0.9906 0.9929 0.9946 0.9960 0.9970 0.9978 0.9984 0.9783 0.9830 0.9868 0.9898 0.9922 0.9941 0.9956 0.9967 0.9976 0.9982 0.9788 0.9834 0.9871 0.9901 0.9925 0.9943 0.9957 0.9968 0.9977 0.9983 0.9988 0.9992 0.9994 0.9996 0.9997 0.9987 0.9991 0.9993 0.9995 0.9997 0.9989 0.9992 0.9994 0.9996 0.9997 0.9989 0.9992 0.9994 0.9996 0.9997 0.9987 0.9990 0.9993 0.9995 0.9997 0.9987 0.9991 0.9994 0.9995 0.9997 0.8577 0.8790 0.8980 0.9147 0.9292 0.9418 0.9525 0.9616 0.9693 0.9756 0.9808 0.9850 0.9884 0.9911 0.9932 0.9949 0.9962 0.9972 0.9979 0.9985 0.9989 0.9992 0.9995 0.9996 0.9997 0.08 0.09 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761 0.9812 0.9854 0.9887 0.9913 0.9934 0.9951 0.9963 0.9973 0.9980 0.9986 0.9990 0.9993 0.9995 0.9996 0.9997 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 0.9857 0.9890 0.9916 0.9936 0.9952 0.9964 0.9974 0.9981 0.9986 0.9990 0.9993 0.9995 0.9997 0.9998 YPositive Z 514 APPENDIX C. DISTRIBUTION TABLES C.3 ttt-Probability Table A ttt-probability table may be used to ﬁnd tail areas of a t-distribution using a T-score, or viceversa. Such a table lists T-scores and the corresponding percentiles. A partial ttt-table is shown in Figure C.3, and the complete table starts on page 516. Each row in the t-table represents a t-distribution with diﬀerent degrees of freedom. The columns correspond to tail probabilities. For instance, if we know we are working with the t-distribution with df = 18, we can examine row 18, which is highlighted in Figure C.3. If we want the value in this
row that identiﬁes the T-score (cutoﬀ) for an upper tail of 10%, we can look in the column where one tail is 0.100. This cutoﬀ is 1.33. If we had wanted the cutoﬀ for the lower 10%, we would use -1.33. Just like the normal distribution, all t-distributions are symmetric. one tail two tails 1 df 2 3 ... 17 18 19 20 ... 400 500 ∞ 0.100 0.200 3.08 1.89 1.64 ... 1.33 1.33 1.33 1.33 ... 1.28 1.28 1.28 0.050 0.100 6.31 2.92 2.35 ... 1.74 1.73 1.73 1.72 ... 1.65 1.65 1.64 0.025 0.050 12.71 4.30 3.18 ... 2.11 2.10 2.09 2.09 ... 1.97 1.96 1.96 0.010 0.020 31.82 6.96 4.54 ... 2.57 2.55 2.54 2.53 ... 2.34 2.33 2.33 0.005 0.010 63.66 9.92 5.84 2.90 2.88 2.86 2.85 2.59 2.59 2.58 Figure C.3: An abbreviated look at the t-table. Each row represents a diﬀerent t-distribution. The columns describe the cutoﬀs for speciﬁc tail areas. The row with df = 18 has been highlighted. EXAMPLE C.4 What proportion of the t-distribution with 18 degrees of freedom falls below -2.10? Just like a normal probability problem, we ﬁrst draw the picture and shade the area below -2.10: To ﬁnd this area, we ﬁrst identify the appropriate row: df = 18. Then we identify the column containing the absolute value of -2.10; it is the third column. Because we are looking for just one tail, we examine the top line of the table, which shows that a one tail area for a value in the third row corresponds to 0.025. That is, 2.5% of the distribution falls below -2.10. In the next example we encounter a case where the exact T-score is not listed in the table. −4−2024 515 EXAMPLE C.5 A t-distribution with 20 degrees of freedom is shown in the left panel of Figure C.4. Estimate the proportion of the distribution falling above 1.65. We identify the row in the t-table using the degrees of freedom: df = 20. Then we look for 1.65; it is not listed. It falls between the ﬁrst and second columns. Since these values bound 1.65, their tail areas will bound the tail area corresponding to 1.65. We identify the one tail area of the ﬁrst and second columns, 0.050 and 0.10, and we conclude that between 5% and 10% of the distribution is more than 1.65 standard deviations above the mean. If we like, we can identify the precise area using statistical software: 0.0573. Figure C.4: Left: The t-distribution with 20 degrees of freedom, with the area above 1.65 shaded. Right: The t-distribution with 475 degrees of freedom, with the area further than 2 units from 0 shaded. EXAMPLE C.6 A t-distribution with 475 degrees of freedom is shown in the right panel of Figure C.4. Estimate the proportion of the distribution falling more than 2 units from the mean (above or below). As before, ﬁrst identify the appropriate row: df = 475. This row does not exist! When this happens, we use the next smaller row, which in this case is df = 400. Next, ﬁnd the columns that capture 2.00; because 1.97 < 3 < 2.34, we use the third and fourth columns. Finally, we ﬁnd bounds for the tail areas by looking at the two tail values: 0.02 and 0.05. We use the two tail values because we are looking for two symmetric tails in the t-distribution. GUIDED PRACTICE C.7 What proportion of the t-distribution with 19 degrees of freedom falls above -1.79 units?1 EXAMPLE C.8 Find the value of t of freedom where 95% of the distribution lies between -t 18 using the t-table, where t 18 and +t 18. 18 is the cutoﬀ for the t-distribution with 18 degrees For a 95% conﬁdence interval, we want to ﬁnd the cutoﬀ t 18 such that 95% of the t-distribution is between -t 18; this is the same as where the two tails have a total area of 0.05. We look in the t-table on page 514, ﬁnd the column with area totaling 0.05 in the two tails (third column), and then the row with 18 degrees of freedom: t 18 and t 18 = 2.10. 1We ﬁnd the shaded area above -1.79 (we leave the picture to you). The small left tail is between 0.025 and 0.05, so the larger upper region must have an area between 0.95 and 0.975. −4−2024−4−2024 516 APPENDIX C. DISTRIBUTION TABLES one tail two tails 1 df 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 0.100 0.200 3.08 1.89 1.64 1.53 1.48 1.44 1.41 1.40 1.38 1.37 1.36 1.36 1.35 1.35 1.34 1.34 1.33 1.33 1.33 1.33 1.32 1.32 1.32 1.32 1.32 1.31 1.31 1.31 1.31 1.31 0.050 0.100 6.31 2.92 2.35 2.13 2.02 1.94 1.89 1.86 1.83 1.81 1.80 1.78 1.77 1.76 1.75 1.75 1.74 1.73 1.73 1.72 1.72 1.72 1.71 1.71 1.71 1.71 1.70 1.70 1.70 1.70 0.025 0.050 12.71 4.30 3.18 2.78 2.57 2.45 2.36 2.31 2.26 2.23 2.20 2.18 2.16 2.14 2.13 2.12 2.11 2.10 2.09 2.09 2.08 2.07 2.07 2.06 2.06 2.06 2.05 2.05 2.05 2.04 0.010 0.020 31.82 6.96 4.54 3.75 3.36 3.14 3.00 2.90 2.82 2.76 2.72 2.68 2.65 2.62 2.60 2.58 2.57 2.55 2.54 2.53 2.52 2.51 2.50 2.49 2.49 2.48 2.47 2.47 2.46 2.46 0.005 0.010 63.66 9.92 5.84 4.60 4.03 3.71 3.50 3.36 3.25 3.17 3.11 3.05 3.01 2.98 2.95 2.92 2.90 2.88 2.86 2.85 2.83 2.82 2.81 2.80 2.79 2.78 2.77 2.76 2.76 2.75 −3−2−10123One Tail−3−2−10123One Tail−3−2−10123Two Tails 517 one tail two tails 31 df 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 60 70 80 90 100 150 200 300 400 500 ∞ 0.100 0.200 1.31 1.31 1.31 1.31 1.31 1.31 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.30 1.29 1.29 1.29 1.29 1.29 1.29 1.28 1.28 1.28 1.28 0.050 0.100 1.70 1.69 1.69 1.69 1.69 1.69 1.69 1.69 1.68 1.68 1.68 1.68 1.68 1.68 1.68 1.68 1.68 1.68 1.68 1.68 1.67 1.67 1.66 1.66 1.66 1.66 1.65 1.65 1.65 1.65 1.65 0.025 0.050 2.04 2.04 2.03 2.03 2.03 2.03 2.03 2.02 2.02 2.02 2.02 2.02 2.02 2.02 2.01 2.01 2.01 2.01 2.01 2.01 2.00 1.99 1.99 1.99 1.98 1.98 1.97 1.97 1.97 1.96 1.96 0.010 0.020 2.45 2.45 2.44 2.44 2.44 2.43 2.43 2.43 2.43 2.42 2.42 2.42 2.42 2.41 2.41 2.41 2.41 2.41 2.40 2.40 2.39 2.38 2.37 2.37 2.36 2.35 2.35 2.34 2.34 2.33 2.33 0.005 0.010 2.74 2.74 2.73 2.73 2.72 2.72 2.72 2.71 2.71 2.70 2.70 2.70 2.70 2.69 2.69 2.69 2.68 2.68 2.68 2.68 2.66 2.65 2.64 2.63 2.63 2.61 2.60 2.59 2.59 2.59 2.58 −3−2−10123One Tail−3−2−10123One Tail−3−2−10123Two Tails 518 APPENDIX C. DISTRIBUTION TABLES C.4 Chi-Square Probability Table A chi-square probability table may be used to ﬁnd tail areas of a chi-square distribution. The chi-square table is partially shown in Figure C.5, and the complete table may be found on page 519. When using a chi-square table, we examine a particular row for distributions with diﬀerent degrees of freedom, and we identify a range for the area (e.g. 0.025 to 0.05). Note that the chi-square table provides upper tail values, which is diﬀerent than the normal and t-distribution tables. Upper tail 2 df 3 4 5 6 7 0.3 2.41 3.66 4.88 6.06 7.23 8.38 0.2 3.22 4.64 5.99 7.29 8.56 9.80 0.1 4.61 6.25 7.78 9.24 10.64 12.02 0.05 5.99 7.81 9.49 11.07 12.59 14.07 0.02 7.82 9.84 11.67 13.39 15.03 16.62 0.01 9.21 11.34 13.28 15.09 16.81 18.48 0.005 10.60 12.84 14.86 16.75 18.55 20.28 0.001 13.82 16.27 18.47 20.52 22.46 24.32 Figure C.5: A section of the chi-square table. A complete table is in Appendix C.4. EXAMPLE C.9 Figure C.6(a) shows a chi-square distribution with 3 degrees of freedom and an upper shaded tail starting at 6.25. Use Figure C.5 to estimate the shaded area. This distribution has three degrees of freedom, so only the row with 3 degrees of freedom (df) is relevant. This row has been italicized in the table. Next, we see that the value – 6.25 – falls in the column with upper tail area 0.1. That is, the shaded upper tail of Figure C.6(a) has area 0.1. This example was unusual, in that we observed the exact value in the table. In the next examples, we encounter situations where we cannot precisely estimate the tail area and must instead provide a range of values. EXAMPLE C.10 Figure C.6(b) shows the upper tail of a chi-square distribution with 2 degrees of freedom. The area above value 4.3 has been shaded; ﬁnd this tail area. The cutoﬀ 4.3 falls between the second and third columns in the 2 degrees of freedom row. Because these columns correspond to tail areas of 0.2 and 0.1, we can be certain that the area shaded in Figure C.6(b) is between 0.1 and 0.2. EXAMPLE C.11 Figure C.6(c) shows an upper tail for a chi-square distribution with 5 degrees of freedom and a cutoﬀ of 5.1. Find the tail area. Looking in the row with 5 df, 5.1 falls below the smallest cutoﬀ for this row (6.06). That means we can only say that the area is greater than 0.3. EXAMPLE C.12 Figure C.6(d) shows a cutoﬀ of 11.7 on a chi-square distribution with 7 degrees of freedom. Find the area of the upper tail. The value 11.7 falls between 9.80 and 12.02 in the 7 df row. Thus, the area is between 0.1 and 0.2. 519 (a) (c) (b) (d) Figure C.6: (a) Chi-square distribution with 3 degrees of freedom, area above 6.25 shaded. (b) 2 degrees of freedom, area above 4.3 shaded. (c) 5 degrees of freedom, area above 5.1 shaded. (d) 7 degrees of freedom, area above 11.7 shaded. Upper tail 1 df 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 40 50 0.3 1.07 2.41 3.66 4.88 6.06 7.23 8.38 9.52 10.66 11.78 12.90 14.01 15.12 16.22 17.32 18.42 19.51 20.60 21.69 22.77 28.17 33.53 44.16 54.72 0.2 1.64 3.22 4.64 5.99 7.29 8.56 9.80 11.03 12.24 13.44 14.63 15.81 16.98 18.15 19.31 20.47 21.61 22.76 23.90 25.04 30.68 36.25 47.27 58.16 0.1 2.71 4.61 6.25 7.78 9.24 10.64 12.02 13.36 14.68 15.99 17.28 18.55 19.81 21.06 22.31 23.54 24.77 25.99 27.20 28.41 34.38 40.26 51.81 63.17 0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51 16.92 18.31 19.68 21.03 22.36 23.68 25.00 26.30 27.59 28.87 30.14 31.41 37.65 43.77 55.76 67.50 0.02 5.41 7.82 9.84 11.67 13.39 15.03 16.62 18.17 19.68 21.16 22.62 24.05 25.47 26.87 28.26 29.63 31.00 32.35 33.69 35.02 41.57 47.96 60.44 72.61 0.01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 21.67 23.21 24.72 26.22 27.69 29.14 30.58 32.00 33.41 34.81 36.19 37.57 44.31 50.89 63.69 76.15 0.005 7.88 10.60 0.001 10.83 13.82 12.84 14.86 16.75 18.55 20.28 21.95 23.59 25.19 26.76 28.30 29.82 31.32 32.80 34.27 35.72 37.16 38.58 40.00 46.93 53.67 66.77 79.49 16.27 18.47 20.52 22.46 24.32 26.12 27.88 29.59 31.26 32.91 34.53 36.12 37.70 39.25 40.79
42.31 43.82 45.31 52.62 59.70 73.40 86.66 0510152025051015202505101520250510152025 520 Index 5-number summary, 95 accurate, 259 Addition Rule of disjoint outcomes, 140 alternative hypothesis, 274, 284 anecdotal evidence, 28 ask, 506 associated, 20, 22 average, 27 bar chart segmented, 119 side-by-side, 119 bar chart, 116 Bayes’ Theorem, 167, 164–167 Bayesian statistics, 167 bias, 255 bimodal, 69 binomial coeﬃcient, 201, 212 binomial distribution, 212 binomial formula, 201, 212 bivariate, 59, 69 blind, 47 blocked experiment, 48, 48–50, 51 blocking, 48, 53 blocks, 48 box plot, 80, 95 side-by-side box plot, 91 case, 16 categorical, 19, 22, 131 census, 37 center, 95 Central Limit Theorem, 222, 234, 234–236, 239 proportion, 222 chi-square distribution, 328 χ2 goodness of ﬁt test, 326–338 chi-square probability table, 518 χ2-statistic, 338 chi-square table, 329, 518 χ2 test for homogeneity, 343–346, 353 χ2 test for independence, 347–350, 353 cluster random sample, 43 clusters, 40, 43 code comment, 256 cohort, 30 collections, 140 column totals, 116 combining random variables, 190 comparing distributions, 95 complement, 143, 147, 147 completely randomized experiment, 48, 48–50, 51 condition, 154 conditional probability, 154, 154–155, 167, 168 conditions are met, 284 conﬁdence interval, 253, 263, 263–269, 270 interpretation, 269 conﬁdence level, 266, 270 confounded, 34 confounder, 33 confounding factor, 33, 43 confounding variable, 33 contingency table, 116 column proportion, 117 column totals, 116 row proportions, 117 row totals, 116 continuous, 19, 22 control, 47 control group, 12, 14, 46 convenience sample, 36 correlation, 428, 428–429, 431 county, 503, 504, 508 county complete, 503, 508 cpr, 506 critical value, 267 data, 11, 503–508 approval ratings, 347 baby smoke, 402–404 Congress approval rating, 300–301 county, 17–21, 29, 89–92, 456 CPR and blood thinner, 312 dolphins and mercury, 369 email, 116–140, 143 email50, 59–92 health care, 316 loan50, 16–17, 59 loans, 116–121 malaria vaccine, 125–128 medical consultant, 273–278 midterm elections, 469–472 521 photo classify, 152–155 possum, 422–426 racial make-up of jury, 328, 332 run17samp, 230 SAT prep company, 387, 390 search algorithm, 344 smallpox, 156–163 stem cells, heart function, 406 stroke, 12–14, 19 supreme court, 295 textbooks, 383–385 data density, 65 data matrix, 17 decision errors, 284 deck of cards, 141 degrees of freedom (df) t-distribution, 363 chi-square, 328, 338 dependent, 20, 22, 29, 168 descriptive statistics, 69 deviation, 76 df, see degrees of freedom (df) direct control, 47 discrete, 19, 22, 223 discrete probability distribution, 190, 215 disjoint, 139, 139–141, 147 distribution, 61, 69 Bernoulli, 193, 193–194 binomial, 207, 200–211 normal approximation, 208–211 geometric, 195, 196, 195–197 normal, 101, 219 dot plot, 63, 69 double-blind, 47 ebola survey, 505 eﬀect size, 290 elmhurst, 508 email, 504, 505 email50, 504 empirical rule, 77, 108 error, 221, 232, 255 estimate, 254 event, 140, 140–141 E(X), 180 expectation, 180–181 expected count, 353 expected value, 180 experiment, 30 explained variance, 443, 451 explanatory variable, 29, 423, 431 exponentially, 195 extrapolation, 443, 451 face card, 141 factor, 46, 51 factorial, 201 failure, 193, 200 false negative, 165 false positive, 165 family college, 504 ﬁnite population correction factor, 222 ﬁrst quartile, 80 ﬁsh oil 18, 506 ﬁve-number summary, 80 frequency, 63 frequency histogram, 64 frequency table, 64 gambler’s fallacy, 160 gear company, 506 General Addition Rule, 142, 147 General Multiplication Rule, 157, 168 geometric distribution, 198 Greek beta (β), 463 mu (µ), 74 sigma (σ), 77 gsearch, 506 healthcare law survey, 506 heterogeneous, 40 high leverage, 448 histogram, 64 cumulative frequency, 66 hollow histogram, 91 homogeneous, 40 hypotheses, 284 hypothesis test, 253, 275, 284 logic of, 284 hypothesis testing, 273–282 decision errors, 280–281 p-value, 277 signiﬁcance level, 277, 282 statistically signiﬁcant, 277 independent, 21, 22, 29, 144, 147, 168, 190, 198, 212, 215 independent and identically distributed (iid), 195 indicator variable, 450 inference, 254, 260 inferential statistics, 69 inﬂuential point, 448, 451 intensity map, 92, 92 interquartile range (IQR), 80, 81, 95 joint probability, 153, 153–154 jury, 506 Law of Large Numbers, 138, 147, 175 leaf, 62 least squares regression, 437 extrapolation, 442–443 interpreting parameters, 441 R-squared (R2), 443, 443–444 least squares regression line, 451 left skewed, 69, 95 levels, 19, 46, 51 522 INDEX linear combination, 185 linear regression, 419 linear transformations of data, 95 loan50, 503, 504, 508 loans full schema, 503, 504 logic of a hypothesis test, 284 lower variability, 260 lurking variable, 33 machine learning (ML), 152 malaria, 504 margin of error, 268, 270, 300, 300–301, 307, 373 marginal probability, 153, 153–154 mariokart, 508 matched pairs, 50, 48–50 matched-pairs experiment, 51 mean, 27, 30, 73, 95, 190 average, 73 weighted mean, 74 median, 75, 95 midterm election, 469 midterms house, 508 minimum sample size, 307 modality bimodal, 68 multimodal, 68 unimodal, 68 mode, 68, 95 mosaic plot, 120 multimodal, 69 Multiplication Rule for independent events, 145 mutually exclusive, 139, 139–141, 147, 168 n choose x, 201 nba players 19, 504 ncbirths, 507 negative association, 21, 60, 69 nominal, 19 non-response, 37 non-response bias, 37 nonlinear, 60 normal, 307 normal approximation, 227 normal curve, 101 normal distribution, 101, 113, 219 standard, 102 normal probability plot, 109 normal probability table, 510 null hypothesis, 274, 284 null value, 275 numerical, 19, 22 outcome, 138 outcome of interest, 154 outlier, 63, 69, 81, 86, 95, 451 p-value, 276, 277, 284 paired, 69, 383 paired t-interval, see also t-interval for a mean of diﬀerences, 388–391 paired t-test, see also t-test for a mean of diﬀer- ences, 384–388 paired data, 59, 383–391 parameter, 27, 30, 102, 254, 274, 26–463 patients, 46 percentile, 104, 511 pew energy 2018, 505 pie chart, 121 placebo, 30, 47 placebo eﬀect, 47 playing cards, 504 point estimate, 254, 260, 274, 307, 254–307 single proportion, 295 poker, 505 pooled sample proportion, 317, 322 population, 26, 30, 26–38 population mean, 255 positive association, 21, 60, 69 possum, 504, 508 power, 282, 285 power analysis, 282 practically signiﬁcant, 283 precise, 259 prediction, 431 primary, 163 probability, 138, 147, 136–167, 254 probability distribution, 178 probability of a success, 193, 200 probability of failure 1 − p, 198 probability of success p , 198 probability sample, see sample probability table, 104 proportion, 27, 30 prospective study, 35 quartile ﬁrst quartile (Q1), 80 second quartile (Q2, median), 80 third quartile (Q3), 80 R, 256 R-squared, 451 random assignment, 53 random noise, 126 random numbers, 172 observational study, 30 observational unit, 16 one-sample t-interval, see t-interval for a mean one-sample t-test, see t-test for a mean one-sided, 275 ordinal, 19 psuedo-random numbers, 172 random process, 138, 138–139 random sampling, 53 random variable, 180, 177–189 combine, 190 randomization, 127 523 randomized experiment, 30 range, 76, 95 relative frequency, 67, 138, 147, 175 replication, 51 representative, 37 residual, 424, 424–427, 431 residual plot, 425, 431 response bias, 37 response variable, 29, 423, 431 resume, 507 retrospective studies, 35 right skewed, 69, 95 robust estimates, 87 row totals, 116 rule of complements, 147 run17, 505 run17samp, 505, 507 sample, 26, 30, 26–38 cluster sampling, 40 convenience sample, 36 multistage cluster sampling, 40 multistage sampling, 40 non-response, 37 non-response bias, 37 random sample, 35–38 simple random sampling, 38 strata, 40 stratiﬁed sampling, 40 systematic sampling, 38 sample mean, 255 sample proportion, 194, 270 sample space, 143 sample statistic, 86 sampling distribution, 219, 231, 239, 256 sampling error, 255 sat improve, 507 scatterplot, 20, 59, 69, 431 SD, see standard deviation SE, see standard error second quartile, 80 secondary, 163 sets, 140 shape, 67 shape of the sampling distribution, 307 side-by-side box plot, 91 signiﬁcance level, 277, 277, 282, 284 signiﬁcant, 14 simple random sample, 36, 43 simulated scatter, 508 simulation, 127, 172, 172–175, 279 single-blind, 47 skew left skewed, 67 right skewed, 67 strongly skewed guideline, 236 symmetric, 67 slope, 451 smallpox, 504 spread, 80, 95 standard deviation, 76, 95, 182, 190 standard deviation of the residuals, 425 standard error, 258, 260, 307 single mean, 233 single proportion, 294 standard normal distribution, 102 standard units, 79 statistic, 27, 30, 26–30 statistically signiﬁcant, 14, 128, 277, 284 stem, 62 stem-and-leaf plot, 62, 69 split stem-and-leaf plot, 62 stem cells, 507 stent30, 503 stent365, 503 stocks 18, 504 strata, 40, 43 stratiﬁed random sample, 43 stratifying, 53 study participants, 46 success, 193, 200 success-failure condition, 222, 227, 294 suits, 141 summary statistic, 13, 14, 20, 86 symmetric, 67, 69 systematic random sample, 43 t-distribution, 363–366 t-interval for a diﬀerence of means, 399, 397–402, 409 for a mean, 370, 367–372, 379 for a mean of diﬀerences, 389, 388–391, 392 for a slope, 467, 477, 465–477 t-probability table, 514 T-statistic, 375 t-table, 364, 514 t-test for a diﬀerence of means, 405, 402–408, 409 for a mean, 376, 374–378, 379 for a mean of diﬀerences, 386, 384–388, 392 for a slope, 473, 469–475, 477 table proportions, 153 tail, 67 test statistic, 104, 284 textbooks, 507, 508 the ﬁrst success on the xth trial, 198 third quartile, 80 time series, 464 time series data, 236 transform, 458, 460 transformation, 456, 458, 460 transplant, 505 treatment, 50 treatment group, 12, 14, 46 tree diagram, 163, 163–167 trial, 193, 200 524 INDEX two-proportion Z-interval, see Z-interval for a dif- ference of proportions two-proportion Z-test, see Z-test for a diﬀerence of proportions two-sample t-interval, see t-interval for a diﬀer- ence of means two-sampl
e t-test, see t-test for a diﬀerence of means two-sided, 275 two-way table, 353 Type I Error, 281, 284 Type II Error, 281, 284 ucla textbooks f18, 507, 508 unbiased, 223, 260 unconditional probability, 168 undercoverage bias, 36 unimodal, 69 unit of observation, 16 univariate, 61, 69 variability, 76, 80 variable, 16, 27 variance, 76, 182 Venn diagrams, 141 volunteer sample, 37 volunteers, 46 weighted mean, 74 whiskers, 81 with replacement, 159 without replacement, 158, 168 y-intercept, 451 Z-interval for a diﬀerence of proportions, 313, 312–315, 322 for a proportion, 297, 295–299, 307 Z-score, 79, 95, 113, 227 Z-test for a diﬀerence of proportions, 319, 316–321, 322 for a proportion, 304, 302–306, 307 525 Appendix D Technology reference, Formulas, and Inference guide D.1 Technology reference Instructions for the TI-83/84 and the Casio fx-9750GII, and their associated videos. Summarizing 1-variable statistics Entering data Calculating summary statistics. Drawing a box plot Finding normal probabilities Finding area under the normal curve Finding a Z-score that corresponds to a percentile Binomial probabilities Computing the binomial coeﬃcient Computing the binomial formula Computing cumulative binomial probabilities Inference for a single proportion 1-proportion Z-interval 1-proportion Z-test Inference for a diﬀerence of proportions 2-proportion Z-interval 2-proportion Z-test Chi-square for one-way tables Finding area under chi-square curve Chi-square goodness of ﬁt test Chi-square for two-way tables Entering data in a two-way table Chi-square test for homogeneity and independence Finding the expected counts Inference for a single mean 1-sample t-test 1-sample t-interval Inference for a mean of diﬀerences 1-sample t-test with paired data 1-sample t-interval with paired data Inference for a diﬀerence of means 2-sample t-interval 2-sample t-test The least squares regression line Finding the y-intercept, slope, r, and R2 What to do if you get Dim Mismatch Inference for the slope of a regression line page 83 page 84 page 84 page 110 page 112 page 205 page 206 page 206 page 299 page 306 page 315 page 321 page 331 page 337 page 351 page 351 page 351 page 378 page 372 page 388 page 391 page 401 page 407 page 445 page 446 t-interval for the slope t-test for the slope see Graphing Calculator Guides at openintro.org/ahss see Graphing Calculator Guides at openintro.org/ahss D.2 Formulas Descriptive Statistics xi n xi = ¯x = 1 n sx = 1 n − 1 (xi − ¯x)2 r = 1 n − 1 xi − ¯x yi − ¯y sx sy ˆy = a + bx ¯y = a + b¯x b = r sy sx s = (yi − ˆyi)2 n − 2 Inferential Statistics Probability P (A ∪ B) = P (A) + P (B) − P (A ∩ B) P (A\|B) = P (A ∩ B) P (B) µX = E(X) = xi · P (xi) σX = (xi − µx)2 · P (xi) µ¯x = µ σ¯x = σ √ n If X has a binomial distribution with parameters n and p, then: P (X = x) = n x px(1 − p)n−x µX = np σX = np(1 − p) µ ˆp = p σ ˆp = p(1 − p) n standardized test statistic: point estimate − null value SE of estimate conﬁdence interval: point estimate ± critical value × SE of estimate parameter point estimate single proportion p ˆp SE of estimate ˆp(1− ˆp) n when H0: p = p0, use p0(1−p0) n diﬀ. of proportions p1 − p2 ˆp1 − ˆp2 ˆp1(1− ˆp1) n1 + ˆp2(1− ˆp2) n2 when H0: p1 = p2, use √ ˆpc (1− ˆpc ) 1 n1 + 1 n2 single mean µ mean of diﬀerences µdiﬀ ¯x ¯xdiﬀ diﬀerence of means µ1 − µ2 ¯x1 − ¯x2 slope of reg. line β b s√ n sdiﬀ√ ndiﬀ s2 1 n1 + s2 2 n2 s √ n−1 sx Chi-square test statistic = (observed−expected)2 expected D.3 Inference Guide Ω INFERENCE GUIDE CONFIDENCE INTERVALS Use confidence intervals to estimate a parameter with a particular confidence level, C. IDENTIFY: Identify the parameter and the confidence level. CHOOSE: Choose and name the appropriate interval. CHECK: Check that conditions for the procedure are met. CALCULATE: 𝐂𝐈: 𝐩𝐨𝐢𝐧𝐭 𝐞𝐬𝐭𝐢𝐦𝐚𝐭𝐞 ± 𝐜𝐫𝐢𝐭𝐢𝐜𝐚𝐥 𝐯𝐚𝐥𝐮𝐞 × 𝑺𝑬 𝐨𝐟 𝐞𝐬𝐭𝐢𝐦𝐚𝐭𝐞 𝑑𝑓 = (if applicable) ( ____ , ____ ) CONCLUDE: We are C% confident that the true [parameter] is between ____ and ____ . (Put the parameter in context.) We have evidence that […], because […]. OR We do not have evidence that […], because […]. When the parameter is: a single proportion p CHOOSE: 1-Proportion Z-Interval to estimate 𝑝, or 1-Proportion Z-Test to test 𝐻0: 𝑝=𝑝0. CHECK: - Data come from a random sample or process. - for CI: 𝑛𝑝̂≥10 and 𝑛(1−𝑝̂)≥10. for Test: 𝑛𝑝0≥10 and 𝑛(1−𝑝0)≥10. CALCULATE: (1-PropZInt or 1-PropZTest) point estimate: sample proportion 𝑝̂ SE of estimate: for CI, use √𝑝̂(1−𝑝̂)𝑛 ; for Test, use √𝑝0(1−𝑝0)𝑛 When the parameter is: a difference of proportions p1-p2 CHOOSE: 2-Proportion Z-Interval to estimate 𝑝1−𝑝2, or 2-Proportion Z-Test to test 𝐻0: 𝑝1=𝑝2. CHECK: - Data come from 2 independent random samples or 2 randomly assigned treatments. - 𝑛1𝑝̂1≥10, 𝑛1(1−𝑝̂1)≥10, 𝑛2𝑝̂2≥10, 𝑛2(1−𝑝̂2)≥10. Note: use 𝑝̂𝑐, the pooled proportion, in place of 𝑝̂1 and 𝑝̂2 when checking condition for the 2-Proportion Z-Test CALCULATE: (2-PropZInt or 2-PropZTest) point estimate: difference of sample proportions 𝑝̂1−𝑝̂2 SE of estimate: CI, use√𝑝̂1(1−𝑝̂1)𝑛1+𝑝̂2(1−𝑝̂2)𝑛2 ; Test, use√𝑝̂𝑐(1−𝑝𝑐̂) √1𝑛1+1𝑛2 HYPOTHESIS TESTS Use hypothesis tests to test 𝐻0 versus 𝐻𝐴 at a particular significance level, α. IDENTIFY: Identify the hypotheses and the significance level. CHOOSE: Choose and name the appropriate test. CHECK: Check that conditions for the procedure are met. CALCULATE: 𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝𝐢𝐳𝐞𝐝 𝐭𝐞𝐬𝐭 𝐬𝐭𝐚𝐭𝐢𝐬𝐭𝐢𝐜=𝐩𝐨𝐢𝐧𝐭 𝐞𝐬𝐭𝐢𝐦𝐚𝐭𝐞− 𝐧𝐮𝐥𝐥 𝐯𝐚𝐥𝐮𝐞𝑺𝑬 𝐨𝐟 𝐞𝐬𝐭𝐢𝐦𝐚𝐭𝐞 𝑑𝑓 = (if applicable) p-value = CONCLUDE: p-value < α, so we reject 𝐻0. We have evidence that [𝐻𝐴]. (Put 𝐻𝐴 in context.) OR p-value > α, so we do NOT reject 𝐻0. We do NOT have evidence that [𝐻𝐴]. (Put 𝐻𝐴 in context.) When the parameter is: a single mean μ CHOOSE: 1-Sample T-Interval to estimate 𝜇, or 1-Sample T-Test to test 𝐻0: 𝜇=𝜇0. CHECK: - Data come from a random sample or process. - 𝑛≥30, OR population known to be nearly normal, OR population could to be nearly normal because data has no excessive skew or outliers (draw graph). CALCULATE: (TInterval or T-Test) point estimate: sample mean 𝑥̅ SE of estimate: 𝑠√𝑛 𝑑𝑓=𝑛−1 When the parameter is: a difference of means μ1-μ2 CHOOSE: 2-Sample T-Interval to estimate 𝜇1−𝜇2, or 2-Sample T-Test to test 𝐻0: 𝜇1=𝜇2. CHECK: - Data come from 2 independent random samples or 2 randomly assigned treatments. - 𝑛1≥30 and 𝑛2≥30, OR both populations known to be nearly normal, OR both populations could be nearly normal because both data sets have no excessive skew or outliers (draw 2 graphs). CALCULATE: (2-SampTInt or 2-SampTTest) point estimate: difference of sample means 𝑥̅1−𝑥̅2 SE of estimate: √𝑠12𝑛1+𝑠22𝑛2 𝑑𝑓: use technology When the parameter is: a mean of differences μdiff CHOOSE: 1-Sample T-Interval to estimate 𝜇abcc, or 1-Sample T-Test to test 𝐻<: 𝜇abcc = 0.CHECK: -There is paired data from a random sample or matchedpairs experiment.-𝑛abcc ≥ 30, OR population of differences known to benearly normal, OR population of differences could benearly normal because observed differences have noexcessive skew or outliers (draw graph of differences).CALCULATE: (TInterval or T-Test) point estimate: mean of sample difference 𝑥̅abcc SE of estimate: _deffVNdeff𝑑𝑓=𝑛abcc−1 When the parameter is: the slope β of a regression lineCHOOSE: T-Interval for the slope to estimate 𝛽, or T-Test for the slope to test 𝐻<: 𝛽 = 0.CHECK: -There is (x, y) data from a random sample or experiment.-The residual plot shows no pattern making a linearmodel reasonable. (More specifically, the residualsshould be independent, nearly normal, and haveconstant standard deviation.)CALCULATE: (LinRegTInt or LinRegTTest) point estimate: sample slope 𝑏 SE of estimate: SE of slope (from computer output) 𝑑𝑓=𝑛−2 The χ2 tests for categorical variables: chi-square statistic =∑(𝐨𝐛𝐬𝐞𝐫𝐯𝐞𝐝 M 𝐞𝐱𝐩𝐞𝐜𝐭𝐞𝐝)𝟐𝐞𝐱𝐩𝐞𝐜𝐭𝐞𝐝 When comparing the distribution of one categorical variable to a fixed/specified population distribution CHOOSE: χ2 Goodness of Fit Test CHECK: -Data come from a random sample or process.-All expected counts ≥ 5. (To calculate expected counts for each category, multiply the sample size by the expected proportion under 𝐻<.) CALCULATE: (χ2GOF-Test) 𝜒P = 𝑑𝑓= # of categories – 1 When comparing the distribution of a categorical variable across 2 or more populations/treatments CHOOSE: χ2 Test for Homogeneity CHECK: -Data come from 2 or more independent random samples or 2 or more randomly assigned treatments.-All expected counts ≥ 5. (Calculate expected counts and verify this to be true.) CALCULATE: (χ2-Test, then 2ND MATRIX,EDIT,2:[B] to find expected counts) 𝜒P = 𝑑𝑓 = (# of rows – 1)(# of cols – 1) When looking for association or dependence between two categorical variables CHOOSE: χ2 Test for Independence CHECK: -Data come from a random sample or process.-All expected counts ≥ 5. (Calculate expected counts and verify this to be true.) CALCULATE: (χ2-Test, then 2ND MATRIX,EDIT,2:[B] to find expected counts) 𝜒P = 𝑑𝑓 = (# of rows – 1)(# of cols – 1) ______________________________________________________________________________________________________________
−3, −2, −1, The set of rational numbers is written as ⎧ positive integers zero 0, 1, 2, 3, ⋯ ⎨m ⎬. Notice from the definition that rational n \|m and n are integers and n ≠ 0 numbers are fractions (or quotients) containing integers in both the numerator and the denominator, and the denominator is never 0. We can also see that every natural number, whole number, and integer is a rational number with a denominator of 1. ⎭ ⎫ ⎩ Because they are fractions, any rational number can also be expressed in decimal form. Any rational number can be represented as either: 1. a terminating decimal: 15 8 = 1.875, or This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 11 2. a repeating decimal: 4 11 = 0.36363636 … = 0.36 We use a line drawn over the repeating block of numbers instead of writing the group multiple times. Example 1.1 Writing Integers as Rational Numbers Write each of the following as a rational number. a. 7 b. 0 c. –8 Solution Write a fraction with the integer in the numerator and 1 in the denominator. a. b. −8 = − 8 1 1.1 Write each of the following as a rational number. a. 11 b. 3 c. –4 Example 1.2 Identifying Rational Numbers Write each of the following rational numbers as either a terminating or repeating decimal. a. − 5 7 b. c. 15 5 13 25 Solution Write each fraction as a decimal by dividing the numerator by the denominator. a. − 5 7 ——— = −0.714285 , a repeating decimal 12 Chapter 1 Prerequisites b. c. 15 5 13 25 = 3 (or 3.0), a terminating decimal = 0.52, a terminating decimal 1.2 Write each of the following rational numbers as either a terminating or repeating decimal. a. b. 68 17 8 13 c. − 17 20 Irrational Numbers At some point in the ancient past, someone discovered that not all numbers are rational numbers. A builder, for instance, may have found that the diagonal of a square with unit sides was not 2 or even 3 , but was something else. Or a garment 2 maker might have observed that the ratio of the circumference to the diameter of a roll of cloth was a little bit more than 3, but still not a rational number. Such numbers are said to be irrational because they cannot be written as fractions. These numbers make up the set of irrational numbers. Irrational numbers cannot be expressed as a fraction of two integers. It is impossible to describe this set of numbers by a single rule except to say that a number is irrational if it is not rational. So we write this as shown. {h\|h is not a rational number} Example 1.3 Differentiating Rational and Irrational Numbers Determine whether each of the following numbers is rational or irrational. If it is rational, determine whether it is a terminating or repeating decimal. a. b. c. d. e. 25 33 9 11 17 34 0.3033033303333 … Solution a. b. 25 : This can be simplified as 25 = 5. Therefore, 25 is rational. 33 9 : Because it is a fraction, 33 9 is a rational number. Next, simplify and divide. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 13 33 9 11 = 33 9 3 ¯ = 3. 6 = 11 3 So, 33 9 is rational and a repeating decimal. c. d. 11 : This cannot be simplified any further. Therefore, 11 is an irrational number. 17 34 : Because it is a fraction, 17 34 is a rational number. Simplify and divide. 17 34 1 = 17 34 2 = 1 2 = 0.5 So, 17 34 is rational and a terminating decimal. e. 0.3033033303333 … is not a terminating decimal. Also note that there is no repeating pattern because the group of 3s increases each time. Therefore it is neither a terminating nor a repeating decimal and, hence, not a rational number. It is an irrational number. Determine whether each of the following numbers is rational or irrational. If it is rational, determine 1.3 whether it is a terminating or repeating decimal. a. b. c. d. e. 7 77 81 4.27027002700027 … 91 13 39 Real Numbers Given any number n, we know that n is either rational or irrational. It cannot be both. The sets of rational and irrational numbers together make up the set of real numbers. As we saw with integers, the real numbers can be divided into three subsets: negative real numbers, zero, and positive real numbers. Each subset includes fractions, decimals, and irrational numbers according to their algebraic sign (+ or –). Zero is considered neither positive nor negative. The real numbers can be visualized on a horizontal number line with an arbitrary point chosen as 0, with negative numbers to the left of 0 and positive numbers to the right of 0. A fixed unit distance is then used to mark off each integer (or other basic value) on either side of 0. Any real number corresponds to a unique position on the number line.The converse is also true: Each location on the number line corresponds to exactly one real number. This is known as a one-to-one correspondence. We refer to this as the real number line as shown in Figure 1.2. Figure 1.2 The real number line Example 1.4 14 Chapter 1 Prerequisites Classifying Real Numbers Classify each number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 0 on the number line? a. − 10 3 b. 5 c. − 289 d. −6π e. 0.615384615384 … Solution a. − 10 3 is negative and rational. It lies to the left of 0 on the number line. b. 5 is positive and irrational. It lies to the right of 0. c. − 289 = − 172 = −17 is negative and rational. It lies to the left of 0. d. −6π is negative and irrational. It lies to the left of 0. e. 0.615384615384 … is a repeating decimal so it is rational and positive. It lies to the right of 0. Classify each number as either positive or negative and as either rational or irrational. Does the number lie 1.4 to the left or the right of 0 on the number line? a. 73 b. −11.411411411 … c. 47 19 d. − 5 2 e. 6.210735 Sets of Numbers as Subsets Beginning with the natural numbers, we have expanded each set to form a larger set, meaning that there is a subset relationship between the sets of numbers we have encountered so far. These relationships become more obvious when seen as a diagram, such as Figure 1.3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 15 Figure 1.3 Sets of numbers N: the set of natural numbers W: the set of whole numbers I: the set of integers Q: the set of rational numbers Q´: the set of irrational numbers Sets of Numbers The set of natural numbers includes the numbers used for counting: {1, 2, 3, ...}. The set of whole numbers is the set of natural numbers plus zero: {0, 1, 2, 3, ...}. The set of integers adds the negative natural numbers to the set of whole numbers: {..., −3, −2, −1, 0, 1, 2, 3, ...}. The set of rational numbers includes fractions written as ⎧ ⎨m n \|m and n are integers and n ≠ 0 ⎩ ⎫ ⎬. ⎭ The set of irrational numbers is the set of numbers that are not rational, are nonrepeating, and are nonterminating: {h\|h is not a rational number}. Example 1.5 Differentiating the Sets of Numbers Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q′). a. b. c. 36 8 3 73 d. −6 e. 3.2121121112 … Solution 16 Chapter 1 Prerequisites a. 36 = 6 ¯ = 2. 6 b. 8 3 c. 73 d. –6 e. 3.2121121112... ′ X X Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), 1.5 and/or irrational number (Q′). a. − 35 7 b. 0 c. d. e. 169 24 4.763763763 … Performing Calculations Using the Order of Operations When we multiply a number by itself, we square it or raise it to a power of 2. For example, 42 = 4 ⋅ 4 = 16. We can raise any number to any power. In general, the exponential notation an means that the number or variable a is used as a factor n times. an = a ⋅ a ⋅ a ⋅ … ⋅ a In this notation, an is read as the nth power of a, where a is called the base and n is called the exponent. A term in exponential notation may be part of a mathematical expression, which is a combination of numbers and operations. For example, 24 + 6 ⋅ 2 3 − 42 is a mathematical expression. n factors To evaluate a mathematical expression, we perform the various operations. However, we do not perform them in any random order. We use the order of operations. This is a sequence of rules for evaluating such expressions. Recall that in mathematics we use parentheses ( ), brackets [ ], and braces { } to group numbers and expressions so that anything appearing within the symbols is treated as a unit. Additionally, fraction bars, radicals, and absolute value bars are treated as grouping symbols. When evaluating a mathematical expression, begin by simplifying expressions within grouping symbols. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 17 The next step is to address any exponents or radicals. Afterward, perform multiplication and division from left to right and finally addition and subtraction from left to right. Let’s take a look at the expression provided. There are no grouping symbols, so we move on to exponents or radicals. The number 4 is raised to a power of 2, so simplify 42 as 16. 24 + 6 ⋅ 2 3 − 42 Next, perform multiplication or division, left to right. Lastly, perform addition or subtraction, left to right. 24 + 6 ⋅ 2 3 24 + 6 ⋅ 2 3 − 42 − 16 24 + 6 ⋅ 2 3 − 16 24 + 4 − 16 24 + 4 − 16 28 − 16 12 Therefore, 24 + 6 ⋅ 2 3 − 42 = 12. For some complicated expressions, several passes through the order of operations will be needed. For instance, there may be a radical expression inside parentheses that must be simplified before the parentheses are evaluated. Following the order of operations ensures that anyone simplifying the same mathematical expression will get the same result. Order of Operations Operations in mathematical expressions must be evaluated in a systematic order, which can be simplified using the acronym PEMDAS: P(arentheses) E(xponents) M(ultiplication) and D(ivision) A(ddition) and S(ubtraction) Given a mathematical expression, simplify
it using the order of operations. 1. Simplify any expressions within grouping symbols. 2. Simplify any expressions containing exponents or radicals. 3. Perform any multiplication and division in order, from left to right. 4. Perform any addition and subtraction in order, from left to right. Example 1.6 Using the Order of Operations Use the order of operations to evaluate each of the following expressions. 18 Chapter 1 Prerequisites a. b. c. d. e. (3 ⋅ 2)2 − 4(6 + 2) 52 − 4 7 − 11 − 2 6 − \|5 − 8\| + 3(4 − 1) 14 − 3 ⋅ 2 2 ⋅ 5 − 32 ⎡ ⎣(6 − 3) − 42⎤ 7(5 ⋅ 3) − 2 ⎦ + 1 Solution (3 ⋅ 2)2 − 4(6 + 2) = (6)2 − 4(8) = 36 − 4(8) = 36 − 32 = 4 Simplify parentheses Simplify exponent Simplify multiplication Simplify subtraction 52 7 − 11 − 2 = 52 − 4 7 = 52 − 4 7 = 25 − 4 7 − 3 = 21 7 = 3 − 3 = 0 − 9 Simplify grouping symbols (radical) − 3 − 3 Simplify radical Simplify exponent Simplify subtraction in numerator Simplify division Simplify subtraction Note that in the first step, the radical is treated as a grouping symbol, like parentheses. Also, in the third step, the fraction bar is considered a grouping symbol so the numerator is considered to be grouped. 6 − \|5 − 8\| + 3(4 − 1) = 6 − \|−3\| + 3(3) = 6 − 3 + 3(3 = 12 Simplify inside grouping symbols Simplify absolute value Simplify multiplication Simplify subtraction Simplify addition a. b. c. d. 14 − 3 ⋅ 2 2 ⋅ 5 − 32 = 14 − 3 ⋅ 2 2 ⋅ 5 − 9 = 14 − 6 10 − 9 = 8 1 = 8 Simplify exponent Simplify products Simplify diffe ences Simplify quotient In this example, the fraction bar separates the numerator and denominator, which we simplify separately until the last step. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 19 e. ⎡ ⎣(6 − 3) − 42⎤ 7(5 ⋅ 3) − 2 ⎡ ⎣(3) − 42⎤ ⎦ + 1 = 7(15) − 2 ⎦ + 1 Simplify inside parentheses = 7(15) − 2(3 − 16) + 1 = 7(15) − 2(−13) + 1 = 105 + 26 + 1 = 132 Simplify exponent Subtract Multiply Add 1.6 Use the order of operations to evaluate each of the following expressions. a. 52 − 42 + 7(5 − 4)2 b. d. e. \|1.8 − 4.3\| + 0.4 15 + 10 ⎡ ⎣5 ⋅ 32 − 72⎤ 1 ⋅ 92 2 ⎦ + 1 3 ⎤ ⎡ ⎣(3 − 8)2 − 4 ⎦ − (3 − 8) Using Properties of Real Numbers For some activities we perform, the order of certain operations does not matter, but the order of other operations does. For example, it does not make a difference if we put on the right shoe before the left or vice-versa. However, it does matter whether we put on shoes or socks first. The same thing is true for operations in mathematics. Commutative Properties The commutative property of addition states that numbers may be added in any order without affecting the sum. We can better see this relationship when using real numbers. a + b = b + a (−2) + 7 = 5 and 7 + (−2) = 5 Similarly, the commutative property of multiplication states that numbers may be multiplied in any order without affecting the product. Again, consider an example with real numbers. a ⋅ b = b ⋅ a (−11) ⋅ (−4) = 44 and (−4) ⋅ (−11) = 44 It is important to note that neither subtraction nor division is commutative. For example, 17 − 5 is not the same as 5 − 17. Similarly, 20 ÷ 5 ≠ 5 ÷ 20. Associative Properties The associative property of multiplication tells us that it does not matter how we group numbers when multiplying. We can move the grouping symbols to make the calculation easier, and the product remains the same. 20 Chapter 1 Prerequisites Consider this example. a(bc) = (ab)c (3 ⋅ 4) ⋅ 5 = 60 and 3 ⋅ (4 ⋅ 5) = 60 The associative property of addition tells us that numbers may be grouped differently without affecting the sum. This property can be especially helpful when dealing with negative integers. Consider this example. a + (b + c) = (a + b) + c ⎡ ⎣15 + (−9)⎤ ⎦ + 23 = 29 and 15 + ⎡ ⎣(−9) + 23⎤ ⎦ = 29 Are subtraction and division associative? Review these examples. 8 − (3 − 15) =? (8 − 3) − 15 8 − ( − 12) = 5 − 15 64 ÷ (8 ÷ 4) =? 64 ÷ 2 =? (64 ÷ 8) ÷ 4 8 ÷ 4 20 ≠ 20 − 10 32 ≠ 2 As we can see, neither subtraction nor division is associative. Distributive Property The distributive property states that the product of a factor times a sum is the sum of the factor times each term in the sum. a ⋅ (b + c) = a ⋅ b + a ⋅ c This property combines both addition and multiplication (and is the only property to do so). Let us consider an example. Note that 4 is outside the grouping symbols, so we distribute the 4 by multiplying it by 12, multiplying it by –7, and adding the products. To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example. 6 + (3 ⋅ 5) =? 6 + (15) =? (6 + 3) ⋅ (6 + 5) (9) ⋅ (11) 21 ≠ 99 Multiplication does not distribute over subtraction, and division distributes over neither addition nor subtraction. A special case of the distributive property occurs when a sum of terms is subtracted. a − b = a + (−b) For example, consider the difference 12 − (5 + 3). We can rewrite the difference of the two terms 12 and (5 + 3) by turning the subtraction expression into addition of the opposite. So instead of subtracting (5 + 3), we add the opposite. Now, distribute −1 and simplify the result. 12 + (−1) ⋅ (5 + 3) 12 − (5 + 3) = 12 + (−1) ⋅ (5 + 3) = 12 + [(−1) ⋅ 5 + (−1) ⋅ 3] = 12 + (−8) = 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 21 This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign of each term and add the results. With this in mind, we can rewrite the last example. 12 − (5 + 3) = 12 + (−5 − 3) = 12 + (−8) = 4 Identity Properties The identity property of addition states that there is a unique number, called the additive identity (0) that, when added to a number, results in the original number. a + 0 = a The identity property of multiplication states that there is a unique number, called the multiplicative identity (1) that, when multiplied by a number, results in the original number. For example, we have (−6) + 0 = −6 and 23 ⋅ 1 = 23. There are no exceptions for these properties; they work for every real number, including 0 and 1. a ⋅ 1 = a Inverse Properties The inverse property of addition states that, for every real number a, there is a unique number, called the additive inverse (or opposite), denoted−a, that, when added to the original number, results in the additive identity, 0. For example, if a = −8, the additive inverse is 8, since (−8) + 8 = 0. a + (−a) = 0 The inverse property of multiplication holds for all real numbers except 0 because the reciprocal of 0 is not defined. The property states that, for every real number a, there is a unique number, called the multiplicative inverse (or reciprocal), denoted 1 a, that, when multiplied by the original number, results in the multiplicative identity, 1. a ⋅ 1 a = 1 For example, if a = − 2 3 , the reciprocal, denoted 1 a, is − 3 2 because a ⋅ 1 a = ⎛ ⎝− 2 3 ⎞ ⎠ ⋅ ⎛ ⎝− 3 2 ⎞ ⎠ = 1 Properties of Real Numbers The following properties hold for real numbers a, b, and c. 22 Chapter 1 Prerequisites Commutative Property Associative Property Distributive Property Identity Property Inverse Property Addition Multiplication b + c) = (a + b) + c a(bc) = (ab)c a ⋅ (b + c) = a ⋅ b + a ⋅ c There exists a unique real number called the additive identity, 0, such that, for any real number a There exists a unique real number called the multiplicative identity, 1, such that, for any real number Every real number a has an additive inverse, or opposite, denoted –a, such that a + (−a) = 0 Every nonzero real number a has a multiplicative inverse, or reciprocal, denoted 1 a, such that ⎞ 1 ⎠ = 1 a a ⋅ ⎛ ⎝ Example 1.7 Using Properties of Real Numbers Use the properties of real numbers to rewrite and simplify each expression. State which properties apply. a. b. c. d. e. 3 ⋅ 6 + 3 ⋅ 4 (5 + 8) + (−8) 6 − (15 + 9 100 ⋅ ⎡ ⎣0.75 + (−2.38)⎤ ⎦ Solution a6 + 4) = 3 ⋅ 10 = 30 Distributive property Simplify Simplify This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 23 b. c. d. e. (5 + 8) + (−8) = 5 + [8 + (−8)] = 5 + 0 = 5 Associative property of addition Inverse property of addition Identity property of addition 6 − (15 + 9) = 6 + [(−15) + (−9)] = 6 + (−24) = −18 Distributive property Simplify Simplify Commutative property of multiplication Associative property of multiplication Inverse property of multiplication Identity property of multiplication 100 ⋅ [0.75 + ( − 2.38)] = 100 ⋅ 0.75 + 100 ⋅ (−2.38) = 75 + (−238) = −163 Distributive property Simplify Simplify 1.7 Use the properties of real numbers to rewrite and simplify each expression. State which properties apply. a. b. c. d. e. ⎛ ⎝− 23 5 ⎞ ⎠ ⋅ ⎡ ⎣11 ⋅ ⎛ ⎝− 5 23 ⎤ ⎞ ⎦ ⎠ 5 ⋅ (6.2 + 0.4) 18 − (7−15) 17 18 + ⋅ ⎡ ⎣ 4 9 + ⎛ ⎝− 17 18 ⎤ ⎞ ⎦ ⎠ 6 ⋅ (−3) + 6 ⋅ 3 Evaluating Algebraic Expressions So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see expressions such as x + 5, 4 πr 3, or 2m3 n2. In the expression x + 5, 5 is called a constant because it does not vary 3 and x is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division. We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are treated the same way. (−3)5 = (−3) ⋅ (−3) ⋅ (−3) ⋅ (−3) ⋅ (−3) (2 ⋅ 7)3 = (2 ⋅ 7) ⋅ (2 ⋅ 7) ⋅ (2 ⋅ 7) x5 = yz)3 = (yz) ⋅ (yz) ⋅ (yz) In each case, the exponent tells us how many factors of the base to use, whether the base consists of constants or variables. 24 Chapter 1 Prerequisites Any variable in an algebraic expression may take
on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value, then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before. Example 1.8 Describing Algebraic Expressions List the constants and variables for each algebraic expression. a. x + 5 b. c. πr 3 4 3 2m3 n2 Solution Constants Variables a. x + 5 5 πr 3 b. 2m3 n2 2 m, n 1.8 List the constants and variables for each algebraic expression. a. 2πr(r + h) b. 2(L + W) c. 4y3 + y Example 1.9 Evaluating an Algebraic Expression at Different Values Evaluate the expression 2x − 7 for each value for x. a. b. c This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 25 d. x = −4 Solution a. Substitute 0 for x. b. Substitute 1 for x. c. Substitute 1 2 for x. d. Substitute −4 for x. 2x − 7 = 2(0) − 7 = 0 − 7 = −7 2x − 7 = 2(1) − 7 = 2 − 7 = −5 2x − 6 2x − 7 = 2( − 4) − 7 = −8 − 7 = −15 1.9 Evaluate the expression 11 − 3y for each value for y. a. b. c. d5 Example 1.10 Evaluating Algebraic Expressions Evaluate each expression for the given values. a. b. c. d. e. x + 5 for x = −5 t 2t−1 for t = 10 πr 3 for r = 5 4 3 a + ab + b for a = 11, b = −8 2m3 n2 for m = 2, n = 3 26 Chapter 1 Prerequisites Solution a. Substitute −5 for x. b. Substitute 10 for t. x + 5 = (−5) + 5 = 0 t 2t − 1 = = (10) 2(10) − 1 10 20 − 1 = 10 19 c. Substitute 5 for r. 4 3 d. Substitute 11 for a and –8 for b. π(5)3 πr 3 = 4 3 = 4 π(125) 3 = 500 3 π e. Substitute 2 for m and 3 for n. a + ab + b = (11) + (11)(−8) + (−8) = 11 − 88 − 8 = −85 2m3 n2 = 2(2)3 (3)2 = 2(8)(9) = 144 = 12 1.10 Evaluate each expression for the given values. y + 3 y − 3 for y = 5 7 − 2t for t = −2 πr 2 for r = 11 1 3 3 ⎝p2 q⎞ ⎛ ⎠ for p = −2, q = 3 4(m − n) − 5(n − m) for . b. c. d. e. Formulas An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation 2x + 1 = 7 has the unique solution x = 3 because when we substitute 3 for x in the equation, we obtain the true statement 2(3) + 1 = 7. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 27 A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area A of a circle in terms of the radius r of the circle: A = πr 2. For any value of r, the area A can be found by evaluating the expression πr 2. Example 1.11 Using a Formula A right circular cylinder with radius r and height h has the surface area S (in square units) given by the formula S = 2πr(r + h). See Figure 1.4. Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of π. Figure 1.4 Right circular cylinder Solution Evaluate the expression 2πr(r + h) for r = 6 and h = 9. S = 2πr(r + h) = 2π(6)[(6) + (9)] = 2π(6)(15) = 180π The surface area is 180π square inches. 28 Chapter 1 Prerequisites 1.11 A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm2) is found to be A = (L + 16)(W + 16) − L ⋅ W. See Figure 1.5. Find the area of a matte for a photograph with length 32 cm and width 24 cm. Figure 1.5 Simplifying Algebraic Expressions Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions. Example 1.12 Simplifying Algebraic Expressions Simplify each algebraic expression. a. b. c. d. 3x − 2y + x − 3y − 7 2r − 5(3 − r) + 4 ⎛ ⎝4t − 5 4 s⎞ ⎠ − ⎛ ⎝ 2 3 t + 2s⎞ ⎠ 2mn − 5m + 3mn + n Solution a. b. 3x − 2y + x − 3y − 7 = 3x + x − 2y − 3y − 7 = 4x − 5y − 7 Commutative property of addition Simplify 2r − 5(3 − r) + 4 = 2r − 15 + 5r + 4 = 2r + 5y − 15 + 4 = 7r − 11 Distributive property Commutative property of addition Simplify This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 29 ⎛ ⎝t − 5 4t − 4 4 s⎞ ⎠ − ⎛ ⎝ 2 3 c. d. t + 2s⎞ t − 2s ⎠ = 4t − 5 4 = 4t − 2 3 t − 13 4 = 10 3 s − 2 3 t − 5 s − 2s 4 s Distributive property Commutative property of addition Simplify mn − 5m + 3mn + n = 2mn + 3mn − 5m + n = 5mn − 5m + n Commutative property of addition Simplify 1.12 Simplify each algebraic expression. a. b. c 4p⎛ ⎝q − 1⎞ ⎠ + q⎛ ⎝1 − p⎞ ⎠ d. 9r − (s + 2r) + (6 − s) Example 1.13 Simplifying a Formula A rectangle with length L and width W has a perimeter P given by P = L + W + L + W. Simplify this expression. Solution = 2L + 2W P = 2(L + W) Commutative property of addition Simplify Distributive property 1.13 If the amount P is deposited into an account paying simple interest r for time t, the total value of the deposit A is given by A = P + Prt. Simplify the expression. (This formula will be explored in more detail later in the course.) Access these online resources for additional instruction and practice with real numbers. • Simplify an Expression (http://openstaxcollege.org/l/simexpress) • Evaluate an Expression1 (http://openstaxcollege.org/l/ordofoper1) • Evaluate an Expression2 (http://openstaxcollege.org/l/ordofoper2) 30 Chapter 1 Prerequisites 1.1 EXERCISES Verbal 1. Is 2 an example of a rational terminating, rational repeating, or irrational number? Tell why it fits that category. 2. What is the order of operations? What acronym is used to describe the order of operations, and what does it stand for? What do the Associative Properties allow us to do when 3. following the order of operations? Explain your answer. Numeric 22. 23. (12 ÷ 3 × 3)2 25 ÷ 52 − 7 24. (15 − 7) × (3 − 7) 25. 2 × 4 − 9(−1) 26. 42 − 25 × 1 5 27. 12(3 − 1) ÷ 6 For the following exercises, simplify the given expression. Algebraic 4. 10 + 2 × (5 − 3) 5. 6. 7. 8. 9. 10. 11. 12. 6 ÷ 2 − ⎛ ⎝81 ÷ 32⎞ ⎠ 18 + (6 − 8)3 −2 × ⎡ ⎣16 ÷ (8 − 4)2(5 − 8) 4 + 6 − 10 ÷ 2 12 ÷ (36 ÷ 9) + 6 (4 + 5)2 ÷ 3 13. 3 − 12 × 2 + 19 14. 2 + 8 × 7 ÷ 4 15. 16. 5 + (6 + 4) − 11 9 − 18 ÷ 32 17. 14 × 3 ÷ 7 − 6 18. 9 − (3 + 11) × 2 19. 6 + 2 × 2 − 1 20. 64 ÷ (8 + 4 × 2) 21. ⎝22⎞ ⎛ 9 + 4 ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, solve for the variable. 28. 8(x + 3) = 64 29. 4y + 8 = 2y 30. (11a + 3) − 18a = −4 31. 4z − 2z(1 + 4) = 36 32. 33. 4y(7 − 2)2 = −200 −(2x)2 + 1 = −3 34. 8(2 + 4) − 15b = b 35. 2(11c − 4) = 36 36. 4(3 − 1)x = 4 37. 1 4 ⎛ ⎝8w − 42⎞ ⎠ = 0 For the following exercises, simplify the expression. 38. 4x + x(13 − 7) 39. 40. 2y − (4)2 y − 11 a 23(64) − 12a ÷ 6 41. 8b − 4b(3) + 1 42. 5l ÷ 3l × (9 − 6) 7z − 3 + z × 62 43. 44. Chapter 1 Prerequisites 4 × 3 + 18x ÷ 9 − 12 45. ⎝y + 8⎞ 9⎛ ⎠ − 27 46. ⎞ t − 4 ⎠2 ⎛ ⎝ 9 6 47. 6 + 12b − 3 × 6b 48. 49. 18y − 2⎛ ⎝1 + 7y⎞ ⎠ 2 ⎞ ⎠ ⎛ ⎝ 4 9 × 27x 50. 8(3 − m) + 1(−8) 51. 9x + 4x(2 + 3) − 4(2x + 3x) 52. 52 − 4(3x) For the following exercise, solve the given problem. 31 Ramon runs the marketing department at his company. 59. His department gets a budget every year, and every year, he must spend the entire budget without going over. If he spends less than the budget, then his department gets a smaller budget the following year. At the beginning of this year, Ramon got $2.5 million for the annual marketing budget. He must that 2,500,000 − x = 0. What property of addition tells us what the value of x must be? budget spend such the Technology For the following exercises, use a graphing calculator to solve for x. Round the answers to the nearest hundredth. 60. 61. 0.5(12.3)2 − 48x = 3 5 (0.25 − 0.75)2 x − 7.2 = 9.9 Real-World Applications Extensions If a whole number is not a natural number, what must 62. the number be? Determine whether the statement is true or false: The 63. multiplicative inverse of a rational number is also rational. Determine whether the statement is true or false: The 64. product of a rational and irrational number is always irrational. Determine whether 65. rational or irrational: −18 − 4(5)(−1). the simplified expression is Determine whether 66. rational or irrational: −16 + 4(5) + 5. the simplified expression is The division of two whole numbers will always result 67. in what type of number? What property of real numbers would simplify the 68. following expression: 4 + 7(x − 1) ? For the following exercises, consider this scenario: Fred earns $40 mowing lawns. He spends $10 on mp3s, puts half of what is left in a savings account, and gets another $5 for washing his neighbor’s car. Write the expression that represents the number of in his savings 53. dollars Fred keeps (and does not put account). Remember the order of operations. 54. How much money does Fred keep? For the following exercises, solve the given problem. According to the U.S. Mint, the diameter of a quarter is 55. 0.955 inches. The circumference of the quarter would be the diameter multiplied by π. Is the circumference of a quarter a whole number, a rational number, or an irrational number? Jessica and her roommate, Adriana, have decided to 56. share a change jar for joint expenses. Jessica put her loose change in the jar first, and then Adriana put her change in the jar. We know that it does not matter in which order the change was added to the jar. What property of addition desc
ribes this fact? For the following exercises, consider this scenario: There is a mound of g pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel is added to the mound. Two orders of 600 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,200 pounds of gravel. 57. Write the equation that describes the situation. 58. Solve for g. 32 Chapter 1 Prerequisites 1.2 \| Exponents and Scientific Notation Learning Objectives In this section students will: 1.2.1 Use the product rule of exponents. 1.2.2 Use the quotient rule of exponents. 1.2.3 Use the power rule of exponents. 1.2.4 Use the zero exponent rule of exponents. 1.2.5 Use the negative rule of exponents. 1.2.6 Find the power of a product and a quotient. 1.2.7 Simplify exponential expressions. 1.2.8 Use scientific notation. Mathematicians, scientists, and economists commonly encounter very large and very small numbers. But it may not be obvious how common such figures are in everyday life. For instance, a pixel is the smallest unit of light that can be perceived and recorded by a digital camera. A particular camera might record an image that is 2,048 pixels by 1,536 pixels, which is a very high resolution picture. It can also perceive a color depth (gradations in colors) of up to 48 bits per frame, and can shoot the equivalent of 24 frames per second. The maximum possible number of bits of information used to film a one-hour (3,600-second) digital film is then an extremely large number. Using a calculator, we enter 2,048 × 1,536 × 48 × 24 × 3,600 and press ENTER. The calculator displays 1.304596316E13. What does this mean? The “E13” portion of the result represents the exponent 13 of ten, so there are a maximum of approximately 1.3 × 1013 bits of data in that one-hour film. In this section, we review rules of exponents first and then apply them to calculations involving very large or small numbers. Using the Product Rule of Exponents Consider the product x3 ⋅ x4. Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression. 3 factors x3 ⋅ x4 = x ⋅ x ⋅ x 4 factors ⋅ = x7 7 factors The result is that x3 ⋅ x4 = x3 + 4 = x7. Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents. Now consider an example with real numbers. am ⋅ an = am + n 23 ⋅ 24 = 23 + 4 = 27 We can always check that this is true by simplifying each exponential expression. We find that 23 is 8, 24 is 16, and 27 is 128. The product 8 ⋅ 16 equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents. The Product Rule of Exponents For any real number a and natural numbers m and n, the product rule of exponents states that This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites Example 1.14 Using the Product Rule am ⋅ an = am + n 33 (1.1) Write each of the following products with a single base. Do not simplify further. a. b. c. t 5 ⋅ t 3 (−3)5 ⋅ (−3) x2 ⋅ x5 ⋅ x3 Solution Use the product rule to simplify each expression. a. b. c (−3)5 ⋅ (−3) = (−3)5 ⋅ (−3)1 = (−3)5 + 1 = (−3)6 x2 ⋅ x5 ⋅ x3 At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two. x2 ⋅ x5 ⋅ x3 = ⎝x2 ⋅ x5⎞ ⎛ ⎠ ⋅ x3 = ⎛ ⎝x2 + 5⎞ ⎠ ⋅ x3 = x7 ⋅ x3 = x7 + 3 = x10 Notice we get the same result by adding the three exponents in one step. x2 ⋅ x5 ⋅ x3 = x2 + 5 + 3 = x10 1.14 Write each of the following products with a single base. Do not simplify further. a. b. c Using the Quotient Rule of Exponents The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as ym yn , where m > n. Consider the example y9 y5. Perform the division by canceling common factors. 34 Chapter 1 Prerequisites y9 y5 = = = = y4 Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend. am an = am − n In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents. y9 y5 = y9 − 5 = y4 For the time being, we must be aware of the condition m > n. Otherwise, the difference m − n could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers. The Quotient Rule of Exponents For any real number a and natural numbers m and n, such that m > n, the quotient rule of exponents states that am an = am − n (1.2) Example 1.15 Using the Quotient Rule Write each of the following products with a single base. Do not simplify further. a. b. c. (−2)14 (−2)9 t 23 t 15 5 ⎛ ⎝z 2⎞ ⎠ z 2 Solution Use the quotient rule to simplify each expression. a. b. (−2)14 (−2)9 = (−2)14 − 9 = (−2)5 t 23 t 15 = t 23 − 15 = t 8 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 35 c. 5 ⎛ ⎝z 2⎞ ⎠ z 2 = ⎛ ⎝z 2⎞ ⎠ 5 − 1 = ⎛ ⎝z 2⎞ ⎠ 4 1.15 Write each of the following products with a single base. Do not simplify further. a. b. c. s75 s68 (−3)6 −3 5 3 ⎝e f 2⎞ ⎛ ⎠ ⎝e f 2⎞ ⎛ ⎠ Using the Power Rule of Exponents Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression ⎛ ⎝x2⎞ ⎠ 3 . The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3. 3 ⎛ ⎝x2⎞ ⎠ = 3 factors ⎛ ⎝x2⎞ ⎛ ⎝x2⎞ ⎠ ⋅ ⎠ ⋅ ⎛ ⎝x2⎞ ⎠ 3 factors ⎛ ⎧⎩⎨2 factors⎞ ⎜ x ⋅ x ⎟ ⋅ ⎝ ⎠ ⎛ ⎧⎩⎨2 factors⎞ ⎜ x ⋅ x ⎟ ⎝ ⎠ = ⎧⎩⎨2 factors = x6 The exponent of the answer is the product of the exponents: ⎛ ⎝x2⎞ ⎠ 3 = x2 ⋅ 3 = x6. In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents. (am ) = am ⋅ n n Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents. Product Rule 53 ⋅ 54 = 53 + 4 x5 ⋅ x2 = x5 + 2 (3a)7 ⋅ (3a)10 = (3a)7 + 10 = 57 = x7 = (3a)17 but but but (53)4 (x5)2 ((3a)7)10 Power Rule = = = = 512 = x10 53 ⋅ 4 x5 ⋅ 2 (3a)7 ⋅ 10 = (3a)70 The Power Rule of Exponents For any real number a and positive integers m and n, (am ) n the power rule of exponents states that = am ⋅ n (1.3) 36 Chapter 1 Prerequisites Example 1.16 Using the Power Rule Write each of the following products with a single base. Do not simplify further. a. b. c. 7 ⎝x2⎞ ⎛ ⎠ 3 ⎝(2t)5⎞ ⎛ ⎠ 11 ⎝(−3)5⎞ ⎛ ⎠ Solution Use the power rule to simplify each expression. a. b. c. 7 ⎝x2⎞ ⎛ ⎠ = x2 ⋅ 7 = x14 3 ⎝(2t)5⎞ ⎛ ⎠ = (2t)5 ⋅ 3 = (2t)15 11 ⎛ ⎝(−3)5⎞ ⎠ = (−3)5 ⋅ 11 = (−3)55 1.16 Write each of the following products with a single base. Do not simplify further. a. b. c. 3 ⎛ ⎛ ⎝3y⎞ ⎝ ⎠ 8⎞ ⎠ 7 ⎝t 5⎞ ⎛ ⎠ 4 ⎛ ⎝(−g)4⎞ ⎠ Using the Zero Exponent Rule of Exponents Return to the quotient rule. We made the condition that m > n so that the difference m − n would never be zero or negative. What would happen if m = n ? In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example If we were to simplify the original expression using the quotient rule, we would have If we equate the two answers, the result is t 0 = 1. This is true for any nonzero real number, or any variable representing a real number. a0 = 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 37 The sole exception is the expression 00. This appears later in more advanced courses, but for now, we will consider the value to be undefined. The Zero Exponent Rule of Exponents For any nonzero real number a, the zero exponent rule of exponents states that a0 = 1 (1.4) Example 1.17 Using the Zero Exponent Rule Simplify each expression using the zero exponent rule of exponents. a. c3 c3 b. −3x5 x5 c. d. 4 ⎛ ⎝j2 k⎞ ⎠ ⎛ ⎝j2 k⎞ ⎠ ⋅ ⎛ ⎝j2 k⎞ ⎠ 3 2 5 ⎛ ⎝rs2⎞ ⎠ 2 ⎛ ⎝rs2⎞ ⎠ Solution Use the zero exponent and other rules to simplify each expression. a. b. c3 c3 = c3 − 3 = c3 − 3 = c3 − 3 −3x5 x5 = −3 ⋅ x5 x5 = −3 ⋅ x5 − 5 = −3 ⋅ x0 = −3 ⋅ 1 = −3 38 Chapter 1 Prerequisites c. d. 4 ⎝j2 k⎞ ⎛ ⎠ ⎛ ⎝j2 k⎞ ⎠ ⋅ ⎝j2 k⎞ ⎛ ⎠ 3 = 4 ⎝j2 k⎞ ⎛ ⎠ 1 + 3 ⎝j2 k⎞ ⎛ ⎠ Use the product rule in the denominator. 4 4 ⎝j2 k⎞ ⎛ ⎠ ⎝j2 k⎞ ⎛ ⎠ 4 − 4 ⎝j2 k⎞ ⎛ ⎠ 0 ⎝j2 k⎞ ⎛ ⎠ = = = = 1 Simplify. Use the quotient rule. Simplify. 2 − 2 ⎝rs2⎞ ⎛ ⎠ Use the quotient rule. 2 5 ⎝rs2⎞ ⎛ ⎠ 2 = 5 ⎛ ⎝rs2⎞ ⎠ 0 = 5 ⎛ ⎝rs2⎞ ⎠ = 5 ⋅ 1 = 5 Simplify. Use the zero exponent rule. Simplify. 1.17 Simplify each expression using the zero exponent rule of exponents. a. b. t 7 t 7 11 11 ⎛ ⎝de2⎞ ⎠ ⎛ ⎝de2⎞ ⎠ 2 c. w4 ⋅ w2 w6 d Using the Negative Rule of Exponents Another useful result occurs if we relax the condition that m > n in the quotient rule even further. For example, can we simplify h3 h5 ? When m < n —that is, where the difference m − n is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, h3 h5. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequi
sites 39 h3 h5 = = = h2 If we were to simplify the original expression using the quotient rule, we would have h3 h5 = h3 − 5 h−2 = Putting the answers together, we have h−2 = 1 h2. This is true for any nonzero real number, or any variable representing a nonzero real number. A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa. an and an We have shown that the exponential expression an is defined when n is a natural number, 0, or the negative of a natural number. That means that an is defined for any integer n. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer n. = 1 a−n = 1 a−n The Negative Rule of Exponents For any nonzero real number a and natural number n, the negative rule of exponents states that a−n = 1 an (1.5) Example 1.18 Using the Negative Exponent Rule Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents. a. b. c. θ 3 θ 10 z2 ⋅ z z4 4 8 ⎝−5t 3⎞ ⎛ ⎠ ⎝−5t 3⎞ ⎛ ⎠ Solution 40 Chapter 1 Prerequisites a. b. c. θ 3 θ 10 = θ 3 − 10 = θ −7 = 1 θ 7 z2 + 1 z2 ⋅ z z4 = z4 = z3 z4 = z3 − 4 = z−1 = 1 z 4 ⎝−5t 3⎞ ⎛ ⎠ 8 = ⎝−5t 3⎞ ⎛ ⎠ ⎝−5t 3⎞ ⎛ ⎠ 4 − 8 ⎝−5t 3⎞ ⎛ ⎠ = −4 = 1 ⎝−5t 3⎞ ⎛ ⎠ 4 Write each of the following quotients with a single base. Do not simplify further. Write answers with 1.18 positive exponents. a. b. c. (−3t)2 (−3t)8 f 47 f 49 ⋅ f 2k 4 5k 7 Example 1.19 Using the Product and Quotient Rules Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents. a. b. c. b2 ⋅ b−8 (−x)5 ⋅ (−x)−5 −7z (−7z)5 Solution a. b. c. b2 ⋅ b−8 = b2 − 8 = b−6 = 1 b6 (−x)5 ⋅ (−x)−5 = (−x)5 − 5 = (−x)0 = 1 −7z (−7z)5 = (−7z)1 (−7z)5 = (−7z)1 − 5 = (−7z)−4 = 1 (−7z)4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 41 1.19 Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents. a. b. t −11 ⋅ t 6 2512 2513 Finding the Power of a Product To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider (pq)3. We begin by using the associative and commutative properties of multiplication to regroup the factors. 3 factors (pq)3 = (pq) ⋅ (pq) ⋅ (pq factors ⋅ q ⋅ q ⋅ q 3 factors = p ⋅ p ⋅ p = p3 ⋅ q3 In other words, (pq)3 = p3 ⋅ q3. The Power of a Product Rule of Exponents For any real numbers a and b and any integer n, the power of a product rule of exponents states that (ab) n = an bn (1.6) Example 1.20 Using the Power of a Product Rule Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents. a. b. c. d. e. 3 ⎛ ⎝ab2⎞ ⎠ (2t)15 3 ⎝−2w3⎞ ⎛ ⎠ 1 (−7z)4 7 ⎝e−2 f 2⎞ ⎛ ⎠ Solution Use the product and quotient rules and the new definitions to simplify each expression. 42 Chapter 1 Prerequisites a. b. c. d. e. 3 ⎝ab2⎞ ⎛ ⎠ = (a)3 ⋅ ⎝b2⎞ ⎛ ⎠ 3 = a1 ⋅ 3 ⋅ b2 ⋅ 3 = a3 b6 2t 15 = (2)15 ⋅ (t)15 = 215 t 15 = 32, 768t 15 3 ⎝−2w3⎞ ⎛ ⎠ = (−2)3 ⋅ ⎝w3⎞ ⎛ ⎠ 3 = −8 ⋅ w3 ⋅ 3 = −8w9 1 (−7z)4 = 1 (−7)4 ⋅ (z)4 = 1 2, 401z4 ⎝e−2 f 2⎞ ⎛ ⎠ 7 ⎝e−2⎞ ⎛ ⎠ = 7 7 ⎝ f 2⎞ ⎛ ⎠ ⋅ = e−−14 f 14 = f 14 e14 Simplify each of the following products as much as possible using the power of a product rule. Write 1.20 answers with positive exponents. a. b. c. d. 5 ⎝g2 h3⎞ ⎛ ⎠ (5t)3 3 ⎝−3y5⎞ ⎛ ⎠ 1 ⎝a6 b7⎞ ⎛ ⎠ 3 e. 4 ⎝r 3 s−2⎞ ⎛ ⎠ Finding the Power of a Quotient To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example. Let’s rewrite the original problem differently and look at the result. ⎝e−2 f 2⎞ ⎛ ⎠ 7 = f 14 e14 7 ⎝e−2 f 2 ⎜ e2 ⎝ f 14 e14 = = It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 43 7 ⎝e−2 f 2⎞ ⎛ ⎠ 7 ⎛ ⎞ f 2 ⎜ ⎟ e2 ⎝ ⎠ ( f 2)7 (e2)7 f 2 ⋅ 7 e2 ⋅ 7 f 14 e14 = = = = The Power of a Quotient Rule of Exponents For any real numbers a and b and any integer n, the power of a quotient rule of exponents states that n ⎛ ⎝ a b ⎞ ⎠ = an bn (1.7) Example 1.21 Using the Power of a Quotient Rule Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents. 3 ⎞ ⎠ 6 a. ⎛ 4 ⎝ z11 b. ⎛ ⎜ ⎝ p q3 ⎞ ⎟ ⎠ c. ⎛ −1 ⎝ t 2 27 ⎞ ⎠ d. e. 4 ⎝j3 k −2⎞ ⎛ ⎠ 3 ⎛ ⎝m−2 n−2⎞ ⎠ Solution a. ⎛ 4 ⎝ z11 3 ⎞ ⎠ = (4)3 ⎝z11⎞ ⎛ ⎠ 3 = 64 z11 ⋅ 3 = 64 z33 b. 6 ⎛ ⎜ ⎝ p q3 ⎞ ⎟ ⎠ = (p)6 ⎝q3⎞ ⎛ ⎠ 6 = p1 ⋅ 6 q3 ⋅ 6 = p6 q18 44 Chapter 1 Prerequisites c. ⎛ −1 ⎝ t 2 27 ⎞ ⎠ = (−1)27 ⎝t 2⎞ ⎛ ⎠ 27 = −1 t 2 ⋅ 27 = −1 t 54 = − 1 t 54 d. ⎝j3 k −2⎞ ⎛ ⎠ 4 = 4 ⎛ ⎜ ⎝ j3 j3⎞ ⎛ ⎠ ⎝k 2⎞ ⎛ ⎠ j3 ⋅ 4 k 2 ⋅ 4 = j12 k 8 e. ⎛ ⎝m−2 n−2⎞ ⎠ 3 = ⎛ 1 ⎝ m2 n2 3 ⎞ ⎠ = (1)3 ⎛ ⎝m2 n2⎞ ⎠ 3 = 1 3 ⎛ ⎝n2⎞ ⎠ 3 = ⎛ ⎝m2⎞ ⎠ 1 m2 ⋅ 3 ⋅ n2 ⋅ 3 = 1 m6 n6 Simplify each of the following quotients as much as possible using the power of a quotient rule. Write 1.21 answers with positive exponents. a. 3 b5 c ⎛ ⎝ ⎞ ⎠ b. ⎛ 5 ⎝ u8 4 ⎞ ⎠ c. ⎛ −1 ⎝ w3 35 ⎞ ⎠ d. e. 8 ⎝p−4 q3⎞ ⎛ ⎠ 4 ⎝c−5 d −3⎞ ⎛ ⎠ Simplifying Exponential Expressions Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions. Example 1.22 Simplifying Exponential Expressions Simplify each expression and write the answer with positive exponents only. a. b. 3 ⎝6m2 n−1⎞ ⎛ ⎠ 175 ⋅ 17−4 ⋅ 17−3 c. ⎛ u−1 v ⎝ v−1 2 ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 45 d. e. f. ⎝−2a3 b−1⎞ ⎛ ⎛ ⎝5a−2 b2⎞ ⎠ ⎠ ⎛ ⎝x2 2 ⎞ ⎠ 4 ⎛ ⎞ ⎝x2 2 ⎠ −4 5 ⎝3w2⎞ ⎛ ⎠ ⎝6w−2⎞ ⎛ ⎠ 2 Solution a. 3 ⎛ ⎝6m2 n−1⎞ ⎠ 3 3 ⎛ ⎝n−1⎞ ⎝m2⎞ = (6)3 ⎛ ⎠ ⎠ = 63 m2 ⋅ 3 n−1 ⋅ 3 = 216m6 n−3 = 216m6 n3 175 ⋅ 17−4 ⋅ 17−3 = 175 − 4 − 3 = 17−2 = 1 172 or 1 289 ⎛ u−1 v ⎝ v−1 2 ⎞ ⎠ = = (u−1 v)2 (v−1)2 u−2 v2 v−2 = u−2 v2 − (−2) = u−2 v4 v4 u2 = b. c. d. e. The power of a product rule The power rule Simplify. The negative exponent rule The product rule Simplify. The negative exponent rule The power of a quotient rule The power of a product rule The quotient rule Simplify. The negative exponent rule ⎛ ⎛ ⎝−2a3 b−1⎞ ⎝5a−2 b2⎞ ⎠ ⎠ = −2 ⋅ 5 ⋅ a3 ⋅ a−2 ⋅ b−1 ⋅ b2 = −10 ⋅ a3 − 2 ⋅ b−1 + 2 = −10ab ⎛ ⎝x2 2 ⎞ ⎠ 4 ⎛ ⎞ ⎝x2 2 ⎠ −4 = ⎛ ⎞ ⎝x2 2 ⎠ 4 − 4 = ⎛ ⎞ ⎝x2 2 ⎠ 0 = 1 Commutative and associative laws of multiplication The product rule Simplify. The product rule Simplify. The zero exponent rule 46 Chapter 1 Prerequisites f. (3w2)5 (6w−2)2 = (3)5 ⋅ (w2)5 (6)2 ⋅ (w−2)2 = 35 w2 ⋅ 5 62 w−2 ⋅ 2 = 243w10 36w−4 = 27w10 − (−4) 4 = 27w14 4 The power of a product rule The power rule Simplify. The quotient rule and reduce fraction Simplify. 1.22 Simplify each expression and write the answer with positive exponents only. a. b. c. d. e. f. −3 ⎛ ⎝2uv−2⎞ ⎠ x8 ⋅ x−12 ⋅ x 2 ⎛ ⎜ ⎝ e2 f −3 f −1 ⎞ ⎟ ⎠ ⎛ ⎝9r −5 s3⎞ ⎝3r 6 s−4⎞ ⎛ ⎠ ⎠ tw−2⎞ ⎠ ⎛ ⎝ 4 9 −3 ⎛ ⎝ 4 9 3 tw−2⎞ ⎠ 4 ⎛ ⎝2h2 k⎞ ⎠ ⎛ ⎝7h−1 k 2⎞ ⎠ 2 Using Scientific Notation Recall at the beginning of the section that we found the number 1.3 × 1013 when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these? A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number, n is positive. If you moved the decimal right as in a small large number, n is negative. For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 47 We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number. 2.780418 × 106 Working with small numbers is similar. Take, for example, the radius of an electron, 0.00000000000047 m. Perform the same series of steps as above, except move the decimal point to the right. Be careful not to include the leading 0 in your count. We move the decimal point 13 places to the right, so the exponent of 10 is 13. The exponent is negative because we moved the decimal point to the right. This is what we should expect for a small number. 4.7 × 10−13 Scientific Notation A number is written in scientific notation if it is written in the form a × 10 n , where 1 ≤ \|a\| < 10 and n is an integer. Example 1.23 Converting Standard Notation to Scientific Notation Write each number in scientific notation. a. Distance to Andromeda Galaxy from Earth: 24,000,000,000,000,000,000,000 m b. Diameter of Andromeda Galaxy: 1,300,000,000,000,000,000,000 m c. Number of stars in Andromeda Galaxy: 1,000,000,000,000 d. Diameter of electron: 0.00000000000094 m e. Probability of being struck by lightning in any single year: 0.00000143 Solution a. b. c. 24,000,000,000,000,000,000,000 m 24,000,000,000,000,000,000,000 m ← 22 places 2.4 × 1022 m
1,300,000,000,000,000,000,000 m 1,300,000,000,000,000,000,000 m ← 21 places 1.3 × 1021 m 1,000,000,000,000 1,000,000,000,000 ← 12 places 1 × 1012 48 Chapter 1 Prerequisites d. e. 0.00000000000094 m 0.00000000000094 m → 6 places 9.4 × 10−13 m 0.00000143 0.00000143 → 6 places 1.43 × 10−6 Analysis Observe that, if the given number is greater than 1, as in examples a–c, the exponent of 10 is positive; and if the number is less than 1, as in examples d–e, the exponent is negative. 1.23 Write each number in scientific notation. a. U.S. national debt per taxpayer (April 2014): $152,000 b. World population (April 2014): 7,158,000,000 c. World gross national income (April 2014): $85,500,000,000,000 d. Time for light to travel 1 m: 0.00000000334 s e. Probability of winning lottery (match 6 of 49 possible numbers): 0.0000000715 Converting from Scientific to Standard Notation To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal n places to the right if n is positive or n places to the left if n is negative and add zeros as needed. Remember, if n is positive, the value of the number is greater than 1, and if n is negative, the value of the number is less than one. Example 1.24 Converting Scientific Notation to Standard Notation Convert each number in scientific notation to standard notation. a. 3.547 × 1014 b. −2 × 106 c. 7.91 × 10−7 d. −8.05 × 10−12 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 49 a. b. c. d. 3.547 × 1014 3.54700000000000 → 14 places 354,700,000,000,000 −2 × 106 −2.000000 → 6 places −2,000,000 7.91 × 10−7 0000007.91 → 7 places 0.000000791 −8.05 × 10−12 −000000000008.05 → 12 places −0.00000000000805 1.24 Convert each number in scientific notation to standard notation. a. 7.03 × 105 b. −8.16 × 1011 c. −3.9 × 10−13 d. 8 × 10−6 Using Scientific Notation in Applications Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around 1.32 × 1021 molecules of water and 1 L of water holds about 1.22 × 104 average drops. Therefore, there are approximately ⎠ ≈ 4.83 × 1025 atoms in 1 L of water. We simply multiply the decimal terms and add the 3 ⋅ exponents. Imagine having to perform the calculation without using scientific notation! ⎝1.32 × 1021⎞ ⎛ ⎠ ⋅ ⎝1.22 × 104⎞ ⎛ When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product ⎛ ⎠ = 35 × 1010. The answer is not in proper scientific notation because ⎝5 × 106⎞ ⎛ 35 is greater than 10. Consider 35 as 3.5 × 10. That adds a ten to the exponent of the answer. ⎝7 × 104⎞ ⎠ ⋅ (35) × 1010 = (3.5 × 10) × 1010 = 3.5 × ⎛ ⎝10 × 1010⎞ ⎠ = 3.5 × 1011 50 Chapter 1 Prerequisites Example 1.25 Using Scientific Notation Perform the operations and write the answer in scientific notation. a. b. c. d. e. ⎝6.5 × 1010⎞ ⎛ ⎝8.14 × 10−7⎞ ⎛ ⎠ ⎠ ⎝4 × 105⎞ ⎛ ⎠ ÷ ⎛ ⎝−1.52 × 109⎞ ⎠ ⎝2.7 × 105⎞ ⎛ ⎝6.04 × 1013⎞ ⎛ ⎠ ⎠ ⎝1.2 × 108⎞ ⎛ ⎠ ÷ ⎛ ⎝9.6 × 105⎞ ⎠ ⎝5.62 × 105⎞ ⎛ ⎛ ⎝3.33 × 104⎞ ⎝−1.05 × 107⎞ ⎛ ⎠ ⎠ ⎠ Solution a. b. c. d. e. ⎝6.5 × 1010⎞ ⎛ ⎝8.14 × 10−7⎞ ⎛ ⎠ ⎝10−7 × 1010⎞ ⎛ ⎠ = (8.14 × 6.5) ⎠ ⎝103⎞ ⎛ = (52.91) ⎠ = 5.291 × 104 ⎝4 × 105⎞ ⎛ ⎠ ÷ ⎛ ⎝−1.52 × 109⎞ ⎠ = ⎛ ⎝ 4 −1.52 ⎛ ⎞ 105 ⎞ ⎠ ⎝ ⎠ 109 ⎛ ⎝10−4⎞ ≈ (−2.63) ⎠ = −2.63 × 10−4 ⎝2.7 × 105⎞ ⎛ ⎝6.04 × 1013⎞ ⎛ ⎠ ⎝105 × 1013⎞ ⎛ ⎠ = (2.7 × 6.04) ⎠ ⎝1018⎞ ⎛ = (16.308) ⎠ = 1.6308 × 1019 ⎝1.2 × 108⎞ ⎛ ⎠ ÷ ⎛ ⎝9.6 × 105⎞ ⎠ = ⎛ ⎝ 1.2 9.6 ⎞ ⎛ 108 ⎞ ⎠ ⎠ ⎝ 105 ⎝103⎞ ⎛ = (0.125) ⎠ = 1.25 × 102 Commutative and associative properties of multiplication Product rule of exponents Scientific n tation Commutative and associative properties of multiplication Quotient rule of exponents Scientific n tation Commutative and associative properties of multiplication Product rule of exponents Scientific n tation Commutative and associative properties of multiplication Quotient rule of exponents Scientific n tation ⎝5.62 × 105⎞ ⎛ ⎝3.33 × 104⎞ ⎛ ⎝−1.05 × 107⎞ ⎛ ⎠ ⎠ ⎝104 × 107 × 105⎞ ⎛ ⎠ = [3.33 × (−1.05) × 5.62] ⎠ ⎝1016⎞ ⎛ ≈ (−19.65) ⎠ = −1.965 × 1017 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 51 1.25 Perform the operations and write the answer in scientific notation. a. b. c. d. e. ⎝−7.5 × 108⎞ ⎛ ⎛ ⎝1.13 × 10−2⎞ ⎠ ⎠ ⎝1.24 × 1011⎞ ⎛ ⎠ ÷ ⎝1.55 × 1018⎞ ⎛ ⎠ ⎛ ⎝3.72 × 109⎞ ⎝8 × 103⎞ ⎛ ⎠ ⎠ ⎝9.933 × 1023⎞ ⎛ ⎠ ÷ ⎛ ⎝−2.31 × 1017⎞ ⎠ ⎝−6.04 × 109⎞ ⎛ ⎝−2.81 × 102⎞ ⎛ ⎝7.3 × 102⎞ ⎛ ⎠ ⎠ ⎠ Example 1.26 Applying Scientific Notation to Solve Problems In April 2014, the population of the United States was about 308,000,000 people. The national debt was about $17,547,000,000,000. Write each number in scientific notation, rounding figures to two decimal places, and find the amount of the debt per U.S. citizen. Write the answer in both scientific and standard notations. Solution The population was 308,000,000 = 3.08 × 108. The national debt was $17,547,000,000,000 ≈ $1.75 × 1013. To find the amount of debt per citizen, divide the national debt by the number of citizens. ⎝1.75 × 1013⎞ ⎛ ⎠ ÷ ⎛ ⎝3.08 × 108⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⋅ ⎛ 1013 1.75 ⎝ 108 3.08 ≈ 0.57 × 105 = 5.7 × 104 ⎞ ⎠ The debt per citizen at the time was about $5.7 × 104, or $57,000. 1.26 An average human body contains around 30,000,000,000,000 red blood cells. Each cell measures approximately 0.000008 m long. Write each number in scientific notation and find the total length if the cells were laid end-to-end. Write the answer in both scientific and standard notations. Access these online resources for additional instruction and practice with exponents and scientific notation. • Exponential Notation (http://openstaxcollege.org/l/exponnot) • Properties of Exponents (http://openstaxcollege.org/l/exponprops) • Zero Exponent (http://openstaxcollege.org/l/zeroexponent) • Simplify Exponent Expressions (http://openstaxcollege.org/l/exponexpres) • Quotient Rule for Exponents (http://openstaxcollege.org/l/quotofexpon) • Scientific Notation (http://openstaxcollege.org/l/scientificnota) • Converting to Decimal Notation (http://openstaxcollege.org/l/decimalnota) 52 Chapter 1 Prerequisites 1.2 EXERCISES Verbal 69. Is 23 the same as 32 ? Explain. 70. When can you add two exponents? 5 ⎝33 ÷ 34⎞ ⎛ ⎠ the following exercises, express the decimal For scientific notation. in 71. What is the purpose of scientific notation? 72. Explain what a negative exponent does. 89. 0.0000314 90. 148,000,000 Numeric For the following exercises, simplify the given expression. Write answers with positive exponents. 73. 92 74. 15−2 75. 76. 77. 32 × 33 44 ÷ 4 −2 ⎛ ⎝22⎞ ⎠ 78. (5 − 8)0 79. 80. 81. 113 ÷ 114 65 × 6−7 2 ⎝80⎞ ⎛ ⎠ 82. 5−2 ÷ 52 For the following exercises, write each expression with a single base. Do not simplify further. Write answers with positive exponents. 42 × 43 ÷ 4−4 612 69 ⎛ ⎝123 × 12 ⎞ ⎠ 10 106 ÷ ⎝1010⎞ ⎛ ⎠ −2 7−6 × 7−3 83. 84. 85. 86. 87. 88. This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, convert each number in scientific notation to standard notation. 91. 92. 1.6 × 1010 9.8 × 10−9 Algebraic For the following exercises, simplify the given expression. Write answers with positive exponents. 93. 94. a3 a2 a mn2 m−2 95. 2 ⎝b3 c4⎞ ⎛ ⎠ 96. 97. 98. 99. −5 ⎛ ⎜x−3 y2 ⎝ ⎞ ⎟ ⎠ ab2 ÷ d −3 −1 ⎝w0 x5⎞ ⎛ ⎠ m4 n0 100. 2 y−4 ⎛ ⎝y2⎞ ⎠ 101. p−4 q2 p2 q−3 102. (l × w)2 103. 3 ⎝y7⎞ ⎛ ⎠ ÷ x14 104. 53 ⎛ 123 m33 ⎝ 4−3 2 ⎞ ⎠ 120. 173 ÷ 152 x3 Extensions For the following exercises, simplify the given expression. Write answers with positive exponents. 121. 122. 123. 124. 125. 2 −2 ⎛ ⎞ ⎝ ⎠ ⎛ 32 ⎝ a3 ⎞ ⎠ a4 22 ⎞ ⎛ ⎝62 −24 ⎠ −5 2 ⎛ ⎝ x y ⎞ ⎠ ÷ m2 n3 a2 c−3 ⋅ a−7 n−2 m2 c4 10 ⎛ ⎜ ⎝ x6 y3 x3 y−3 ⋅ y−7 x−3 ⎞ ⎟ ⎠ −3 ⎛ ⎜ ⎜ ⎝ ⎛ ⎝ab2 c⎞ ⎠ b−3 2 ⎞ ⎟ ⎟ ⎠ Avogadro’s constant is used to calculate the number 126. of particles in a mole. A mole is a basic unit in chemistry to is measure the amount of a substance. The constant 6.0221413 × 1023. Write Avogadro’s in standard notation. constant Planck’s constant is an important unit of measure in 127. quantum physics. It describes the relationship between energy and frequency. The constant is written as 6.62606957 × 10−34. Write Planck’s constant in standard notation. Chapter 1 Prerequisites 2 ⎞ ⎠ ⎛ a ⎝ 23 105. 52 m ÷ 50 m 106. 107. 108. (16 x)2 y−1 23 (3a)−2 2 ⎝ma6⎞ ⎛ ⎠ 1 m3 a2 109. 3 ⎝b−3 c⎞ ⎛ ⎠ 110. ⎝x2 y13 ÷ y0⎞ ⎛ ⎠ 2 111. ⎝9z3⎞ ⎛ ⎠ −2 y Real-World Applications To reach escape velocity, a rocket must travel at the 112. rate of 2.2 × 106 ft/min. Rewrite the rate in standard notation. A dime is the thinnest coin in U.S. currency. A dime’s 113. thickness measures 2.2 × 106 m. Rewrite the number in standard notation. The average distance between Earth and the Sun is 114. 92,960,000 mi. Rewrite the distance using scientific notation. A 115. 1,099,500,000,000 bytes. Rewrite in scientific notation. terabyte approximately made of is The Gross Domestic Product (GDP) for the United 116. States in the first quarter of 2014 was $1.71496 × 1013. Rewrite the GDP in standard notation. 117. One picometer is approximately 3.397 × 10−11 in. Rewrite this length using standard notation. The value of the services sector of the U.S. economy 118. in the first quarter of 2012 was $10,633.6 billion. Rewrite this amount in scientific notation. Technology For the following exercises, use a graphing calculator to simplify. Round the answers to the nearest hundredth. 119. 54 Chapter 1 Prerequisites 1.3 \| Radicals and Rational Expressions Learning Objectives In this section students will: 1.3.1 Evaluate square roots. 1.3.2 Use the product rule to simplify square roots. 1.3.3 Use the quotient rule to simplify square roots. 1.3.4 Add and subtract square roots. 1.3.5 Rationalize denominators. 1.3.6 Use ratio
nal roots. A hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in Figure 1.6, and use the Pythagorean Theorem. Figure 1.6 a2 + b2 = c2 52 + 122 = c2 169 = c2 Now, we need to find out the length that, when squared, is 169, to determine which ladder to choose. In other words, we need to find a square root. In this section, we will investigate methods of finding solutions to problems such as this one. Evaluating Square Roots When the square root of a number is squared, the result is the original number. Since 42 = 16, the square root of 16 is 4. The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root. In general terms, if a is a positive real number, then the square root of a is a number that, when multiplied by itself, gives a. The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals a. The square root obtained using a calculator is the principal square root. The principal square root of a is written as a. The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 55 Principal Square Root The principal square root of a is the nonnegative number that, when multiplied by itself, equals a. It is written as a radical expression, with a symbol called a radical over the term called the radicand: a. Does 25 = ± 5 ? No. Although both 52 and (−5)2 are 25, square root. The principal square root of 25 is 25 = 5. the radical symbol implies only a nonnegative root, the principal Example 1.27 Evaluating Square Roots Evaluate each expression. a. b. c. d. 100 16 25 + 144 49 − 81 Solution a. b. c. d. 100 = 10 because 102 = 100 16 = 4 = 2 because 42 = 16 and 22 = 4 25 + 144 = 169 = 13 because 132 = 169 49 − 81 = 7 − 9 = −2 because 72 = 49 and 92 = 81 For 25 + 144, can we find the square roots before adding? No. 25 + 144 = 5 + 12 = 17. This is not equivalent to 25 + 144 = 13. The order of operations requires us to add the terms in the radicand before finding the square root. 1.27 Evaluate each expression. a. b. c. d. 225 81 25 − 9 36 + 121 56 Chapter 1 Prerequisites Using the Product Rule to Simplify Square Roots To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite 15 as 3 ⋅ 5. We can also use the product rule to express the product of multiple radical expressions as a single radical expression. The Product Rule for Simplifying Square Roots If a and b are nonnegative, the square root of the product ab is equal to the product of the square roots of a and b. ab = a ⋅ b Given a square root radical expression, use the product rule to simplify it. 1. Factor any perfect squares from the radicand. 2. Write the radical expression as a product of radical expressions. 3. Simplify. Example 1.28 Using the Product Rule to Simplify Square Roots Simplify the radical expression. a. b. 300 162a5 b4 Solution a. b. 100 ⋅ 3 100 ⋅ 3 10 3 Factor perfect square from radicand. Write radical expression as product of radical expressions. Simplify. 81a4 b4 ⋅ 2a 81a4 b4 ⋅ 2a 9a2 b2 2a Factor perfect square from radicand. Write radical expression as product of radical expressions. Simplify. 1.28 Simplify 50x2 y3 z. Given the product of multiple radical expressions, use the product rule to combine them into one radical expression. 1. Express the product of multiple radical expressions as a single radical expression. 2. Simplify. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 57 Example 1.29 Using the Product Rule to Simplify the Product of Multiple Square Roots Simplify the radical expression. 12 ⋅ 3 Solution 12 ⋅ 3 Express the product as a single radical expression. 36 6 Simplify. 1.29 Simplify 50x ⋅ 2x assuming x > 0. Using the Quotient Rule to Simplify Square Roots Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. We can rewrite 5 2 . as 5 2 The Quotient Rule for Simplifying Square Roots The square root of the quotient a b is equal to the quotient of the square roots of a and b, where b ≠ 0. a b = a b Given a radical expression, use the quotient rule to simplify it. 1. Write the radical expression as the quotient of two radical expressions. 2. Simplify the numerator and denominator. Example 1.30 Using the Quotient Rule to Simplify Square Roots Simplify the radical expression. 5 36 Solution 58 Chapter 1 Prerequisites 5 36 5 6 Write as quotient of two radical expressions. Simplify denominator. 1.30 Simplify 2x2 9y4. Example 1.31 Using the Quotient Rule to Simplify an Expression with Two Square Roots Simplify the radical expression. 234x11 y 26x7 y Solution 234x11 y 26x7 y 9x4 3x2 Combine numerator and denominator into one radical expression. Simplify fraction. Simplify square root. 1.31 Simplify 9a5 b14 3a4 b5 . Adding and Subtracting Square Roots We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of 2 and 3 2 is 4 2. However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression 18 can be written with a 2 in the radicand, as 3 2, so 2 + 18 = 2 + 3 2 = 4 2. Given a radical expression requiring addition or subtraction of square roots, solve. 1. Simplify each radical expression. 2. Add or subtract expressions with equal radicands. Example 1.32 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 59 Adding Square Roots Add 5 12 + 2 3. Solution We can rewrite 5 12 as 5 4 · 3. According the product rule, this becomes 5 4 3. The square root of 4 is 2, so the expression becomes 5(2) 3, which is 10 3. Now we can the terms have the same radicand so we can add. 10 3 + 2 3 = 12 3 1.32 Add 5 + 6 20. Example 1.33 Subtracting Square Roots Subtract 20 72a3 b4 c − 14 8a3 b4 c. Solution Rewrite each term so they have equal radicands. 2 c ⎝b2⎞ ⎠ 20 72a3 b4 c = 20 9 4 2 a a2 ⎛ = 20(3)(2)\|a\|b2 2ac = 120\|a\|b2 2ac 14 8a3 b4 c = 14 2 4 a a2 ⎛ = 14(2)\|a\|b2 2ac = 28\|a\|b2 2ac 2 c ⎝b2⎞ ⎠ Now the terms have the same radicand so we can subtract. 120\|a\|b2 2ac − 28\|a\|b2 2ac = 92\|a\|b2 2ac 1.33 Subtract 3 80x − 4 45x. Rationalizing Denominators When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator. 60 Chapter 1 Prerequisites We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical. For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is b c, multiply by c c. For a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the then the conjugate is a − b c. denominator. If the denominator is a + b c, Given an expression with a single square root radical term in the denominator, rationalize the denominator. a. Multiply the numerator and denominator by the radical in the denominator. b. Simplify. Example 1.34 Rationalizing a Denominator Containing a Single Term in simplest form. Write 2 3 3 10 Solution The radical in the denominator is 10. So multiply the fraction by 10 10 . Then simplify. ⋅ 10 10 2 3 3 10 2 30 30 30 15 1.34 Write 12 3 2 in simplest form. Given an expression with a radical term and a constant in the denominator, rationalize the denominator. 1. Find the conjugate of the denominator. 2. Multiply the numerator and denominator by the conjugate. 3. Use the distributive property. 4. Simplify. Example 1.35 Rationalizing a Denominator Containing Two Terms This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 61 Write 4 1 + 5 in simplest form. Solution Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of 1 + 5 is 1 − 5. Then multiply the fraction by 4 5 − 1 Use the distributive property. Simplify. 1.35 Write 7 2 + 3 in simplest form. Using Rational Roots Although square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain num
ber. Understanding nth Roots Suppose we know that a3 = 8. We want to find what number raised to the 3rd power is equal to 8. Since 23 = 8, we say that 2 is the cube root of 8. The nth root of a is a number that, when raised to the nth power, gives a. For example, −3 is the 5th root of −243 because (−3)5 = −243. If a is a real number with at least one nth root, then the principal nth root of a is the number with the same sign as a that, when raised to the nth power, equals a. The principal nth root of a is written as an expression, n is called the index of the radical. , where n is a positive integer greater than or equal to 2. In the radical Principal nth Root If a is a real number with at least one nth root, then the principal nth root of a, written as an , the same sign as a that, when raised to the nth power, equals a. The index of the radical is n. is the number with Example 1.36 Simplifying nth Roots Simplify each of the following: a. 5 −32 62 Chapter 1 Prerequisites b. 4 4 4 ⋅ 1, 024 3 c. − 8x6 125 d. 4 8 3 4 − 48 Solution 5 a. −32 = −2 because (−2)5 = −32 b. First, express the product as a single radical expression. 4,096 4 = 8 because 84 = 4,096 c. d. 3 − 8x6 3 125 −2x2 Write as quotient of two radical expressions. Simplify. Simplify to get equal radicands. Add. 1.36 Simplify. a. b. c. 3 −216 4 3 80 4 5 3 6 9, 000 3 + 7 576 Using Rational Exponents Radical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index n is even, then a cannot be negative. = an We can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an nth root. The numerator tells us the power and the denominator tells us the root. a 1 n All of the properties of exponents that we learned for integer exponents also hold for rational exponents. m n a = ( an ) m = amn Rational Exponents Rational exponents are another way to express principal nth roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is = amn = ( an ) m n a m This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 63 Given an expression with a rational exponent, write the expression as a radical. 1. Determine the power by looking at the numerator of the exponent. 2. Determine the root by looking at the denominator of the exponent. 3. Using the base as the radicand, raise the radicand to the power and use the root as the index. Example 1.37 Writing Rational Exponents as Radicals Write 343 2 3 as a radical. Simplify. Solution The 2 tells us the power and the 3 tells us the root. 2 3 = 2 ⎛ 3 ⎝ 343 ⎞ ⎠ 343 3 = 3432 3 We know that 343 root before squaring for this problem. In general, it is easier to find the root first and then raise it to a power. = 7 because 73 = 343. Because the cube root is easy to find, it is easiest to find the cube 2 3 = 2 ⎛ 3 ⎝ 343 ⎞ ⎠ 343 = 72 = 49 1.37 5 2 as a radical. Simplify. Write 9 Example 1.38 Writing Radicals as Rational Exponents using a rational exponent. Write 4 a2 7 Solution The power is 2 and the root is 7, so the rational exponent will be 2 7 . We get 4 2 7 a . Using properties of exponents, −2 7 . = 4a we get 4 a2 7 1.38 Write x (5y)9 using a rational exponent. 64 Chapter 1 Prerequisites Example 1.39 Simplifying Rational Exponents Simplify: a. b. ⎛ ⎜2x ⎝ 3 4 ⎛ ⎞ ⎜3x ⎟ ⎠ ⎝ 1 5 ⎞ ⎟ ⎠ 5 − 1 2 ⎛ ⎝ 16 9 ⎞ ⎠ Solution a. 30x 30x 30x b Multiply the coefficien . Use properties of exponents. 19 20 Simplify. 1 2 ⎛ ⎝ ⎞ ⎠ 9 16 9 16 9 16 3 4 Use definition of ne ative exponents. Rewrite as a radical. Use the quotient rule. Simplify. 1.39 Simplify (8x) 1 3 ⎛ ⎜14x ⎝ 6 5 ⎞ ⎟. ⎠ Access these online resources for additional instruction and practice with radicals and rational exponents. • Radicals (http://openstaxcollege.org/l/introradical) • Rational Exponents (http://openstaxcollege.org/l/rationexpon) • Simplify Radicals (http://openstaxcollege.org/l/simpradical) • Rationalize Denominator (http://openstaxcollege.org/l/rationdenom) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 65 1.3 EXERCISES Verbal What does it mean when a radical does not have an 128. index? Is the expression equal to the radicand? Explain. Where would radicals come in the order of 129. operations? Explain why. Every number will have two square roots. What is the 130. principal square root? Can a radical with a negative radicand have a real 131. square root? Why or why not? Numeric For the following exercises, simplify each expression. 132. 133. 134. 135. 136. 137. 138. 139. 140. 141. 142. 143. 144. 145. 146. 147. 148. 256 256 4(9 + 16) 289 − 121 196 1 98 27 64 81 5 800 169 + 144 8 50 18 162 192 14 6 − 6 24 15 5 + 7 45 150 149. 150. 151. 152. 153. 154. 155. 156. 96 100 ( 42)⎛ ⎝ 30⎞ ⎠ 12 3 − 4 75 4 225 405 324 360 361 5 1 + 3 8 1 − 17 157. 4 16 158. 3 128 3 + 3 2 159. 5 −32 243 160. 4 15 125 4 5 161. 3 3 −432 3 + 16 Algebraic For the following exercises, simplify each expression. 162. 163. 400x4 4y2 164. 49p 165. 1 2 ⎝144p2 q6⎞ ⎛ ⎠ 166. 5 2 289 m 167. 9 3m2 + 27 66 168. 169. 170. 171. 172. 173. 174. 175. 176. 177. 178. 179. 180. 181. 182. 183. 184. 185. 186. 3 ab2 − b a 4 2n 16n4 225x3 49x 3 44z + 99z 50y8 490bc2 32 14d 3 2 63p q 8 1 − 3x 20 121d 4 3 2 32 − w 3 2 50 w 108x4 + 27x4 12x 2 + 2 3 147k 3 125n10 42q 36q3 81m 361m2 72c − 2 2c 144 324d 2 187. 3 24x6 3 + 81x6 188. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 4 162x6 16x4 189. 3 64y 190. 3 128z3 3 − −16z3 191. 5 1,024c10 Real-World Applications 192. A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So the length of the guy wire can be found by evaluating 90,000 + 160,000. What is the length of the guy wire? 193. A car accelerates at a rate of 6 − 4 t m/s2 where t is the time in seconds after the car moves from rest. Simplify the expression. Extensions For the following exercises, simplify each expression. 194. 195. 196. 1 2 − 2 8 − 16 4 − 2 4 3 2 3 2 − 16 1 3 8 mn3 a2 c−3 ⋅ a−7 n−2 m2 c4 197. a a − c 198. 199. 200. x 64y + 4 y 128y ⎛ ⎜ 250x2 100b3 ⎝ ⎞ ⎟⎛ ⎝ ⎠ 7 b 125x ⎞ ⎠ 3 4 + 256 64 64 + 256 Chapter 1 Prerequisites 67 1.4 \| Polynomials Learning Objectives In this section students will: 1.4.1 Identify the degree and leading coefficient of polynomials. 1.4.2 Add and subtract polynomials. 1.4.3 Multiply polynomials. 1.4.4 Use FOIL to multiply binomials. 1.4.5 Perform operations with polynomials of several variables. Earl is building a doghouse, whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which the dog can enter and exit the house. Earl wants to find the area of the front of the doghouse so that he can purchase the correct amount of paint. Using the measurements of the front of the house, shown in Figure 1.7, we can create an expression that combines several variable terms, allowing us to solve this problem and others like it. Figure 1.7 First find the area of the square in square feet. Then find the area of the triangle in square feet. A = s2 = (2x)2 = 4x2 bh 2x) ⎝ ⎞ ⎠ 3 2 x Next find the area of the rectangular door in square feet. A = lw = x ⋅ 1 = x The area of the front of the doghouse can be found by adding the areas of the square and the triangle, and then subtracting the area of the rectangle. When we do this, we get 4x2 + 3 2 x − x ft2, or 4x2 + 1 2 x ft2. In this section, we will examine expressions such as this one, which combine several variable terms. 68 Chapter 1 Prerequisites Identifying the Degree and Leading Coefficient of Polynomials The formula just found is an example of a polynomial, which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as 384π, is known as a coefficient. Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Each product ai xi is a term of a polynomial. If a term does not contain a variable, it is called a constant. , such as 384πw, A polynomial containing only one term, such as 5x4, 2x − 9, is called a binomial. A polynomial containing three terms, such as −3x2 + 8x − 7, is called a trinomial. is called a monomial. A polynomial containing two terms, such as We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term because it is usually written first. The coefficient of the leading term is called the leading coefficient. When a polynomial is written so that the powers are descending, we say that it is in standard form. Polynomials A polynomial is an expression that can be written in the form an xn + ... + a2 x2 + a1 x + a0 Each real number ai is called a coefficient. The number a0 that is not multiplied by a variable is called a constant. Each product ai xi is a term of a polynomial. The highest power of the variable that occurs in the polynomial is called the degree of a polynomial. The leading term is the term with the highest power, and its coefficient is called the leading coefficient. Given a polynomial expression, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree. 2. 3. Identify the term containing the highest power of x to find the leading term. Identify the coefficient of the leading term. Example 1.40 Identifying the Degree and Leading Coefficient of a Polynomial For the following p
olynomials, identify the degree, the leading term, and the leading coefficient. a. b. c. 3 + 2x2 − 4x3 5t 5 − 2t 3 + 7t 6p − p3 − 2 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 69 a. The highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4x3. The leading coefficient is the coefficient of that term, −4. b. The highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t 5. The leading coefficient is the coefficient of that term, 5. c. The highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, − p3, The leading coefficient is the coefficient of that term, −1. 1.40 Identify the degree, leading term, and leading coefficient of the polynomial 4x2 − x6 + 2x − 6. Adding and Subtracting Polynomials We can add and subtract polynomials by combining like terms, which are terms that contain the same variables raised to the same exponents. For example, 5x2 and −2x2 are like terms, and can be added to get 3x2, but 3x and 3x2 are not like terms, and therefore cannot be added. Given multiple polynomials, add or subtract them to simplify the expressions. 1. Combine like terms. 2. Simplify and write in standard form. Example 1.41 Adding Polynomials Find the sum. ⎛ ⎞ ⎝12x2 + 9x − 21 ⎠ + ⎞ ⎛ ⎝4x3 + 8x2 − 5x + 20 ⎠ Solution ⎝12x2 + 8x2⎞ ⎛ 4x3 + 4x3 + 20x2 + 4x − 1 ⎠ + (9x − 5x) + (−21 + 20) Combine like terms. Simplify. Analysis We can check our answers to these types of problems using a graphing calculator. To check, graph the problem as given along with the simplified answer. The two graphs should be equivalent. Be sure to use the same window to compare the graphs. Using different windows can make the expressions seem equivalent when they are not. 1.41 Find the sum. ⎛ ⎞ ⎝2x3 + 5x2 − x + 1 ⎠ + ⎞ ⎛ ⎝2x2 − 3x − 4 ⎠ 70 Chapter 1 Prerequisites Example 1.42 Subtracting Polynomials Find the difference. ⎞ ⎛ ⎝7x4 − x2 + 6x + 1 ⎠ − ⎞ ⎛ ⎝5x3 − 2x2 + 3x + 2 ⎠ Solution ⎝−x2 + 2x2⎞ ⎛ 7x4 − 5x3 + 7x4 − 5x3 + x2 + 3x − 1 ⎠ + (6x − 3x) + (1 − 2) Combine like terms. Simplify. Analysis Note that finding the difference between two polynomials is the same as adding the opposite of the second polynomial to the first. 1.42 Find the difference. ⎞ ⎛ ⎝−7x3 − 7x2 + 6x − 2 ⎠ − ⎞ ⎛ ⎝4x3 − 6x2 − x + 7 ⎠ Multiplying Polynomials Multiplying polynomials is a bit more challenging than adding and subtracting polynomials. We must use the distributive property to multiply each term in the first polynomial by each term in the second polynomial. We then combine like terms. We can also use a shortcut called the FOIL method when multiplying binomials. Certain special products follow patterns that we can memorize and use instead of multiplying the polynomials by hand each time. We will look at a variety of ways to multiply polynomials. Multiplying Polynomials Using the Distributive Property To multiply a number by a polynomial, we use the distributive property. The number must be distributed to each term of the polynomial. We can distribute the 2 in 2(x + 7) to obtain the equivalent expression 2x + 14. When multiplying polynomials, the distributive property allows us to multiply each term of the first polynomial by each term of the second. We then add the products together and combine like terms to simplify. Given the multiplication of two polynomials, use the distributive property to simplify the expression. 1. Multiply each term of the first polynomial by each term of the second. 2. Combine like terms. 3. Simplify. Example 1.43 Multiplying Polynomials Using the Distributive Property Find the product. ⎞ ⎛ ⎝3x2 − x + 4 (2x + 1) ⎠ Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 71 2x⎛ ⎞ ⎛ ⎞ ⎝3x2 − x + 4 ⎝3x2 − x + 4 ⎠ ⎠ + 1 ⎝6x3 − 2x2 + 8x⎞ ⎛ ⎞ ⎛ ⎝3x2 − x + 4 ⎠ + ⎠ ⎝−2x2 + 3x2⎞ ⎛ 6x3 + 6x3 + x2 + 7x + 4 ⎠ + (8x − x) + 4 Use the distributive property. Multiply. Combine like terms. Simplify. Analysis We can use a table to keep track of our work, as shown in Table 1.2. Write one polynomial across the top and the other down the side. For each box in the table, multiply the term for that row by the term for that column. Then add all of the terms together, combine like terms, and simplify. 3x2 −x 2x 6x3 −2x2 +1 3x2 −x +4 8x 4 Table 1.2 1.43 Find the product. ⎛ ⎝x3 − 4x2 + 7 (3x + 2) ⎞ ⎠ Using FOIL to Multiply Binomials A shortcut called FOIL is sometimes used to find the product of two binomials. It is called FOIL because we multiply the first terms, the outer terms, the inner terms, and then the last terms of each binomial. The FOIL method arises out of the distributive property. We are simply multiplying each term of the first binomial by each term of the second binomial, and then combining like terms. Given two binomials, use FOIL to simplify the expression. 1. Multiply the first terms of each binomial. 2. Multiply the outer terms of the binomials. 3. Multiply the inner terms of the binomials. 4. Multiply the last terms of each binomial. 5. Add the products. 6. Combine like terms and simplify. 72 Chapter 1 Prerequisites Example 1.44 Using FOIL to Multiply Binomials Use FOIL to find the product. (2x - 10)(3x + 3) Solution Find the product of the first terms. Find the product of the outer terms. Find the product of the inner terms. Find the product of the last terms. 6x2 + 6x − 54x − 54 6x2 + (6x − 54x) − 54 6x2 − 48x − 54 Add the products. Combine like terms. Simplify. 1.44 Use FOIL to find the product. (x + 7)(3x − 5) Perfect Square Trinomials Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster. Let’s look at a few perfect square trinomials to familiarize ourselves with the form. (x + 5)2 = x2 + 10x + 25 (x − 3)2 = x2 − 6x + 9 (4x − 1)2 = 4x2 − 8x + 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 73 Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial. Perfect Square Trinomials When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared. (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2 (1.8) Given a binomial, square it using the formula for perfect square trinomials. 1. Square the first term of the binomial. 2. Square the last term of the binomial. 3. For the middle term of the trinomial, double the product of the two terms. 4. Add and simplify. Example 1.45 Expanding Perfect Squares Expand (3x − 8)2. Solution Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms. Simplify (3x)2 − 2(3x)(8) + (−8)2 9x2 − 48x + 64. (1.9) 1.45 Expand (4x − 1)2. Difference of Squares Another special product is called the difference of squares, which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply (x + 1)(x − 1) using the FOIL method. (x + 1)(x − 1) = x2 − x + x − 1 = x2 − 1 The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples. 74 Chapter 1 Prerequisites (x + 5)(x − 5) = x2 − 25 (x + 11)(x − 11) = x2 − 121 (2x + 3)(2x − 3) = 4x2 − 9 Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term. Is there a special form for the sum of squares? No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares. Difference of Squares When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term. (a + b)(a − b) = a2 − b2 (1.10) Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares. 1. Square the first term of the binomials. 2. Square the last term of the binomials. 3. Subtract the square of the last term from the square of the first term. Example 1.46 Multiplying Binomials Resulting in a Difference of Squares Multiply (9x + 4)(9x − 4). Solution Square the first term to get (9x)2 = 81x2. Square the last term to get 42 = 16. Subtract the square of the last term from the square of the first term to find the product of 81x2 − 16. 1.46 Multiply (2x + 7)(2x − 7). Performing Operations with Polynomials of Several Variables We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example: (a + 2b)(4a − b − c) a(4a − b − c) + 2b(4a − b − c) 4a2 − ab − ac + 8ab − 2b2 − 2bc 4a2 + ( − ab + 8ab) − ac − 2b2 − 2bc 4a2 + 7ab − ac − 2bc − 2b2 Use the distributive property. Multiply. Combine like terms. Simplify. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 75 Example 1.47 Multiplying Polynomials Containing Several Variables Multiply (x + 4)(3x − 2y + 5). Solution Follow the same steps that we used to multiply polynomials containing only one variable. x(3x − 2y + 5) + 4(3x − 2y + 5) 3x2 − 2xy + 5x + 12x − 8y + 20 3x2 − 2xy + (5x + 12x)
− 8y + 20 3x2 − 2xy + 17x − 8y + 20 Use the distributive property. Multiply. Combine like terms. Simplify. 1.47 Multiply (3x − 1)(2x + 7y − 9). Access these online resources for additional instruction and practice with polynomials. • Adding and Subtracting Polynomials (http://openstaxcollege.org/l/addsubpoly) • Multiplying Polynomials (http://openstaxcollege.org/l/multiplpoly) • Special Products of Polynomials (http://openstaxcollege.org/l/specialpolyprod) 76 Chapter 1 Prerequisites 1.4 EXERCISES Verbal 201. Evaluate the following statement: The degree of a polynomial in standard form is the exponent of the leading term. Explain why the statement is true or false. 202. Many times, multiplying two binomials with two variables results in a trinomial. This is not the case when there is a difference of two squares. Explain why the product in this case is also a binomial. 203. You can multiply polynomials with any number of terms and any number of variables using four basic steps over and over until you reach the expanded polynomial. What are the four steps? 204. State whether the following statement is true and explain why or why not: A trinomial is always a higher degree than a monomial. 217. (4x + 2)(6x − 4) 218. ⎝2c2 − 3c⎞ ⎛ ⎝14c2 + 4c⎞ ⎛ ⎠ ⎠ 219. ⎛ ⎝6b2 − 6 ⎞ ⎛ ⎝4b2 − 4 ⎠ ⎞ ⎠ 220. (3d − 5)(2d + 9) 221. (9v − 11)(11v − 9) 222. ⎛ ⎝4t 2 + 7t⎞ ⎛ ⎞ ⎝−3t 2 + 4 ⎠ ⎠ 223. ⎞ ⎛ ⎝n2 + 9 (8n − 4) ⎠ For the following exercises, expand the binomial. Algebraic 224. (4x + 5)2 For the following exercises, identify the degree of the polynomial. 225. (3y − 7)2 205. 7x − 2x2 + 13 206. 207. 208. 14m3 + m2 − 16m + 8 −625a8 + 16b4 200p − 30p2 m + 40m3 209. x2 + 4x + 4 210. 6y4 − y5 + 3y − 4 226. (12 − 4x)2 227. ⎛ ⎝4p + 9⎞ ⎠ 2 228. (2m − 3)2 229. (3y − 6)2 230. (9b + 1)2 For the following exercises, multiply the binomials. For the following exercises, find the sum or difference. 231. (4c + 1)(4c − 1) 211. ⎝12x2 + 3x⎞ ⎛ ⎠ − ⎞ ⎛ ⎝8x2 −19 ⎠ 212. ⎝4z3 + 8z2 − z⎞ ⎛ ⎠ + ⎛ ⎝−2z2 + z + 6 ⎞ ⎠ 213. ⎞ ⎛ ⎝6w2 + 24w + 24 ⎠ − (3w − 6w + 3) 232. (9a − 4)(9a + 4) 233. (15n − 6)(15n + 6) 234. (25b + 2)(25b − 2) 235. (4 + 4m)(4 − 4m) 236. (14p + 7)(14p − 7) ⎞ ⎛ ⎝7a3 + 6a2 − 4a − 13 ⎠ + ⎛ ⎞ ⎝−3a3 − 4a2 + 6a + 17 ⎠ 214. 215. ⎞ ⎛ ⎝11b4 − 6b3 + 18b2 − 4b + 8 ⎠ − ⎝3b3 + 6b2 + 3b⎞ ⎛ ⎠ 237. (11q − 10)(11q + 10) 216. ⎛ ⎞ ⎝49p2 − 25 ⎠ + ⎛ ⎞ ⎝16p4 − 32p2 + 16 ⎠ For the following exercises, multiply the polynomials. For the following exercises, find the product. 238. ⎛ ⎞ ⎝2x2 + 2x + 1 ⎠(4x − 1) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 77 ⎝a2 − 4c2⎞ ⎛ ⎝a2 + 4ac + 4c2⎞ ⎛ ⎠ ⎠ 239. ⎞ ⎛ ⎛ ⎝4t 2 − 1 ⎝4t 2 + t − 7 ⎠ ⎞ ⎠ 240. ⎞ ⎛ ⎝x2 − 2x + 1 (x − 1) ⎠ 241. ⎛ ⎝y2 − 4y − 9 (y − 2) ⎞ ⎠ 242. ⎞ ⎛ ⎝6k 2 + 5k − 1 (6k − 5) ⎠ 243. ⎞ ⎛ ⎝3p2 + 2p − 10 ⎠(p − 1) 244. ⎛ ⎞ ⎝2m2 − 7m + 9 (4m − 13) ⎠ 245. (a + b)(a − b) 246. (4x − 6y)(6x − 4y) 247. (4t − 5u)2 248. (9m + 4n − 1)(2m + 8) 249. (4t − x)(t − x + 1) 250. ⎛ ⎞ ⎛ ⎝a2 + 2ab + b2⎞ ⎝b2 − 1 ⎠ ⎠ 251. (4r − d)(6r + 7d) 252. ⎛ ⎝x2 − xy + y2⎞ (x + y) ⎠ Real-World Applications A developer wants to purchase a plot of land to build 253. a house. The area of the plot can be described by the following is expression: (4x + 1)(8x − 3) where x measured in meters. Multiply the binomials to find the area of the plot in standard form. A prospective buyer wants to know how much grain a 254. specific silo can hold. The area of the floor of the silo is (2x + 9)2. The height of the silo is 10x + 10, where x is measured in feet. Expand the square and multiply by the height to find the expression that shows how much grain the silo can hold. Extensions For the following exercises, perform the given operations. 255. (4t − 7)2(2t + 1) − ⎞ ⎛ ⎝4t 2 + 2t + 11 ⎠ 256. ⎞ ⎛ ⎝9b2 − 36 (3b + 6)(3b − 6) ⎠ 257. 78 Chapter 1 Prerequisites 1.5 \| Factoring Polynomials Learning Objectives In this section students will: 1.5.1 Factor the greatest common factor of a polynomial. 1.5.2 Factor a trinomial. 1.5.3 Factor by grouping. 1.5.4 Factor a perfect square trinomial. 1.5.5 Factor a difference of squares. 1.5.6 Factor the sum and difference of cubes. 1.5.7 Factor expressions using fractional or negative exponents. Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in Figure 1.8. Figure 1.8 The area of the entire region can be found using the formula for the area of a rectangle. A = lw = 10x ⋅ 6x = 60x2 units2 The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of A = s2 = 42 = 16 units2. The other rectangular region has one side of length 10x − 8 and one side of length 4, giving an area of A = lw = 4(10x − 8) = 40x − 32 units2. So the region that must be subtracted has an area of 2(16) + 40x − 32 = 40x units2. The area of the region that requires grass seed is found by subtracting 60x2 − 40x units2. This area can also be expressed in factored form as 20x(3x − 2) units2. We can confirm that this is an equivalent expression by multiplying. Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions. Factoring the Greatest Common Factor of a Polynomial When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 4 is the GCF of 16 and 20 because it is the largest number that divides evenly into both 16 and 20 The GCF of polynomials works the same way: 4x is the GCF of 16x and 20x2 because it is the largest polynomial that divides evenly into both 16x and 20x2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 79 When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables. Greatest Common Factor The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials. Given a polynomial expression, factor out the greatest common factor. 1. 2. Identify the GCF of the coefficients. Identify the GCF of the variables. 3. Combine to find the GCF of the expression. 4. Determine what the GCF needs to be multiplied by to obtain each term in the expression. 5. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by. Example 1.48 Factoring the Greatest Common Factor Factor 6x3 y3 + 45x2 y2 + 21xy. Solution First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of x3, x2, and x is x. (Note that the GCF of a set of expressions in the form xn will always be the exponent of lowest degree.) And the GCF of y3, y2, and y is y. Combine these to find the GCF of the polynomial, 3xy. Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that 3xy⎛ ⎠ = 6x3 y3, 3xy(15xy) = 45x2 y2, and 3xy(7) = 21xy. ⎝2x2 y2⎞ Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by. ⎛ ⎞ ⎝2x2 y2 + 15xy + 7 (3xy) ⎠ Analysis After factoring, we can check our work by multiplying. Use the distributive property to confirm that ⎞ ⎛ ⎠ = 6x3 y3 + 45x2 y2 + 21xy. ⎝2x2 y2 + 15xy + 7 (3xy) 1.48 Factor x⎛ ⎝b2 − a⎞ ⎝b2 − a⎞ ⎛ ⎠ + 6 ⎠ by pulling out the GCF. Factoring a Trinomial with Leading Coefficient 1 Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial x2 + 5x + 6 has a GCF of 1, but it can be written as the product of the factors (x + 2) and (x + 3). 80 Chapter 1 Prerequisites Trinomials of the form x2 + bx + c can be factored by finding two numbers with a product of c and a sum of b. The trinomial x2 + 10x + 16, for example, can be factored using the numbers 2 and 8 because the product of those numbers is 16 and their sum is 10. The trinomial can be rewritten as the product of (x + 2) and (x + 8). Factoring a Trinomial with Leading Coefficient 1 A trinomial of p + q = b. the form x2 + bx + c can be written in factored form as (x + p)(x + q) where pq = c and Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime. Given a trinomial in the form x2 + bx + c, factor it. 1. List factors of c. 2. Find p and q, a pair of factors of c with a sum of b. 3. Write the factored expression (x + p)(x + q). Example 1.49 Factoring a Trinomial with Leading Coefficient 1 Factor x2 + 2x − 15. Solution We have a trinomial with leading coefficient 1, b = 2, and c = −15. We need to find two numbers with a product of −15 and a sum of 2. In Table 1.3, we list factors until we find a pair with the desired sum. Factors of −15 Sum of Factors 1, −15 −14 −1, 15 3, −5 −3, 5 Table 1.3 14 −2 2 Now that we have identified p and q as −3 and 5, write the factored form as (x − 3)(x + 5). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 81 Analysis We can check our work by multiplying. Use FOIL to confirm that (x − 3)(x + 5) = x2 + 2x − 15. Does the order of the factors matter? No. Multiplication is commutative, so the order of the factors does not matter. 1.49 Factor x2 − 7x + 6. Factoring by Grouping Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial 2x2 + 5x + 3 can be rewritten as (2x + 3)(x + 1) using this process. We begin by rewriting the original expression as 2x2 + 2x + 3x + 3 and then factor each portion of the expression to obtain 2
x(x + 1) + 3(x + 1). We then pull out the GCF of (x + 1) to find the factored expression. Factor by Grouping To factor a trinomial in the form ax2 + bx + c by grouping, we find two numbers with a product of ac and a sum of b. We use these numbers to divide the x term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression. Given a trinomial in the form ax2 + bx + c, factor by grouping. 1. List factors of ac. 2. Find p and q, a pair of factors of ac with a sum of b. 3. Rewrite the original expression as ax2 + px + qx + c. 4. Pull out the GCF of ax2 + px. 5. Pull out the GCF of qx + c. 6. Factor out the GCF of the expression. Example 1.50 Factoring a Trinomial by Grouping Factor 5x2 + 7x − 6 by grouping. Solution 82 Chapter 1 Prerequisites We have a trinomial with a = 5, b = 7, and c = −6. First, determine ac = −30. We need to find two numbers with a product of −30 and a sum of 7. In Table 1.4, we list factors until we find a pair with the desired sum. Factors of −30 Sum of Factors 1, −30 −1, 30 2, −15 −2, 15 3, −10 −3, 10 Table 1.4 −29 29 −13 13 −7 7 So p = −3 and q = 10. 5x2 − 3x + 10x − 6 x(5x − 3) + 2(5x − 3) (5x − 3)(x + 2) Rewrite the original expression as ax2 + px + qx + c. Factor out the GCF of each part. Factor out the GCF of the expression. Analysis We can check our work by multiplying. Use FOIL to confirm that (5x − 3)(x + 2) = 5x2 + 7x − 6. 1.50 Factor a. 2x2 + 9x + 9 b. 6x2 + x − 1 Factoring a Perfect Square Trinomial A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term. a2 + 2ab + b2 = (a + b)2 and a2 − 2ab + b2 = (a − b)2 We can use this equation to factor any perfect square trinomial. Perfect Square Trinomials A perfect square trinomial can be written as the square of a binomial: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites a2 + 2ab + b2 = (a + b)2 Given a perfect square trinomial, factor it into the square of a binomial. 1. Confirm that the first and last term are perfect squares. 2. Confirm that the middle term is twice the product of ab. 3. Write the factored form as (a + b)2. 83 (1.11) Example 1.51 Factoring a Perfect Square Trinomial Factor 25x2 + 20x + 4. Solution Notice that 25x2 and 4 are perfect squares because 25x2 = (5x)2 and 4 = 22. Then check to see if the middle term is twice the product of 5x and 2. The middle term is, indeed, twice the product: 2(5x)(2) = 20x. Therefore, the trinomial is a perfect square trinomial and can be written as (5x + 2)2. 1.51 Factor 49x2 − 14x + 1. Factoring a Difference of Squares A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied. We can use this equation to factor any differences of squares. a2 − b2 = (a + b)(a − b) Differences of Squares A difference of squares can be rewritten as two factors containing the same terms but opposite signs. a2 − b2 = (a + b)(a − b) (1.12) Given a difference of squares, factor it into binomials. 1. Confirm that the first and last term are perfect squares. 2. Write the factored form as (a + b)(a − b). Example 1.52 84 Chapter 1 Prerequisites Factoring a Difference of Squares Factor 9x2 − 25. Solution Notice that 9x2 and 25 are perfect squares because 9x2 = (3x)2 and 25 = 52. The polynomial represents a difference of squares and can be rewritten as (3x + 5)(3x − 5). 1.52 Factor 81y2 − 100. Is there a formula to factor the sum of squares? No. A sum of squares cannot be factored. Factoring the Sum and Difference of Cubes Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial. ⎛ ⎝a2 − ab + b2⎞ a3 + b3 = (a + b) ⎠ Similarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs. ⎛ ⎝a2 + ab + b2⎞ a3 − b3 = (a − b) ⎠ We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example. ⎛ x3 − 23 = (x − 2) ⎝x2 + 2x + 4 ⎞ ⎠ The sign of the first 2 is the same as the sign between x3 − 23. The sign of the 2x term is opposite the sign between x3 − 23. And the sign of the last term, 4, is always positive. Sum and Difference of Cubes We can factor the sum of two cubes as ⎝a2 − ab + b2⎞ ⎛ a3 + b3 = (a + b) ⎠ We can factor the difference of two cubes as ⎝a2 + ab + b2⎞ ⎛ a3 − b3 = (a − b) ⎠ (1.13) (1.14) Given a sum of cubes or difference of cubes, factor it. 1. Confirm that the first and last term are cubes, a3 + b3 or a3 − b3. ⎝a2 − ab + b2⎞ ⎛ 2. For a sum of cubes, write the factored form as (a + b) ⎠. For a difference of cubes, write the ⎝a2 + ab + b2⎞ ⎛ factored form as (a − b) ⎠. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 85 Example 1.53 Factoring a Sum of Cubes Factor x3 + 512. Solution ⎞ ⎛ Notice that x3 and 512 are cubes because 83 = 512. Rewrite the sum of cubes as (x + 8) ⎝x2 − 8x + 64 ⎠. Analysis After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check. 1.53 Factor the sum of cubes: 216a3 + b3. Example 1.54 Factoring a Difference of Cubes Factor 8x3 − 125. Solution Notice that 8x3 and 125 are cubes because 8x3 = (2x)3 and 125 = 53. Write the difference of cubes as ⎛ ⎝4x2 + 10x + 25 (2x − 5) ⎞ ⎠. Analysis Just as with the sum of cubes, we will not be able to further factor the trinomial portion. 1.54 Factor the difference of cubes: 1,000x3 − 1. Factoring Expressions with Fractional or Negative Exponents Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, 2x 1 4 + 5x 3 4 can be factored by pulling out x 1 4 and being rewritten as x 1 4 ⎛ ⎜2 + 5x ⎝ 1 2 ⎞ ⎟. ⎠ Example 1.55 86 Chapter 1 Prerequisites Factoring an Expression with Fractional or Negative Exponents Factor 3x(x + 2) −1 3 + 4(x + 2) 2 3. Solution Factor out the term with the lowest value of the exponent. In this case, that would be (x + 2) − 1 3. − 1 3(3x + 4(x + 2)) − 1 3(3x + 4x + 8) (x + 2) (x + 2) − 1 3(7x + 8) (x + 2) Factor out the GCF. Simplify. 1.55 Factor 2(5a − 1) 3 4 + 7a(5a − 1) − 1 4. Access these online resources for additional instruction and practice with factoring polynomials. • Identify GCF (http://openstaxcollege.org/l/findgcftofact) • Factor Trinomials when a Equals 1 (http://openstaxcollege.org/l/facttrinom1) • Factor Trinomials when a is not equal to 1 (http://openstaxcollege.org/l/facttrinom2) • Factor Sum or Difference of Cubes (http://openstaxcollege.org/l/sumdifcube) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 87 1.5 EXERCISES Verbal 90v2 −181v + 90 If the terms of a polynomial do not have a GCF, does 258. that mean it is not factorable? Explain. 278. 12t 2 + t − 13 259. A polynomial is factorable, but it is not a perfect square trinomial or a difference of two squares. Can you factor the polynomial without finding the GCF? 260. How do you factor by grouping? Algebraic For the following exercises, find the greatest common factor. 261. 14x + 4xy − 18xy2 262. 263. 264. 265. 266. 49mb2 − 35m2 ba + 77ma2 30x3 y − 45x2 y2 + 135xy3 200p3 m3 − 30p2 m3 + 40m3 36 j4 k 2 − 18 j3 k 3 + 54 j2 k 4 6y4 − 2y3 + 3y2 − y For the following exercises, factor by grouping. 267. 6x2 + 5x − 4 268. 2a2 + 9a − 18 269. 6c2 + 41c + 63 270. 271. 6n2 − 19n − 11 20w2 − 47w + 24 272. 2p2 − 5p − 7 For the following exercises, factor the polynomial. 273. 7x2 + 48x − 7 274. 10h2 − 9h − 9 275. 2b2 − 25b − 247 276. 9d 2 −73d + 8 277. 279. 2n2 − n − 15 280. 16x2 − 100 281. 25y2 − 196 282. 121p2 − 169 283. 4m2 − 9 284. 361d 2 − 81 285. 324x2 − 121 286. 144b2 − 25c2 287. 16a2 − 8a + 1 288. 289. 290. 291. 292. 293. 49n2 + 168n + 144 121x2 − 88x + 16 225y2 + 120y + 16 m2 − 20m + 100 m2 − 20m + 100 36q2 + 60q + 25 For the following exercises, factor the polynomials. 294. x3 + 216 295. 27y3 − 8 296. 125a3 + 343 297. b3 − 8d 3 298. 64x3 −125 299. 729q3 + 1331 300. 125r 3 + 1,728s3 Chapter 1 Prerequisites Find the length of the base of the flagpole by 311. factoring. Extensions the following exercises, For completely. factor the polynomials 312. 16x4 − 200x2 + 625 313. 81y4 − 256 314. 16z4 − 2,401a4 315. 316. 5x(3x + 2) − 2 4 + (12x + 8) 3 2 ⎞ ⎛ ⎝32x3 + 48x2 − 162x − 243 ⎠ −1 88 301. 302. 303. 304. 305. 306. 307. 4x(x − 1) − 2 3 + 3(x − 1) 1 3 3c(2c + 3) − 1 4 − 5(2c + 3) 3 4 3t(10t + 3) 1 3 + 7(10t + 3) 4 3 14x(x + 2) − 2 5 + 5(x + 2) 3 5 9y(3y − 13) 1 5 − 2(3y − 13) 6 5 5z(2z − 9) − 3 2 + 11(2z − 9) 6d(2d + 3) − 1 6 + 5(2d + 3) − 1 2 5 6 Real-World Applications For the following exercises, consider this scenario: Charlotte has appointed a chairperson to lead a city beautification project. The first act is to install statues and fountains in one of the city’s parks. The park is a rectangle with an area of 98x2 + 105x − 27 m2, as shown in the figure below. The length and width of the park are perfect factors of the area. Factor by grouping to find the length and width of the 308. park. A statue is to be placed in the center of the park. The 309. area of the base of the statue is 4x2 + 12x + 9m2. Factor
the area to find the lengths of the sides of the statue. At the northwest corner of the park, the city is going 310. to install a fountain. The area of the base of the fountain is 9x2 − 25m2. Factor the area to find the lengths of the sides of the fountain. For the following exercise, consider the following scenario: A school is installing a flagpole in the central plaza. The plaza is a square with side length 100 yd. as shown in the figure below. The flagpole will take up a square plot with area x2 − 6x + 9 yd2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 89 1.6 \| Rational Expressions Learning Objectives In this section students will: 1.6.1 Simplify rational expressions. 1.6.2 Multiply rational expressions. 1.6.3 Divide rational expressions. 1.6.4 Add and subtract rational expressions. 1.6.5 Simplify complex rational expressions. A pastry shop has fixed costs of $280 per week and variable costs of $9 per box of pastries. The shop’s costs per week in terms of x, the number of boxes made, is 280 + 9x. We can divide the costs per week by the number of boxes made to determine the cost per box of pastries. 280 + 9x x Notice that the result is a polynomial expression divided by a second polynomial expression. In this section, we will explore quotients of polynomial expressions. Simplifying Rational Expressions The quotient of two polynomial expressions is called a rational expression. We can apply the properties of fractions to rational expressions, such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown. We can factor the numerator and denominator to rewrite the expression. x2 + 8x + 16 x2 + 11x + 28 (x + 4)2 (x + 4)(x + 7) Then we can simplify that expression by canceling the common factor (x + 4). x + 4 x + 7 Given a rational expression, simplify it. 1. Factor the numerator and denominator. 2. Cancel any common factors. Example 1.56 Simplifying Rational Expressions Simplify x2 − 9 x2 + 4x + 3 . Solution 90 Chapter 1 Prerequisites (x + 3)(x − 3) (x + 3)(x + 1) x − 3 x + 1 Factor the numerator and the denominator. Cancel common factor (x + 3). Analysis We can cancel the common factor because any expression divided by itself is equal to 1. Can the x2 term be cancelled in Example 1.56? No. A factor is an expression that is multiplied by another expression. The x2 term is not a factor of the numerator or the denominator. 1.56 Simplify x − 6 x2 − 36 . Multiplying Rational Expressions Multiplication of rational expressions works the same way as multiplication of any other fractions. We multiply the numerators to find the numerator of the product, and then multiply the denominators to find the denominator of the product. Before multiplying, it is helpful to factor the numerators and denominators just as we did when simplifying rational expressions. We are often able to simplify the product of rational expressions. Given two rational expressions, multiply them. 1. Factor the numerator and denominator. 2. Multiply the numerators. 3. Multiply the denominators. 4. Simplify. Example 1.57 Multiplying Rational Expressions Multiply the rational expressions and show the product in simplest form: (x + 5)(x − 1) 3(x + 6) ⋅ (2x − 1) (x + 5) (x + 5)(x − 1)(2x − 1) 3(x + 6)(x + 5) (x + 5)(x − 1)(2x − 1) 3(x + 6)(x + 5) (x − 1)(2x − 1) 3(x + 6) Factor the numerator and denominator. Multiply numerators and denominators. Cancel common factors to simplify. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 91 (x + 5)(x − 1) 3(x + 6) ⋅ (2x − 1) (x + 5) (x + 5)(x − 1)(2x − 1) 3(x + 6)(x + 5) (x + 5)(x − 1)(2x − 1) 3(x + 6)(x + 5) (x − 1)(2x − 1) 3(x + 6) Factor the numerator and denominator. Multiply numerators and denominators. Cancel common factors to simplify. 1.57 Multiply the rational expressions and show the product in simplest form: x2 + 11x + 30 x2 + 5x + 6 ⋅ x2 + 7x + 12 x2 + 8x + 16 Dividing Rational Expressions Division of rational expressions works the same way as division of other fractions. To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. Using this approach, we would x ÷ rewrite 1 x2 3 can multiply as we did before. as the product 1 x ⋅ 3 x2. Once the division expression has been rewritten as a multiplication expression, we x ⋅ 3 1 x2 = 3 x3 Given two rational expressions, divide them. 1. Rewrite as the first rational expression multiplied by the reciprocal of the second. 2. Factor the numerators and denominators. 3. Multiply the numerators. 4. Multiply the denominators. 5. Simplify. Example 1.58 Dividing Rational Expressions Divide the rational expressions and express the quotient in simplest form: Solution 2x2 + x − 6 x2 − 1 ÷ x2 − 4 x2 + 2x + 1 9x2 − 16 3x2 + 17x − 28 ÷ 3x2 − 2x − 8 x2 + 5x − 14 92 Chapter 1 Prerequisites 1.58 Divide the rational expressions and express the quotient in simplest form: 9x2 − 16 3x2 + 17x − 28 ÷ 3x2 − 2x − 8 x2 + 5x − 14 Adding and Subtracting Rational Expressions Adding and subtracting rational expressions works just like adding and subtracting numerical fractions. To add fractions, we need to find a common denominator. Let’s look at an example of fraction addition. 5 24 + 1 40 + 3 120 = 25 120 = 28 120 = 7 30 We have to rewrite the fractions so they share a common denominator before we are able to add. We must do the same thing when adding or subtracting rational expressions. The easiest common denominator to use will be the least common denominator, or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were (x + 3)(x + 4) and (x + 4)(x + 5), then the LCD would be (x + 3)(x + 4)(x + 5). Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. We would need to multiply the expression with a denominator of (x + 3)(x + 4) by x + 5 x + 5 and the expression with a denominator of (x + 4)(x + 5) by x + 3 x + 3 . Given two rational expressions, add or subtract them. 1. Factor the numerator and denominator. 2. Find the LCD of the expressions. 3. Multiply the expressions by a form of 1 that changes the denominators to the LCD. 4. Add or subtract the numerators. 5. Simplify. Example 1.59 Adding Rational Expressions Add the rational expressions: 5 x + 6 y Solution First, we have to find the LCD. In this case, the LCD will be xy. We then multiply each expression by the appropriate form of 1 to obtain xy as the denominator for each fraction ⋅ 5y xy + 6x xy This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 93 Now that the expressions have the same denominator, we simply add the numerators to find the sum. 6x + 5y xy Analysis Multiplying by y y or x x does not change the value of the original expression because any number divided by itself is 1, and multiplying an expression by 1 gives the original expression. Example 1.60 Subtracting Rational Expressions Subtract the rational expressions: Solution 2 (x + 2)(x − 2x + 2)2 − 6 (x + 2)2 ⋅ 6(x − 2) (x + 2)2(x − 2) 6x − 12 − (2x + 4) (x + 2)2(x − 2) − 2 (x + 2)(x − 2) 2(x + 2) (x + 2)2(x − 2) 4x − 16 (x + 2)2(x − 2) 4(x − 4) (x + 2)2(x − 2) 6 x2 + 4x + 4 − 2 x2 −4 Factor. ⋅ x + 2 x + 2 Multiply each fraction to get LCD as denominator. Multiply. Apply distributive property. Subtract. Simplify. Do we have to use the LCD to add or subtract rational expressions? No. Any common denominator will work, but it is easiest to use the LCD. 1.59 Subtract the rational expressions: 3 x + 5 − 1 x−3 . Simplifying Complex Rational Expressions A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator as single rational expressions and dividing. The complex rational expression a can be simplified by rewriting the numerator 1 b + c 94 Chapter 1 Prerequisites as the fraction a 1 and combining the expressions in the denominator as 1 + bc b . We can then rewrite the expression as a multiplication problem using the reciprocal of the denominator. We get a 1 ⋅ Given a complex rational expression, simplify it. b 1 + bc, which is equal to ab 1 + bc. 1. Combine the expressions in the numerator into a single rational expression by adding or subtracting. 2. Combine the expressions in the denominator into a single rational expression by adding or subtracting. 3. Rewrite as the numerator divided by the denominator. 4. Rewrite as multiplication. 5. Multiply. 6. Simplify. Example 1.61 Simplifying Complex Rational Expressions Simplify: y + 1 x x y . Solution Begin by combining the expressions in the numerator into one expression. x x + 1 x y ⋅ xy x + 1 x xy + 1 x Multiply by x x to get LCD as denominator. Add numerators. Now the numerator is a single rational expression and the denominator is a single rational expression. We can rewrite this as division, and then multiplication. xy + 1 x x y ÷ xy + 1 x x y xy + 1 y ⋅ x x y(xy + 1) x2 Rewrite as multiplication. Multiply. 1.60 Simplify: y x x y − y Can a complex rational expression always be simplified? Yes. We can always rewrite a complex rational expression as a simplified rational expression. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 95 Access these online resources for additional instruction and practice with rational expressions. • Simplify Rational Expressions (http://openstaxcollege.org/l/simpratexpress) • Multiply and Divide Rational Expressions (http://openstaxcollege.org/
l/multdivratex) • Add and Subtract Rational Expressions (http://openstaxcollege.org/l/addsubratex) • Simplify a Complex Fraction (http://openstaxcollege.org/l/complexfract) 96 Chapter 1 Prerequisites 1.6 EXERCISES Verbal How can you use factoring to simplify rational 317. expressions? How do you use the LCD to combine two rational 318. expressions? 319. Tell whether the following statement is true or false and explain why: You only need to find the LCD when adding or subtracting rational expressions. Algebraic For the expressions. following exercises, simplify the rational 320. 321. 322. 323. 324. 325. 326. 327. 328. 329. x2 − 16 x2 − 5x + 4 y2 + 10y + 25 y2 + 11y + 30 6a2 − 24a + 24 6a2 − 24 9b2 + 18b + 9 3b + 3 m − 12 m2 − 144 2x2 + 7x − 4 4x2 + 2x − 2 6x2 + 5x − 4 3x2 + 19x + 20 a2 + 9a + 18 a2 + 3a − 18 3c2 + 25c − 18 3c2 − 23c + 14 12n2 − 29n − 8 28n2 − 5n − 3 the For expressions and express the product in simplest form. following exercises, multiply the rational x2 − x − 6 2x2 + x − 6 ⋅ 2x2 + 7x − 15 x2 − 9 330. 331. This content is available for free at https://cnx.org/content/col11758/1.5 c2 + 2c − 24 c2 + 12c + 36 ⋅ c2 − 10c + 24 c2 − 8c + 16 332. 333. 334. 335. 336. 337. 338. 339. 2d 2 + 9d − 35 d 2 + 10d + 21 ⋅ 3d 2 + 2d − 21 3d 2 + 14d − 49 10h2 − 9h − 9 2h2 − 19h + 24 ⋅ h2 − 16h + 64 5h2 − 37h − 24 6b2 + 13b + 6 4b2 − 9 ⋅ 6b2 + 31b − 30 18b2 − 3b − 10 2d 2 + 15d + 25 4d 2 − 25 ⋅ 2d 2 − 15d + 25 25d 2 − 1 6x2 − 5x − 50 15x2 − 44x − 20 ⋅ 20x2 − 7x − 6 2x2 + 9x + 10 t 2 − 1 t 2 + 4t + 3 ⋅ t 2 + 2t − 15 t 2 − 4t + 3 2n2 − n − 15 6n2 + 13n − 5 ⋅ 12n2 − 13n + 3 4n2 − 15n + 9 36x2 − 25 6x2 + 65x + 50 ⋅ 3x2 + 32x + 20 18x2 + 27x + 10 For the following exercises, divide the rational expressions. 340. 341. 342. 343. 344. 345. 346. 3y2 − 7y − 6 2y2 − 3y − 9 ÷ y2 + y − 2 2y2 + y − 3 6p2 + p − 12 8p2 + 18p + 9 ÷ 6p2 − 11p + 4 2p2 + 11p − 6 q2 − 9 q2 + 6q + 9 ÷ q2 − 2q − 3 q2 + 2q − 3 18d 2 + 77d − 18 27d 2 − 15d + 2 ÷ 3d 2 + 29d − 44 9d 2 − 15d + 4 16x2 + 18x − 55 32x2 − 36x − 11 ÷ 2x2 + 17x + 30 4x2 + 25x + 6 144b2 − 25 72b2 − 6b − 10 ÷ 18b2 − 21b + 5 36b2 − 18b − 10 16a2 − 24a + 9 4a2 + 17a − 15 ÷ 16a2 − 9 4a2 + 11a + 6 Chapter 1 Prerequisites 97 347. 348. 22y2 + 59y + 10 12y2 + 28y − 5 ÷ 11y2 + 46y + 8 24y2 − 10y + 1 9x2 + 3x − 20 3x2 − 7x + 4 ÷ 6x2 + 4x − 10 x2 − 2x + 1 a b − b a a + b ab 364. For the following exercises, add and subtract the rational expressions, and then simplify. 365. 3 + 4x 2x 2c c + 1 2c + 1 c + 1 349. x + 10 4 y 350. 351. 2q − 6 12 3p 4 a + 1 + 5 a − 3 352 353 354. x − 1 x + 1 − 2x + 3 2x + 1 355. 3z z + 1 + 2z + 5 z − 2 356. 4p p + 1 − p + 1 4p 357 the For expression. following exercises 2b 3a 358. 359. 360. 361. 362. 363. 366 Real-World Applications Brenda is placing tile on her bathroom floor. The area 367. of the floor is 15x2 − 8x − 7 ft2. The area of one tile is x2 − 2x + 1ft2. To find the number of tiles needed, simplify the rational expression: 15x2 − 8x − 7 x2 − 2x + 1 . The area of Sandy’s yard is 25x2 − 625 ft2. A patch 368. of sod has an area of x2 − 10x + 25 ft2. Divide the two areas and simplify to find how many pieces of sod Sandy needs to cover her yard. Aaron wants to mulch his garden. His garden is 369. x2 + 18x + 81 ft2. One bag of mulch covers x2 − 81 ft2. Divide the expressions and simplify to find how many bags of mulch Aaron needs to mulch his garden. Extensions For the following exercises, perform the given operations and simplify. x2 + x − 6 x2 − 2x − 3 ⋅ 2x2 − 3x − 9 x2 − x − 2 ÷ 10x2 + 27x + 18 x2 + 2x + 1 3y2 − 10y + 3 3y2 + 5y − 2 ⋅ 2y2 − 3y − 20 2y2 − y − 15 y − 4 370. 371. 372. simplify the rational 98 4a + 1 2a − 3 + 2a − 3 2a + 3 4a2 + 9 a 373. x2 + 7x + 12 x2 + x − 6 ÷ 3x2 + 19x + 28 8x2 − 4x − 24 ÷ 2x2 + x − 3 3x2 + 4x − 7 Chapter 1 Prerequisites This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 99 CHAPTER 1 REVIEW KEY TERMS algebraic expression constants and variables combined using addition, subtraction, multiplication, and division associative property of addition the sum of three numbers may be grouped differently without affecting the result; in symbols, a + (b + c) = (a + b) + c associative property of multiplication the product of three numbers may be grouped differently without affecting the result; in symbols, a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c base in exponential notation, the expression that is being multiplied binomial a polynomial containing two terms coefficient any real number ai in a polynomial in the form an xn + ... + a2 x2 + a1 x + a0 commutative property of addition two numbers may be added in either order without affecting the result; in symbols, a + b = b + a commutative property of multiplication two numbers may be multiplied in any order without affecting the result; in symbols, a ⋅ b = b ⋅ a constant a quantity that does not change value degree the highest power of the variable that occurs in a polynomial difference of squares opposite sign the binomial that results when a binomial is multiplied by a binomial with the same terms, but the distributive property the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols, a ⋅ (b + c) = a ⋅ b + a ⋅ c equation a mathematical statement indicating that two expressions are equal exponent in exponential notation, the raised number or variable that indicates how many times the base is being multiplied exponential notation a shorthand method of writing products of the same factor factor by grouping a method for factoring a trinomial in the form ax2 + bx + c by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression formula an equation expressing a relationship between constant and variable quantities greatest common factor the largest polynomial that divides evenly into each polynomial identity property of addition there is a unique number, called the additive identity, 0, which, when added to a number, results in the original number; in symbols, a + 0 = a identity property of multiplication there is a unique number, called the multiplicative identity, 1, which, when multiplied by a number, results in the original number; in symbols, a ⋅ 1 = a index the number above the radical sign indicating the nth root integers the set consisting of the natural numbers, their opposites, and 0: { … , −3, −2, −1, 0, 1, 2, 3,…} inverse property of addition there is a unique number, called the additive inverse (or opposite), denoted − a, which, when added to the original number, results in the additive identity, 0; in symbols, a + (−a) = 0 for every real number a, 100 Chapter 1 Prerequisites inverse property of multiplication for every non-zero real number a, there is a unique number, called the multiplicative inverse (or reciprocal), denoted 1 a, which, when multiplied by the original number, results in the multiplicative identity, 1; in symbols, a ⋅ 1 a = 1 irrational numbers the set of all numbers that are not rational; they cannot be written as either a terminating or repeating decimal; they cannot be expressed as a fraction of two integers leading coefficient the coefficient of the leading term leading term the term containing the highest degree least common denominator the smallest multiple that two denominators have in common monomial a polynomial containing one term natural numbers the set of counting numbers: {1, 2, 3,…} order of operations operations a set of rules governing how mathematical expressions are to be evaluated, assigning priorities to perfect square trinomial the trinomial that results when a binomial is squared polynomial a sum of terms each consisting of a variable raised to a nonnegative integer power principal nth root the number with the same sign as a that when raised to the nth power equals a principal square root the nonnegative square root of a number a that, when multiplied by itself, equals a radical the symbol used to indicate a root radical expression an expression containing a radical symbol radicand the number under the radical symbol rational expression the quotient of two polynomial expressions rational numbers the set of all numbers of the form m n , where m and n are integers and n ≠ 0. Any rational number may be written as a fraction or a terminating or repeating decimal. real number line a horizontal line used to represent the real numbers. An arbitrary fixed point is chosen to represent 0; positive numbers lie to the right of 0 and negative numbers to the left. real numbers the sets of rational numbers and irrational numbers taken together scientific notation a shorthand notation for writing very large or very small numbers in the form a × 10 n where 1 ≤ \|a\| < 10 and n is an integer term of a polynomial any ai xi of a polynomial in the form an xn + ... + a2 x2 + a1 x + a0 trinomial a polynomial containing three terms variable a quantity that may change value whole numbers the set consisting of 0 plus the natural numbers: {0, 1, 2, 3,…} KEY EQUATIONS This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 101 Rules of Exponents For nonzero real numbers a and b and integers m and n Product rule Quotient rule Power rule Zero exponent rule Negative rule am ⋅ an = am + n am an = am − n (am ) n = am ⋅ n a0 = 1 a−n = 1 an Power of a product rule (a ⋅ b) n = an ⋅ bn Power of a quotient rule n ⎛ ⎝ a b ⎞ ⎠ = an bn perfect square trinomial (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2 difference of squares (a + b)(a − b) = a2 − b2 difference of squares a2 − b2 = (a + b)(a − b) perfect square trinomial a2 + 2ab + b2 = (a + b)2 sum of cubes ⎛ ⎝a2 − ab + b2⎞ a3 + b3 = (a + b) ⎠ difference of cubes ⎝a2 + ab + b2⎞ ⎛ a3 − b3 = (a − b) ⎠ KEY CONCEPTS 1.1 Real Numbers: Algebra Essentials • Rational numbers may be written as fractions or terminating or repeating decimals. See Example 1.1 and Example 1.2. • Determine whether a number is rational or irrational by
writing it as a decimal. See Example 1.3. • The rational numbers and irrational numbers make up the set of real numbers. See Example 1.4. A number can be classified as natural, whole, integer, rational, or irrational. See Example 1.5. 102 Chapter 1 Prerequisites • The order of operations is used to evaluate expressions. See Example 1.6. • The real numbers under the operations of addition and multiplication obey basic rules, known as the properties of real numbers. These are the commutative properties, the associative properties, the distributive property, the identity properties, and the inverse properties. See Example 1.7. • Algebraic expressions are composed of constants and variables that are combined using addition, subtraction, multiplication, and division. See Example 1.8. They take on a numerical value when evaluated by replacing variables with constants. See Example 1.9, Example 1.10, and Example 1.12 • Formulas are equations in which one quantity is represented in terms of other quantities. They may be simplified or evaluated as any mathematical expression. See Example 1.11 and Example 1.13. 1.2 Exponents and Scientific Notation • Products of exponential expressions with the same base can be simplified by adding exponents. See Example 1.14. • Quotients of exponential expressions with the same base can be simplified by subtracting exponents. See Example 1.15. • Powers of exponential expressions with the same base can be simplified by multiplying exponents. See Example 1.16. • An expression with exponent zero is defined as 1. See Example 1.17. • An expression with a negative exponent is defined as a reciprocal. See Example 1.18 and Example 1.19. • The power of a product of factors is the same as the product of the powers of the same factors. See Example 1.20. • The power of a quotient of factors is the same as the quotient of the powers of the same factors. See Example 1.21. • The rules for exponential expressions can be combined to simplify more complicated expressions. See Example 1.22. • Scientific notation uses powers of 10 to simplify very large or very small numbers. See Example 1.23 and Example 1.24. • Scientific notation may be used to simplify calculations with very large or very small numbers. See Example 1.25 and Example 1.26. 1.3 Radicals and Rational Expressions • The principal square root of a number a is the nonnegative number that when multiplied by itself equals a. See Example 1.27. • • If a and b are nonnegative, the square root of the product ab is equal to the product of the square roots of a and b See Example 1.28 and Example 1.29. If a and b are nonnegative, the square root of the quotient a b is equal to the quotient of the square roots of a and b See Example 1.30 and Example 1.31. • We can add and subtract radical expressions if they have the same radicand and the same index. See Example 1.32 and Example 1.33. • Radical expressions written in simplest form do not contain a radical in the denominator. To eliminate the square root radical from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. See Example 1.34 and Example 1.35. • The principal nth root of a is the number with the same sign as a that when raised to the nth power equals a. These roots have the same properties as square roots. See Example 1.36. • Radicals can be rewritten as rational exponents and rational exponents can be rewritten as radicals. See Example 1.37 and Example 1.38. • The properties of exponents apply to rational exponents. See Example 1.39. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 103 1.4 Polynomials • A polynomial is a sum of terms each consisting of a variable raised to a non-negative integer power. The degree is the highest power of the variable that occurs in the polynomial. The leading term is the term containing the highest degree, and the leading coefficient is the coefficient of that term. See Example 1.40. • We can add and subtract polynomials by combining like terms. See Example 1.41 and Example 1.42. • To multiply polynomials, use the distributive property to multiply each term in the first polynomial by each term in the second. Then add the products. See Example 1.43. • FOIL (First, Outer, Inner, Last) is a shortcut that can be used to multiply binomials. See Example 1.44. • Perfect square trinomials and difference of squares are special products. See Example 1.45 and Example 1.46. • Follow the same rules to work with polynomials containing several variables. See Example 1.47. 1.5 Factoring Polynomials • The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem. See Example 1.48. • Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term. See Example 1.49. • Trinomials can be factored using a process called factoring by grouping. See Example 1.50. • Perfect square trinomials and the difference of squares are special products and can be factored using equations. See Example 1.51 and Example 1.52. • The sum of cubes and the difference of cubes can be factored using equations. See Example 1.53 and Example 1.54. • Polynomials containing fractional and negative exponents can be factored by pulling out a GCF. See Example 1.55. 1.6 Rational Expressions • Rational expressions can be simplified by cancelling common factors in the numerator and denominator. See Example 1.56. • We can multiply rational expressions by multiplying the numerators and multiplying the denominators. See Example 1.57. • To divide rational expressions, multiply by the reciprocal of the second expression. See Example 1.58. • Adding or subtracting rational expressions requires finding a common denominator. See Example 1.59 and Example 1.60. • Complex rational expressions have fractions in the numerator or the denominator. These expressions can be simplified. See Example 1.61. CHAPTER 1 REVIEW EXERCISES Real Numbers: Algebra Essentials 377. 5x + 9 = −11 For the following exercises, perform the given operations. 374. (5 − 3 ⋅ 2)2 − 6 375. 64 ÷ (2 ⋅ 8) + 14 ÷ 7 376. 2 ⋅ 52 + 6 ÷ 2 For the following exercises, solve the equation. 378. 2y + 42 = 64 For the following exercises, simplify the expression. 379. 9⎛ ⎝y + 2⎞ ⎠ ÷ 3 ⋅ 2 + 1 380. 3m(4 + 7) − m 104 Chapter 1 Prerequisites For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer. 381. 11 382. 0 383. 5 6 384. 11 Exponents and Scientific Notation For the following exercises, simplify the expression. 385. 22 ⋅ 24 386. 45 43 387. ⎛ a2 ⎝ b3 4 ⎞ ⎠ 388. 6a2 ⋅ a0 2a−4 389. (xy)4 y3 ⋅ 2 x5 390. 4−2 x3 y−3 2x0 −2 391. ⎛ ⎝ 2x2 y ⎞ ⎠ 396. 196 397. 361 398. 75 399. 162 400. 401. 32 25 80 81 402. 49 1250 403. 2 4 + 2 404. 4 3 + 6 3 405. 12 5 − 13 5 406. 5 −243 407. 3 250 3 −8 Polynomials For the following exercises, perform the given operations and simplify. 408. ⎞ ⎛ ⎝3x3 + 2x − 1 ⎠ + ⎞ ⎛ ⎝4x2 − 2x + 7 ⎠ 392. ⎛ 16a3 ⎝ b2 ⎞ ⎝4ab−1⎞ ⎛ ⎠ ⎠ −2 409. ⎛ ⎝2y + 1⎞ ⎠ − ⎞ ⎛ ⎝2y2 − 2y − 5 ⎠ the number in standard notation: 410. ⎛ ⎞ ⎝2x2 + 3x − 6 ⎠ + ⎛ ⎞ ⎝3x2 − 4x + 9 ⎠ 393. Write 2.1314 × 10−6 394. Write the number in scientific notation: 16,340,000 Radicals and Rational Expressions For the following exercises, find the principal square root. 395. 121 411. ⎞ ⎛ ⎝6a2 + 3a + 10 ⎠ − ⎛ ⎝6a2 −3a + 5 ⎞ ⎠ 412. (k + 3)(k − 6) 413. (2h + 1)(3h − 2) 414. ⎞ ⎛ ⎝x2 + 1 (x + 1) ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 1 Prerequisites 105 Rational Expressions For the following exercises, simplify the expression. 434. x2 − x − 12 x2 − 8x + 16 435. 4y2 − 25 4y2 − 20y + 25 436. 2a2 − a − 3 2a2 − 6a − 8 ⋅ 5a2 − 19a − 4 10a2 − 13a − 3 437 − 16 438. m2 + 5m + 6 2m2 − 5m − 3 ÷ 2m2 + 3m − 9 4m2 − 4m − 3 439. 4d 2 − 7d − 2 6d 2 − 17d + 10 ÷ 8d 2 + 6d + 1 6d 2 + 7d − 10 440. 10 x + 6 y 441. 12 a2 + 2a + 1 − 3 a2 −1 442. d + 2 1 c 6c + 12d dc 443. 3 x − 7 y 2 x 415. ⎞ ⎛ ⎝m2 + 2m − 3 (m − 2) ⎠ 416. (a + 2b)(3a − b) 417. (x + y)(x − y) Factoring Polynomials For the following exercises, find the greatest common factor. 418. 81p + 9pq − 27p2 q2 419. 12x2 y + 4xy2 −18xy 420. 88a3 b + 4a2 b − 144a2 For the following exercises, factor the polynomial. 421. 2x2 − 9x − 18 422. 8a2 + 30a − 27 423. d 2 − 5d − 66 424. x2 + 10x + 25 425. y2 − 6y + 9 426. 4h2 − 12hk + 9k 2 427. 361x2 − 121 428. p3 + 216 429. 8x3 − 125 430. 64q3 − 27p3 431. 4x(x − 1) − 1 4 + 3(x − 1) 3 4 432. 3p⎛ ⎝p + 3⎞ ⎠ 1 3 −8⎛ ⎝p + 3⎞ ⎠ 4 3 433. 4r(2r − 1) − 2 3 − 5(2r − 1) 1 3 106 Chapter 1 Prerequisites CHAPTER 1 PRACTICE TEST For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer. 461. 3 4 −8 625 444. −13 445. 2 For the following exercises, evaluate the equations. 446. 2(x + 3) − 12 = 18 447. y(3 + 3)2 − 26 = 10 462. ⎞ ⎛ ⎝13q3 + 2q2 − 3 ⎠ − ⎞ ⎛ ⎝6q2 + 5q − 3 ⎠ 463. ⎛ ⎞ ⎝6p2 + 2p + 1 ⎠ + ⎛ ⎞ ⎝9p2 −1 ⎠ 464. ⎞ ⎛ ⎝n2 − 4n + 4 (n − 2) ⎠ 465. (a − 2b)(2a + b) 448. Write the number in standard notation: 3.1415 × 106 For the following exercises, factor the polynomial. 466. 16x2 − 81 467. y2 + 12y + 36 468. 27c3 − 1331 469. 3x(x − 6) − 1 4 + 2(x − 6) 3 4 For the following exercises, simplify the expression. 470. 2z2 + 7z + 3 z2 − 9 ⋅ 4z2 − 15z + 9 4z2 − 1 471. x y + 2 x 472. a 2b − 2b 9a 3a − 2b 6a 449. Write 0.0000000212. the number in scientific notation: For the following exercises, simplify the expression. 450. −2 ⋅ (2 + 3 ⋅ 2)2 + 144 451. 4(x + 3) − (6x + 2) 452. 35 ⋅ 3−3 453. 3 ⎞ ⎠ ⎛ ⎝ 2 3 454. 8x3 (2x)2 455. ⎝16y0⎞ ⎛ ⎠2y−2 456. 441 457. 490 458. 9x 16 459. 121b2 1 + b 460. 6 24 + 7 54 − 12 6 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 107 2 \| EQUATIONS AND INEQUALITIES Figure 2.1 Chapter Outline 2.1 The Rectangular Coordinate Systems and Graphs 2.2 Line
ar Equations in One Variable 2.3 Models and Applications 2.4 Complex Numbers 2.5 Quadratic Equations 2.6 Other Types of Equations 2.7 Linear Inequalities and Absolute Value Inequalities Introduction For most people, the term territorial possession indicates restrictions, usually dealing with trespassing or rite of passage and takes place in some foreign location. What most Americans do not realize is that from September through December, territorial possession dominates our lifestyles while watching the NFL. In this area, territorial possession is governed by the referees who make their decisions based on what the chains reveal. If the ball is at point A (x1, y1), then it is up to the quarterback to decide which route to point B (x2, y2), the end zone, is most feasible. 108 Chapter 2 Equations and Inequalities 2.1 \| The Rectangular Coordinate Systems and Graphs Learning Objectives In this section you will: 2.1.1 Plot ordered pairs in a Cartesian coordinate system. 2.1.2 Graph equations by plotting points. 2.1.3 Graph equations with a graphing utility. 2.1.4 Find x -intercepts and y -intercepts. 2.1.5 Use the distance formula. 2.1.6 Use the midpoint formula. Figure 2.2 Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 2.2. Laying a rectangular coordinate grid over the map, we can see that each stop aligns with an intersection of grid lines. In this section, we will learn how to use grid lines to describe locations and changes in locations. Plotting Ordered Pairs in the Cartesian Coordinate System An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis. While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis. The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure 2.3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 109 Figure 2.3 The center of the plane is the point at which the two axes cross. It is known as the origin, or point (0, 0). From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in Figure 2.4. Figure 2.4 Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form (x, y). An ordered pair is also known as a coordinate pair because it consists of x- and ycoordinates. For example, we can represent the point (3, −1) in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See Figure 2.5. 110 Chapter 2 Equations and Inequalities Figure 2.5 When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2, or 10, or 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities. Cartesian Coordinate System A two-dimensional plane where the • x-axis is the horizontal axis • y-axis is the vertical axis A point in the plane is defined as an ordered pair, (x, y), such that x is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin. Example 2.1 Plotting Points in a Rectangular Coordinate System Plot the points (−2, 4), (3, 3), and (0, −3) in the plane. Solution To plot the point (−2, 4), begin at the origin. The x-coordinate is –2, so move two units to the left. The ycoordinate is 4, so then move four units up in the positive y direction. To plot the point (3, 3), begin again at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y direction. To plot the point (0, −3), begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y direction. See the graph in Figure 2.6. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 111 Figure 2.6 Analysis Note that when either coordinate is zero, the point must be on an axis. If the x-coordinate is zero, the point is on the y-axis. If the y-coordinate is zero, the point is on the x-axis. Graphing Equations by Plotting Points We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a two-dimensional plane is a graph in two variables. Suppose we want to graph the equation y = 2x − 1. We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of x- and y-values is an ordered pair that can be plotted. Table 2.1 lists values of x from –3 to 3 and the resulting values for y. 112 Chapter 2 Equations and Inequalities x y = 2x − 1 (x, y) −3 −2 −1 0 1 2 3 y = 2(−3) − 1 = −7 (−3, −7) y = 2(−2) − 1 = −5 (−2, −5) y = 2(−1) − 1 = −3 (−1, −3) y = 2(0) − 1 = −1 (0, −1) y = 2(1) − 1 = 1 (1, 1) y = 2(2) − 1 = 3 (2, 3) y = 2(3) − 1 = 5 (3, 5) Table 2.1 We can plot the points in the table. The points for this particular equation form a line, so we can connect them. See Figure 2.7. This is not true for all equations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 113 Figure 2.7 Note that the x-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph. Given an equation, graph by plotting points. 1. Make a table with one column labeled x, a second column labeled with the equation, and a third column listing the resulting ordered pairs. 2. Enter x-values down the first column using positive and negative values. Selecting the x-values in numerical order will make the graphing simpler. 3. Select x-values that will yield y-values with little effort, preferably ones that can be calculated mentally. 4. Plot the ordered pairs. 5. Connect the points if they form a line. Example 2.2 Graphing an Equation in Two Variables by Plotting Points Graph the equation y = − x + 2 by plotting points. 114 Chapter 2 Equations and Inequalities Solution First, we construct a table similar to Table 2.2. Choose x values and calculate y. x y = − x + 2 (x, y) −5 −3 −1 0 1 3 5 y = − (−5) + 2 = 7 (−5, 7) y = − (−3) + 2 = 5 (−3, 5) y = − (−1) + 2 = 3 (−1, 3) y = − (0) + 2 = 2 (0, 2) y = − (1) + 2 = 1 (1, 1) y = − (3) + 2 = −1 (3, −1) y = − (5) + 2 = −3 (5, −3) Table 2.2 Now, plot the points. Connect them if they form a line. See Figure 2.8 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 115 Figure 2.8 2.1 Construct a table and graph the equation by plotting points: y = 1 2 x + 2. Graphing Equations with a Graphing Utility Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style y = _____. The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen. For example, the equation y = 2x − 20 has been entered in the TI-84 Plus shown in Figure 2.9a. In Figure 2.9b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows −10 ≤ x ≤ 10, and −10 ≤ y ≤ 10. See Figure 2.9c. Figure 2.9 a. Enter the equation. b. This is the graph in the original window. c. These are the original settings. 116 Chapter 2 Equations and Inequalities By changing
the window to show more of the positive x-axis and more of the negative y-axis, we have a much better view of the graph and the x- and y-intercepts. See Figure 2.10a and Figure 2.10b. Figure 2.10 a. This screen shows the new window settings. b. We can clearly view the intercepts in the new window. Example 2.3 Using a Graphing Utility to Graph an Equation Use a graphing utility to graph the equation . Solution Enter the equation in the y= function of the calculator. Set the window settings so that both the x- and y- intercepts are showing in the window. See Figure 2.11. Figure 2.11 Finding x-intercepts and y-intercepts The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero. To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. For example, lets find the intercepts of the equation y = 3x − 1. To find the x-intercept, set y = 0. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 117 To find the y-intercept, set x = 0. y = 3x − 1 0 = 3x − 1 1 = 3x = 3x − 1 y = 3(0) − 1 y = −1 (0, −1) x−intercept y−intercept We can confirm that our results make sense by observing a graph of the equation as in Figure 2.12. Notice that the graph crosses the axes where we predicted it would. Figure 2.12 Given an equation, find the intercepts. 1. Find the x-intercept by setting y = 0 and solving for x. 2. Find the y-intercept by setting x = 0 and solving for y. Example 2.4 Finding the Intercepts of the Given Equation Find the intercepts of the equation y = −3x − 4. Then sketch the graph using only the intercepts. Solution Set y = 0 to find the x-intercept. 118 Chapter 2 Equations and Inequalities y = −3x − 4 0 = −3x − 4 4 = −3x = x − 4 3 ⎛ ⎝− 4 3 ⎞ , 0 ⎠ x−intercept Set x = 0 to find the y-intercept. y = −3x − 4 y = −3(0) − 4 y = −4 (0, −4) y−intercept Plot both points, and draw a line passing through them as in Figure 2.13. Figure 2.13 2.2 Find the intercepts of the equation and sketch the graph: y = − 3 4 x + 3. Using the Distance Formula Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, a2 + b2 = c2, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. See Figure 2.14. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 119 Figure 2.14 The relationship of sides \|x2 − x1\| and \|y2 − y1\| to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, \|−3\| = 3. ) The symbols \|x2 − x1\| and \|y2 − y1\| indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem. It follows that the distance formula is given as c2 = a2 + b2 → c = a2 + b2 d 2 = (x2 − x1)2 + (y2 − y1)2 → d = (x2 − x1)2 + (y2 − y1)2 We do not have to use the absolute value symbols in this definition because any number squared is positive. The Distance Formula Given endpoints (x1, y1) and (x2, y2), the distance between two points is given by d = (x2 − x1)2 + (y2 − y1)2 Example 2.5 Finding the Distance between Two Points Find the distance between the points (−3, −1) and (2, 3). Solution Let us first look at the graph of the two points. Connect the points to form a right triangle as in Figure 2.15. 120 Chapter 2 Equations and Inequalities Figure 2.15 Then, calculate the length of d using the distance formula. d = (x2 − x1)2 + (y2 − y1)2 d = (2 − (−3))2 + (3 − (−1))2 = (5)2 + (4)2 = 25 + 16 = 41 2.3 Find the distance between two points: (1, 4) and (11, 9). Example 2.6 Finding the Distance between Two Locations Let’s return to the situation introduced at the beginning of this section. Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 2.2. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions. Solution The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at (1, 1). The next stop is 5 blocks to the east, so it is at (5, 1). After that, she traveled 3 blocks east and 2 blocks north to (8, 3). Lastly, she traveled 4 blocks north to (8, 7). We can label these points on the grid as in Figure 2.16. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 121 Figure 2.16 Next, we can calculate the distance. Note that each grid unit represents 1,000 feet. • From her starting location to her first stop at (1, 1), Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop. • Her second stop is at (5, 1). So from (1, 1) to (5, 1), Tracie drove east 4,000 feet. • Her third stop is at (8, 3). There are a number of routes from (5, 1) to (8, 3). Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet. • Tracie’s final stop is at (8, 7). This is a straight drive north from (8, 3) for a total of 4,000 feet. Next, we will add the distances listed in Table 2.3. 122 Chapter 2 Equations and Inequalities From/To Number of Feet Driven (0, 0) to (1, 1) (1, 1) to (5, 1) (5, 1) to (8, 3) (8, 3) to (8, 7) 2,000 4,000 5,000 4,000 Total 15,000 Table 2.3 The total distance Tracie drove is 15,000 feet, or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points (0, 0) and (8, 7). d = (8 − 0)2 + (7 − 0)2 = 64 + 49 = 113 = 10.63 units At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point (8, 7). Perhaps you have heard the saying “as the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways. Using the Midpoint Formula When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment, (x1, y1) and (x2, y2), the midpoint formula states how to find the coordinates of the midpoint M. M = ⎛ ⎝ x1 + x2 2 , y1 + y2 2 ⎞ ⎠ A graphical view of a midpoint is shown in Figure 2.17. Notice that the line segments on either side of the midpoint are congruent. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 123 Figure 2.17 Example 2.7 Finding the Midpoint of the Line Segment Find the midpoint of the line segment with the endpoints (7, −2) and (9, 5). Solution Use the formula to find the midpoint of the line segment. ⎛ ⎝ x1 + x2 2 , y1 + y2 2 ⎞ ⎛ ⎠ = ⎝ ⎛ ⎝82 + 5 2 ⎞ ⎠ 2.4 Find the midpoint of the line segment with endpoints (−2, −1) and (−8, 6). Example 2.8 Finding the Center of a Circle The diameter of a circle has endpoints (−1, −4) and (5, −4). Find the center of the circle. Solution The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point. , x1 + x2 ⎛ ⎝ 2 , −4 − 4 2 ⎞ ⎠ = y1 + y2 ⎞ ⎠ 2, −4) ⎛ ⎝ −1 + 5 2 124 Chapter 2 Equations and Inequalities Access these online resources for additional instruction and practice with the Cartesian coordinate system. • Plotting points on the coordinate plane (http://Openstaxcollege.org/l/coordplotpnts) • Find x and y intercepts based on the graph of a line (http://Openstaxcollege.org/l/ xyintsgraph) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 125 2.1 EXERCISES Verbal Is it possible for a point plotted in the Cartesian 1. coordinate system to not lie in one of the four quadrants? Explain. 17. (−4, 1) and (3, −4) 18. (2, −5) and (7, 4) 19. (5, 0) and (5, 6) Describe the process for finding the x-intercept and the 2. y-intercept of a graph algebraically. 20. (−4, 3) and (10, 3) Describe in your own words what the y-intercept of a 3. graph is. using When the formula 4. d = (x2 − x1)2 + (y2 − y1)2, explain the correct order of operations that are to be performed to obtain the correct answer. distance Algebraic 21. Find the distance between the two points given using your calculator, and round your answer to the nearest hundredth. (19, 12) and (41, 71) For each of the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points. For each of the following exercises, find the x-intercept and the y-intercept without graphing. Write the coordinates of each intercept. 22. (−5, −6) and (4, 2) 23. (−1, 1) and (7, −4) 24. (−5, −3) and (−2, −8) 25. (0, 7) and (4, −9) 26. (−43, 17) and (23, −34) Graphical For each of information requested. the following exercises, identify the 27. What are the coordinates of the orig
in? If a point is located on the y-axis, what is the x- 28. coordinate? If a point is located on the x-axis, what is the y- 29. coordinate? For each of the following exercises, plot the three points on the given coordinate plane. State whether the three points you plotted appear to be collinear (on the same line). 30. (4, 1)(−2, −3)(5, 0) 5. 6. 7. 8. 9. y = −3x + 6 4y = 2x − 1 3x − 2y = 6 4x − 3 = 2y 3x + 8y = 9 10. 2x − 2 3 = 3 4 y + 3 For each of the following exercises, solve the equation for y in terms of x. 11. 4x + 2y = 8 12. 3x − 2y = 6 13. 2x = 5 − 3y 14. x − 2y = 7 15. 5y + 4 = 10x 16. 5x + 2y = 0 For each of the following exercises, find the distance between the two points. Simplify your answers, and write the exact answer in simplest radical form for irrational answers. 126 Chapter 2 Equations and Inequalities 31. (−1, 2)(0, 4)(2, 1) 33. Name the coordinates of the points graphed. 32. (−3, 0)(−3, 4)(−3, −3) Name the quadrant 34. in which the following points would be located. If the point is on an axis, name the axis. a.(−3, −4) b.(−5, 0) c.(1, −4) d.(−2, 7) e.(0, −3) For each of the following exercises, construct a table and graph the equation by plotting at least three points. 35. y = 1 3 x + 2 36. y = −3x + 1 37. 2y = x + 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 127 Numeric For each of the following exercises, find and plot the x- and y-intercepts, and graph the straight line based on those two points. 38. 4x − 3y = 12 39. x − 2y = 8 40. y − 5 = 5x 41. 3y = −2x + 6 42. y = x − 3 2 For each of the following exercises, use the graph in the figure below. Find the distance between the two endpoints using the 43. distance formula. Round to three decimal places. x and it will display the y value for any x value you input. Use this and plug in x = 0, thus finding the y-intercept, for each of the following graphs. 48. Y1 = −2x + 5 49. 50. Y1 = 3x − 8 4 Y1 = x + 5 2 For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. After graphing it, use the 2nd CALC button and 2:zero button, hit enter. At the lower part of the screen you will see “left bound?” and a blinking cursor on the graph of the line. Move this cursor to the left of the x-intercept, hit ENTER. Now it says “right bound?” Move the cursor to the right of the x-intercept, hit enter. Now it says “guess?” Move your cursor to the left somewhere in between the left and right bound near the x-intercept. Hit enter. At the bottom of your screen it will display the coordinates of the x-intercept or the “zero” to the y-value. Use this to find the x-intercept. Note: With linear/straight line functions the zero is not really a “guess,” but it is necessary to enter a “guess” so it will search and find the exact x-intercept between your right and left boundaries. With other types of functions (more than one x-intercept), they may be irrational numbers so “guess” is more appropriate to give it the correct limits to find a very close approximation between the left and right boundaries. 51. Y1 = −8x + 6 52. Y1 = 4x − 7 53. Y1 = 3x + 5 4 thousandth. Round your answer to the nearest Find the coordinates of the midpoint of the line 44. segment connecting the two points. Extensions 45. Find the distance that (−3, 4) is from the origin. 46. Find the distance that (5, 2) is from the origin. Round to three decimal places. 47. Which point is closer to the origin? Technology For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. After graphing it, use the 2nd CALC button and 1:value button, hit enter. At the lower part of the screen you will see “x=” and a blinking cursor. You may enter any number for A man drove 10 mi directly east from his home, made a 54. left turn at an intersection, and then traveled 5 mi north to his place of work. If a road was made directly from his home to his place of work, what would its distance be to the nearest tenth of a mile? If the road was made in the previous exercise, how 55. much shorter would the man’s one-way trip be every day? Given 56. A(1, 3), B(−3, 5), C(4, 7), and D(5, −4), find these four points: the coordinates of the midpoint of line segments AB and CD. 57. 128 Chapter 2 Equations and Inequalities After finding the two midpoints in the previous exercise, find the distance between the two midpoints to the nearest thousandth. 58. Given the graph of the rectangle shown and the coordinates of its vertices, prove that the diagonals of the rectangle are of equal length. If we rent a truck and pay a $75/day fee plus $.20 for every mile we travel, write a linear equation that would express the total cost y, using x to represent the number of miles we travel. Graph this function on your graphing calculator and find the total cost for one day if we travel 70 mi. In the previous exercise, find the coordinates of the 59. midpoint for each diagonal. Real-World Applications The coordinates on a map for San Francisco are 60. (53, 17) and those for Sacramento are (123, 78). Note that coordinates represent miles. Find the distance between the cities to the nearest mile. 61. If San Jose’s coordinates are (76, −12), where the coordinates represent miles, find the distance between San Jose and San Francisco to the nearest mile. the boat A small craft in Lake Ontario sends out a distress 62. signal. The coordinates of in trouble were (49, 64). One rescue boat is at the coordinates (60, 82) and a second Coast Guard craft is at coordinates (58, 47). Assuming both rescue craft travel at the same rate, which one would get to the distressed boat the fastest? A man on the top of a building wants to have a guy 63. wire extend to a point on the ground 20 ft from the building. To the nearest foot, how long will the wire have to be if the building is 50 ft tall? 64. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 129 2.2 \| Linear Equations in One Variable Learning Objectives In this section you will: 2.2.1 Solve equations in one variable algebraically. 2.2.2 Solve a rational equation. 2.2.3 Find a linear equation. 2.2.4 Given the equations of perpendicular. 2.2.5 Write the equation of a line parallel or perpendicular to a given line. two lines, determine whether their graphs are parallel or Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 2.18. Figure 2.18 Solving Linear Equations in One Variable A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form ax + b = 0 and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation. 3x = 2x + x The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x will make the equation true. A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5x + 2 = 3x − 6, we have the following: The solution set consists of one number: {−4}. It is the only solution and, therefore, we have solved a conditional equation. 5x + 2 = 3x − 6 2x = −8 x = −4 130 Chapter 2 Equations and Inequalities An inconsistent equation results in a false statement. For example, if we are to solve 5x − 15 = 5(x − 4), we have the following: 5x − 15 = 5x − 20 5x − 15 − 5x = 5x − 20 − 5x −15 ≠ −20 Subtract 5x from both sides. False statement Indeed, −15 ≠ −20. There is no solution because this is an inconsistent equation. Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows. Linear Equation in One Variable A linear equation in one variable can be written in the form ax + b = 0 where a and b are real numbers, a ≠ 0. Given a linear equation in one variable, use algebra to solve it. The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = _________, if x is the unknown. There is no set order, as the steps used depend on what is given: 1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero. 2. Apply the distributive property as needed: a(b + c) = ab + ac. 3. Isolate the variable on one side of the equation. 4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient. Example 2.9 Solving an Equation in One Variable Solve the following equation: 2x + 7 = 19. Solution This equation can be written in the form ax + b = 0 by subtracting 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations. 2x + 7 = 19 2x = 12 x = 6 Subtract 7 from both sides. Multiply both sides by 1 2 or divide by 2. The solution is x = 6. 2.5 Solve the linear equation in one variable: 2x + 1 = −9. This
content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 131 Example 2.10 Solving an Equation Algebraically When the Variable Appears on Both Sides Solve the following equation: 4(x−3) + 12 = 15−5(x + 6). Solution Apply standard algebraic properties. 4(x − 3) + 12 = 15 − 5(x + 6) 4x − 12 + 12 = 15 − 5x − 30 4x = −15 − 5x 9x = −15 x = − 15 9 x = − 5 3 Apply the distributive property. Combine like terms. Place x − terms on one side and simplify. Multiply both sides by 1 9 , the reciprocal of 9. Analysis This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . 2.6 Solve the equation in one variable: −2(3x − 1) + x = 14 − x. Solving a Rational Equation In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation. Recall that a rational number is the ratio of two numbers, such as 2 3 or 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples. x + 1 x2 − 4 , 1 x − 3 , or 4 x2 + x − 2 Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD). Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out. Example 2.11 Solving a Rational Equation Solve the rational equation: 7 2x − 5 3x = 22 3 . 132 Chapter 2 Equations and Inequalities Solution We have three denominators; 2x, 3x, and 3. The LCD must contain 2x, 3x, and 3. An LCD of 6x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6x. ⎛ (6x) ⎝ ⎛ (6x ) ⎝ 7 2x ⎡ (6x) ⎣ ⎛ ⎞ 7 ⎠ − (6x) ⎝ 2x ⎛ ⎞ ⎠ − (6x ) ⎝ ⎤ ⎦(6x) ⎞ ⎠(6x) ⎞ ⎠(6 x) ⎡ ⎣ ⎛ ⎝ ⎛ ⎝ 22 3 22 3 22 3 ⎤ 2x − 5 7 ⎦ = 3x ⎞ 5 ⎠ = 3x ⎞ 5 ⎠ = 3x 3(7) − 2(5) = 22(2x) 21 − 10 = 44x 11 = 44x 11 = x 44 1 4 = x Use the distributive property. Cancel out the common factors. Multiply remaining factors by each numerator. A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as (x + 1). Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x, x − 1, and 3x − 3. (x − 1), and 3(x − 1) as the denominators. (Note the parentheses placed First, factor all denominators. We then have x, around the second denominator.) Only the last two denominators have a common factor of (x − 1). The x in the first denominator is separate from the x in the (x − 1) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x, one factor of (x − 1), and the 3. Thus, the LCD is the following: x(x − 1)3 = 3x(x − 1) So, both sides of the equation would be multiplied by 3x(x − 1). Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out. Another example is a problem with two denominators, such as x and x2 + 2x. Once the second denominator is factored as x2 + 2x = x(x + 2), there is a common factor of x in both denominators and the LCD is x(x + 2). Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation. a b = c d We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign. Multiply a(d) and b(c), which results in ad = bc. Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities. Rational Equations A rational equation contains at least one rational expression where the variable appears in at least one of the denominators. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 133 Given a rational equation, solve it. 1. Factor all denominators in the equation. 2. Find and exclude values that set each denominator equal to zero. 3. Find the LCD. 4. Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left. 5. Solve the remaining equation. 6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator Example 2.12 Solving a Rational Equation without Factoring Solve the following rational equation: x − 3 2 2 = 7 2x Solution We have three denominators: x, 2, and 2x. No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2x. Only one value is excluded from a solution set, x = 0. Next, multiply the whole equation (both sides of the equal sign) by 2x. 2 x ⎛ ⎝ 2 x 7 2x 7 2x ⎞ ⎡ ⎤ 2x⎛ 3 ⎠ = ⎝ ⎝ 2 2(2) − 3x = 7 4 − 3x = 7 −3x = 3 ⎤ ⎦2x ⎞ ⎠2x Distribute 2x. Denominators cancel out. x = −1 or {−1} The proposed solution is x = −1, which is not an excluded value, so the solution set contains one number, x = −1, or {−1} written in set notation. 2.7 Solve the rational equation: 2 3x = 1 4 − 1 6x. Example 2.13 Solving a Rational Equation by Factoring the Denominator Solve the following rational equation: 1 x = 1 10 − 3 4x. 134 Chapter 2 Equations and Inequalities Solution find the common denominator. The three denominators in factored form are x, 10 = 2 ⋅ 5, and First 4x = 2 ⋅ 2 ⋅ x. The smallest expression that is divisible by each one of the denominators is 20x. Only x = 0 is an excluded value. Multiply the whole equation by 20x. ⎞ ⎠20x 1 10 ⎞ ⎠ = ⎛ 20x⎛ − 3 1 x ⎝ ⎝ 4x 20 = 2x − 15 35 = 2x 35 = x 2 The solution is x = 35 2 . 2.8 Solve the rational equation: − 5 2x + 3 4x = − 7 4 . Example 2.14 Solving Rational Equations with a Binomial in the Denominator Solve the following rational equations and state the excluded values: a. b. c Solution a. The denominators x and x − 6 have nothing in common. Therefore, the LCD is the product x(x − 6). However, for this problem, we can cross-multiply. 3 x − 6 = 5 x 3x = 5(x − 6) 3x = 5x − 30 −2x = −30 x = 15 Distribute. The solution is x = 15. The excluded values are x = 6 and x = 0. b. The LCD is 2(x − 3). Multiply both sides of the equation by 2(x − 3). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 135 ⎡ 2(x − 3) ⎣ x x − 3 2 (x − 3(x − 3)5 x − 3 ⎤ ⎦2(x − 3) − 2(x − 3) 2 2x = 10 − (x − 3) 2x = 10 − x + 3 2x = 13 − x 3x = 13 x = 13 3 . The excluded value is x = 3. The solution is x = 13 3 c. The least common denominator is 2(x − 2). Multiply both sides of the equation by x(x − 2). ⎡ 2(x − 2) ⎣ x x − 2 ⎤ ⎦2(x − 2 2x = 10 − (x − 2) 2x = 12 − x 3x = 12 x = 4 The solution is x = 4. The excluded value is x = 2. 2.9 Solve −3 2x + 1 = 4 3x + 1 . State the excluded values. Example 2.15 Solving a Rational Equation with Factored Denominators and Stating Excluded Values Solve the rational equation after factoring the denominators = 2x x2 − 1 . State the excluded values. Solution We must factor the denominator x2 −1. We recognize this as the difference of squares, and factor it as (x − 1)(x + 1). Thus, the LCD that contains each denominator is (x − 1)(x + 1). Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation. ⎡ (x − 1)(x + 1(x − 1) − 1(x + 1) = 2x 2x − 2 − x − 1 = 2x −3 − x = 0 −3 = x 2x (x − 1)(x + 1) ⎤ ⎦(x − 1)(x + 1) Distribute the negative sign. The solution is x = −3. The excluded values are x = 1 and x = −1. 136 Chapter 2 Equations and Inequalities 2.10 Solve the rational equation x2 − x − 2 . Finding a Linear Equation Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = mx + b, where m = slope and b = y−intercept. Let us begin with the slope. The Slope of a Line The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run. m = y2 − y1 x2 − x1 If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2.19. The lines indicate the following slopes: m = −3, m = 2, and m = 1 3 . Figure 2.19 The Slope of a Line The slope of a line, m, represents the change in y over the change in x. Given two points, (x1, y1) and (x2, y2), following formula determines the slope of a line containing these points: the m = y2 − y1 x2 − x1 Example 2.16 Finding the Slope of a Line Given Two Points Find the slope of a line that passes through the points (2, −1) and (−5, 3). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 137 Solution We substitute the y-values and the x-values into the formula. m = 3 − (−1) −5 − 2 = 4 −7 = − 4 7 The slope is − 4 7 . Analysis It does not matter which point is called (x1, y1) or (x2, y2). As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result. 2.11 Find the slope of the line that passes through the points (−2, 6) and (1, 4)
. Example 2.17 Identifying the Slope and y-intercept of a Line Given an Equation Identify the slope and y-intercept, given the equation y = − 3 4 x − 4. Solution As the line is in y = mx + b form, the given line has a slope of m = − 3 4 . The y-intercept is b = −4. Analysis The y-intercept is the point at which the line crosses the y-axis. On the y-axis, x = 0. We can always identify the y-intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 and solve for y. The Point-Slope Formula Given the slope and one point on a line, we can find the equation of the line using the point-slope formula. y − y1 = m(x − x1) This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form. The Point-Slope Formula Given one point and the slope, the point-slope formula will lead to the equation of a line: y − y1 = m(x − x1) 138 Chapter 2 Equations and Inequalities Example 2.18 Finding the Equation of a Line Given the Slope and One Point Write the equation of the line with slope m = −3 and passing through the point (4, 8). Write the final equation in slope-intercept form. Solution Using the point-slope formula, substitute −3 for m and the point (4, 8) for (x1, y1). y − y1 = m(x − x1) y − 8 = −3(x − 4) y − 8 = −3x + 12 y = −3x + 20 Analysis Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained. 2.12 Given m = 4, find the equation of the line in slope-intercept form passing through the point (2, 5). Example 2.19 Finding the Equation of a Line Passing Through Two Given Points Find the equation of the line passing through the points (3, 4) and (0, −3). Write the final equation in slopeintercept form. Solution First, we calculate the slope using the slope formula and two points. m = −3 − 4 0 − 3 = −7 −3 = 7 3 Next, we use the point-slope formula with the slope of 7 3 , and either point. Let’s pick the point (3, 4) for (x1, y1). x − 3) x − 7 x − 3 Distribute the 7 3 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 139 In slope-intercept form, the equation is written as y = 7 3 x − 3. Analysis To prove that either point can be used, let us use the second point (0, −3) and see if we get the same equationx − 0) x x − 3 We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope. Standard Form of a Line Another way that we can represent the equation of a line is in standard form. Standard form is given as Ax + By = C where A, B, and C are integers. The x- and y-terms are on one side of the equal sign and the constant term is on the other side. Example 2.20 Finding the Equation of a Line and Writing It in Standard Form Find the equation of the line with m = −6 and passing through the point ⎛ ⎝ form. 1 4 ⎞ ⎠. Write the equation in standard , −2 Solution We begin using the point-slope formula. ⎛ ⎝x − 1 y − (−2) = −6 4 y + 2 = −6x + 3 2 ⎞ ⎠ From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right. 2(y + 2) = ⎛ ⎝−6x + 3 2 2y + 4 = −12x + 3 ⎞ ⎠2 This equation is now written in standard form. 12x + 2y = −1 2.13 Find the equation of the line in standard form with slope m = − 1 3 and passing through the point ⎛ ⎝1, 1 3 ⎞ ⎠. 140 Chapter 2 Equations and Inequalities Vertical and Horizontal Lines The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as x = c where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c. Suppose that we want to find the equation of a line containing the following points: (−3, −5), (−3, 1), (−3, 3), and (−3, 5). First, we will find the slope. m = 5 − 3 −3 − (−3) = 2 0 Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through x = −3. See Figure 2.20. The equation of a horizontal line is given as y = c where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c. Suppose we want to find the equation of a line that contains the following set of points: (−2, −2), (0, −2), (3, −2), and (5, −2). We can use the point-slope formula. First, we find the slope using any two points on the line. m = −2 − (−2) 0 − (−2) = 0 2 = 0 Use any point for (x1, y1) in the formula, or use the y-intercept. y − (−2) = 0(x − 3) y + 2 = 0 y = −2 The graph is a horizontal line through y = −2. Notice that all of the y-coordinates are the same. See Figure 2.20. Figure 2.20 The line x = −3 is a vertical line. The line y = −2 is a horizontal line. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 141 Example 2.21 Finding the Equation of a Line Passing Through the Given Points Find the equation of the line passing through the given points: (1, −3) and (1, 4). Solution The x-coordinate of both points is 1. Therefore, we have a vertical line, x = 1. 2.14 Find the equation of the line passing through (−5, 2) and (2, 2). Determining Whether Graphs of Lines are Parallel or Perpendicular Parallel lines have the same slope and different y-intercepts. Lines that are parallel to each other will never intersect. For example, Figure 2.21 shows the graphs of various lines with the same slope, m = 2. Figure 2.21 Parallel lines All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts. Lines that are perpendicular intersect to form a 90° -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is −1 : m1 ⋅ m2 = −1. For example, Figure 2.22 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . 142 Chapter 2 Equations and Inequalities m1 ⋅ m2 = −1 ⎞ ⎠ = −1 ⎛ ⎝− 1 3 3 ⋅ Figure 2.22 Perpendicular lines Example 2.22 Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3y = − 4x + 3 and 3x − 4y = 8. Solution The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form. First equation: Second equation: See the graph of both lines in Figure 2.23 3y = −4x + 3 y = − 4 3 x + 1 3x − 4y = 8 −4y = −3x + 8 y = 3 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 143 Figure 2.23 From the graph, we can see that the lines appear perpendicular, but we must compare the slopes. m1 = − 4 3 m2 = 3 4 ⎛ ⎝− 4 3 m1 ⋅ m2 = ⎞ ⎛ ⎝ ⎠ 3 4 ⎞ ⎠ = −1 The slopes are negative reciprocals of each other, confirming that the lines are perpendicular. 2.15 Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2y − x = 10 and 2y = x + 4. Writing the Equations of Lines Parallel or Perpendicular to a Given Line As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line. Given an equation for a line, write the equation of a line parallel or perpendicular to it. 1. Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form. 2. Use the slope and the given point with the point-slope formula. 3. Simplify the line to slope-intercept form and compare the equation to the given line. Example 2.23 Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point 144 Chapter 2 Equations and Inequalities Write the equation of line parallel to a 5x + 3y = 1 and passing through the point (3, 5). Solution First, we will write the equation in slope-intercept form to find the slope. 5x + 3y = 1 3y = 5x + The slope is m = − 5 3 . The y-intercept is 1 3 , but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formulax − 3) x + 5 x + 10 The equation of the line is y = − 5 3 x + 10. See Figure 2.24. Figure 2.24 2.16 Find the equation of the line parallel to 5x = 7 + y and passing through the point (−1, −2). Example 2.24 Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point Find the equation of the line perpendicular to 5x − 3y + 4 = 0 (−4, 1). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 145 Solution The first step is to write the equation in slope-intercept form. 5x − 3y + 4 = 0 −3y = −5x − 4 x + 4 3 y = 5 3 We see that the slope is m = 5 3 . This means that the slope of the line perpendicular to the given line is the negative reciprocal, or − 3 5 . Next, we use the point-slope formula with this new slope and the given pointx − (−4)⎞ ⎠ x − 1
2 5 x − 12 5 x − 7 5 + 5 5 Access these online resources for additional instruction and practice with linear equations. • Solving rational equations (http://openstaxcollege.org/l/rationaleqs) • Equation of a line given two points (http://openstaxcollege.org/l/twopointsline) • Finding the equation of a line perpendicular to another line through a given point (http://openstaxcollege.org/l/findperpline) • Finding the equation of a line parallel to another line through a given point (http://openstaxcollege.org/l/findparaline) 146 Chapter 2 Equations and Inequalities 2.2 EXERCISES Verbal What does it mean when we say that two lines are 65. parallel? What 66. is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)? 67. How do we recognize when an equation, for example y = 4x + 3, will be a straight line (linear) when graphed? What does it mean when we say that a linear equation 68. is inconsistent? 69. When solving the following equation explain why we must possible solutions from the solution set. exclude x = 5 and x = −1 as Algebraic For the following exercises, solve the equation for x. 70. 7x + 2 = 3x − 9 71. 4x − 3 = 5 72. 3(x + 2) − 12 = 5(x + 1) 73. 12 − 5(x + 3) = 2x − 5 74. 75. 76 = 2x + 3 12 2 3 x + 1 2 = 31 6 77. 3(2x − 1) + x = 5x + 3 78. 79. 2x 3 − 3 4 = x 6 + 21 For the following exercises, solve each rational equation for x. State all x-values that are excluded from the solution set. 80. 3 x − 1 3 = 1 6 This content is available for free at https://cnx.org/content/col11758/1.5 81. 82. 83. 84. 85x − 1)(x − 2) 3x 6 x2 − 2x − 3 x = 1 1 5 + 3 2x For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form. 86. 87. (0, 3) with a slope of 2 3 (1, 2) with a slope of −4 5 88. x-intercept is 1, and (−2, 6) 89. y-intercept is 2, and (4, −1) 90. (−3, 10) and (5, −6) 91. (1, 3) and (5, 5) parallel to y = 2x + 5 and passes through the point 92. (4, 3) perpendicular to 3y = x − 4 and passes through the 93. point (−2, 1) . For the following exercises, find the equation of the line using the given information. 94. (−2, 0) and (−2, 5) 95. (1, 7) and (3, 7) The slope is undefined and it passes through the point 96. (2, 3). The slope equals zero and it passes through the point 97. (1, −4). 98. The slope is 3 4 and it passes through the point (1,4). Chapter 2 Equations and Inequalities 147 99. (−1, 3) and (4, −5) 112. 4,500x − 200y = 9,528 Graphical For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither. 113. 200 − 30y x = 70 Extensions 100. y = 2x + 7 y = −1 2 x − 4 101. 3x − 2y = 5 6y − 9x = 6 102. y = 3x + 1 y = 3x + 2 4 103. x = 4 y = −3 Numeric For the following exercises, find the slope of the line that passes through the given points. 104. (5, 4) and (7, 9) 105. (−3, 2) and (4, −7) 106. (−5, 4) and (2, 4) 107. (−1, −2) and (3, 4) 108. (3, −2) and (3, −2) For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular. 109. 110. (−1, 3) and (5, 1) (−2, 3) and (0, 9) (2, 5) and (5, 9) (−1, −1) and (2, 3) Technology For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y-intercept occurs. State your ymin and ymax values. 111. 0.537x − 2.19y = 100 with Starting formula the 114. y − y1 = m(x − x1), solve this expression for x in terms of x1, y, y1, and m. point-slope Starting with the standard form of an equation 115. Ax + By = C, solve this expression for y in terms of A, B, C, and x. Then put the expression in slopeintercept form. Use the above derived formula to put the following 116. standard equation in slope intercept form: 7x − 5y = 25. Given that the following coordinates are the vertices 117. of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular. (−1, 1), (2, 0), (3, 3), and (0, 4) Find the slopes of the diagonals in the previous 118. exercise. Are they perpendicular? Real-World Applications The slope for a wheelchair ramp for a home has to be . If the vertical distance from the ground to the door 119. 1 12 bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope. If the profit equation for a small business selling x item two is item one and y number of the y value 120. number of p = 3x + 4y, find p = $453 and x = 75. when For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25x, where x is the number of miles traveled. 121. What is your cost if you travel 50 mi? 122. If your cost were $63.75, how many miles were you charged for traveling? 123. 148 Chapter 2 Equations and Inequalities Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 149 2.3 \| Models and Applications Learning Objectives In this section you will: 2.3.1 Set up a linear equation to solve a real-world application. 2.3.2 Use a formula to solve a real-world application. Figure 2.25 Credit: Kevin Dooley Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer. Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems. Setting up a Linear Equation to Solve a Real-World Application To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write 0.10x. This expression represents a variable cost because it changes according to the number of miles driven. If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost C. When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 2.4 lists some common verbal expressions and their equivalent mathematical expressions. C = 0.10x + 50 150 Verbal One number exceeds another by a Twice a number One number is a more than another number One number is a less than twice another number The product of a number and a, decreased by b Chapter 2 Equations and Inequalities Translation to Math Operations x, x + a 2x x, x + a x, 2x − a ax − b The quotient of a number and the number plus a is three times the number x x + a = 3x The product of three times a number and the number decreased by b is c 3x(x − b) = c Table 2.4 Given a real-world problem, model a linear equation to fit it. 1. Identify known quantities. 2. Assign a variable to represent the unknown quantity. 3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first. 4. Write an equation interpreting the words as mathematical operations. 5. Solve the equation. Be sure the solution can be explained in words, including the units of measure. Example 2.25 Modeling a Linear Equation to Solve an Unknown Number Problem Find a linear equation to solve for the following unknown quantities: One number exceeds another number by 17 and their sum is 31. Find the two numbers. Solution Let x equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as x + 17. The sum of the two numbers is 31. We usually interpret the word is as an equal sign. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 151 Simplify and solve. x + (x + 17) = 31 2x + 17 = 31 2x = 14 x = 7 x + 17 = 7 + 17 = 24 The two numbers are 7 and 24. Find a linear equation to solve for the following unknown quantities: One number is three more than 2.17 twice another number. If the sum of the two numbers is 36, find the numbers. Example 2.26 Setting Up a Linear Equation to Solve a Real-World Application There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time. a. Write a linear equation that models the packages offered by both companies. b. c. If the average number of minutes used each month is 1,160, which company offers the better plan? If the average number of minutes used each month is 420, which company offers the better plan? d. How many minutes of talk-time would yield equal monthly st
atements from both companies? Solution a. The model for Company A can be written as A = 0.05x + 34. This includes the variable cost of 0.05x plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of $40, but a lower variable cost of 0.04x. Company B’s model can be written as B = 0.04x + $40. b. If the average number of minutes used each month is 1,160, we have the following: Company A = 0.05(1.160) + 34 = 58 + 34 = 92 Company B = 0.04(1, 1600) + 40 = 46.4 + 40 = 86.4 So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160. c. If the average number of minutes used each month is 420, we have the following: Company A = 0.05(420) + 34 = 21 + 34 = 55 Company B = 0.04(420) + 40 = 16.8 + 40 = 56.8 152 Chapter 2 Equations and Inequalities If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of $56.80. d. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of (x, y) coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x. Check the x-value in each equation. 0.05x + 34 = 0.04x + 40 0.01x = 6 x = 600 0.05(600) + 34 = 64 0.04(600) + 40 = 64 Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2.26 Figure 2.26 2.18 Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses? Using a Formula to Solve a Real-World Application Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, A = LW; the perimeter of a rectangle, P = 2L + 2W; and the volume of a rectangular solid, V = LWH. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time. Example 2.27 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 153 Solving an Application Using a Formula It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work? Solution This is a distance problem, so we can use the formula d = rt, where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution. First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or 1 2 h at rate r. His drive home takes 40 min, or 2 3 h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance d. A table, such as Table 2.5, is often helpful for keeping track of information in these types of problems. d d d r r r − 10 t 1 2 2 3 To Work To Home Table 2.5 Write two equations, one for each trip. d = rr − 10) ⎝ ⎞ ⎠ 2 3 To work To home As both equations equal the same distance, we set them equal to each other and solve for r. ⎞ ⎠ 2 3 ⎛ r⎛ ⎞ 1 ⎠ = (r − 10) ⎝ ⎝ 2 r − 20 2r = 2 1 3 3 r = − 20 3 r = − 20 3 r = − 20 3 r = 40 (−6 We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/ h. Now we can answer the question. Substitute the rate back into either equation and solve for d. The distance between home and work is 20 mi. d = 40 ⎞ ⎠ ⎛ ⎝ 1 2 = 20 154 Chapter 2 Equations and Inequalities Analysis Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for r. r⎛ ⎝ 6 × rr − 10r − 10) ⎝ 2 3r = 4(r − 10) 3r = 4r − 40 −r = −40 r = 40 ⎞ ⎠ 2 3 2.19 On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday? Example 2.28 Solving a Perimeter Problem The perimeter of a rectangular outdoor patio is 54 ft. The length is 3 ft greater than the width. What are the dimensions of the patio? Solution The perimeter formula is standard: P = 2L + 2W. We have two unknown quantities, length and width. However, we can write the length in terms of the width as L = W + 3. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure 2.27. Figure 2.27 Now we can solve for the width and then calculate the length. P = 2L + 2W 54 = 2(W + 3) + 2W 54 = 2W + 6 + 2W 54 = 4W + 6 48 = 4W 12 = W (12 + 3) = L 15 = L The dimensions are L = 15 ft and W = 12 ft. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 155 Find the dimensions of a rectangle given that the perimeter is 110 cm and the length is 1 cm more than 2.20 twice the width. Example 2.29 Solving an Area Problem The perimeter of a tablet of graph paper is 48 in.2. The length is 6 in. more than the width. Find the area of the graph paper. Solution The standard formula for area is A = LW; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one. We know that the length is 6 in. more than the width, so we can write length as L = W + 6. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length. P = 2L + 2W 48 = 2(W + 6) + 2W 48 = 2W + 12 + 2W 48 = 4W + 12 36 = 4W 9 = W (9 + 6) = L 15 = L Now, we find the area given the dimensions of L = 15 in. and W = 9 in. A = LW A = 15(9) = 135 in.2 The area is 135 in.2. 2.21 A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft2 of new carpeting should be ordered? Example 2.30 Solving a Volume Problem Find the dimensions of a shipping box given that the length is twice the width, the height is 8 inches, and the volume is 1,600 in.3. Solution 156 Chapter 2 Equations and Inequalities The formula for the volume of a box is given as V = LWH, given that L = 2W, and H = 8. The volume is 1,600 cubic inches. the product of length, width, and height. We are V = LWH 1, 600 = (2W)W(8) 1, 600 = 16W 2 100 = W 2 10 = W The dimensions are L = 20 in., W = 10 in., and H = 8 in. Analysis Note that the square root of W 2 would result in a positive and a negative value. However, because we are describing width, we can use only the positive result. Access these online resources for additional instruction and practice with models and applications of linear equations. • Problem solving using linear equations (http://openstaxcollege.org/l/lineqprobsolve) • Problem solving using equations (http://openstaxcollege.org/l/equationprsolve) • Finding the dimensions of area given the perimeter (http://openstaxcollege.org/l/ permareasolve) • Find the distance between the cities using the distance = rate * time formula (http://openstaxcollege.org/l/ratetimesolve) • Linear equation application (Write a cost equation) (http://openstaxcollege.org/l/ lineqappl) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 157 2.3 EXERCISES Verbal To set up a model linear equation to fit real-world 124. applications, what should always be the first step? Use your own words to describe this equation where n 125. is a number: 5(n + 3) = 2n 126. If the total amount of money you had to invest was $2,000 and you deposit x amount in one investment, how can you represent the remaining amount? If a man sawed a 10-ft board into two sections and long, how long would the other 127. one section was n ft section be in terms of n ? If Bill was traveling v mi/h, how would you 128. represent Daemon’s speed if he was traveling 10 mi/h faster? Real-World Applications For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked. 129. Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113? Beth and Ann are joking that their combined ages 130. equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages? Ben originally filled out 8 more applications than 131. Henry. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out? For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls. Find the model of the total cost of Company A’s plan, 132. using m for the minutes. Find the model of the total cost of Company B’s plan, 133. using m for the minutes. Find out how many
minutes of calling would make 134. the two plans equal. 135. If the person makes a monthly average of 200 min of calls, which plan should for the person choose? the following plans For the following exercises, use this scenario: A wireless that a person is carrier offers considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 5 lines, and data charges of $40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use P for the number of devices that need data plans as part of their cost. 136. Find the model of the total cost of the Family Plan. Find the model of the total cost of the Mobile Share 137. Plan. Assuming they stay under their data limit, find the 138. number of devices that would make the two plans equal in cost. If a family has 3 smart phones, which plan should 139. they choose? For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%. If we let x be the amount the woman invests in the 140. 15% bond, how much will she be able to invest in the CD? Set up and solve the equation for how much the 141. woman should invest in each option to sustain a $6,000 annual return. Two planes fly in opposite directions. One travels 450 142. mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart? Ben starts walking along a path at 4 mi/h. One and a 143. half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben? 144. Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h? 145. A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result? 158 Chapter 2 Equations and Inequalities 146. Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000? For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/ wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/ mi driven. Write the model equation for the cost of renting a 147. truck with plan A. Write the model equation for the cost of renting a 148. truck with plan B. Use the formula from the previous question to find f when p = 8 and q = 13. Solve 159. y = mx + b for m in the slope-intercept formula: Use the formula from the previous question to find the point are (4, 7) and 160. m when the coordinates of b = 12. 161. A = 1 2 The area h⎛ ⎝b1 + b2 a of trapezoid by ⎠. Use the formula to find the area of a given is ⎞ trapezoid with h = 6, b1 = 14, and b2 = 8. Find the number of miles that would generate the 149. same cost for both plans. 162. Solve for h: A = 1 2 h⎛ ⎝b1 + b2 ⎞ ⎠ If Tim knows he has to travel 300 mi, which plan 150. should he choose? For the following exercises, use the given formulas to answer the questions. 151. A = P(1 + rt) is used to find the principal amount Pdeposited, earning r% interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if A = $8,000. 152. The formula F = mv2 R relates force (F), velocity (v), mass (m), and resistance (R). Find R when m = 45, v = 7, and F = 245. F = ma indicates that force (F) equals mass (m) 153. times acceleration (a). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it. Use the formula from the previous question to find the with 163. height A = 150, b1 = 19, and b2 = 11. trapezoid of a Find the dimensions of an American football field. 164. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula P = 2L + 2W. Distance equals rate times time, d = rt. Find the 165. distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h. Using the formula in the previous exercise, find the 166. distance that Susan travels if she is moving at a rate of 60 mi/h for 6.75 h. What is the total distance that two people travel in 3 h 167. if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h? 154. Sum = 1 1 − r is the formula for an infinite series 168. sum. If the sum is 5, find r. If the area model for a triangle is A = 1 2 bh, find the area of a triangle with a height of 16 in. and a base of 11 in. For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question. 169. Solve for h: A = 1 2 bh 155. Solve for W: P = 2L + 2W Use the formula from the previous question to find the 156. width, W, of a rectangle whose length is 15 and whose perimeter is 58. Solve for 157. 158. Use the formula from the previous question to find the 170. height to the nearest tenth of a triangle with a base of 15 and an area of 215. 171. The volume formula for a cylinder is V = πr 2 h. Using the symbol π in your answer, find the volume of a cylinder with a radius, r, of 4 cm and a height of 14 cm. Solve for h: V = πr 2 h 172. 173. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 159 Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of 16π 174. Solve for r: V = πr 2 h Use the formula from the previous question to find the 175. radius of a cylinder with a height of 36 and a volume of 324π. The formula for the circumference of a circle is 176. C = 2πr. Find the circumference of a circle with a diameter of 12 in. (diameter = 2r). Use the symbol π in your final answer. Solve the formula from the previous question for π. 177. Notice why π is sometimes defined as the ratio of the circumference to its diameter. 160 Chapter 2 Equations and Inequalities 2.4 \| Complex Numbers Learning Objectives In this section you will: 2.4.1 Add and subtract complex numbers. 2.4.2 Multiply and divide complex numbers. 2.4.3 Solve quadratic equations with complex numbers Figure 2.28 Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple. In order to better understand it, we need to become familiar with a new set of numbers. Keep in mind that the study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it. Expressing Square Roots of Negative Numbers as Multiples of i We know how to find the square root of any positive real number. In a similar way, we can find the square root of any negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number i is defined as the square root of −1. So, using properties of radicals, −1 = i i2 = ( −1)2 = −1 We can write the square root of any negative number as a multiple of i. Consider the square root of −49. We use 7i and not −7i because the principal root of 49 is the positive root. −49 = 49 ⋅ (−1) = 49 −1 = 7i This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 161 A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a + bi where a is the real part and b is the imaginary part. For example, 5 + 2i is a complex number. So, too, is 3 + 4i 3. Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers. Imaginary and Complex Numbers A complex number is a number of the form a + bi where • a is the real part of the complex number. • b is the imaginary part of the complex number. If b = 0, imaginary number. An imaginary number is an even root of a negative number. then a + bi is a real number. If a = 0 and b is not equal to 0, the complex number is called a pure Given an imaginary number, express it in the standard form of a complex number. 1. Write −a as a −1. 2. Express −1 as i. 3. Write a ⋅ i in simplest form. Example 2.31 Expressing an Imaginary Number in Standard Form Express −9 in standard form. Solution In standard form, this is 0 + 3i. −9 = 9 −1 = 3i 2.22 Express −24 in standard form. Plotting a Complex Number on the Complex Plane We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number, we need to address the two components of the number. We use
the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs (a, b), where a represents the coordinate for the horizontal axis and b represents the coordinate for the vertical axis. 162 Chapter 2 Equations and Inequalities Let’s consider the number −2 + 3i. The real part of the complex number is −2 and the imaginary part is 3. We plot the ordered pair (−2, 3) to represent the complex number −2 + 3i, as shown in Figure 2.29. Figure 2.29 Complex Plane In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis, as shown in Figure 2.30. Figure 2.30 Given a complex number, represent its components on the complex plane. 1. Determine the real part and the imaginary part of the complex number. 2. Move along the horizontal axis to show the real part of the number. 3. Move parallel to the vertical axis to show the imaginary part of the number. 4. Plot the point. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 163 Example 2.32 Plotting a Complex Number on the Complex Plane Plot the complex number 3 − 4i on the complex plane. Solution The real part of the complex number is 3, and the imaginary part is –4. We plot the ordered pair (3, −4) as shown in Figure 2.31. Figure 2.31 2.23 Plot the complex number −4 − i on the complex plane. Adding and Subtracting Complex Numbers Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and then combine the imaginary parts. Complex Numbers: Addition and Subtraction Adding complex numbers: Subtracting complex numbers: (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i 164 Chapter 2 Equations and Inequalities Given two complex numbers, find the sum or difference. 1. Identify the real and imaginary parts of each number. 2. Add or subtract the real parts. 3. Add or subtract the imaginary parts. Example 2.33 Adding and Subtracting Complex Numbers Add or subtract as indicated. 1. 2. (3 − 4i) + (2 + 5i) (−5 + 7i) − (−11 + 2i) Solution We add the real parts and add the imaginary parts. 1. 2. (3 − 4i) + (2 + 5i) = 3 − 4i + 2 + 5i = 3 + 2 + (−4i) + 5i = (3 + 2) + (−4 + 5)i = 5 + i (−5 + 7i) − (−11 + 2i) = −5 + 7i + 11 − 2i = −5 + 11 + 7i − 2i = (−5 + 11) + (7 − 2)i = 6 + 5i 2.24 Subtract 2 + 5i from 3 – 4i. Multiplying Complex Numbers Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately. Multiplying a Complex Number by a Real Number Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example, 3(6 + 2i) : Given a complex number and a real number, multiply to find the product. 1. Use the distributive property. 2. Simplify. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 165 Example 2.34 Multiplying a Complex Number by a Real Number Find the product 4(2 + 5i). Solution Distribute the 4. 2.25 Find the product: 1 2 (5 − 2i). 4(2 + 5i) = (4 ⋅ 2) + (4 ⋅ 5i) = 8 + 20i Multiplying Complex Numbers Together Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. The difference with complex numbers is that when we get a squared term, i2, it equals −1. (a + bi)(c + di) = ac + adi + bci + bdi2 = ac + adi + bci − bd = (ac − bd) + (ad + bc)i i2 = −1 Group real terms and imaginary terms. Given two complex numbers, multiply to find the product. 1. Use the distributive property or the FOIL method. 2. Remember that i2 = −1. 3. Group together the real terms and the imaginary terms Example 2.35 Multiplying a Complex Number by a Complex Number Multiply: (4 + 3i)(2 − 5i). Solution (4 + 3i)(2 − 5i) = 4(2) − 4(5i) + 3i(2) − (3i)(5i) ⎝i2⎞ ⎛ = 8 − 20i + 6i − 15 ⎠ = (8 + 15) + (−20 + 6)i = 23 − 14i 2.26 Multiply: (3 − 4i)(2 + 3i). 166 Chapter 2 Equations and Inequalities Dividing Complex Numbers Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form a + bi. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of a + bi is a − bi. For example, the product of a + bi and a − bi is (a + bi)(a − bi) = a2 − abi + abi − b2 i2 = a2 + b2 The result is a real number. Note that complex conjugates have an opposite relationship: The complex conjugate of a + bi is a − bi, and the complex conjugate of a − bi is a + bi. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another. Suppose we want to divide c + di by a + bi, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply. c + di a + bi where a ≠ 0 and b ≠ 0 Multiply the numerator and denominator by the complex conjugate of the denominator. (c + di) (a + bi) ⋅ (a − bi) (a − bi) = (c + di)(a − bi) (a + bi)(a − bi) Apply the distributive property. Simplify, remembering that i2 = −1. = ca − cbi + adi − bdi2 a2 − abi + abi − b2 i2 = = ca − cbi + adi − bd(−1) a2 − abi + abi − b2(−1) (ca + bd) + (ad − cb)i a2 + b2 The Complex Conjugate The complex conjugate of a complex number a + bi is a − bi. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged. • When a complex number is multiplied by its complex conjugate, the result is a real number. • When a complex number is added to its complex conjugate, the result is a real number. Example 2.36 Finding Complex Conjugates Find the complex conjugate of each number. 1. 2 + i 5 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 167 2. − 1 2 i Solution 1. The number is already in the form a + bi. The complex conjugate is a − bi, or 2 − i 5. 2. We can rewrite this number in the form a + bi as 0 − 1 2 i. The complex conjugate is a − bi, or 0 + 1 2 i. This can be written simply as 1 2 i. Analysis Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i. 2.27 Find the complex conjugate of −3 + 4i. Given two complex numbers, divide one by the other. 1. Write the division problem as a fraction. 2. Determine the complex conjugate of the denominator. 3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. 4. Simplify. Example 2.37 Dividing Complex Numbers Divide: (2 + 5i) by (4 − i). Solution We begin by writing the problem as a fraction. (2 + 5i) (4 − i) Then we multiply the numerator and denominator by the complex conjugate of the denominator. (2 + 5i) (4 − i) ⋅ (4 + i) (4 + i) To multiply two complex numbers, we expand the product as we would with polynomials (using FOIL). 168 Chapter 2 Equations and Inequalities (2 + 5i) (4 − i) ⋅ (4 + i) (4 + i) = 8 + 2i + 20i + 5i2 16 + 4i − 4i − i2 8 + 2i + 20i + 5(−1) 16 + 4i − 4i − (−1) = Because i2 = −1. = 3 + 22i 17 = 3 + 22 17 17 i Separate real and imaginary parts. Note that this expresses the quotient in standard form. Simplifying Powers of i The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers. i1 = i i2 = −1 i3 = i2 ⋅ i = −1 ⋅ i = −i i4 = i3 ⋅ i = −i ⋅ i = −i2 = − (−1) = 1 i5 = i4 ⋅ i = 1 ⋅ i = i We can see that when we get to the fifth power of i, increasing powers, we will see a cycle of four. Let’s examine the next four powers of i. it is equal to the first power. As we continue to multiply i by i6 = i5 ⋅ i = i ⋅ i = i2 = −1 i7 = i6 ⋅ i = i2 ⋅ i = i3 = −i i8 = i7 ⋅ i = i3 ⋅ i = i4 = 1 i9 = i8 ⋅ i = i4 ⋅ i = i5 = i The cycle is repeated continuously: i, −1, − i, 1, every four powers. Example 2.38 Simplifying Powers of i Evaluate: i35. Solution Since i4 = 1, we can simplify the problem by factoring out as many factors of i4 as possible. To do so, first determine how many times 4 goes into 35: 35 = 4 ⋅ 8 + 3. i35 = i4 ⋅ 8 + 3 = i4 ⋅ 8 ⋅ i3 = 8 ⎝i4⎞ ⎛ ⎠ ⋅ i3 = 18 ⋅ i3 = i3 = − i 2.28 Evaluate: i18 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 169 Can we write i35 in other helpful ways? As we saw in Example 2.38, we reduced i35 to i3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of i35 may be more useful. Table 2.6 shows some other possible factorizations. Factorization of i35 i34 ⋅ i i33 ⋅ i2 i31 ⋅ i4 i19 ⋅ i16 Reduced form 17 ⎝i2⎞ ⎛ ⎠ ⋅ i i33 ⋅ (−1) i31 ⋅ 1 4 i19 ⋅ ⎝i4⎞ ⎛ ⎠ Simplified form (−1)17 ⋅ i −i33 i31 i19 Table 2.6 Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method. Access these online resources for additional instruction and practice with complex numbers. • Adding and Subtracting Complex Numbers (http://openstaxcollege.org/l/addsubcomplex) • Multiply Compl
ex Numbers (http://openstaxcollege.org/l/multiplycomplex) • Multiplying Complex Conjugates (http://openstaxcollege.org/l/multcompconj) • Raising i to Powers (http://openstaxcollege.org/l/raisingi) 170 Chapter 2 Equations and Inequalities 2.4 EXERCISES Verbal 178. Explain how to add complex numbers. 179. What complex numbers? is the basic principle in multiplication of Give an example to show that the product of two 180. imaginary numbers is not always imaginary. What is a characteristic of the plot of a real number in 181. the complex plane? Algebraic the following exercises, evaluate the algebraic For expressions. 182. 183. 184. 185. If y = x2 + x − 4, evaluate y given x = 2i. If y = x3 − 2, evaluate y given x = i. If y = x2 + 3x + 5, evaluate y given x = 2 + i. If y = 2x2 + x − 3, evaluate y given x = 2 − 3i. 186. If y = x + 1 2 − x, evaluate y given x = 5i. 187. If y = 1 + 2x x + 3 , evaluate y given x = 4i. Graphical For the following exercises, plot the complex numbers on the complex plane. 188. 1 − 2i 189. −2 + 3i 190. i 191. −3 − 4i Numeric For the following exercises, perform the indicated operation and express the result as a simplified complex number. 192. (3 + 2i) + (5 − 3i) 193. (−2 − 4i) + (1 + 6i) 194. (−5 + 3i) − (6 − i) This content is available for free at https://cnx.org/content/col11758/1.5 195. (2 − 3i) − (3 + 2i) 196. (−4 + 4i) − (−6 + 9i) 197. (2 + 3i)(4i) 198. (5 − 2i)(3i) 199. (6 − 2i)(5) 200. (−2 + 4i)(8) 201. (2 + 3i)(4 − i) 202. (−1 + 2i)(−2 + 3i) 203. (4 − 2i)(4 + 2i) 204. (3 + 4i)(3 − 4i) 205. 206. 3 + 4i 2 6 − 2i 3 207. −5 + 3i 2i 208. 209. 210. 211. 212. 213. 214. 215. 6 + 4i i 2 − 3i 4 + 3i 3 + 4i 2 − i 2 + 3i 2 − 3i −9 + 3 −16 − −4 − 4 −25 2 + −12 2 4 + −20 2 216. i8 217. i15 Chapter 2 Equations and Inequalities 171 218. i22 Technology For the following exercises, use a calculator to help answer the questions. 219. Evaluate (1 + i) k for k = 4, 8, and 12. Predict the value if k = 16. 220. Evaluate (1 − i) k for k = 2, 6, and 10. Predict the value if k = 14. 221. Evaluate (l + i) Predict the value for k = 16. − (l − i) k k for k = 4, 8, and 12. 222. 223. Show that a solution of x6 + 1 = 0 is 3 2 + 1 2 i. Show that a solution of x8 −1 = 0 is 2 2 + 2 2 i. Extensions For the following exercises, evaluate the expressions, writing the result as a simplified complex number. 224. 225. i + 4 1 i3 1 i11 − 1 i21 226. i7 ⎛ ⎝1 + i2⎞ ⎠ 227. i−3 + 5i7 228. (2 + i)(4 − 2i) (1 + i) 229. (1 + 3i)(2 − 4i) (1 + 2i) 230. (3 + i)2 (1 + 2i)2 231. 3 + 2i 2 + i + (4 + 3i) 232. 4 + i i + 3 − 4i 1 − i 233. 3 + 2i 1 + 2i − 2 − 3i 3 + i 172 Chapter 2 Equations and Inequalities 2.5 \| Quadratic Equations Learning Objectives In this section you will: 2.5.1 Solve quadratic equations by factoring. 2.5.2 Solve quadratic equations by the square root property. 2.5.3 Solve quadratic equations by completing the square. 2.5.4 Solve quadratic equations by using the quadratic formula. Figure 2.32 The computer monitor on the left in Figure 2.32 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods. Solving Quadratic Equations by Factoring An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as 2x2 + 3x − 1 = 0 and x2 − 4 = 0 are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics. Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if a ⋅ b = 0, then a = 0 or b = 0, where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero. Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression (x − 2)(x + 3) by multiplying the two factors together. (x − 2)(x + 3) = x2 + 3x − 2x − 6 = x2 + x − 6 The product is a quadratic expression. Set equal to zero, x2 + x − 6 = 0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 173 The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, ax2 + bx + c = 0, where a, b, and c are real numbers, and a ≠ 0. The equation x2 + x − 6 = 0 is in standard form. We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section. The Zero-Product Property and Quadratic Equations The zero-product property states If a ⋅ b = 0, then a = 0 or b = 0, where a and b are real numbers or algebraic expressions. A quadratic equation is an equation containing a second-degree polynomial; for example where a, b, and c are real numbers, and if a ≠ 0, it is in standard form. ax2 + bx + c = 0 Solving Quadratics with a Leading Coefficient of 1 In the quadratic equation x2 + x − 6 = 0, factoring quadratic equations in this form. the leading coefficient, or the coefficient of x2, is 1. We have one method of Given a quadratic equation with the leading coefficient of 1, factor it. 1. Find two numbers whose product equals c and whose sum equals b. 2. Use those numbers to write two factors of the form (x + k) or (x − k), where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2, the factors are (x + 1)(x − 2). 3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable. Example 2.39 Factoring and Solving a Quadratic with Leading Coefficient of 1 Factor and solve the equation: x2 + x − 6 = 0. Solution To factor x2 + x − 6 = 0, we look for two numbers whose product equals −6 and whose sum equals 1. Begin by looking at the possible factors of −6. 1 ⋅ (−6) (−6) ⋅ 1 2 ⋅ (−3) 3 ⋅ (−2) The last pair, 3 ⋅ (−2) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors. 174 Chapter 2 Equations and Inequalities To solve this equation, we use the zero-product property. Set each factor equal to zero and solve. (x − 2)(x + 3) = 0 (x − 2)(x + 3) = 0 (x − 2) = 0 x = 2 (x + 3) = 0 x = −3 The two solutions are x = 2 and x = −3. We can see how the solutions relate to the graph in Figure 2.33. The solutions are the x-intercepts of x2 + x − 6 = 0. Figure 2.33 2.29 Factor and solve the quadratic equation: x2 − 5x − 6 = 0. Example 2.40 Solve the Quadratic Equation by Factoring Solve the quadratic equation by factoring: x2 + 8x + 15 = 0. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 175 Solution Find two numbers whose product equals 15 and whose sum equals 8. List the factors of 15. 1 ⋅ 15 3 ⋅ 5 (−1) ⋅ (−15) (−3) ⋅ (−5) The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve. (x + 3)(x + 5) = 0 (x + 3) = 0 x = −3 (x + 5) = 0 x = −5 The solutions are x = −3 and x = −5. 2.30 Solve the quadratic equation by factoring: x2 − 4x − 21 = 0. Example 2.41 Using the Zero-Product Property to Solve a Quadratic Equation Written as the Difference of Squares Solve the difference of squares equation using the zero-product property: x2 − 9 = 0. Solution Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property. x2 − 9 = 0 (x − 3)(x + 3) = 0 (x − 3) = 0 x = 3 (x + 3) = 0 x = −3 The solutions are x = 3 and x = −3. 2.31 Solve by factoring: x2 − 25 = 0. 176 Chapter 2 Equations and Inequalities Factoring and Solving a Quadratic Equation of Higher Order When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures: 1. With the quadratic in standard form, ax2 + bx + c = 0, multiply a ⋅ c. 2. Find two numbers whose product equals ac and whose sum equals b. 3. Rewrite the equation replacing the bx term with two terms using the numbers found in step 1 as coefficients of x. 4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping. 5. Factor out the expression in parentheses. 6. Set the expressions equal to zero and solve for the variable. Example 2.42 Solving a Quadratic Equation Using Grouping Use grouping to factor and solve the quadratic equation: 4x2 + 15x + 9 = 0. Solution First, multiply ac : 4(9) = 36. Then list the factors of 36. 1 ⋅ 36 2 ⋅ 18 3 ⋅ 12 4 ⋅ 9 6 ⋅ 6 The only pair of factors that sums to 15 is 3 + 12. Rewrite the equation replacing the b term, 15x, with two terms using 3 and 12 as coefficients of x. Factor the first two terms, and then
factor the last two terms. Solve using the zero-product property. 4x2 + 3x + 12x + 9 = 0 x(4x + 3) + 3(4x + 3) = 0 (4x + 3)(x + 3) = 0 (4x + 3)(x + 3) = 0 (4x + 3) = 0 x = − 3 4 (x + 3) = 0 x = −3 The solutions are x = − 3 4 , x = −3. See Figure 2.34. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 177 Figure 2.34 2.32 Solve using factoring by grouping: 12x2 + 11x + 2 = 0. Example 2.43 Solving a Higher Degree Quadratic Equation by Factoring Solve the equation by factoring: −3x3 − 5x2 − 2x = 0. Solution This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out − x from all of the terms and then proceed with grouping. −3x3 − 5x2 − 2x = 0 ⎞ − x⎛ ⎝3x2 + 5x + 2 ⎠ = 0 Use grouping on the expression in parentheses. − x⎛ ⎞ ⎝3x2 + 3x + 2x + 2 ⎠ = 0 −x[3x(x + 1) + 2(x + 1)] = 0 −x(3x + 2)(x + 1) = 0 Now, we use the zero-product property. Notice that we have three factors. 178 Chapter 2 Equations and Inequalities −x = 0 x = 0 3x + 1 The solutions are x = 0, x = − 2 3 , and x = −1. 2.33 Solve by factoring: x3 + 11x2 + 10x = 0. Using the Square Root Property When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the x2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x2 term so that the square root property can be used. The Square Root Property With the x2 term isolated, the square root property states that: where k is a nonzero real number. if x2 = k, then x = ± k Given a quadratic equation with an x2 term but no x term, use the square root property to solve it. 1. Isolate the x2 term on one side of the equal sign. 2. Take the square root of both sides of the equation, putting a ± sign before the expression on the side opposite the squared term. 3. Simplify the numbers on the side with the ± sign. Example 2.44 Solving a Simple Quadratic Equation Using the Square Root Property Solve the quadratic using the square root property: x2 = 8. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 179 Take the square root of both sides, and then simplify the radical. Remember to use a ± sign before the radical symbol. x2 = 8 x = ± 8 = ±2 2 The solutions are x = 2 2, x = −2 2. Example 2.45 Solving a Quadratic Equation Using the Square Root Property Solve the quadratic equation: 4x2 + 1 = 7. Solution First, isolate the x2 term. Then take the square root of both sides. 4x2 + 1 = 7 4x2 = 6 x2 = 6 4 x = ± 6 2 The solutions are .34 Solve the quadratic equation using the square root property: 3(x − 4)2 = 15. Completing the Square Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square. We will use the example x2 + 4x + 1 = 0 to illustrate each step. 1. Given a quadratic equation that cannot be factored, and with a = 1, first add or subtract the constant term to the right sign of the equal sign. 2. Multiply the b term by 1 2 and square it. x2 + 4x = −1 180 Chapter 2 Equations and Inequalities 1 2 (4) = 2 22 = 4 3. Add ⎛ ⎝ 2 b⎞ ⎠ 1 2 to both sides of the equal sign and simplify the right side. We have 4. The left side of the equation can now be factored as a perfect square. x2 + 4x + 4 = −1 + 4 x2 + 4x + 4 = 3 5. Use the square root property and solve. x2 + 4x + 4 = 3 (x + 2)2 = 3 (x + 2)2 ± 3 6. The solutions are x = −2 + 3, x = −2 − 3. Example 2.46 Solving a Quadratic by Completing the Square Solve the quadratic equation by completing the square: x2 − 3x − 5 = 0. Solution First, move the constant term to the right side of the equal sign. x2 − 3x = 5 Then, take 1 2 of the b term and square it. 1 2 ⎛ ⎝− 3 2 (−3 Add the result to both sides of the equal sign. x2 − 3x + 2 ⎞ ⎛ ⎝− 3 ⎠ 2 x2 − 3x + 9 4 = 5 + 2 ⎛ ⎝− 3 2 ⎞ ⎠ = 5 + 9 4 Factor the left side as a perfect square and simplify the right side. 2 ⎛ ⎝x − 3 2 ⎞ ⎠ = 29 4 Use the square root property and solve. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 181 2 ⎛ ⎞ = ± 29 ⎝x − 3 ⎠ 4 2 ⎞ ⎛ ⎠ = ± 29 ⎝x − 3 2 2 ± 29 x = 3 2 2 The solutions are x = 3 2 + 29 2 , x = 3 2 − 29 2 . 2.35 Solve by completing the square: x2 − 6x = 13. Using the Quadratic Formula The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number. We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 and obtain a positive a. Given ax2 + bx + c = 0, a ≠ 0, we will complete the square as follows: 1. First, move the constant term to the right side of the equal sign: ax2 + bx = − c 2. As we want the leading coefficient to equal 1, divide through by a: x2 + b ax = − c a 3. Then, find 1 2 of the middle term, and add ⎛ ⎝ 2 b a ⎞ ⎠ 1 2 = b2 4a2 to both sides of the equal sign: x2 + b ax + b2 4a2 = b2 4a2 − c a 4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction: 5. Now, use the square root property, which gives ⎛ ⎝x + 2 ⎞ ⎠ b 2a = b2 − 4ac 4a2 x + x + b2 − 4ac b 2a = ± 4a2 2a = ± b2 − 4ac 2a b 6. Finally, add − b 2a to both sides of the equation and combine the terms on the right side. Thus, x = −b ± b2 − 4ac 2a 182 Chapter 2 Equations and Inequalities The Quadratic Formula Written in standard form, ax2 + bx + c = 0, any quadratic equation can be solved using the quadratic formula: x = −b ± b2 − 4ac 2a (2.1) where a, b, and c are real numbers and a ≠ 0. Given a quadratic equation, solve it using the quadratic formula 1. Make sure the equation is in standard form: ax2 + bx + c = 0. 2. Make note of the values of the coefficients and constant term, a, b, and c. 3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula. 4. Calculate and solve. Example 2.47 Solve the Quadratic Equation Using the Quadratic Formula Solve the quadratic equation: x2 + 5x + 1 = 0. Solution Identify the coefficients: a = 1, b = 5, c = 1. Then use the quadratic formula. x = −(5) ± (5)2 − 4(1)(1) 2(1) = −5 ± 25 − 4 2 = −5 ± 21 2 Example 2.48 Solving a Quadratic Equation with the Quadratic Formula Use the quadratic formula to solve x2 + x + 2 = 0. Solution First, we identify the coefficients: a = 1, b = 1, and c = 2. Substitute these values into the quadratic formula. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 183 x = −b ± b2 − 4ac 2a = −(1) ± (1)2 − (4) ⋅ (1) ⋅ (2) 2 ⋅ 1 = −1 ± 1 − 8 2 2 = −1 ± −7 = −1 ± i 7 2 The solutions to the equation are x = −1 + i 7 2 and x = −1 − i 7 2 or x = −1 2 + i 7 2 and x = −1 2 − i 7 2 . 2.36 Solve the quadratic equation using the quadratic formula: 9x2 + 3x − 2 = 0. The Discriminant The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, b2 − 4ac. The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 2.7 relates the value of the discriminant to the solutions of a quadratic equation. Value of Discriminant Results b2 − 4ac = 0 One rational solution (double solution) b2 − 4ac > 0, perfect square Two rational solutions b2 − 4ac > 0, not a perfect square Two irrational solutions b2 − 4ac < 0 Table 2.7 The Discriminant Two complex solutions For ax2 + bx + c = 0, where a, b, and c are real numbers, the discriminant is the expression under the radical in the quadratic formula: b2 − 4ac. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect. 184 Chapter 2 Equations and Inequalities Example 2.49 Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation Use the discriminant to find the nature of the solutions to the following quadratic equations: a. b. c. d. x2 + 4x + 4 = 0 8x2 + 14x + 3 = 0 3x2 − 5x − 2 = 0 3x2 − 10x + 15 = 0 Solution Calculate the discriminant b2 − 4ac for each equation and state the expected type of solutions. a. x2 + 4x + 4 = 0 b2 − 4ac = (4)2 − 4(1)(4) = 0. There will be one rational double solution. b. 8x2 + 14x + 3 = 0 b2 − 4ac = (14)2 − 4(8)(3) = 100. As 100 is a perfect square, there will be two rational solutions. c. 3x2 − 5x − 2 = 0 b2 − 4ac = (−5)2 − 4(3)(−2) = 49. As 49 is a perfect square, there will be two rational solutions. d. 3x2 −10x + 15 = 0 b2 − 4ac = (−10)2 − 4(3)(15) = −80. There will be two complex solutions. Using the Pythagorean Theorem One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as a2 +
b2 = c2, where a and b refer to the legs of a right triangle adjacent to the 90° angle, and c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications. We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side. The Pythagorean Theorem is given as a2 + b2 = c2 where a and b refer to the legs of a right triangle adjacent to the 90∘ angle, and c refers to the hypotenuse, as shown in Figure 2.35. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 185 Figure 2.35 Example 2.50 Finding the Length of the Missing Side of a Right Triangle Find the length of the missing side of the right triangle in Figure 2.36. Figure 2.36 Solution As we have measurements for side b and the hypotenuse, the missing side is a. a2 + b2 = c2 a2 + (4)2 = (12)2 a2 + 16 = 144 a2 = 128 a = 128 = 8 2 Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg 2.37 b measures 3 units. Find the length of the hypotenuse. Access these online resources for additional instruction and practice with quadratic equations. • Solving Quadratic Equations by Factoring (http://openstaxcollege.org/l/quadreqfactor) • The Zero-Product Property (http://openstaxcollege.org/l/zeroprodprop) • Completing the Square (http://openstaxcollege.org/l/complthesqr) • Quadratic Formula with Two Rational Solutions (http://openstaxcollege.org/l/ quadrformrat) • Length of a leg of a right triangle (http://openstaxcollege.org/l/leglengthtri) 186 Chapter 2 Equations and Inequalities 2.5 EXERCISES Verbal 234. How do we recognize when an equation is quadratic? For the following exercises, solve the quadratic equation by using the square root property. 235. When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why form when quadratic ax2 + bx + c = 0 we may equation y = ax2 + bx + c and have no zeroes (x-intercepts). equation graph in the solving the a 236. When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side? 237. In the quadratic formula, what is the name of the expression under the radical sign b2 − 4ac, and how does it determine the number of and nature of our solutions? Describe two scenarios where using the square root 238. property to solve a quadratic equation would be the most efficient method. Algebraic 252. x2 = 36 253. x2 = 49 254. (x − 1)2 = 25 255. (x − 3)2 = 7 256. (2x + 1)2 = 9 257. (x − 5)2 = 4 For the following exercises, solve the quadratic equation by completing the square. Show each step. 258. 259. x2 − 9x − 22 = 0 2x2 − 8x − 5 = 0 For the following exercises, solve the quadratic equation by factoring. 260. x2 − 6x = 13 239. 240. 241. 242. 243. x2 + 4x − 21 = 0 x2 − 9x + 18 = 0 2x2 + 9x − 5 = 0 6x2 + 17x + 5 = 0 4x2 − 12x + 8 = 0 244. 3x2 − 75 = 0 245. 8x2 + 6x − 9 = 0 246. 4x2 = 9 247. 2x2 + 14x = 36 248. 5x2 = 5x + 30 249. 4x2 = 5x 250. 7x2 + 3x = 0 251. x 3 − 9 x = 2 This content is available for free at https://cnx.org/content/col11758/1.5 261. x2 + 2 3 x − 1 3 = 0 262. 2 + z = 6z2 263. 6p2 + 7p − 20 = 0 264. 2x2 − 3x − 1 = 0 For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve. 265. 266. 267. 268. 269. 270. 2x2 − 6x + 7 = 0 x2 + 4x + 7 = 0 3x2 + 5x − 8 = 0 9x2 − 30x + 25 = 0 2x2 − 3x − 7 = 0 6x2 − x − 2 = 0 For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution. Chapter 2 Equations and Inequalities 187 271. 2x2 + 5x + 3 = 0 272. x2 + x = 4 273. 274. 275. 276. 2x2 − 8x − 5 = 0 3x2 − 5x + 1 = 0 x2 + 4x + 2 = 0 4 + 1 x − 1 x2 = 0 Technology For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the x-intercepts) by using 2nd CALC 2:zero. Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth. 277. 278. 279. 280. Y1 = 4x2 + 3x − 2 Y1 = −3x2 + 8x − 1 Y1 = 0.5x2 + x − 7 To solve the quadratic equation x2 + 5x − 7 = 4, we can graph these two equations Y1 = x2 + 5x − 7 Y2 = 4 and find the points of intersection. Recall 2nd CALC 5:intersection. Do this and find the solutions to the nearest tenth. To the 281. 0.3x2 + 2x − 4 = 2, we can graph these two equations quadratic solve equation Y1 = 0.3x2 + 2x − 4 Y2 = 2 and find the points of intersection. Recall 2nd CALC 5:intersection. Do this and find the solutions to the nearest tenth. Extensions Beginning with the general form of a quadratic 282. equation, ax2 + bx + c = 0, solve for x by using the completing the square method, thus deriving the quadratic formula. Show that 283. quadratic equation is −b a . the sum of the two solutions to the A person has a garden that has a length 10 feet longer 284. than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft.2. Solve the quadratic equation to find the length and width. Abercrombie and Fitch stock had a price given as 285. P = 0.2t 2 − 5.6t + 50.2, where t is the time in months from 1999 to 2001. ( t = 1 is January 1999). Find the two months in which the price of the stock was $30. an that Suppose equation given the is 286. p = −2x2 + 280x − 1000, where x represents number of items sold at an auction and p is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2nd CALC maximum. To obtain a good window for the curve, set x [0,200] and y [0,10000]. Real-World Applications 287. man A formula for the normal systolic blood pressure for a as in mmHg, age A, measured given is P = 0.006A2 − 0.02A + 120. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg. The cost function for a revenue certain company is 288. C = 60x + 300 and by R = 100x − 0.5x2. Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300. given the is A falling object travels a distance given by the 289. formula d = 5t + 16t 2 ft, where t is measured in seconds. How long will it take for the object to traveled 74 ft? A vacant lot is being converted into a community 290. garden. The garden and the walkway around its perimeter have an area of 378 ft2. Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long. 188 Chapter 2 Equations and Inequalities 291. An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P, who contracted the flu t days after is given by the model P = − t 2 + 13t + 130, where 1 ≤ t ≤ 6. Find the day that 160 students had the flu. Recall that the restriction on t is at most 6. it broke out This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 189 2.6 \| Other Types of Equations Learning Objectives In this section you will: 2.6.1 Solve equations involving rational exponents. 2.6.2 Solve equations using factoring. 2.6.3 Solve radical equations. 2.6.4 Solve absolute value equations. 2.6.5 Solve other types of equations. We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes. Solving Equations Involving Rational Exponents Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, 16 exponents is a useful skill, as it is highly applicable in calculus. 1 2 is another way of writing 16; 8 1 3 3 is another way of writing 8 . The ability to work with rational We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, 2 3 ⎞ ⎠ = 1, and so on. Rational Exponents A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent: m n a = m ⎛ ⎜am ) = amn = ( an ) m Example 2.51 Evaluating a Number Raised to a Rational Exponent Evaluate 8 2 3. Solution 190 Chapter 2 Equations and Inequalities Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite 8 ⎛ 2 3 as ⎜.38 Evaluate 64 − 1 3. Example 2.52 2 ⎛ ⎜8 ⎝ 1 3 ⎞ ⎟ ⎠ = (2)2 = 4 Solve the Equation Including a Variable Raised to a Rational Exponent Solve the equation in which a variable is raised to a rational exponent: x 5 4 = 32. Solution The way to remove the
exponent on x is by raising both sides of the equation to a power that is the reciprocal of 5 4 , which is 4 5 . 5 4 = 32 4 5 = (32) 4 5 x ⎛ ⎜2)4 = 16 The fi th root of 32 is 2. 2.39 Solve the equation x 3 2 = 125. Example 2.53 Solving an Equation Involving Rational Exponents and Factoring Solve 3x 3 4 = x 1 2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 191 Solution This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero. 3 4 − 3x ⎛ ⎜ Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with 3 4 − x 3x the lowest exponent. Rewrite x 1 2 as x 2 4. Then, factor out x 2 4 from both terms on the left ⎠ 3x ⎛ ⎜3x ⎝ 2 4 x Where did x 1 4 come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply x before the factoring, which is what should happen. We need an exponent such that when added to 2 4 2 4 back in using the distributive property, we get the expression we had equals 3 4 . Thus, the exponent on x in the parentheses is 1 4 . Let us continue. Now we have two factors and can use the zero factor theorem. 2 4 x ⎛ ⎜3x ⎝ ⎞ 3x 3x x 1 4 ⎛ ⎜ 81 Divide both sides by 3. 4 Raise both sides to the reciprocal of 1 4 . The two solutions are x = 0, x = 1 81 . 2.40 Solve: (x + 5) 3 2 = 8. 192 Chapter 2 Equations and Inequalities Solving Equations Using Factoring We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring. Polynomial Equations A polynomial of degree n is an expression of the type an xn + an − 1 xn − 1 + ⋅ ⋅ ⋅ + a2 x2 + a1 x + a0 where n is a positive integer and an, … , a0 are real numbers and an ≠ 0. Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n. Example 2.54 Solving a Polynomial by Factoring Solve the polynomial by factoring: 5x4 = 80x2. Solution First, set the equation equal to zero. Then factor out what is common to both terms, the GCF. 5x4 − 80x2 = 0 5x2⎛ ⎞ ⎝x2 − 16 ⎠ = 0 Notice that we have the difference of squares in the factor x2 − 16, which we will continue to factor and obtain two solutions. The first term, 5x2, generates, technically, two solutions as the exponent is 2, but they are the same solution. 5x2 = 0 x = 0 x2 − 16 = 0 (x − 4)(x + 4) = 0 x = 4 x = −4 The solutions are x = 0 (double solution), x = 4, and x = −4. Analysis We can see the solutions on the graph in Figure 2.37. The x-coordinates of the points where the graph crosses the x-axis are the solutions—the x-intercepts. Notice on the graph that at the solution x = 0, the graph touches the x-axis and bounces back. It does not cross the x-axis. This is typical of double solutions. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 193 Figure 2.37 2.41 Solve by factoring: 12x4 = 3x2. Example 2.55 Solve a Polynomial by Grouping Solve a polynomial by grouping: x3 + x2 − 9x − 9 = 0. Solution This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested. x3 + x2 − 9x − 9 = 0 x2(x + 1) − 9(x + 1) = 0 ⎞ ⎛ ⎝x2 − 9 ⎠(x + 1) = 0 The grouping process ends here, as we can factor x2 − 9 using the difference of squares formula. ⎞ ⎛ ⎝x2 − 9 ⎠(x + 1) = 0 (x − 3)(x + 3)(x + 1) = 0 x = 3 x = −3 x = −1 The solutions are x = 3, x = −3, and x = −1. Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in Figure 2.38. 194 Chapter 2 Equations and Inequalities Figure 2.38 Analysis We looked at solving quadratic equations by factoring when the leading coefficient is 1. When the leading coefficient is not 1, we solved by grouping. Grouping requires four terms, which we obtained by splitting the linear term of quadratic equations. We can also use grouping for some polynomials of degree higher than 2, as we saw here, since there were already four terms. Solving Radical Equations Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as 3x + 18 = Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions. Radical Equations An equation containing terms with a variable in the radicand is called a radical equation. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 195 Given a radical equation, solve it. 1. 2. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side. If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an nth root radical, raise both sides to the nth power. Doing so eliminates the radical symbol. 3. Solve the remaining equation. 4. If a radical term still remains, repeat steps 1–2. 5. Confirm solutions by substituting them into the original equation. Example 2.56 Solving an Equation with One Radical Solve 15 − 2x = x. Solution The radical is already isolated on the left side of the equal side, so proceed to square both sides. 15 − 2x = x ⎛ ⎝ 15 − 2x⎞ ⎠ 2 = (x)2 We see that the remaining equation is a quadratic. Set it equal to zero and solve. 15 − 2x = x2 0 = x2 + 2x − 15 = (x + 5)(x − 3) x = −5 x = 3 The proposed solutions are x = −5 and x = 3. Let us check each solution back in the original equation. First, check x = −5. 15 − 2x = x 15 − 2( − 5) = −5 25 = −5 5 ≠ −5 This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation. Check x = 3. The solution is x = 3. 15 − 2x = x 15 − 2(3.42 Solve the radical equation: x + 3 = 3x − 1 196 Chapter 2 Equations and Inequalities Example 2.57 Solving a Radical Equation Containing Two Radicals Solve 2x + 3 + x − 2 = 4. Solution As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical. 2x + 3 + x − 2 = 4 2x + 3 = 4 − x − 2 ⎝4 − x − 2⎞ 2 = ⎛ ⎝ 2x + 3⎞ ⎠ ⎠ ⎛ 2 Subtract x − 2 from both sides. Square both sides. Use the perfect square formula to expand the right side: (a − b)2 = a2 −2ab + b2. ⎝ x − 2⎞ ⎠ 2x + 3 = (4)2 − 2(4) x − 2 + ⎛ 2x + 3 = 16 − 8 x − 2 + (x − 2) 2x + 3 = 14 + x − 8 x − 2 x − 11 = −8 x − 2 (x − 11)2 = ⎛ ⎝−8 x − 2⎞ x2 − 22x + 121 = 64(x − 2) 2 ⎠ 2 Combine like terms. Isolate the second radical. Square both sides. Now that both radicals have been eliminated, set the quadratic equal to zero and solve. x2 − 22x + 121 = 64x − 128 x2 − 86x + 249 = 0 (x − 3)(x − 83) = 0 x = 3 x = 83 Factor and solve. The proposed solutions are x = 3 and x = 83. Check each solution in the original equation. One solution is x = 3. Check x = 83. 2x + 3 + x − 2 = 4 2x + 3 = 4 − x − 2 2(3) + 3 = 4 − (3 2x + 3 + x − 2 = 4 2x + 3 = 4 − x − 2 2(83) + 3 = 4 − (83 − 2) 169 = 4 − 81 13 ≠ −5 The only solution is x = 3. We see that x = 83 is an extraneous solution. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 197 2.43 Solve the equation with two radicals: 3x + 7 + x + 2 = 1. Solving an Absolute Value Equation Next, we will learn how to solve an absolute value equation. To solve an equation such as \|2x − 6\| = 8, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is 8 or −8. This leads to two different equations we can solve independently. 2x − 6 = 8 2x = 14 x = 7 or 2x − 6 = −8 2x = −2 x = −1 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point. Absolute Value Equations The absolute value of x is written as \|x\|. It has the following properties: If x ≥ 0, then \|x\| = x. If x < 0, then \|x\| = −x. For real numbers A and B, an equation of the form \|A\| = B, with B ≥ 0, will have solutions when A = B or A = − B. If B < 0, the equation \|A\| = B has no solution. An absolute value equation in the form \|ax + b\| = c has the following properties: If c < 0, \|ax + b\| = c has no solution. If c = 0, \|ax + b\| = c has one solution. If c > 0, \|ax + b\| = c has two solutions. Given an absolute value equation, solve it. 1. 2. Isolate the absolute value expression on one side of the equal sign. If c > 0, write and solve two equations: ax + b = c and ax + b = − c. Example 2.58 Solving Absolute Value Equations Solve the following absolute value equations: (a) \|6x + 4\| = 8 (b) \|3x + 4\| = −9 (c) \|3x − 5\| − 4 = 6 (d) \|−5x + 10\| = 0 Solution (a) \|6x + 4\| = 8 198 Chapter 2 Equations and Inequalities Write two equations and solve each: 6x + 4 = 8 6x = 4 x = 2 3 6x + 4 = −8 6x = −12 x = −2 The two solutions are x = 2 3 , x = −2. (b) \|3x + 4\| = −9 There is no so
lution as an absolute value cannot be negative. (c) \|3x − 5\| − 4 = 6 Isolate the absolute value expression and then write two equations. \|3x − 5\| − 4 = 6 \|3x − 5\| = 10 3x − 5 = 10 3x = 15 x = 5 3x − 5 = −10 3x = −5 x = − 5 3 There are two solutions: x = 5, x = − 5 3 . (d) \|−5x + 10\| = 0 The equation is set equal to zero, so we have to write only one equation. −5x + 10 = 0 −5x = −10 x = 2 There is one solution: x = 2. 2.44 Solve the absolute value equation: \|1 − 4x\| + 8 = 13. Solving Other Types of Equations There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic. Solving Equations in Quadratic Form Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include x4 − 5x2 + 4 = 0, x6 + 7x3 − 8 = 0, and 2 3 + 4x 1 3 + 2 = 0. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We x can solve these equations by substituting a variable for the middle term. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 199 Quadratic Form If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form, which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form. Given an equation quadratic in form, solve it. 1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term. 2. If it is, substitute a variable, such as u, for the variable portion of the middle term. 3. Rewrite the equation so that it takes on the standard form of a quadratic. 4. Solve using one of the usual methods for solving a quadratic. 5. Replace the substitution variable with the original term. 6. Solve the remaining equation. Example 2.59 Solving a Fourth-degree Equation in Quadratic Form Solve this fourth-degree equation: 3x4 − 2x2 − 1 = 0. Solution This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let u = x2. Rewrite the equation in u. Now solve the quadratic. 3u2 − 2u − 1 = 0 3u2 − 2u − 1 = 0 (3u + 1)(u − 1) = 0 Solve each factor and replace the original term for u. 3u + 1 = 0 3u = −1 u = − 1 3 x2 = − 1 3 x = ± x2 = 1 x = ±1 The solutions are x = ± i 1 3 and x = ± 1. 200 Chapter 2 Equations and Inequalities 2.45 Solve using substitution: x4 − 8x2 − 9 = 0. Example 2.60 Solving an Equation in Quadratic Form Containing a Binomial Solve the equation in quadratic form: (x + 2)2 + 11(x + 2) − 12 = 0. Solution This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting u = x + 2. Then rewrite the equation in u. Solve using the zero-factor property and then replace u with the original expression. u2 + 11u − 12 = 0 (u + 12)(u − 1) = 0 u + 12 = 0 u = −12 x + 2 = −12 x = −14 1 The second factor results in We have two solutions: x = −14, x = −1. 2.46 Solve: (x − 5)2 − 4(x − 5) − 21 = 0. Solving Rational Equations Resulting in a Quadratic Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution. Example 2.61 Solving a Rational Equation Leading to a Quadratic Solve the following rational equation: −4x 8 x2 − 1 . Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 201 We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, x2 −1 = (x + 1)(x − 1). Then, the LCD is (x + 1)(x − 1). Next, we multiply the whole equation by the LCD. ⎡ (x + 1)(x − 1) ⎣ −4x 8 (x + 1)(x − 1) ⎤ ⎦(x + 1)(x − 1) −4x(x + 1) + 4(x − 1) = −8 −4x2 − 4x + 4x − 4 = −8 −4x2 + 4 = 0 ⎞ ⎛ ⎝x2 − 1 ⎠ = 0 −4 −4(x + 1)(x − 1) = 0 x = −1 x = 1 In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution. 2.47 Solve 3x + 2 x − 2 + 1 x = −2 x2 − 2x . Access these online resources for additional instruction and practice with different types of equations. • Rational Equation with no Solution (http://openstaxcollege.org/l/rateqnosoln) • Solving equations with rational exponents using reciprocal powers (http://openstaxcollege.org/l/ratexprecpexp) • Solving radical equations part 1 of 2 (http://openstaxcollege.org/l/radeqsolvepart1) • Solving radical equations part 2 of 2 (http://openstaxcollege.org/l/radeqsolvepart2) 202 Chapter 2 Equations and Inequalities 2.6 EXERCISES Verbal In a radical equation, what does it mean if a number is 292. an extraneous solution? Explain why possible solutions must be checked in 293. radical equations. 294. 3 2 and Your friend tries to calculate the value − 9 keeps getting an ERROR message. What mistake is he or she probably making? 295. Explain why \|2x + 5\| = −7 has no solutions. Explain how to change a rational exponent into the 296. correct radical expression. Algebraic For the following exercises, solve the rational exponent equation. Use factoring where necessary. 297. 298. 299. 300. 301. 302. 303. 2 3 = 16 x 3 4 = 27 x 1 2 − x 1 4 = 0 2x (x − 1) 3 4 = 8 (x + 1) 2 3 = 4 2 3 − 5x 1 3 + 6 = 0 x 7 3 − 3x 4 3 − 4x 1 3 = 0 x For the following exercises, solve the following polynomial equations by grouping and factoring. 304. 305. x3 + 2x2 − x − 2 = 0 3x3 − 6x2 − 27x + 54 = 0 306. 4y3 − 9y = 0 307. 308. x3 + 3x2 − 25x − 75 = 0 m3 + m2 − m − 1 = 0 This content is available for free at https://cnx.org/content/col11758/1.5 309. 310. 2x5 −14x3 = 0 5x3 + 45x = 2x2 + 18 For the following exercises, solve the radical equation. Be sure to check all to eliminate extraneous solutions. solutions 311. 312. 313. 314. 315. 316. 317. 318. 319. 3x − 3t + 12 − x = x 2x + 3 − x + 2 = 2 3x + 7 + x + 2 = 1 2x + 3 − x + 1 = 1 For the following exercises, solve the equation involving absolute value. 320. \|3x − 4\| = 8 321. \|2x − 3\| = −2 322. \|1 − 4x\| − 1 = 5 323. \|4x + 1\| − 3 = 6 324. \|2x − 1\| − 7 = −2 325. \|2x + 1\| − 2 = −3 326. \|x + 5\| = 0 327. −\|2x + 1\| = −3 the following exercises, solve the equation by For identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. 328. 329. x4 − 10x2 + 9 = 0 4(t − 1)2 − 9(t − 1) = −2 Chapter 2 Equations and Inequalities 203 330. 2 ⎞ ⎛ ⎝x2 − 1 ⎠ ⎞ ⎛ ⎝x2 − 1 ⎠ − 12 = 0 + 331. (x + 1)2 − 8(x + 1) − 9 = 0 332. (x − 3)2 − 4 = 0 Extensions the following exercises, solve for For variable. the unknown 333. 334. 335. 336. x−2 − x−1 − 12 = 0 \|x\|2 = x t 25 − t 5 + 1 = 0 \|x2 + 2x − 36\| = 12 Real-World Applications For the following exercises, use the model for the period of a pendulum, T, such that T = 2π L g, where the length of the pendulum is L and the acceleration due to gravity is g. If the acceleration due to gravity is 9.8 m/s2 and the 337. period equals 1 s, find the length to the nearest cm (100 cm = 1 m). If the gravity is 32 ft/s2 and the period equals 1 s, find 338. the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in. For the following exercises, use a model for body surface area, BSA, such that BSA = kg and h = height in cm. wh 3600 , where w = weight in Find the height of a 72-kg female to the nearest cm 339. whose BSA = 1.8. Find the weight of a 177-cm male to the nearest kg 340. whose BSA = 2.1. 204 Chapter 2 Equations and Inequalities 2.7 \| Linear Inequalities and Absolute Value Inequalities Learning Objectives In this section you will: 2.7.1 Use interval notation. 2.7.2 Use properties of inequalities. 2.7.3 Solve inequalities in one variable algebraically. 2.7.4 Solve absolute value inequalities. Figure 2.39 It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities. Using Interval Notation Indicating the solution to an inequality such as x ≥ 4 can be achieved in several ways. We can use a number line as shown in Figure 2.40. The blue ray begins at x = 4 and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4. Figure 2.40 We can use set-builder notation: {x\|x ≥ 4}, which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set. The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions to x ≥ 4 are represented as [4, ∞). This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses. The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore,
cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, are ⎡ ⎣−2, 6), or all numbers between −2 and 6, including −2, but not including 6; (−1, 0), all real numbers between, but not including −1 and 0; and (−∞, 1], all real numbers less than and including 1. Table 2.8 outlines the possibilities. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 205 Set Indicated Set-Builder Notation Interval Notation All real numbers between a and b, but not including a or b {x\|a < x < b} All real numbers greater than a, but not including a All real numbers less than b, but not including b All real numbers greater than a, including a All real numbers less than b, including b All real numbers between a and b, including a All real numbers between a and b, including b All real numbers between a and b, including a and b {x\|x > a} {x\|x < b} {x\|x ≥ a} {x\|x ≤ b} {x\|a ≤ x < b} {x\|a < x ≤ b} {x\|a ≤ x ≤ b} (a, b) (a, ∞) (−∞, b) [a, ∞) (−∞, b⎤ ⎦ ⎡ ⎣a, b) (a, b⎤ ⎦ ⎡ ⎣a, b⎤ ⎦ All real numbers less than a or greater than b {x\|x < a and x > b} (−∞, a) ∪ (b, ∞) All real numbers Table 2.8 Example 2.62 {x\|x is all real numbers} (−∞, ∞) Using Interval Notation to Express All Real Numbers Greater Than or Equal to a Use interval notation to indicate all real numbers greater than or equal to −2. Solution Use a bracket on the left of −2 and parentheses after infinity: [−2, ∞). The bracket indicates that −2 is included in the set with all real numbers greater than −2 to infinity. 2.48 Use interval notation to indicate all real numbers between and including −3 and 5. 206 Chapter 2 Equations and Inequalities Example 2.63 Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b Write the interval expressing all real numbers less than or equal to −1 or greater than or equal to 1. Solution We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at − ∞ and ends at −1, which is written as (−∞, −1]. The second interval must show all real numbers greater than or equal to 1, which is written as [1, ∞). However, we want to combine these two sets. We accomplish this by inserting the union symbol, ∪ , between the two intervals. (−∞, −1] ∪ [1, ∞) 2.49 Express all real numbers less than −2 or greater than or equal to 3 in interval notation. Using the Properties of Inequalities When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol. Properties of Inequalities Addition Property If a < b, then a + c < b + c. Multiplication Property If a < b and c > 0, then ac < bc. If a < b and c < 0, then ac > bc. These properties also apply to a ≤ b, a > b, and a ≥ b. Example 2.64 Demonstrating the Addition Property Illustrate the addition property for inequalities by solving each of the following: (a) x − 15 < 4 (b) 6 ≥ x − 1 (c) x + 7 > 9 Solution The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 207 a. b. c. x − 15 < 4 x − 15 + 15 < 4 + 15 x < 19 Add 15 to both sides Add 1 to both sides. Subtract 7 from both sides. 2.50 Solve: 3x−2 < 1. Example 2.65 Demonstrating the Multiplication Property Illustrate the multiplication property for inequalities by solving each of the following: a. 3x < 6 b. −2x − 1 ≥ 5 c. 5 − x > 10 Solution a. b. c. 1 3 3x < 6 (3x) < (6)1 3 x < 2 −2x − 1 ≥ 5 −2x ≥ 6 ⎞ ⎛ ⎝− 1 ⎠( − 2x) ≥ (6) 2 x ≤ − 3 ⎛ ⎝− 1 2 ⎞ ⎠ 5 − x > 10 −x > 5 ( − 1)( − x) > (5)( − 1) x < − 5 Multiply by − 1 2 . Reverse the inequality. Multiply by − 1. Reverse the inequality. 208 Chapter 2 Equations and Inequalities 2.51 Solve: 4x + 7 ≥ 2x − 3. Solving Inequalities in One Variable Algebraically As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable. Example 2.66 Solving an Inequality Algebraically Solve the inequality: 13 − 7x ≥ 10x − 4. Solution Solving this inequality is similar to solving an equation up until the last step. 13 − 7x ≥ 10x − 4 13 − 17x ≥ −4 −17x ≥ −17 x ≤ 1 Move variable terms to one side of the inequality. Isolate the variable term. Dividing both sides by −17 reverses the inequality. The solution set is given by the interval (−∞, 1], or all real numbers less than and including 1. 2.52 Solve the inequality and write the answer using interval notation. Example 2.67 Solving an Inequality with Fractions Solve the following inequality and write the answer in interval notation. Solution We begin solving in the same way we do when solving an equation 12 − 17 12 − 9 12 ⎝− 12 8 17 Put variable terms on one side. Write fractions with common denominator. ⎞ ⎠ Multiplying by a negative number reverses the inequality. x ≤ 15 34 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 209 The solution set is the interval ⎛ ⎝−∞, 15 34 ⎤ ⎦. 2.53 Solve the inequality and write the answer in interval notation. Understanding Compound Inequalities A compound inequality includes two inequalities in one statement. A statement such as 4 < x ≤ 6 means 4 < x and x ≤ 6. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods. Example 2.68 Solving a Compound Inequality Solve the compound inequality: 3 ≤ 2x + 2 < 6. Solution The first method is to write two separate inequalities: 3 ≤ 2x + 2 and 2x + 2 < 6. We solve them independently. 3 ≤ 2x + 2 1 ≤ 2x 1 ≤ x 2 and 2x + 2 < 6 2x < 4 x < 2 Then, we can rewrite the solution as a compound inequality, the same way the problem began. In interval notation, the solution is written as ⎡ ⎣ ⎞ , 2 ⎠. 1 2 ≤ x < 2 1 2 The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time. 3 ≤ 2x + 2 < 6 1 ≤ 2x < 4 1 2 ≤ x < 2 We get the same solution: ⎡ ⎣ ⎞ ⎠. , 2 1 2 Isolate the variable term, and subtract 2 from all three parts. Divide through all three parts by 2. 2.54 Solve the compound inequality: 4 < 2x − 8 ≤ 10. 210 Chapter 2 Equations and Inequalities Example 2.69 Solving a Compound Inequality with the Variable in All Three Parts Solve the compound inequality with variables in all three parts: 3 + x > 7x − 2 > 5x − 10. Solution Let's try the first method. Write two inequalities: 3 + x > 7x − 2 3 > 6x − 2 5 > 6x 5 > x 6 x < 5 6 and 7x − 2 > 5x − 10 2x − 2 > −10 2x > −8 x > −4 −4 < x The solution set is −4 < x < 5 6 ⎞ ⎠. Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See Figure 2.41. or in interval notation ⎛ ⎝−4, 5 6 Figure 2.41 2.55 Solve the compound inequality: 3y < 4 − 5y < 5 + 3y. Solving Absolute Value Inequalities As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at (−x, 0) has an absolute value of x, as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero. An absolute value inequality is an equation of the form \|A\| < B, \|A\| ≤ B, \|A\| > B, or \|A\| ≥ B, Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all x -values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values. There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph. Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between x and 600 is less than 200. We represent the distance between x and 600 as \|x − 600\|, and therefore, \|x − 600\| ≤ 200 or −200 ≤ x − 600 ≤ 200 −200 + 600 ≤ x − 600 + 600 ≤ 200 + 600 400 ≤ x ≤ 800 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 211 This means our returns would be between $400 and $800. To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently. Absolute Value Inequalities For an algebraic expression X, and k > 0, an absolute value inequality is an inequality of the form \|X\| < k is equivalent to − k < X < k \|X\| > k is equivalent to X < − k or X > k These statements also apply to \|X\| ≤ k and \|X\| ≥ k. Example 2.70 Determining a Number within a Prescribed Distance Describe all values x within a distance of 4 from the number 5. Solution We want the distance between x and 5 to be less than or equal to 4. We can draw a number line, such as in Figure 2.42, to represent the condition to be satisfied. Figure 2.42 The distance from x to 5 can be represented using an absolute value symbol, \|x − 5\|. Write the values of x that satisfy the condition as an absolute value inequa
lity. We need to write two inequalities as there are always two solutions to an absolute value equation. \|x − 5 and x − 5 ≥ − 4 x ≥ 1 If the solution set is x ≤ 9 and x ≥ 1, including 1 and 9. So \|x − 5\| ≤ 4 is equivalent to [1, 9] in interval notation. then the solution set is an interval including all real numbers between and 2.56 Describe all x-values within a distance of 3 from the number 2. Example 2.71 212 Chapter 2 Equations and Inequalities Solving an Absolute Value Inequality Solve \|x − 1\| ≤ 3 . Solution \|x − 1\| ≤ 3 −3 ≤ x − 1 ≤ 3 −2 ≤ x ≤ 4 [−2, 4] Example 2.72 Using a Graphical Approach to Solve Absolute Value Inequalities Given the equation y = − 1 2\|4x − 5\| + 3, determine the x-values for which the y-values are negative. Solution We are trying to determine where y < 0, which is when − 1 2\|4x − 5\| + 3 < 0. We begin by isolating the Multiply both sides by –2, and reverse the inequality. absolute value. − 1 2\|4x − 5\| < − 3 \|4x − 5\| > 6 Next, we solve for the equality \|4x − 5\| = 6. 4x − 5 = 6 or 4x = 11 x = 11 4 4x − 5 = − 6 4x = − 1 x = − 1 4 Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at x = − 1 4 , and that the graph opens downward. See Figure 2.43. and x = 11 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 213 Figure 2.43 2.57 Solve − 2\|k − 4\| ≤ − 6. Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities. • Interval notation (http://openstaxcollege.org/l/intervalnotn) • How to solve linear inequalities (http://openstaxcollege.org/l/solvelinineq) • How to solve an inequality (http://openstaxcollege.org/l/solveineq) • Absolute value equations (http://openstaxcollege.org/l/absvaleq) • Compound inequalities (http://openstaxcollege.org/l/compndineqs) • Absolute value inequalities (http://openstaxcollege.org/l/absvalineqs) 214 Chapter 2 Equations and Inequalities 2.7 EXERCISES Verbal When solving an inequality, explain what happened 341. from Step 1 to Step 2: Step 1 Step 2 −2x > 6 x < − 3 342. When solving an inequality, we arrive at Explain what our solution set is. When writing our solution in interval notation, how 343. do we represent all the real numbers? 344. When solving an inequality, we arrive at 357. \|3x − 1\| > 11 358. \|2x + 1\| + 1 ≤ 6 359. \|x − 2\| + 4 ≥ 10 360. \|−2x + 7\| ≤ 13 361. \|x − 7\| < −4 362. 363. \|x − 20\| > −1 \|x − 3 4 \| < 2 For the following exercises, describe all the x-values within or including a distance of the given values. 364. Distance of 5 units from the number 7 Explain what our solution set is. 365. Distance of 3 units from the number 9 345. Describe how to graph y = \|x − 3\| Algebraic 366. Distance of10 units from the number 4 367. Distance of 11 units from the number 1 For the following exercises, solve the inequality. Write your final answer in interval notation. For the following exercises, solve the compound inequality. Express your answer using inequality signs, and then write your answer using interval notation. 346. 4x − 7 ≤ 9 347. 3x + 2 ≥ 7x − 1 348. −2x + 3 > x − 5 349. 4(x + 3) ≥ 2x − 1 350. − 1 2 x ≤ −5 4 + 2 5 x 351. −5(x − 1) + 3 > 3x − 4 − 4x 352. −3(2x + 1) > −2(x + 4) 353. 354 10 For the following exercises, solve the inequality involving absolute value. Write your final answer in interval notation. 355. \|x + 9\| ≥ −6 356. \|2x + 3\| < 7 This content is available for free at https://cnx.org/content/col11758/1.5 368. −4 < 3x + 2 ≤ 18 369. 3x + 1 > 2x − 5 > x − 7 370. 3y < 5 − 2y < 7 + y 371. 2x − 5 < −11 or 5x + 1 ≥ 6 372. x + 7 < x + 2 Graphical For the following exercises, graph the function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation. 373. \|x − 1\| > 2 374. \|x + 3\| ≥ 5 375. \|x + 7\| ≤ 4 376. \|x − 2\| < 7 377. \|x − 2\| < 0 Chapter 2 Equations and Inequalities 215 For the following exercises, graph both straight lines (lefthand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the yvalues of the lines. the points of intersection, recall (2nd CALC 5:intersection, 1st curve, enter, 2nd curve, enter, guess, enter). Copy a sketch of the graph and shade the x-axis for your solution set to the inequality. Write final answers in interval notation. 378. x + 3 < 3x − 4 379. x − 2 > 2x + 1 380. x + 1 > x + 4 381. 382 4x + 1 < 1 2 x + 3 Numeric 394. \|x + 2\| − 5 < 2 395. −1 2 \|x + 2\| < 4 396. \|4x + 1\| − 3 > 2 397. \|x − 4\| < 3 398. \|x + 2\| ≥ 5 Extensions For the following exercises, write the set notation. in interval 399. Solve \|3x + 1\| = \|2x + 3\| 383. {x\|−1 < x < 3} 384. {x\|x ≥ 7} 385. {x\|x < 4} 386. { x\| x is all real numbers} 400. Solve x2 − x > 12 401. x − 5 x + 7 ≤ 0, x ≠ −7 402. p = − x2 + 130x − 3000 is a profit formula for a small business. Find the set of x-values that will keep this profit positive. For the following exercises, write the interval in set-builder notation. Real-World Applications 387. (−∞, 6) 388. (4, + ∞) 389. [−3, 5) 390. [−4, 1] ∪ [9, ∞) For the following exercises, write the set of numbers represented on the number line in interval notation. In chemistry the volume for a certain gas is given by 403. V = 20T, where V is measured in cc and T is temperature in ºC. If the temperature varies between 80ºC and 120ºC, find the set of volume values. A basic cellular package costs $20/mo. for 60 min of 404. calling, with an additional charge of $.30/min beyond that time.. The cost formula would be C = $20 + .30(x − 60). If you have to keep your bill lower than $50, what is the maximum calling minutes you can use? 391. 392. 393. Technology For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter y2 = the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, 1:abs(. Find 216 Chapter 2 Equations and Inequalities CHAPTER 2 REVIEW KEY TERMS absolute value equation an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression area in square units, the area formula used in this section is used to find the area of any two-dimensional rectangular region: A = LW Cartesian coordinate system a grid system designed with perpendicular axes invented by René Descartes completing the square a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square complex conjugate a complex number containing the same terms as another complex number, but with the opposite operator. Multiplying a complex number by its conjugate yields a real number. complex number the sum of a real number and an imaginary number; the standard form is a + bi, where a is the real part and b is the complex part. complex plane the coordinate plane in which the horizontal axis represents the real component of a complex number, and the vertical axis represents the imaginary component, labeled i. compound inequality a problem or a statement that includes two inequalities conditional equation an equation that is true for some values of the variable discriminant the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots. distance formula a formula that can be used to find the length of a line segment if the endpoints are known equation in two variables a mathematical statement, typically written in x and y, in which two expressions are equal equations in quadratic form equations with a power other than 2 but with a middle term with an exponent that is one- half the exponent of the leading term extraneous solutions any solutions obtained that are not valid in the original equation graph in two variables the graph of an equation in two variables, which is always shown in two variables in the two- dimensional plane identity equation an equation that is true for all values of the variable imaginary number the square root of −1 : i = −1. inconsistent equation an equation producing a false result intercepts the points at which the graph of an equation crosses the x-axis and the y-axis interval an interval describes a set of numbers within which a solution falls interval notation a mathematical statement that describes a solution set and uses parentheses or brackets to indicate where an interval begins and ends linear equation an algebraic equation in which each term is either a constant or the product of a constant and the first power of a variable linear inequality similar to a linear equation except that the solutions will include sets of numbers midpoint formula a formula to find the point that divides a line segment into two parts of equal length ordered pair a pair of numbers indicating horizontal displacement and vertical displacement from the origin; also known as a coordinate pair, (x, y) origin the point where the two axes cross in the center of the plane, described by the ordered pair (0, 0) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 217 perimeter in linear units, the perimeter formula is used to find the linear measurement, or outside length and width, around a two-dimensional regular object; for a rectangle: P = 2L + 2W polynomial equation an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents Pythagorean Theorem a theorem that states the relationship among the lengths of the sides of a right trian
gle, used to solve right triangle problems quadrant one quarter of the coordinate plane, created when the axes divide the plane into four sections quadratic equation an equation containing a second-degree polynomial; can be solved using multiple methods quadratic formula a formula that will solve all quadratic equations radical equation an equation containing at least one radical term where the variable is part of the radicand rational equation an equation consisting of a fraction of polynomials slope the change in y-values over the change in x-values solution set the set of all solutions to an equation square root property one of the methods used to solve a quadratic equation, in which the x2 term is isolated so that the square root of both sides of the equation can be taken to solve for x volume in cubic units, the volume measurement includes length, width, and depth: V = LWH x-axis the common name of the horizontal axis on a coordinate plane; a number line increasing from left to right x-coordinate the first coordinate of an ordered pair, representing the horizontal displacement and direction from the origin x-intercept the point where a graph intersects the x-axis; an ordered pair with a y-coordinate of zero y-axis the common name of the vertical axis on a coordinate plane; a number line increasing from bottom to top y-coordinate the second coordinate of an ordered pair, representing the vertical displacement and direction from the origin y-intercept a point where a graph intercepts the y-axis; an ordered pair with an x-coordinate of zero zero-product property the property that formally states that multiplication by zero is zero, so that each factor of a quadratic equation can be set equal to zero to solve equations KEY EQUATIONS quadratic formula x = −b ± b2 − 4ac 2a KEY CONCEPTS 2.1 The Rectangular Coordinate Systems and Graphs • We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x-axis and displacement from the y-axis. See Example 2.1. • An equation can be graphed in the plane by creating a table of values and plotting points. See Example 2.2. • Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y=_____. See Example 2.3. • Finding the x- and y-intercepts can define the graph of a line. These are the points where the graph crosses the axes. See Example 2.4. • The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment. See Example 2.5 and Example 2.6. 218 Chapter 2 Equations and Inequalities • The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x- coordinates and the sum of the y-coordinates of the endpoints by 2. See Example 2.7 and Example 2.8. 2.2 Linear Equations in One Variable • We can solve linear equations in one variable in the form ax + b = 0 using standard algebraic properties. See Example 2.9 and Example 2.10. • A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See Example 2.11 and Example 2.12. • All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See Example 2.13 and Example 2.14. • Given two points, we can find the slope of a line using the slope formula. See Example 2.15. • We can identify the slope and y-intercept of an equation in slope-intercept form. See Example 2.16. • We can find the equation of a line given the slope and a point. See Example 2.17. • We can also find the equation of a line given two points. Find the slope and use the point-slope formula. See Example 2.19. • The standard form of a line has no fractions. See Example 2.20. • Horizontal lines have a slope of zero and are defined as y = c, where c is a constant. • Vertical lines have an undefined slope (zero in the denominator), and are defined as x = c, where c is a constant. See Example 2.21. • Parallel lines have the same slope and different y-intercepts. See Example 2.23. • Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical. See Example 2.24. 2.3 Models and Applications • A linear equation can be used to solve for an unknown in a number problem. See Example 2.25. • Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities. See Example 2.26. • There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the d = rt formula. See Example 2.27. • Many geometry problems are solved using the perimeter formula P = 2L + 2W, the area formula A = LW, or the volume formula V = LWH. See Example 2.28, Example 2.29, and Example 2.30. 2.4 Complex Numbers • The square root of any negative number can be written as a multiple of i. See Example 2.31. • To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. See Example 2.32. • Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See Example 2.33. • Complex numbers can be multiplied and divided. ◦ To multiply complex numbers, distribute just as with polynomials. See Example 2.34 and Example 2.35. ◦ To divide complex numbers, multiply both numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See Example 2.36 and Example 2.37. • The powers of i are cyclic, repeating every fourth one. See Example 2.38. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 219 2.5 Quadratic Equations • Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions. See Example 2.39, Example 2.40, and Example 2.41. • Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method. See Example 2.42 and Example 2.43. • Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. See Example 2.44 and Example 2.45. • Completing the square is a method of solving quadratic equations when the equation cannot be factored. See Example 2.46. • A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See Example 2.47. • The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. See Example 2.48. • The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. See Example 2.49. 2.6 Other Types of Equations • Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See Example 2.51, Example 2.52, and Example 2.53. • Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See Example 2.54 and Example 2.55. • We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See Example 2.56 and Example 2.57. • To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See Example 2.58. • Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See Example 2.59 and Example 2.60. • Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See Example 2.61. 2.7 Linear Inequalities and Absolute Value Inequalities • Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See Table 2.8 and Example 2.63. • Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See Example 2.64, Example 2.65, Example 2.66, and Example 2.67. • Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See Example 2.68 and Example 2.69. • Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See Example 2.70 and Example 2.71. 220 Chapter 2 Equations and Inequalities • Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See Example 2.72. CHAPTER 2 REVIEW EXERCISES The Rectangular Coordinate Systems and Graphs 418. 7x − 3 = 5 For the following exercises, find the x-intercept and the y
intercept without graphing. 419. 12 − 5(x + 1) = 2x − 5 405. 4x − 3y = 12 406. 2y − 4 = 3x For the following exercises, solve for y in terms of x, putting the equation in slope–intercept form. 407. 5x = 3y − 12 408. 2x − 5y = 7 For the following exercises, find the distance between the two points. 409. (−2, 5)(4, −1) 410. (−12, −3)(−1, 5) 411. Find the distance between the two points (−71,432) and (511,218) using your calculator, and round your answer to the nearest thousandth. For the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points. 420. 2x 3 − 3 4 = x 6 + 21 4 For the following exercises, solve for x. State all x-values that are excluded from the solution set. 421. x x2 − 9 + 4 x + 3 = 3 x2 − 9 x ≠ 3, −3 422. 1 2 + 2 x = 3 4 For the following exercises, find the equation of the line using the point-slope formula. 423. Passes through these two points: (−2, 1),(4, 2). 424. Passes through the point (−3, 4) and has a slope of −1 3 . 425. Passes through the point (−3, 4) and is parallel to the graph y = 2 3 x + 5. 412. (−1, 5) and (4, 6) 426. Passes through these two points: (5, 1),(5, 7). 413. (−13, 5) and (17, 18) Models and Applications For the following exercises, construct a table and graph the equation by plotting at least three points. 414. y = 1 2 x + 4 415. 4x − 3y = 6 Linear Equations in One Variable For the following exercises, solve for x. 416. 5x + 2 = 7x − 8 417. 3(x + 2) − 10 = x + 4 For the following exercises, write and solve an equation to answer each question. 427. The number of males in the classroom is five more than three times the number of females. If the total number of students is 73, how many of each gender are in the class? 428. A man has 72 ft. of fencing to put around a rectangular garden. If the length is 3 times the width, find the dimensions of his garden. 429. A truck rental is $25 plus $.30/mi. Find out how many miles Ken traveled if his bill was $50.20. Complex Numbers For the following exercises, use the quadratic equation to solve. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 221 430. x2 − 5x + 9 = 0 431. 2x2 + 3x + 7 = 0 448. x2 = 49 449. (x − 4)2 = 36 the For component and the vertical component. following exercises, name the horizontal For the following exercises, solve the quadratic equation by completing the square. 432. 4 − 3i 433. −2 − i the following exercises, perform the operations For indicated. 434. (9 − i) − (4 − 7i) 435. (2 + 3i) − (−5 − 8i) 436. 2 −75 + 3 25 437. −16 + 4 −9 438. −6i(i − 5) 439. (3 − 5i)2 440. −4 · −12 441. −2⎛ ⎝ −8 − 5⎞ ⎠ 442. 2 5 − 3i 443. 3 + 7i i 450. x2 + 8x − 5 = 0 451. 4x2 + 2x − 1 = 0 For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No real solution. 452. 2x2 − 5x + 1 = 0 453. 15x2 − x − 2 = 0 For the following exercises, solve the quadratic equation by the method of your choice. 454. (x − 2)2 = 16 455. x2 = 10x + 3 Other Types of Equations For the following exercises, solve the equations. 456. 3 2 = 27 x 457. 1 2 − 4x 1 4 = 0 x 458. 4x3 + 8x2 − 9x − 18 = 0 Quadratic Equations 459. 3x5 − 6x3 = 0 For the following exercises, solve the quadratic equation by factoring. 460. x + 9 = x − 3 444. 2x2 − 7x − 4 = 0 445. 3x2 + 18x + 15 = 0 446. 25x2 − 9 = 0 447. 7x2 − 9x = 0 For the following exercises, solve the quadratic equation by using the square-root property. 461. 3x + 7 + x + 2 = 1 462. \|3x − 7\| = 5 463. \|2x + 3\| − 5 = 9 222 Chapter 2 Equations and Inequalities Linear Inequalities and Absolute Value Inequalities For the following exercises, solve the inequality. Write your final answer in interval notation. 464. 5x − 8 ≤ 12 465. −2x + 5 > x − 7 466 467. \|3x + 2\| + 1 ≤ 9 468. \|5x − 1\| > 14 469. \|x − 3\| < −4 For the following exercises, solve the compound inequality. Write your answer in interval notation. 470. −4 < 3x + 2 ≤ 18 471. 3y < 1 − 2y < 5 + y For the following exercises, graph as described. 472. Graph the absolute value function and graph the constant function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation. \|x + 3\| ≥ 5 473. Graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines. See the interval where the inequality is true. x + 3 < 3x − 4 CHAPTER 2 PRACTICE TEST 474. Graph the following: 2y = 3x + 4. 475. Find the x- and y-intercepts for the following: 476. Find the x- and y-intercepts of this equation, and sketch the graph of the line using just the intercepts plotted. Find the exact distance between (5, −3) and (−2, 8). Find the coordinates of the midpoint of the line segment joining the two points. 477. Write the interval notation for the set of numbers represented by {x\|x ≤ 9}. 3x − 4y = 12 478. Solve for x: 5x + 8 = 3x − 10. 479. Solve for x: 3(2x − 5) − 3(x − 7) = 2x − 9. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 2 Equations and Inequalities 223 480. Solve for x: x 2 + 1 = 4 x 481. Solve for x . 497. Solve: 4x2 − 4x − 1 = 0 498. Solve: x − 7 = x − 7 482. The perimeter of a triangle is 30 in. The longest side is 2 less than 3 times the shortest side and the other side is 2 more than twice the shortest side. Find the length of each side. 499. Solve: 2 + 12 − 2x = x 500. Solve: (x − 1) 2 3 = 9 For the following exercises, find the real solutions of each equation by factoring. 501. 2x3 − x2 − 8x + 4 = 0 502. (x + 5)2 − 3(x + 5) − 4 = 0 483. Solve for x. Write the answer in simplest radical form. x2 3 − x = −1 2 484. Solve: 3x − 8 ≤ 4. 485. Solve: \|2x + 3\| < 5. 486. Solve: \|3x − 2\| ≥ 4. For the following exercises, find the equation of the line with the given information. 487. Passes through the points (−4, 2) and (5, −3). 488. Has an undefined slope and passes through the point (4, 3). 489. Passes through the point (2, 1) and is perpendicular to y = −2 5 x + 3. 490. Add these complex numbers: (3 − 2i) + (4 − i). 491. Simplify: −4 + 3 −16. 492. Multiply: 5i(5 − 3i). 493. Divide: 4 − i 2 + 3i. 494. Solve this quadratic equation and write the two complex roots in a + bi form: x2 − 4x + 7 = 0. 495. Solve: (3x − 1)2 − 1 = 24. 496. Solve: x2 − 6x = 13. 224 Chapter 2 Equations and Inequalities This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 225 3 \| FUNCTIONS Figure 3.1 Standard and Poor’s Index with dividends reinvested (credit "bull": modification of work by Prayitno Hadinata; credit "graph": modification of work by MeasuringWorth) Chapter Outline 3.1 Functions and Function Notation 3.2 Domain and Range 3.3 Rates of Change and Behavior of Graphs 3.4 Composition of Functions 3.5 Transformation of Functions 3.6 Absolute Value Functions 3.7 Inverse Functions Introduction Toward the end of the twentieth century, the values of stocks of Internet and technology companies rose dramatically. As a result, the Standard and Poor’s stock market average rose as well. Figure 3.1 tracks the value of that initial investment of just under $100 over the 40 years. It shows that an investment that was worth less than $500 until about 1995 skyrocketed up to about $1100 by the beginning of 2000. That five-year period became known as the “dot-com bubble” because so many Internet startups were formed. As bubbles tend to do, though, the dot-com bubble eventually burst. Many companies grew too fast and then suddenly went out of business. The result caused the sharp decline represented on the graph beginning at the end of 2000. Notice, as we consider this example, that there is a definite relationship between the year and stock market average. For any year we choose, we can determine the corresponding value of the stock market average. In this chapter, we will explore these kinds of relationships and their properties. 226 Chapter 3 Functions 3.1 \| Functions and Function Notation Learning Objectives In this section, you will: 3.1.1 Determine whether a relation represents a function. 3.1.2 Find the value of a function. 3.1.3 Determine whether a function is one-to-one. 3.1.4 Use the vertical line test to identify functions. 3.1.5 Graph the functions listed in the library of functions. A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships. Determining Whether a Relation Represents a Function A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first. The domain is {1, 2, 3, 4, 5}. The range is {2, 4, 6, 8, 10}. {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)} Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter x. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter y. A function f is a relation that assigns a single element in the range to each element in the domain. In other words, no xvalues are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, {1, 2, 3, 4, 5}, is paired with exactly one element in the range, {2, 4, 6, 8
, 10}. Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as {(odd, 1), (even, 2), (odd, 3), (even, 4), (odd, 5)} Notice that each element in the domain, {even, odd} is not paired with exactly one element in the range, {1, 2, 3, 4, 5}. For example, the term “odd” corresponds to three values from the domain, {1, 3, 5} and the term “even” corresponds to two values from the range, {2, 4}. This violates the definition of a function, so this relation is not a function. Figure 3.2 compares relations that are functions and not functions. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 227 Figure 3.2 (a) This relationship is a function because each input is associated with a single output. Note that input q and r both give output n. (b) This relationship is also a function. In this case, each input is associated with a single output. (c) This relationship is not a function because input q is associated with two different outputs. Function A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.” The input values make up the domain, and the output values make up the range. Given a relationship between two quantities, determine whether the relationship is a function. 1. 2. 3. Identify the input values. Identify the output values. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function. Example 3.1 Determining If Menu Price Lists Are Functions The coffee shop menu, shown in Figure 3.3 consists of items and their prices. a. b. Is price a function of the item? Is the item a function of the price? Figure 3.3 Solution a. Let’s begin by considering the input as the items on the menu. The output values are then the prices. Each item on the menu has only one price, so the price is a function of the item. 228 Chapter 3 Functions b. Two items on the menu have the same price. If we consider the prices to be the input values and the items to be the output, then the same input value could have more than one output associated with it. See Figure 3.4. Figure 3.4 Therefore, the item is a not a function of price. Example 3.2 Determining If Class Grade Rules Are Functions In a particular math class, the overall percent grade corresponds to a grade-point average. Is grade-point average a function of the percent grade? Is the percent grade a function of the grade-point average? Table 3.1 shows a possible rule for assigning grade points. 0–56 57–61 62–66 67–71 72–77 78–86 87–91 92–100 0.0 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Percent grade Grade-point average Table 3.1 Solution For any percent grade earned, there is an associated grade-point average, so the grade-point average is a function of the percent grade. In other words, if we input the percent grade, the output is a specific grade point average. In the grading system given, there is a range of percent grades that correspond to the same grade-point average. For example, students who receive a grade point average of 3.0 could have a variety of percent grades ranging from 78 all the way to 86. Thus, percent grade is not a function of grade-point average. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 229 3.1 Table 3.2[1] lists the five greatest baseball players of all time in order of rank. Player Rank Babe Ruth Willie Mays Ty Cobb Walter Johnson Hank Aaron Table 3.2 1 2 3 4 5 a. b. Is the rank a function of the player name? Is the player name a function of the rank? Using Function Notation Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into graphing calculators and computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions. To represent “height is a function of age,” we start by identifying the descriptive variables h for height and a for age. The letters f , g, and h are often used to represent functions just as we use x, y, and z to represent numbers and A, B, and C to represent sets. h is f of a h = f (a) f (a) We name the function f ; height is a function of age. We use parentheses to indicate the function input. We name the function f ; the expression is read as “ f of a.” Remember, we can use any letter to name the function; the notation h(a) shows us that h depends on a. The value a must be put into the function h to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication. We can also give an algebraic expression as the input to a function. For example f (a + b) means “first add a and b, and the result is the input for the function f.” The operations must be performed in this order to obtain the correct result. Function Notation The notation y = f (x) defines a function named f . This is read as “y is a function of x.” The letter x represents the input value, or independent variable. The letter y, or f (x), represents the output value, or dependent variable. Example 3.3 1. http://www.baseball-almanac.com/legendary/lisn100.shtml. Accessed 3/24/2014. 230 Chapter 3 Functions Using Function Notation for Days in a Month Use function notation to represent a function whose input is the name of a month and output is the number of days in that month. Solution The number of days in a month is a function of the name of the month, so if we name the function f , we write days = f (month) or d = f (m). The name of the month is the input to a “rule” that associates a specific number (the output) with each input. For example, f (March) = 31, because March has 31 days. The notation d = f (m) reminds us that the number of days, d (the output), is dependent on the name of the month, m (the input). Analysis Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with in this book will have numbers as inputs and outputs. Example 3.4 Interpreting Function Notation A function N = f (y) gives the number of police officers, N, in a town in year y. What does f (2005) = 300 represent? Solution When we read f (2005) = 300, we see that the input year is 2005. The value for the output, the number of police officers (N), is 300. Remember, N = f (y). The statement f (2005) = 300 tells us that in the year 2005 there were 300 police officers in the town. 3.2 Use function notation to express the weight of a pig in pounds as a function of its age in days d. Instead of a notation such as y = f(x), could we use the same symbol for the output as for the function, such as y = y(x), meaning “y is a function of x?” Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering. However, in exploring math itself we like to maintain a distinction between a function such as f , which is a rule or procedure, and the output y we get by applying f to a particular input x. This is why we usually use notation such as y = f (x), P = W(d), and so on. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 231 Representing Functions Using Tables A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship. Table 3.3 lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function f where D = f (m) identifies months by an integer rather than by name. Month number, m (input) Days in month, D (output) Table 3. 10 11 12 31 28 31 30 31 30 31 31 30 31 30 31 Table 3.4 defines a function Q = g(n). Remember, this notation tells us that g is the name of the function that takes the input n and gives the output Table 3.4 Table 3.5 displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in. Age in years, a (input) 5 5 6 7 8 9 10 Height in inches, h (output) 40 42 44 47 50 52 54 Table 3.5 Given a table of input and output values, determine whether the table represents a function. 1. Identify the input and output values. 2. Check to see if each input value is paired with only one output value. If so, the table represents a function. Example 3.5 Identifying Tables that Represent Functions Which table, Table 3.6, Table 3.7, or Table 3.8, represents a function (if any)? 232 Chapter 3 Functions Input Output 2 5 8 1 3 6 Table 3.6 Input Output –3 0 4 5 1 5 Table 3.7 Input Output 1 5 5 0 2 4 Table 3.8 Solution Table 3.6 and Table 3.7 define functions. In both, each input value corresponds to exactly one output value. Table 3.8 does not define a function because the input value of 5 corresponds to two different output values. When a table represents a function, corresponding input and output values can also be specified using function notation. The function represented by Table 3.6 can be represented by writing Similarly, the statements represent the function in Table 3.7. f (2) = 1, f
(5) = 3, and f (8) = 6 g(−3) = 5, g(0) = 1, and g(4) = 5 Table 3.8 cannot be expressed in a similar way because it does not represent a function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 233 3.3 Does Table 3.9 represent a function? Input Output 1 2 3 10 100 1000 Table 3.9 Finding Input and Output Values of a Function When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value. When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value. Evaluation of Functions in Algebraic Forms When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function f (x) = 5 − 3x2 can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5. Given the formula for a function, evaluate. 1. Replace the input variable in the formula with the value provided. 2. Calculate the result. Example 3.6 Evaluating Functions at Specific Values Evaluate f (x) = x2 + 3x − 4 at a. 2 b. a c. a + h d. f (a + h) − f (a) h Solution Replace the x in the function with each specified value. a. Because the input value is a number, 2, we can use simple algebra to simplify. 234 Chapter 3 Functions f (2) = 22 + 3(2. In this case, the input value is a letter so we cannot simplify the answer any further. c. With an input value of a + h, we must use the distributive property. f (a) = a2 + 3a − 4 f (a + h) = (a + h)2 + 3(a + h) − 4 = a2 + 2ah + h2 + 3a + 3h − 4 d. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that f (a + h) = a2 + 2ah + h2 + 3a + 3h − 4 and we know that Now we combine the results and simplify. f (a) = a2 + 3a − 4 f (a + h) − f (a) h ⎛ ⎝a2 + 2ah + h2 + 3a + 3h − 4 ⎞ ⎠ − ⎛ ⎞ ⎝a2 + 3a − 4 ⎠ = = 2ah + h2 + 3h h h(2a + h + 3) h = h Factor out h. = 2a + h + 3 Simplify. Example 3.7 Evaluating Functions Given the function h(p) = p2 + 2p, evaluate h(4). Solution To evaluate h(4), we substitute the value 4 for the input variable p in the given function. h(p) = p2 + 2p h(4) = (4)2 + 2(4) = 16 + 8 = 24 Therefore, for an input of 4, we have an output of 24. 3.4 Given the function g(m) = m − 4, evaluate g(5). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 235 Example 3.8 Solving Functions Given the function h(p) = p2 + 2p, solve for h(p) = 3. Solution h(p) = 3 p2 + 2p = 3 p2 + 2p − 3 = 0 (p + 3)(p − 1) = 0 ⎠ = 0, either ⎛ ⎝p + 3⎞ Substitute the original function h(p) = p2 + 2p. Subtract 3 from each side. Factor. ⎝p − 1⎞ If ⎛ ⎝p + 3⎞ ⎛ ⎠ ⎝p − 1⎞ ⎠ = 0 or ⎛ ⎠ = 0 (or both of them equal 0). We will set each factor equal to 0 and solve for p in each case. (p + 3) = 0, (p − 1) = 0, p = −3 p = 1 This gives us two solutions. The output h(p) = 3 when the input is either p = 1 or p = − 3. We can also verify by graphing as in Figure 3.5. The graph verifies that h(1) = h(−3) = 3 and h(4) = 24. Figure 3.5 3.5 Given the function g(m) = m − 4, solve g(m) = 2. Evaluating Functions Expressed in Formulas Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation 2n + 6p = 12 expresses a functional relationship between n and p. We can rewrite it to decide if p is a function of n. 236 Chapter 3 Functions Given a function in equation form, write its algebraic formula. 1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable. 2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity. Example 3.9 Finding an Equation of a Function Express the relationship 2n + 6p = 12 as a function p = f (n), if possible. Solution To express the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p = [expression involving n]. 2n + 6p = 12 Subtract 2n from both sides. Divide both sides by 6 and simplify. 6p = 12 − 2n p = 12 − 2n 6 − 2n 6 n p = 12 6 p = 2 − 1 3 Therefore, p as a function of n is written as p = f (n) = 2 − 1 3 n Example 3.10 Expressing the Equation of a Circle as a Function Does the equation x2 + y2 = 1 represent a function with x as input and y as output? If so, express the relationship as a function y = f (x). Solution First we subtract x2 from both sides. We now try to solve for y in this equation. y2 = 1 − x2 y = ± 1 − x2 = + 1 − x2 and − 1 − x2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 237 We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function y = f (x). If we graph both functions on a graphing calculator, we will get the upper and lower semicircles. 3.6 If x − 8y3 = 0, express y as a function of x. Are there relationships expressed by an equation that do represent a function but that still cannot be represented by an algebraic formula? Yes, this can happen. For example, given the equation x = y + 2 y, if we want to express y as a function of x, there is no simple algebraic formula involving only x that equals y. However, each x does determine a unique value for y, and there are mathematical procedures by which y can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for y as a function of x, even though the formula cannot be written explicitly. Evaluating a Function Given in Tabular Form As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours. The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See Table 3.10.[2] Pet Memory span in hours Puppy Adult dog Cat Goldfish Beta fish Table 3.10 0.008 0.083 16 2160 3600 At times, evaluating a function in table form may be more useful than using equations. Here let us call the function P. The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s memory span lasts. We can evaluate the function P at the input value of “goldfish.” We would write P(goldfis ) = 2160. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the 2. http://www.kgbanswers.com/how-long-is-a-dogs-memory-span/4221590. Accessed 3/24/2014. 238 Chapter 3 Functions pertinent row of the table. The tabular form for function P seems ideally suited to this function, more so than writing it in paragraph or function form. Given a function represented by a table, identify specific output and input values. 1. Find the given input in the row (or column) of input values. 2. Identify the corresponding output value paired with that input value. 3. Find the given output values in the row (or column) of output values, noting every time that output value appears. 4. Identify the input value(s) corresponding to the given output value. Example 3.11 Evaluating and Solving a Tabular Function Using Table 3.11, a. Evaluate g(3). b. Solve g(n) = 6. n g(n Table 3.11 Solution a. Evaluating g(3) means determining the output value of the function g for the input value of n = 3. The table output value corresponding to n = 3 is 7, so g(3) = 7. b. Solving g(n) = 6 means identifying the input values, n, that produce an output value of 6. Table 3.12 shows two solutions: n = 2 and n = 4. n g(n Table 3.12 When we input 2 into the function g, our output is 6. When we input 4 into the function g, our output is also 6. 3.7 Using Table 3.11, evaluate g(1). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 239 Finding Function Values from a Graph Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s). Example 3.12 Reading Function Values from a Graph Given the graph in Figure 3.6, a. Evaluate f (2). b. Solve f (x) = 4. Figure 3.6 Solution a. To evaluate f (2), locate the point on the curve where x = 2, then read the y-coordinate of that point. The point has coordinates (2, 1), so f (2) = 1. See Figure 3.7. 240 Chapter 3 Functions Figure 3.7 b. To solve f (x) = 4, we find the output value 4 on the vertical axis. Moving horizontally along the line y = 4, we locate two points of the curve with output value 4: (−1, 4) and (3, 4). These points represent the two solutions to f (x) = 4: x = −1 or x = 3. This means f (−1) = 4 and f (3) = 4, or when the input is −1 or 3, the
output is 4. See Figure 3.8. Figure 3.8 3.8 Using Figure 3.6, solve f (x) = 1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 241 Determining Whether a Function is One-to-One Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in 51260 (https://cnx.org/content/51260/latest/#Figure_01_00_001) at the beginning of this chapter, the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1000. However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in Table 3.13. Letter grade Grade point average A B C D Table 3.13 4.0 3.0 2.0 1.0 This grading system represents a one-to-one function because each letter input yields one particular grade-point average output and each grade-point average corresponds to one input letter. To visualize this concept, let’s look again at the two simple functions sketched in Figure 3.2(a) and Figure 3.2(b). The function in part (a) shows a relationship that is not a one-to-one function because inputs q and r both give output n. The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output. One-to-One Function A one-to-one function is a function in which each output value corresponds to exactly one input value. There are no repeated x- or y-values. Example 3.13 Determining Whether a Relationship Is a One-to-One Function Is the area of a circle a function of its radius? If yes, is the function one-to-one? Solution A circle of radius r has a unique area measure given by A = πr 2, so for any input, r, there is only one output, A. The area is a function of radius r. If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure A is given by the formula A = πr 2. Because areas and radii are positive numbers, there is exactly one solution: r = A π . So the area of a circle is a one-to-one function of the circle’s radius. 242 Chapter 3 Functions 3.9 a. b. c. Is a balance a function of the bank account number? Is a bank account number a function of the balance? Is a balance a one-to-one function of the bank account number? 3.10 a. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade? b. If so, is the function one-to-one? Using the Vertical Line Test As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many inputoutput pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis. The most common graphs name the input value x and the output value y, and we say y is a function of x, or y = f (x) when the function is named f . The graph of the function is the set of all points (x, y) in the plane that satisfies the equation y = f (x). If the function is defined for only a few input values, then the graph of the function consists of only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in Figure 3.9 tell us that f (0) = 2 and f (6) = 1. However, the set of all points (x, y) satisfying y = f (x) is a curve. The curve shown includes (0, 2) and (6, 1) because the curve passes through those points. Figure 3.9 The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See Figure 3.10. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 243 Figure 3.10 Given a graph, use the vertical line test to determine if the graph represents a function. 1. 2. Inspect the graph to see if any vertical line drawn would intersect the curve more than once. If there is any such line, determine that the graph does not represent a function. Example 3.14 Applying the Vertical Line Test Which of the graphs in Figure 3.11 represent(s) a function y = f (x) ? Figure 3.11 Solution If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of Figure 3.11. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in Figure 3.12. 244 Chapter 3 Functions Figure 3.12 3.11 Does the graph in Figure 3.13 represent a function? Figure 3.13 Using the Horizontal Line Test Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function. 1. 2. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once. If there is any such line, determine that the function is not one-to-one. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 245 Example 3.15 Applying the Horizontal Line Test Consider the functions shown in Figure 3.11(a) and Figure 3.11(b). Are either of the functions one-to-one? Solution The function in Figure 3.11(a) is not one-to-one. The horizontal line shown in Figure 3.14 intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.) Figure 3.14 The function in Figure 3.11(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once. 3.12 Is the graph shown in Figure 3.12 one-to-one? Identifying Basic Toolkit Functions In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our “toolkit functions,” which form a set of basic named functions for which we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use x as the input variable and y = f (x) as the output variable. We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in Table 3.14. 246 Chapter 3 Functions Toolkit Functions Name Function Graph Constant f (x) = c, where c is a constant Identity f (x) = x Absolute value f (x) = \|x\| Table 3.14 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 247 Toolkit Functions Name Function Graph Quadratic f (x) = x2 Cubic f (x) = x3 Reciprocal f (x) = 1 x Table 3.14 248 Chapter 3 Functions Toolkit Functions Name Function Graph Reciprocal squared f (x) = 1 x2 Square root f (x) = x Cube root f (x) = x3 Table 3.14 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 249 Access the following online resources for additional instruction and practice with functions. • Determine if a Relation is a Function (http://openstaxcollege.org/l/relationfunction) • Vertical Line Test (http://openstaxcollege.org/l/vertlinetest) • Introduction to Functions (http://openstaxcollege.org/l/introtofunction) • Vertical Line Test on Graph (http://openstaxcollege.org/l/vertlinegraph) • One-to-one Functions (http://openstaxcollege.org/l/onetoone) • Graphs as One-to-one Functions (http://openstaxcollege.org/l/graphonetoone) 250 Chapter 3 Functions 3.1 EXERCISES Verbal 1. What function? is the difference between a relation and a What is the difference between the input and the output 2. of a function? Why does the vertical line test tell us whether the graph 3. of a relation represents a function? How can you determine if a relation is a one-to-one 4. function? Why does the horizontal line test tell us whether the 5. graph of a function is one-to-one? Algebraic For the following exercises, determine whether the relation represents a function. 6. 7. {(a, b), (c, d), (a, c)} {(a, b), (b, c), (c, c)} For the following exercises, determine whether the relation represents y as a function of x. 8. 9. 5x + 2y = 10 y = x2 10. x = y2 3x2 + y = 14 2x + y2 = 6 y = − 2x2 + 40x y = 1 x x = 3y + 5 7y − 1 x = 1 − y2 y = 3x + 5 7x − 1 x2 + y2 = 9 11. 12. 13. 14. 15. 16. 17. 18. 19. This content is available for free at https://cnx.org/content/col11758/1.5 2xy = 1 20. x = y3 21. y = x3 22. 23. 24. 25. y = 1 − x2 y2 = x2 26. y3 = x2 For the following exercises, evaluate the function f at the indicated values f (−3), f (2), f (−a), − f (a), f (a + h). 27. f (x) = 2x − 5 28. 29. 30. f (x) = − 5x2 + 2x − 1 f (x) = 2 − x + 5
f (x) = 6x − 1 5x + 2 31. f (x) = \|x − 1\| − \|x + 1\| 32. Given the function g(x) = 5 − x2, simplify g(x + h) − g(x) h , h ≠ 0. 33. Given g(x) − g(a) x − a the function g(x) = x2 + 2x, simplify , x ≠ a. 34. Given the function k(t) = 2t − 1: a. Evaluate k(2). b. Solve k(t) = 7. 35. Given the function f (x) = 8 − 3x: a. Evaluate f ( − 2). b. Solve f (x) = −1. 36. Given the function p(c) = c2 + c: Chapter 3 Functions 251 a. Evaluate p(−3). b. Solve p(c) = 2. 37. Given the function f (x) = x2 − 3x: a. Evaluate f (5). b. Solve f (x) = 4. 38. Given the function f (x) = x + 2: a. Evaluate f (7). b. Solve f (x) = 4. 39. Consider the relationship 3r + 2t = 18. a. Write the relationship as a function r = f (t). b. Evaluate f (−3). c. Solve f (t) = 2. Graphical For the following exercises, use the vertical line test to determine which graphs show relations that are functions. 40. 41. 42. 43. 44. 45. 252 46. 47. 48. 49. Chapter 3 Functions 50. 51. 52. Given the following graph, • Evaluate f (−1). • Solve for f (x) = 3. 53. Given the following graph, • Evaluate f (0). • Solve for f (x) = −3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 253 54. Given the following graph, • Evaluate f (4). • Solve for f (x) = 1. 58. 59. For the following exercises, determine if the given graph is a one-to-one function. 55. 56. 57. Numeric For the following exercises, determine whether the relation represents a function. 60. 61. 62. {(−1, −1), (−2, −2), (−3, −3)} {(3, 4), (4, 5), (5, 6)} ⎧ ⎨(2, 5), (7, 11), (15, 8), (7, 9)⎫ ⎬ ⎭ ⎩ For the following exercises, determine if the relation represented in table form represents y as a function of x. x y 5 3 10 8 15 14 63. 64. 254 65. x y 5 3 10 15 8 8 x y 5 3 10 8 10 14 For the following exercises, use the function f represented in Table 3.15(x) 74 28 1 53 56 3 36 45 14 47 Table 3.15 66. Evaluate f (3). 67. Solve f (x) = 1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions For the following exercises, evaluate the function f at the values f (−2), f (−1), f (0), f (1), and f (2). 68. f (x) = 4 − 2x 69. f (x) = 8 − 3x 70. 71. 72. f (x) = 8x2 − 7x + 3 f (x) = 3 + x + 3 f (x) = x − 2 x + 3 73. f (x) = 3 x For the following exercises, evaluate the expressions, given functions f , g, and h: • • • f (x) = 3x − 2 g(x) = 5 − x2 h(x) = −2x2 + 3x − 1 74. 3 f (1) − 4g(−2) 75(−2) Technology For the following exercises, graph y = x2 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 76. 77. 78. [ − 0.1, 0.1] [ − 10, 10] [ − 100, 100] For the following exercises, graph y = x3 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 79. 80. 81. [ − 0.1, 0.1] [ − 10, 10] [ − 100, 100] For the following exercises, graph y = x on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 255 Chapter 3 Functions 82. [0, 0.01] 83. [0, 100] 84. [0, 10,000] For the following exercises, graph y = x3 viewing window. Determine the corresponding range for each viewing window. Show each graph. on the given 85. 86. 87. [−0.001, 0.001] [−1000, 1000] [−1,000,000, 1,000,000] Real-World Applications The amount of garbage, G, produced by a city with 88. population p is given by G = f (p). G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function f . b. Explain the meaning of the statement f (5) = 2. The number of cubic yards of dirt, D, needed to cover 89. a garden with area a square feet is given by D = g(a). a. A garden with area 5000 ft2 requires 50 yd3 of dirt. Express this information in terms of the function g. b. Explain the meaning of the statement g(100) = 1. 90. Let f (t) be the number of ducks in a lake t years after 1990. Explain the meaning of each statement: a. b. f (5) = 30 f (10) = 40 91. Let h(t) be the height above ground, in feet, of a rocket t seconds after launching. Explain the meaning of each statement: a. b. h(1) = 200 h(2) = 350 92. Show that the function f (x) = 3(x − 5)2 + 7 is not one-to-one. 256 Chapter 3 Functions 3.2 \| Domain and Range Learning Objectives In this section, you will: 3.2.1 Find the domain of a function defined by an equation. 3.2.2 Graph piecewise-defined functions. If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 3.15 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these. Figure 3.15 Based on data compiled by www.the-numbers.com.[3] Finding the Domain of a Function Defined by an Equation In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0. We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products. See Figure 3.16. 3. The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.thenumbers.com/market/genre/Horror. Accessed 3/24/2014 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 257 Figure 3.16 We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write (0, 100]. We will discuss interval notation in greater detail later. Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative. Before we begin, let us review the conventions of interval notation: • The smallest number from the interval is written first. • The largest number in the interval is written second, following a comma. • Parentheses, ( or ), are used to signify that an endpoint value is not included, called exclusive. • Brackets, [ or ], are used to indicate that an endpoint value is included, called inclusive. See Figure 3.17 for a summary of interval notation. 258 Chapter 3 Functions Figure 3.17 Example 3.16 Finding the Domain of a Function as a Set of Ordered Pairs Find the domain of the following function: {(2, 10), (3, 10), (4, 20), (5, 30), (6, 40)} . Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs. {2, 3, 4, 5, 6} This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 259 3.13 Find the domain of the function: ⎧ ⎨(−5, 4), (0, 0), (5, −4), (10, −8), (15, −12)⎫ ⎬ ⎭ ⎩ Given a function written in equation form, find the domain. 1. 2. Identify the input values. Identify any restrictions on the input and exclude those values from the domain. 3. Write the domain in interval form, if possible. Example 3.17 Finding the Domain of a Function Find the domain of the function f (x) = x2 − 1. Solution The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form, the domain of f is (−∞, ∞). 3.14 Find the domain of the function: f (x) = 5 − x + x3. Given a function written in an equation form that includes a fraction, find the domain. 1. 2. Identify the input values. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x . If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve. 3. Write the domain in interval form, making sure to exclude any restricted values from the domain. Example 3.18 Finding the Domain of a Function Involving a Denominator Find the domain of the function f (x) = x + 1 2 − x. Solution Wh
en there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. 260 Chapter 3 Functions 2 − x = 0 −x = −2 x = 2 Now, we will exclude 2 from the domain. The answers are all real numbers where x < 2 or x > 2 as shown in Figure 3.18. We can use a symbol known as the union, ∪ , to combine the two sets. In interval notation, we write the solution: (−∞, 2) ∪ (2, ∞). Figure 3.18 3.15 Find the domain of the function: f (x) = 1 + 4x 2x − 1 . Given a function written in equation form including an even root, find the domain. 1. Identify the input values. 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 3. The solution(s) are the domain of the function. If possible, write the answer in interval form. Example 3.19 Finding the Domain of a Function with an Even Root Find the domain of the function f (x) = 7 − x. Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 7 − x ≥ 0 −x ≥ −7 x ≤ 7 Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or ( − ∞, 7]. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 261 3.16 Find the domain of the function f (x) = 5 + 2x. Can there be functions in which the domain and range do not intersect at all? Yes. For example, the function f (x) = − 1 x has the set of all positive real numbers as its domain but the set of all negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common. Using Notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, {x\|10 ≤ x < 30} describes the behavior of x in set-builder notation. The braces {} are read as “the set of,” and the vertical bar \| is read as “such that,” so we would read {x\|10 ≤ x < 30} as “the set of x-values such that 10 is less than or equal to x, and x is less than 30.” Figure 3.19 compares inequality notation, set-builder notation, and interval notation. Figure 3.19 To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol, ∪ , to combine two unconnected intervals. For example, the union of the sets {2, 3, 5} and {4, 6} is the set {2, 3, 4, 5, 6}. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending 262 Chapter 3 Functions order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is {x\| \|x\| ≥ 3} = (−∞, − 3] ∪ [3, ∞) Set-Builder Notation and Interval Notation Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form {x\| statement about x} which is read as, “the set of all x such that the statement about x is true.” For example, {x\|4 < x ≤ 12} Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example, (4, 12] Given a line graph, describe the set of values using interval notation. 1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. 2. At the left end of each interval, use [ with each end value to be included in the set (solid dot) or ( for each excluded end value (open dot). 3. At the right end of each interval, use ] with each end value to be included in the set (filled dot) or ) for each excluded end value (open dot). 4. Use the union symbol ∪ to combine all intervals into one set. Example 3.20 Describing Sets on the Real-Number Line Describe the intervals of values shown in Figure 3.20 using inequality notation, set-builder notation, and interval notation. Figure 3.20 Solution To describe the values, x, included in the intervals shown, we would say, “ x is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.” This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 263 Inequality 1 ≤ x ≤ 3 or x > 5 Set-builder notation ⎧ ⎨x\|1 ≤ x ≤ 3 or x > 5⎫ ⎬ ⎭ ⎩ Interval notation [1, 3] ∪ (5, ∞) Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set. 3.17 Given Figure 3.21, specify the graphed set in a. words b. c. set-builder notation interval notation Figure 3.21 Finding Domain and Range from Graphs Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure 3.22. 264 Chapter 3 Functions Figure 3.22 We can observe that the graph extends horizontally from −5 to the right without bound, so the domain is ⎡ ⎣−5, ∞). The ⎦. Note that the domain and range are vertical extent of the graph is all range values 5 and below, so the range is (−∞, 5⎤ always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Example 3.21 Finding Domain and Range from a Graph Find the domain and range of the function f whose graph is shown in Figure 3.23. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 265 Figure 3.23 Solution We can observe that the horizontal extent of the graph is –3 to 1, so the domain of f is (−3, 1]. The vertical extent of the graph is 0 to –4, so the range is [−4, 0). See Figure 3.24. Figure 3.24 Example 3.22 Finding Domain and Range from a Graph of Oil Production Find the domain and range of the function f whose graph is shown in Figure 3.25. 266 Chapter 3 Functions Figure 3.25 (credit: modification of work by the U.S. Energy Information Administration)[4] Solution The input quantity along the horizontal axis is “years,” which we represent with the variable t for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable b for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as 1973 ≤ t ≤ 2008 and the range as approximately 180 ≤ b ≤ 2010. In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines. 3.18 Given Figure 3.26, identify the domain and range using interval notation. Figure 3.26 Can a function’s domain and range be the same? Yes. For example, the domain and range of the cube root function are both the set of all real numbers. Finding Domains and Ranges of the Toolkit Functions We will now return to our set of toolkit functions to determine the domain and range of each. 4. http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPAK2&f=A. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 267 Figure 3.27 For the constant function f (x) = c, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant c, so the range is the set {c} that contains this single element. In interval notation, this is written as [c, c], the interval that both begins and ends with c. Figure 3.28 For the identity function f (x) = x, there is no restriction on x. Both the domain and range are the set of all real numbers. 268 Chapter 3 Functions Figure 3.29 For the absolute value function f (x) = \|x\|, there is no restriction on x. However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. Figure 3.30 For the quadratic function f (x) = x2, the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 269 Figure 3.31 For the cubic function f (x) = x3, the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers. Figure 3.32 For the reciprocal function f (x) = 1 x, we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also wr
ite {x\| x ≠ 0}, the set of all real numbers that are not zero. 270 Chapter 3 Functions Figure 3.33 For the reciprocal squared function f (x) = 1 x2, we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers. Figure 3.34 For the square root function f (x) = x, we cannot take the square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number x is defined to be positive, even though the square of the negative number − x also gives us x. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 271 Figure 3.35 For the cube root function f (x) = x3 , the domain and range include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). Given the formula for a function, determine the domain and range. 1. Exclude from the domain any input values that result in division by zero. 2. Exclude from the domain any input values that have nonreal (or undefined) number outputs. 3. Use the valid input values to determine the range of the output values. 4. Look at the function graph and table values to confirm the actual function behavior. Example 3.23 Finding the Domain and Range Using Toolkit Functions Find the domain and range of f (x) = 2x3 − x. Solution There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result. The domain is (−∞, ∞) and the range is also (−∞, ∞). Example 3.24 Finding the Domain and Range Find the domain and range of f (x) = 2 x + 1 . 272 Chapter 3 Functions Solution We cannot evaluate the function at −1 because division by zero is undefined. The domain is (−∞, −1) ∪ (−1, ∞). Because the function is never zero, we exclude 0 from the range. The range is (−∞, 0) ∪ (0, ∞). Example 3.25 Finding the Domain and Range Find the domain and range of f (x) = 2 x + 4. Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative. The domain of f (x) is [ − 4, ∞). x + 4 ≥ 0 when x ≥ − 4 We then find the range. We know that f (−4) = 0, and the function value increases as x increases without any upper limit. We conclude that the range of f is ⎡ ⎣0, ∞). Analysis Figure 3.36 represents the function f . Figure 3.36 3.19 Find the domain and range of f (x) = − 2 − x. Graphing Piecewise-Defined Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function f (x) = \|x\|. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 273 value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value, the output is the same as the input. f (x) = x if x ≥ 0 If we input a negative value, the output is the opposite of the input. f (x) = − x if x < 0 Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain. We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be 0.1S if S ≤ $10,000 and $1000 + 0.2(S − $10,000) if S > $10,000. Piecewise Function A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this: f (x) = ⎧ ⎨ ⎩ formula 1 if x is in domain 1 formula 2 if x is in domain 2 formula 3 if x is in domain 3 In piecewise notation, the absolute value function is \|x\| = x if x ≥ 0 ⎧ ⎨ −x if x < 0 ⎩ Given a piecewise function, write the formula and identify the domain for each interval. 1. Identify the intervals for which different rules apply. 2. Determine formulas that describe how to calculate an output from an input in each interval. 3. Use braces and if-statements to write the function. Example 3.26 Writing a Piecewise Function A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people, n, to the cost, C. Solution Two different formulas will be needed. For n-values under 10, C = 5n. For values of n that are 10 or greater, C = 50. C(n) = Analysis ⎧ ⎨5n if 0 < n < 10 50 if ⎩ n ≥ 10 274 Chapter 3 Functions The function is represented in Figure 3.37. The graph is a diagonal line from n = 0 to n = 10 and a constant after that. In this example, the two formulas agree at the meeting point where n = 10, but not all piecewise functions have this property. Figure 3.37 Example 3.27 Working with a Piecewise Function A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer. C(g) = ⎧ 25 ⎨ 25 + 10(g − 2) if ⎩ if 0 < g < 2 g ≥ 2 Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data. Solution To find the cost of using 1.5 gigabytes of data, C(1.5), we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula. C(1.5) = $25 To find the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula. C(4) = 25 + 10(4 − 2) = $45 Analysis The function is represented in Figure 3.38. We can see where the function changes from a constant to a shifted and stretched identity at g = 2. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 275 Figure 3.38 Given a piecewise function, sketch a graph. 1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain. 2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function. Example 3.28 Graphing a Piecewise Function Sketch a graph of the function. f (x) = ⎧ ⎨ ⎩ x ≤ 1 x2 if 3 if 1 < x ≤ 2 x if x > 2 Solution Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equalto inequality. Figure 3.39 shows the three components of the piecewise function graphed on separate coordinate systems. 276 Chapter 3 Functions Figure 3.39 (a) f (x) = x2 if x ≤ 1; (b) f (x) = 3 if 1< x ≤ 2; (c) f (x) = x if x > 2 Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure 3.40. Figure 3.40 Analysis Note that the graph does pass the vertical line test even at x = 1 and x = 2 because the points (1, 3) and (2, 2) are not part of the graph of the function, though (1, 1) and (2, 3) are. 3.20 Graph the following piecewise function. f (x) = ⎧ ⎨ ⎩ x3 if −2 if − if x > 4 Can more than one formula from a piecewise function be applied to a value in the domain? No. Each value corresponds to one equation in a piecewise formula. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 277 Access these online resources for additional instruction and practice with domain and range. • Domain and Range of Square Root Functions (http://openstaxcollege.org/l/domainsqroot) • Determining Domain and Range (http://openstaxcollege.org/l/determinedomain) • Find Domain and Range Given the Graph (http://openstaxcollege.org/l/drgraph) • Find Domain and Range Given a Table (http://openstaxcollege.org/l/drtable) • Find Domain and Range Given Points on a Coordinate Plane (http://openstaxcollege.org/ l/drcoordinate) 278 Chapter 3 Functions 3.2 EXERCISES Verbal 93. Why does the domain differ for different functions? How do we determine the domain of a function defined 94. by an equation? 95. Explain why the domain of f (x) = x3 is different from the domain of f (x) = x. 96. When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket? 97. How do you graph a piecewise function? Algebraic For the following exercises, find the domain of each function using interval notation. 98. f (x) = − 2x(x − 1)(x − 2) 99. f (x) = 5 − 2x2 100. 101. 102. 103. 104. 105. 106. 107. 108. f (x) = 3 x − 2 f (x) = 3 − 6 − 2x f (x) = 4 − 3x f (x) = x2 + 4 3 f (x) = 1 − 2x 3 f (x) = x − 1 f (x) = 9 x − 6 f (x) = 3x + 1 4x + 2 f (x) = x + 4 x − 4 1
09. f (x) = x − 3 x2 + 9x − 22 110. f (x) = 1 x2 − x − 6 111. This content is available for free at https://cnx.org/content/col11758/1.5 f (x) = 2x3 − 250 x2 − 2x − 15 112. 113. 114. 5 x − 3 2x + 1 5 − x f (x) = 115. f (x) = 116. f (xx) = x2 − 9x x2 − 81 Find the domain of the function f (x) = 2x3 − 50x 117. 118. by: a. using algebra. b. graphing the radicand and function in the determining intervals on the x-axis for which the radicand is nonnegative. Graphical For the following exercises, write the domain and range of each function using interval notation. 119. 120. 121. Chapter 3 Functions 279 122. 123. 124. 125. 126. 127. 128. 129. 280 Chapter 3 Functions For the following exercises, given each function f , evaluate f (−1), f (0), f (2), and f (4). 141. 142. 143. f (x) = ⎨7x + 3 if x < 0 ⎧ 7x + 6 if x ≥ 0 ⎩ f (x) = ⎧ ⎨ x2 − 2 if x < 2 4 + \|x − 5\| if x ≥ 2 ⎩ f (x) = ⎧ ⎨ ⎩ x < 0 5x if 3 if 0 ≤ x ≤ 3 x2 if x > 3 For the following exercises, write the domain for the piecewise function in interval notation. 144. 145. f (x) = x + 1 if x < − 2 ⎧ ⎨ −2x − 3 if x ≥ − 2 ⎩ f (x) = ⎧ x2 − 2 if x < 1 ⎨ −x2 + 2 if x > 1 ⎩ 146. f (x) = ⎧ ⎨2x − 3 −3x2 ⎩ if x < 0 if x ≥ 2 Technology 147. Graph y = 1 x2 on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine corresponding range for the viewing window. Show the graphs. the 148. Graph y = 1 x on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine the corresponding range for the viewing window. Show the graphs. Extension Suppose the range of a function f is [−5, 8]. What 149. is the range of \| f (x)\| ? Create a function in which the range is all 150. nonnegative real numbers. 151. Create a function in which the domain is x > 2. Real-World Applications The height h of a projectile is a function of the time 152. t it is in the air. The height in feet for t seconds is given by the function h(t) = −16t 2 + 96t. What is the domain of the function? What does the domain mean in the context of the problem? For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation. 130. 131. 132. 133. 134. 135. 136. f (x) = if 2x − 3 if x ≥ − 2 ⎩ f (x) = ⎨2x − 1 if x < 1 ⎧ if x ≥ 1 1 + x ⎩ f (x) = x + 1 if x < 0 ⎧ ⎨ x − 1 if x > 0 ⎩ f (x) = ⎧ ⎨ 3 if x < 0 x if x ≥ 0 ⎩ f (x) = ⎧ ⎨x2 if x < 0 1 − x if x > 0 ⎩ f (x) = ⎧ x2 ⎨ x + 2 ⎩ if x < 0 if x ≥ 0 f (x) = x + 1 if x < 1 ⎧ ⎨ x3 if x ≥ 1 ⎩ ⎧ ⎨\|x\| 1 ⎩ if x < 2 if x ≥ 2 137. f (x) = Numeric For the following exercises, given each function f , evaluate f (−3), f (−2), f (−1), and f (0). 138. 139. 140. f (x) = if 2x − 3 if x ≥ − 2 ⎩ f (x) = ⎧ ⎨1 if x ≤ − 3 0 if x > − 3 ⎩ f (x) = ⎧ ⎨−2x2 + 3 if x ≤ − 1 if x > − 1 ⎩ 5x − 7 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 281 153. The cost in dollars of making x items is given by the function C(x) = 10x + 500. a. The fixed cost is determined when zero items are produced. Find the fixed cost for this item. b. What is the cost of making 25 items? c. Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function, C(x) ? 282 Chapter 3 Functions 3.3 \| Rates of Change and Behavior of Graphs Learning Objectives In this section, you will: 3.3.1 Find the average rate of change of a function. 3.3.2 Use a graph to determine where a function is increasing, decreasing, or constant. 3.3.3 Use a graph to locate local maxima and local minima. 3.3.4 Use a graph to locate the absolute maximum and absolute minimum. Gasoline costs have experienced some wild fluctuations over the last several decades. Table 3.17[5] lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year. y 2005 2006 2007 2008 2009 2010 2011 2012 2.31 2.62 2.84 3.30 2.41 2.84 3.58 3.68 C(y) Table 3.17 If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these. Finding the Average Rate of Change of a Function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table 3.17 did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value. Average rate of change = = = = Change in output Change in input Δy Δx y2 − y1 x2 − x1 ⎠ − f ⎛ f ⎛ ⎝x2 x2 − x1 ⎝x1 ⎞ ⎠ ⎞ The Greek letter Δ (delta) signifies the change in a quantity; we read the ratio as “delta-y over delta-x” or “the change in y divided by the change in x. ” Occasionally we write Δ f instead of Δy, which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function. In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was Δy Δx = $1.37 7 years ≈ 0.196 dollars per year On average, the price of gas increased by about 19.6¢ each year. Other examples of rates of change include: 5. http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 283 • A population of rats increasing by 40 rats per week • A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) • A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) • The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage • The amount of money in a college account decreasing by $4,000 per quarter Rate of Change A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.” The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values. Δy Δx = f (x2) − f (x1) x2 − x1 (3.1) Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values x1 and x2. 1. Calculate the difference y2 − y1 = Δy. 2. Calculate the difference x2 − x1 = Δx. 3. Find the ratio Δy Δx. Example 3.29 Computing an Average Rate of Change Using the data in Table 3.17, find the average rate of change of the price of gasoline between 2007 and 2009. Solution In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is Δy Δx = y2 − y1 x2 − x1 $2.41 − $2.84 2009 − 2007 = = −$0.43 2 years = −$0.22 per year Analysis Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases. 3.21 Using the data in Table 3.17, find the average rate of change between 2005 and 2010. 284 Chapter 3 Functions Example 3.30 Computing Average Rate of Change from a Graph Given the function g(t) shown in Figure 3.41, find the average rate of change on the interval [−1, 2]. Figure 3.41 Solution At t = − 1, Figure 3.42 shows g(−1) = 4. At t = 2, the graph shows g(2) = 1. Figure 3.42 The horizontal change Δt = 3 is shown by the red arrow, and the vertical change Δg(t) = − 3 is shown by the turquoise arrow. The average rate of change is shown by the slope of the orange line segment. The output changes by –3 while the input changes by 3, giving an average rate of change of 1 − 4 2 − (−1) = −3 3 = −1 Analysis Note that the order we choose is very important. If, for example, we use y2 − y1 x1 − x2 , we will not get the correct answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as (x1, y1) and (x2, y2). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 285 Example 3.31 Computing Average Rate of Change from a Table After picking up a friend who lives 10 miles away and leaving on a trip, Anna records her distance from home over time. The values are shown in Table 3.18. Find her average speed over the first 6 hours. t (hours) 0 D(t) (miles) 10 1 55 2 3 4 5 6 7 90 153 214 240 292 300 Table 3.18 Solution Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours. The average speed is 47 miles per hour. 292 − 10 6 − 0 = 282 6 = 47 Analysis Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2,3], the average speed is 63 miles per hour. Example 3.32 Computing Average Rate of Change for a Function Expressed as a Formula Compute the average rate of change of f (x) = x2 − 1 x on the interval [2, 4]. Solution We can start by computing the function values at each endpoint of the interval. f (2) = 22 − 4) = 42 − 1 4 = 16 − 1 4 = 63 4 Now we compute the average rate of change. 286 Chapter 3 Functions Average rate of change = f (4) − f (2) 4 − 2 63 = 4 − 7 2 4 − 2 49 4 2 = 49 8 = 3.22 Find the average rate of change of f (x) = x − 2 x on the interval [1, 9]. Example 3.33 Finding the Average Rate of Change of a Force The electrostatic force F, measured in newtons, between two charged particles can be related to the distance d 2. Find the average rate of change of force if in centimeters, by the formula F(d) = 2 between the particles d, the distance between the particles is increased from 2 cm to 6 cm. Solution We are computing the average rate of change o
f F(d) = 2 d 2 on the interval [2, 6]. Average rate of change = 2 = F(6) − F(2) 6 − 2 62 − 2 22 6 − 2 36 − 2 2 4 − 16 36 4 = − 1 9 = = 4 Simplify. Combine numerator terms. Simplify The average rate of change is − 1 9 newton per centimeter. Example 3.34 Finding an Average Rate of Change as an Expression This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 287 Find the average rate of change of g(t) = t 2 + 3t + 1 on the interval [0, a]. The answer will be an expression involving a in simplest form. Solution We use the average rate of change formula. Average rate of change = = g(a) − g(0) a − 0 ⎛ ⎝a2 + 3a + 1 ⎛ ⎝02 + 3 ⎞ ⎠ − a − 0 ⎛ ⎝0 ⎞ ⎞ ⎠ + 1 ⎠ Evaluate. Simplify. = a2 + 3a + 1 − 1 a a(a + 3) = a = a + 3 Simplify and factor. Divide by the common factor a. This result tells us the average rate of change in terms of a between t = 0 and any other point t = a. For example, on the interval [0, 5], the average rate of change would be 5 + 3 = 8. Find the average rate of change of f (x) = x2 + 2x − 8 on the interval [5, a] in simplest forms in terms 3.23 of a. Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3.43 shows examples of increasing and decreasing intervals on a function. 288 Chapter 3 Functions Figure 3.43 The function f (x) = x3 − 12x is increasing on (−∞, − 2) ∪ (2, ∞) and is decreasing on ( − 2, 2). While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is “local minima.” Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is “extremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term local. Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain. For the function whose graph is shown in Figure 3.44, the local maximum is 16, and it occurs at x = −2. The local minimum is −16 and it occurs at x = 2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 289 Figure 3.44 To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 3.45 illustrates these ideas for a local maximum. Figure 3.45 Definition of a local maximum These observations lead us to a formal definition of local extrema. Local Minima and Local Maxima A function f is an increasing function on an open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a. A function f is a decreasing function on an open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a. A function f has a local maximum at x = b if there exists an interval (a, c) with a < b < c such that, for any x f has a local minimum at x = b if there exists an interval (a, c) with in the interval (a, c), a < b < c such that, for any x in the interval (a, c), f (x) ≤ f (b). Likewise, f (x) ≥ f (b). 290 Chapter 3 Functions Example 3.35 Finding Increasing and Decreasing Intervals on a Graph Given the function p(t) in Figure 3.46, identify the intervals on which the function appears to be increasing. Figure 3.46 Solution We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from t = 1 to t = 3 and from t = 4 on. In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval (4, ∞). Analysis Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at t = 1 , t = 3 , and t = 4 . These points are the local extrema (two minima and a maximum). Example 3.36 Finding Local Extrema from a Graph Graph the function f (x) = 2 x + determine the intervals on which the function is increasing. x 3 . Then use the graph to estimate the local extrema of the function and to Solution Using technology, we find that the graph of the function looks like that in Figure 3.47. It appears there is a low point, or local minimum, between x = 2 and x = 3, and a mirror-image high point, or local maximum, somewhere between x = −3 and x = −2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 291 Figure 3.47 Analysis Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 3.48 provides screen images from two different technologies, showing the estimate for the local maximum and minimum. Figure 3.48 Based on these estimates, the function is increasing on the interval ( − ∞, − 2.449) and (2.449,∞). Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at ± 6, but determining this requires calculus.) 3.24 Graph the function f (x) = x3 − 6x2 − 15x + 20 to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing. Example 3.37 Finding Local Maxima and Minima from a Graph 292 Chapter 3 Functions For the function f whose graph is shown in Figure 3.49, find all local maxima and minima. Figure 3.49 Solution Observe the graph of f . The graph attains a local maximum at x = 1 because it is the highest point in an open interval around x = 1. The local maximum is the y -coordinate at x = 1, which is 2. The graph attains a local minimum at x = −1 because it is the lowest point in an open interval around x = −1. The local minimum is the y-coordinate at x = −1, which is −2. Analyzing the Toolkit Functions for Increasing or Decreasing Intervals We will now return to our toolkit functions and discuss their graphical behavior in Figure 3.50, Figure 3.51, and Figure 3.52. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 293 Figure 3.50 Figure 3.51 294 Chapter 3 Functions Figure 3.52 Use A Graph to Locate the Absolute Maximum and Absolute Minimum There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally) and locating the highest and lowest points on the graph for the entire domain. The y- coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function. See Figure 3.53. Figure 3.53 Not every function has an absolute maximum or minimum value. The toolkit function f (x) = x3 is one such function. Absolute Maxima and Minima The absolute maximum of f at x = c is f (c) where f (c) ≥ f (x) for all x in the domain of f . The absolute minimum of f at x = d is f (d) where f (d) ≤ f (x) for all x in the domain of f . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 295 Example 3.38 Finding Absolute Maxima and Minima from a Graph For the function f shown in Figure 3.54, find all absolute maxima and minima. Figure 3.54 Solution Observe the graph of f . The graph attains an absolute maximum in two locations, x = −2 and x = 2, because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at x = −2 and x = 2, which is 16. The graph attains an absolute minimum at x = 3, because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at x = 3, which is −10. Access this online resource for additional instruction and practice with rates of change. • Average Rate of Change (http://openstaxcollege.org/l/aroc) 296 Chapter 3 Functions 3.3 EXERCISES Verbal Can the average rate of change of a function be 154. constant? a If function f is on (a, b) and 155. decreasing on (b, c), then what can be said about the local extremum of f on (a, c) ? increasing How are the absolute maximum and minimum similar 156. to and different from the local extrema? How does the graph of the absolute value function 157. compare to the graph of the quadratic function, y = x2, in terms of increasing and decreasing intervals? Algebraic For the following exercises, find the ave
rage rate of change of each function on the interval specified for real numbers b or h in simplest form. 158. 159. f (x) = 4x2 − 7 on [1, b] g(x) = 2x2 − 9 on ⎡ ⎣4, b⎤ ⎦ 160. p(x) = 3x + 4 on [2, 2 + h] 161. k(x) = 4x − 2 on [3, 3 + h] 162. 163. 164. 165. 166. 167. 168. f (x) = 2x2 + 1 on [x, x + h] g(x) = 3x2 − 2 on [x, x + h] a(t) = 1 t + 4 on [9, 9 + h] b(x) = 1 x + 3 on [1, 1 + h] j(x) = 3x3 on [1, 1 + h] r(t) = 4t 3 on [2, 2 + h] f (x + h) − f (x) h given f (x) = 2x2 − 3x on [x, x + h] Graphical For the following exercises, consider the graph of f shown in Figure 3.55. This content is available for free at https://cnx.org/content/col11758/1.5 Figure 3.55 Estimate the average rate of change from x = 1 to 169. x = 4. Estimate the average rate of change from x = 2 to 170. x = 5. For the following exercises, use the graph of each function to estimate the intervals on which the function is increasing or decreasing. 171. 172. 173. Chapter 3 Functions 297 Figure 3.57 177. If the complete graph of the function is shown, estimate the intervals where the function is increasing or decreasing. If the complete graph of the function is shown, 178. estimate the absolute maximum and absolute minimum. Numeric 179. Table 3.19 gives the annual sales (in millions of dollars) of a product from 1998 to 2006. What was the average rate of change of annual sales (a) between 2001 and 2002, and (b) between 2001 and 2004? 174. For the following exercises, consider the graph shown in Figure 3.56. Figure 3.56 Estimate the intervals where the function is increasing 175. or decreasing. 176. Estimate the point(s) at which the graph of f has a local maximum or a local minimum. For the following exercises, consider the graph in Figure 3.57. 298 Chapter 3 Functions Year Sales (millions of dollars) Year Population (thousands) 1998 1999 2000 2001 2002 2003 2004 2005 2006 201 219 233 243 249 251 249 243 233 2000 2001 2002 2003 2004 2005 2006 2007 2008 87 84 83 80 77 76 78 81 85 Table 3.19 Table 3.20 180. Table 3.20 gives the population of a town (in thousands) from 2000 to 2008. What was the average rate of change of population (a) between 2002 and 2004, and (b) between 2002 and 2006? For the following exercises, find the average rate of change of each function on the interval specified. 181. 182. 183. 184. 185. 186. 187. f (x) = x2 on [1, 5] h(x) = 5 − 2x2 on [−2, 4] q(x) = x3 on [−4, 2] g(x) = 3x3 − 1 on [−3, 3] y = 1 x on [1, 3] p(t) = ⎞ ⎛ ⎝t 2 − 4 ⎠(t + 1) t 2 + 3 on [−3, 1] k(t) = 6t 2 + 4 t 3 on [−1, 3] Technology For the following exercises, use a graphing utility to estimate the local extrema of each function and to estimate the intervals on which the function is increasing and decreasing. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 299 f (x) = x4 − 4x3 + 5 Real-World Applications At the start of a trip, the odometer on a car read 197. 21,395. At the end of the trip, 13.5 hours later, the odometer read 22,125. Assume the scale on the odometer is in miles. What is the average speed the car traveled during this trip? A driver of a car stopped at a gas station to fill up his 198. gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was full and he noticed that he had pumped 10.7 gallons. What is the average rate of flow of the gasoline into the gas tank? falls is a function of Near the surface of the moon, the distance that an 199. is given by object d(t) = 2.6667t 2, where t is in seconds and d(t) is in feet. If an object is dropped from a certain height, find the average velocity of the object from t = 1 to t = 2. time. It The graph in Figure 3.59 illustrates the decay of a 200. radioactive substance over t days. 188. 189. h(x) = x5 + 5x4 + 10x3 + 10x2 − 1 190. g(t) = t t + 3 191. 192. 193. k(t) = 3t 2 3 − t m(x) = x4 + 2x3 − 12x2 − 10x + 4 n(x) = x4 − 8x3 + 18x2 − 6x + 2 Extension 194. The graph of the function f is shown in Figure 3.58. Figure 3.58 on shot, Based (1.333, 5.185) is which of the following? calculator screen the the point Figure 3.59 Use the graph to estimate the average decay rate from t = 5 to t = 15. A. a relative (local) maximum of the function B. C. the vertex of the function the absolute maximum of the function D. a zero of the function 195. Let f (x) = 1 x. Find a number c such that the average rate of change of the function f on the interval (1, c) is − 1 4 . 196. Let f (x) = 1 x . Find the number b such that the average rate of change of f on the interval (2, b) is − 1 10 . 300 Chapter 3 Functions 3.4 \| Composition of Functions Learning Objectives In this section, you will: 3.4.1 Combine functions using algebraic operations. 3.4.2 Create a new function by composition of functions. 3.4.3 Evaluate composite functions. 3.4.4 Find the domain of a composite function. 3.4.5 Decompose a composite function into its component functions. Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day. Using descriptive variables, we can notate these two functions. The function C(T) gives the cost C of heating a house for a given average daily temperature in T degrees Celsius. The function T(d) gives the average daily temperature on day d of the year. For any given day, Cost = C⎛ ⎠ means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature T(d). For example, we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write C⎛ ⎝T(d)⎞ ⎝T(5)⎞ ⎠. By combining these two relationships into one function, we have performed function composition, which is the focus of this section. Combining Functions Using Algebraic Operations Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function. Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If w(y) is the wife’s income and h(y) is the husband’s income in year y, and we want T to represent the total income, then we can define a new function. T(y) = h(y) + w(y) If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions. T = h + w This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 301 For two functions f (x) and g(x) with real number outputs, we define new functions f + g, f − g, f g, and f relations g by the ( f + g)(x) = f (x) + g(x) ( f − g)(x) = f (x) − g(x) ( f g)(x) = f (x)g(x) f g ⎛ ⎝ ⎞ ⎠(x) = f (x) g(x) where g(x) ≠ 0 Example 3.39 Performing Algebraic Operations on Functions Find and simplify the functions ⎛ ⎝g − f ⎞ ⎠(x) and ⎛ ⎝ g f ⎞ ⎠(x), given f (x) = x − 1 and g(x) = x2 − 1. Are they the same function? Solution Begin by writing the general form, and then substitute the given functions. (g − f )(x) = g(x) − f (x) (g − f )(x) = x2 − 1 − (x − 1) = x2 − x = x(x − 1) ⎛ ⎝ ⎛ ⎝ g f g f ⎞ ⎠(x) = ⎞ ⎠(x) = = g(x) f (x) x2 − 1 x − 1 (x + 1)(x − 1) x − 1 = x + 1 where x ≠ 1 No, the functions are not the same. Note: For ⎛ ⎝ g f ⎞ ⎠(x), the condition x ≠ 1 is necessary because when x = 1, the denominator is equal to 0, which makes the function undefined. 3.25 Find and simplify the functions ⎛ ⎝ f g⎞ ⎠(x) and ⎛ ⎝ f − g⎞ ⎠(x). Are they the same function? f (x) = x − 1 and g(x) = x2 − 1 Create a Function by Composition of Functions Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation: 302 Chapter 3 Functions ⎛ ⎝ f ∘ g⎞ ⎠(x) = f ⎛ ⎝g(x)⎞ ⎠ We read the left-hand side as “ f composed with g at x,” and the right-hand side as “ f of g of x.” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ∘ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in mo
st cases f (g(x)) ≠ f (x)g(x). It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function g takes the input x first and yields an output g(x). Then the function f takes g(x) as an input and yields an output f ⎛ ⎝g(x)⎞ ⎠. In general, f ∘ g and g ∘ f are different functions. In other words, in many cases f ⎛ see that sometimes two functions can be composed only in one specific order. ⎝g(x)⎞ ⎠ ≠ g⎛ ⎝ f (x)⎞ ⎠ for all x. We will also For example, if f (x) = x2 and g(x) = x + 2, then but f (g(x)) = f (x + 2) = (x + 2)2 = x2 + 4x + 4 g( f (x)) = g⎛ ⎝x2⎞ ⎠ = x2 + 2 These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value x = − 1 2 . Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs. Composition of Functions When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input x and functions f and g, this action defines a composite function, which we write as f ∘ g such that ⎛ ⎝ f ∘ g⎞ ⎠(x) = f ⎛ ⎝g(x)⎞ ⎠ (3.2) The domain of the composite function f ∘ g is all x such that x is in the domain of g and g(x) is in the domain of f . It is important to realize that the product of functions f g is not the same as the function composition f ⎛ in general, f (x)g(x) ≠ f ⎛ ⎝g(x)⎞ ⎠. ⎝g(x)⎞ ⎠, because, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 303 Example 3.40 Determining whether Composition of Functions is Commutative Using the functions provided, find f ⎛ commutative. ⎝g(x)⎞ ⎠ and g⎛ ⎝ f (x)⎞ ⎠. Determine whether the composition of the functions is f (x) = 2x + 1 g(x) = 3 − x Solution Let’s begin by substituting g(x) into f (x). Now we can substitute f (x) into g(x). f (g(x)) = 2(3 − x) + 1 = 6 − 2x + 1 = 7 − 2x g( f (x)) = 3 − (2x + 1) = 3 − 2x − 1 = −2x + 2 We find that g( f (x)) ≠ f (g(x)), so the operation of function composition is not commutative. Example 3.41 Interpreting Composite Functions The function c(s) gives the number of calories burned completing s sit-ups, and s(t) gives the number of sit-ups a person can complete in t minutes. Interpret c(s(3)). Solution The inside expression in the composition is s(3). Because the input to the s-function is time, t = 3 represents 3 minutes, and s(3) is the number of sit-ups completed in 3 minutes. Using s(3) as the input to the function c(s) gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups). Example 3.42 Investigating the Order of Function Composition Suppose f (x) gives miles that can be driven in x hours and g(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: f ⎛ ⎠ or g⎛ ⎝g(y)⎞ ⎝ f (x)⎞ ⎠? 304 Chapter 3 Functions Solution The function y = f (x) is a function whose output is the number of miles driven corresponding to the number of hours driven. number of miles = f (number of hours) The function g(y) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means: number of gallons = g (number of miles) The expression g(y) takes miles as the input and a number of gallons as the output. The function f (x) requires a ⎠ number of hours as the input. Trying to input a number of gallons does not make sense. The expression f ⎛ ⎝g(y)⎞ is meaningless. The expression f (x) takes hours as input and a number of miles driven as the output. The function g(y) requires a number of miles as the input. Using f (x) (miles driven) as an input value for g(y), where gallons of gas ⎠ makes sense, and will yield the number of depends on miles driven, does make sense. The expression g⎛ gallons of gas used, g, driving a certain number of miles, f (x), in x hours. ⎝ f (x)⎞ Are there any situations where f(g(y)) and g( f(x)) would both be meaningful or useful expressions? Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order. 3.26 The gravitational force on a planet a distance r from the sun is given by the function G(r). The acceleration of a planet subjected to any force F is given by the function a(F). Form a meaningful composition of these two functions, and explain what it means. Evaluating Composite Functions Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function. Evaluating Composite Functions Using Tables When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function. Example 3.43 Using a Table to Evaluate a Composite Function Using Table 3.21, evaluate f (g(3)) and g( f (3)). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 305 x f(x) g(x) 1 2 3 4 6 8 3 1 Table 3.21 3 5 2 7 Solution To evaluate f (g(3)), we start from the inside with the input value 3. We then evaluate the inside expression g(3) using the table that defines the function g : g(3) = 2. We can then use that result as the input to the function f , so g(3) is replaced by 2 and we get f (2). Then, using the table that defines the function f , we find that f (2) = 8. g(3) = 2 f (g(3)) = f (2) = 8 To evaluate g( f (3)), we first evaluate the inside expression f (3) using the first table: f (3) = 3. Then, using the table for g, we can evaluate Table 3.22 shows the composite functions f ∘ g and g ∘ f as tables. g( f (3)) = g(3) = 2 g(x) f ⎛ ⎝g(x)⎞ ⎠ f (x) ⎝ f (x)⎞ g⎛ ⎠ 2 8 3 2 x 3 Table 3.22 3.27 Using Table 3.21, evaluate f (g(1)) and g( f (4)). Evaluating Composite Functions Using Graphs When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the x- and y- axes of the graphs. 306 Chapter 3 Functions Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs. 1. Locate the given input to the inner function on the x- axis of its graph. 2. Read off the output of the inner function from the y- axis of its graph. 3. Locate the inner function output on the x- axis of the graph of the outer function. 4. Read the output of the outer function from the y- axis of its graph. This is the output of the composite function. Example 3.44 Using a Graph to Evaluate a Composite Function Using Figure 3.60, evaluate f (g(1)). Figure 3.60 Solution To evaluate f (g(1)), we start with the inside evaluation. See Figure 3.61. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 307 Figure 3.61 We evaluate g(1) using the graph of g(x), finding the input of 1 on the x- axis and finding the output value of the graph at that input. Here, g(1) = 3. We use this value as the input to the function f . We can then evaluate the composite function by looking to the graph of f (x), finding the input of 3 on the xaxis and reading the output value of the graph at this input. Here, f (3) = 6, so f (g(1)) = 6. f (g(1)) = f (3) Analysis Figure 3.62 shows how we can mark the graphs with arrows to trace the path from the input value to the output value. 308 Chapter 3 Functions Figure 3.62 3.28 Using Figure 3.60, evaluate g( f (2)). Evaluating Composite Functions Using Formulas When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression. While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that ⎠. To do this, we will extend our idea of function evaluation. Recall that, will calculate the result of a composition f ⎛ when we evaluate a function like f (t) = t 2 − t, we substitute the value inside the parentheses into the formula wherever we see the input variable. ⎝g(x)⎞ Given a formula for a composite function, evaluate the function. 1. Evaluate the inside function using the input value or variable provided. 2. Use the resulting output as the input to the outside function. Example 3.45 Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input Given f (t) = t 2 − t and h(x) = 3x + 2, evaluate f (h(1)). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 309 Solution Because the inside expression is h(1), we start by evaluating h(x) at 1. Then f (h(1)) = f (5), so we evaluate f (t) at an input of 5. h(1) = 3(1) + 2 h(1) = 5 f (h(1)) = f (5) f (h(1)) = 52 − 5 f (h(1)) = 20 Analysis It makes no difference what the input variables t and x were called in this problem because we evaluated for specific numer
ical values. 3.29 Given f (t) = t 2 − t and h(x) = 3x + 2, evaluate a. h( f (2)) b. h( f ( − 2)) Finding the Domain of a Composite Function As we discussed previously, the domain of a composite function such as f ∘ g is dependent on the domain of g and the domain of f . It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as f ∘ g. Let us assume we know the domains of the functions f and g separately. If we write ⎠, we can see right away that x must be a member of the domain of g in the composite function for an input x as f ⎛ order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that g(x) must be a member of the domain of f , otherwise the second function evaluation in f ⎛ ⎠ cannot be completed, and the expression is still undefined. Thus the domain of f ∘ g consists of only those inputs in the domain of g that produce outputs from g belonging to the domain of f . Note that the domain of f composed with g is the set of all x such that x is in the domain of g and g(x) is in the domain of f . ⎝g(x)⎞ ⎝g(x)⎞ Domain of a Composite Function The domain of a composite function f ⎛ domain of f . ⎝g(x)⎞ ⎠ is the set of those inputs x in the domain of g for which g(x) is in the Given a function composition f(g(x)), determine its domain. 1. Find the domain of g. 2. Find the domain of f . 3. Find those inputs x in the domain of g for which g(x) is in the domain of f . That is, exclude those inputs x from the domain of g for which g(x) is not in the domain of f . The resulting set is the domain of f ∘ g. 310 Chapter 3 Functions Example 3.46 Finding the Domain of a Composite Function Find the domain of Solution ⎛ ⎝ f ∘ g⎞ ⎠(x) where f (x) = 5 x − 1 and g(x) = 4 3x − 2 The domain of g(x) consists of all real numbers except x = 2 3 , since that input value would cause us to divide by 0. Likewise, the domain of f consists of all real numbers except 1. So we need to exclude from the domain of g(x) that value of x for which g(x) = 1. 4 3x − 2 = 1 4 = 3x − 2 6 = 3x x = 2 So the domain of f ∘ g is the set of all real numbers except 2 3 and 2. This means that We can write this in interval notation as x ≠ 2 3 or x ≠ 2 ⎛ ⎝−∞, 2, ∞) , 2 Example 3.47 Finding the Domain of a Composite Function Involving Radicals Find the domain of ⎛ ⎝ f ∘ g⎞ ⎠(x) where f (x) = x + 2 and g(x) = 3 − x Solution Because we cannot take the square root of a negative number, the domain of g is (−∞, 3]. Now we check the domain of the composite function ⎛ ⎝ f ∘ g⎞ ⎠(x) = 3 − x + 2 or ⎛ ⎝ f ∘ g⎞ ⎠(x) = 5 − x The domain of this function is (−∞, 5⎤ ⎦. To find the domain of f ∘ g, we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since (−∞, 3] is a proper subset of the domain of f ∘ g. This means the domain of f ∘ g is the same as the domain of g, namely, (−∞, 3]. Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 311 This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of f ∘ g can contain values that are not in the domain of f , though they must be in the domain of g. 3.30 Find the domain of ⎛ ⎝ f ∘ g⎞ ⎠(x) where f (x) = 1 x − 2 and g(x) = x + 4 Decomposing a Composite Function into its Component Functions In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient. Example 3.48 Decomposing a Function Write f (x) = 5 − x2 as the composition of two functions. Solution We are looking for two functions, g and h, so f (x) = g(h(x)). To do this, we look for a function inside a function in the formula for f (x). As one possibility, we might notice that the expression 5 − x2 is the inside of the square root. We could then decompose the function as We can check our answer by recomposing the functions. h(x) = 5 − x2 and g(x) = x g(h(x)) = g⎛ ⎝5 − x2⎞ ⎠ = 5 − x2 3.31 Write f (x) = 4 3 − 4 + x2 as the composition of two functions. Access these online resources for additional instruction and practice with composite functions. • Composite Functions (http://openstaxcollege.org/l/compfunction) • Composite Function Notation Application (http://openstaxcollege.org/l/compfuncnot) • Composite Functions Using Graphs (http://openstaxcollege.org/l/compfuncgraph) • Decompose Functions (http://openstaxcollege.org/l/decompfunction) • Composite Function Values (http://openstaxcollege.org/l/compfuncvalue) 312 Chapter 3 Functions 3.4 EXERCISES Verbal e. ⎛ ⎝ f ∘ f ⎞ ⎠(−2) How does one find the domain of the quotient of two 201. functions, f g ? 202. What is the composition of two functions, f ∘ g ? If the order is reversed when composing two 203. functions, can the result ever be the same as the answer in the original order of the composition? If yes, give an example. If no, explain why not. How do you find the domain for the composition of 204. two functions, f ∘ g ? Algebraic For the following exercises, determine the domain for each function in interval notation. 205. Given f (x) = x2 + 2x and g(x) = 6 − x2, find f + g, f − g, f g, and f g . 206. Given f (x) = − 3x2 + x and g(x) = 5, find f + g, f − g, f g, and f g . 207. Given f (x) = 2x2 + 4x and g(x) = 1 2x, find f + g, f − g, f g, and f g . 208. Given f (x) = 1 x − 4 and g(x) = 1 6 − x, find f + g, f − g, f g, and f g . 209. Given f (x) = 3x2 and g(x) = x − 5, find f + g, f − g, f g, and f g . 210. Given f (x) = x and g(x) = \|x − 3\|, find g f . For the following exercise, find the indicated function 211. given f (x) = 2x2 + 1 and g(x) = 3x − 5. For the following exercises, use each pair of functions to find f ⎛ ⎠. Simplify your answers. ⎠ and g⎛ ⎝ f (x)⎞ ⎝g(x)⎞ 212. 213. f (x) = x2 + 1, g(x) = x + 2 f (x) = x + 2, g(x) = x2 + 3 214. f (x) = \|x\|, g(x) = 5x + 1 215. f (x) = x3 , g(x) = x + 1 x3 216. 217. f (x) = 1 x − 6 , g(x) = 7 x + 6 f (x) = 1 x − 4 , g(x) = 2 x + 4 For the following exercises, use each set of functions to find f ⎛ ⎠. Simplify your answers. ⎝h(x)⎞ ⎝g⎛ ⎞ ⎠ 218. 219. 220. f (x) = x4 + 6, g(x) = x − 6, and h(x) = x f (x) = x2 + 1, g(x) = 1 x, and h(x) = x + 3 Given f (x) = 1 x and g(x) = x − 3, find the following: a. b. c. d. e. ( f ∘ g)(x) the domain of ( f ∘ g)(x) in interval notation (g ∘ f )(x) the domain of (g ∘ f )(x) f g ⎛ ⎝ ⎞ ⎠x 221. Given f (x) = 2 − 4x and g(x) = − 3 x, find the following: a. b. (g ∘ f )(x) the domain of (g ∘ f )(x) in interval notation a. b. c. d. f (g(2)) f (g(x)) g( f (x)) (g ∘ g)(x) Given 222. f (x) = 1 − x x and g(x) = a. b. (g ∘ f )(x) (g ∘ f )(2) This content is available for free at https://cnx.org/content/col11758/1.5 the 1 1 + x2, find the following: functions Chapter 3 Functions 313 223. Given functions p(x) = 1 x and m(x) = x2 − 4, state the domain of each of the following functions using interval notation: 236. h(x) = (5x − 1)3 237. 3 h(x) = x − 1 a. b. p(x) m(x) p(m(x)) c. m(p(x)) 224. Given functions q(x) = 1 x and h(x) = x2 − 9, state the domain of each of the following functions using interval notation. a. b. c. q(x) h(x) q⎛ ⎝h(x)⎞ ⎠ ⎝q(x)⎞ h⎛ ⎠ 225. For f (x) = 1 x and g(x) = x − 1, write the domain of ( f ∘ g)(x) in interval notation. For the following exercises, find functions f (x) and g(x) so the given function can be expressed as h(x) = f ⎛ ⎝g(x)⎞ ⎠. 238. 239. 240. 241. h(x) = \|x2 + 7\| h(x) = 1 (x − 2)3 h(x) = 2 ⎛ ⎝ 1 2x − 3 ⎞ ⎠ h(x) = 2x − 1 3x + 4 Graphical For the following exercises, use the graphs of f , shown in Figure 3.63, and g, shown in Figure 3.64, to evaluate the expressions. 226. h(x) = (x + 2)2 227. h(x) = (x − 5)3 228. h(x) = 3 x − 5 229. h(x) = 4 (x + 2)2 230. h(x) = 4 + x3 231. h(x) = 3 1 2x − 3 232. h(x) = 1 (3x2 − 4)−3 233. 234. 4 h(x) = 3x − 2 x + 5 h(x) = ⎛ 8 + x3 ⎝ 8 − x3 4 ⎞ ⎠ 235. h(x) = 2x + 6 Figure 3.63 Figure 3.64 242. ⎝g(3)⎞ f ⎛ ⎠ 243. ⎝g(1)⎞ f ⎛ ⎠ 244. ⎝ f (1)⎞ g⎛ ⎠ 245. ⎝ f (0)⎞ g⎛ ⎠ 314 246. ⎝ f (5)⎞ f ⎛ ⎠ 247. ⎝ f (4)⎞ f ⎛ ⎠ 248. ⎝g(2)⎞ g⎛ ⎠ 249. ⎝g(0)⎞ g⎛ ⎠ For the following exercises, use graphs of f (x), shown in Figure 3.65, g(x), shown in Figure 3.66, and h(x), shown in Figure 3.67, to evaluate the expressions. Figure 3.65 Figure 3.66 Figure 3.67 250. ⎝ f (1)⎞ g⎛ ⎠ 251. ⎝ f (2)⎞ g⎛ ⎠ 252. ⎝g(4)⎞ f ⎛ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 253. ⎝g(1)⎞ f ⎛ ⎠ 254. ⎝h(2)⎞ f ⎛ ⎠ 255. ⎝ f (2)⎞ h⎛ ⎠ 256. 257. f ⎛ ⎝g⎛ ⎝h(4)⎞ ⎞ ⎠ ⎠ ⎝g⎛ f ⎛ ⎝ f (−2)⎞ ⎞ ⎠ ⎠ Numeric For the following exercises, use the function values for f and g shown in Table 3.23 to evaluate each expression(x) g(x Table 3.23 258. ⎝g(8)⎞ f ⎛ ⎠ 259. ⎝g(5)⎞ f ⎛ ⎠ 260. ⎝ f (5)⎞ g⎛ ⎠ 261. ⎝ f (3)⎞ g⎛ ⎠ 262. Chapter 3 Functions 315 ⎝ f (4)⎞ f ⎛ ⎠ 263. ⎝ f (1)⎞ f ⎛ ⎠ 264. ⎝g(2)⎞ g⎛ ⎠ 265. ⎝g(6)⎞ g⎛ ⎠ For the following exercises, use the function values for f and g shown in Table 3.24 to evaluate the expressions. x f(x) g(x) −3 −2 −1 0 1 2 3 11 9 7 5 3 1 −1 Table 3.24 −8 −3 0 1 0 −3 −8 266. ( f ∘ g)(1) 267. ( f ∘ g)(2) 268. (g ∘ f )(2) 269. (g ∘ f )(3) 270. (g ∘ g)(1) 271. ( f ∘ f )(3) For the following exercises, use each pair of functions to find f ⎛ ⎠ and g⎛ ⎝ f (0)⎞ ⎠. ⎝g(0)⎞ 274. 275. f (x) = x + 4, g(x) = 12 − x3 f (x) = 1 x + 2 , g(x) = 4x + 3 the For following f (x) = 2x2 + 1 and g(x) = 3x + 5 the composite function as indicated. exercises, use the functions to evaluate or find 276. ⎝g(2)⎞ f ⎛ ⎠ 277. ⎝g(x)⎞ f ⎛ ⎠ 278. ⎝ f ( − 3)⎞ g⎛ ⎠ 279. (g ∘ g)(x) Extensions For the following exercises, use f (x) = x3 + 1 and 3 g(x) = x − 1 . 280. Find ( f ∘ g)(x) and (g ∘ f )(x). Compare the two answers. 281. Find ( f ∘ g)(2) and (g ∘ f )(2). 282. What is the domain of (g ∘ f )(x) ? 283. What is the domain of
( f ∘ g)(x) ? 284. Let f (x) = 1 x. a. Find ( f ∘ f )(x). b. Is ( f ∘ f )(x) for any function f the same result as the answer to part (a) for any function? Explain. For the following exercises, let F(x) = (x + 1)5, f (x) = x5, and g(x) = x + 1. 285. True or False: (g ∘ f )(x) = F(x). 286. True or False: ( f ∘ g)(x) = F(x). For the following exercises, find the composition when f (x) = x2 + 2 for all x ≥ 0 and g(x) = x − 2. 287. ( f ∘ g)(6); (g ∘ f )(6) 288. (g ∘ f )(a); ( f ∘ g)(a) f (x) = 4x + 8, g(x) = 7 − x2 272. 273. f (x) = 5x + 7, g(x) = 4 − 2x2 289. ( f ∘ g)(11); (g ∘ f )(11) 316 Chapter 3 Functions Real-World Applications a. Find the composite function r(V(t)). b. Find the exact time when the radius reaches 10 inches. The number of bacteria in a refrigerated food product 297. is given by N(T) = 23T 2 − 56T + 1, 3 < T < 33, where T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by T(t) = 5t + 1.5, where t is the time in hours. a. Find the composite function N(T(t)). b. Find the time (round to two decimal places) when the bacteria count reaches 6752. The function D(p) gives the number of items that 290. will be demanded when the price is p. The production cost C(x) is the cost of producing x items. To determine the cost of production when the price is $6, you would do which of the following? a. Evaluate D⎛ ⎝C(6)⎞ ⎠. b. Evaluate C⎛ ⎝D(6)⎞ ⎠. c. Solve D(C(x)) = 6. d. Solve C⎛ ⎝D(p)⎞ ⎠ = 6. 291. The function A(d) gives the pain level on a scale of 0 to 10 experienced by a patient with d milligrams of a painreducing drug in her system. The milligrams of the drug in the patient’s system after t minutes is modeled by m(t). Which of the following would you do in order to determine when the patient will be at a pain level of 4? a. Evaluate A(m(4)). b. Evaluate m(A(4)). c. Solve A(m(t)) = 4. d. Solve m⎛ ⎝A(d)⎞ ⎠ = 4. 292. A store offers customers a 30% discount on the price x of selected items. Then, the store takes off an additional 15% at the cash register. Write a price function P(x) that computes the final price of the item in terms of the original price x. (Hint: Use function composition to find your answer.) 293. A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to r(t) = 25 t + 2, find the area of the ripple as a function of time. Find the area of the ripple at t = 2. 294. A forest fire leaves behind an area of grass burned in an expanding circular pattern. If the radius of the circle of burning grass is increasing with time according to the formula r(t) = 2t + 1, express the area burned as a function of time, t (minutes). Use the function you found in the previous exercise to 295. find the total area burned after 5 minutes. 296. The radius r, in inches, of a spherical balloon is related to the volume, V, by r(V) = 3V 4π into the balloon, so the volume after t seconds is given by V(t) = 10 + 20t. . Air is pumped 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 317 3.5 \| Transformation of Functions Learning Objectives In this section, you will: 3.5.1 Graph functions using vertical and horizontal shifts. 3.5.2 Graph functions using reflections about the x-axis axis and the y-axis. 3.5.3 Determine whether a function is even, odd, or neither from its graph. 3.5.4 Graph functions using compressions and stretches. 3.5.5 Combine transformations. Figure 3.68 (credit: "Misko"/Flickr) We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations. Graphing Functions Using Vertical and Horizontal Shifts Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve. Identifying Vertical Shifts One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function g(x) = f (x) + k, the function f (x) is shifted vertically k units. See Figure 3.69 for an example. 318 Chapter 3 Functions Figure 3.69 Vertical shift by k = 1 of the cube root function f (x) = x3 . To help you visualize the concept of a vertical shift, consider that y = f (x). Therefore, f (x) + k is equivalent to y + k. Every unit of y is replaced by y + k, so the y-value increases or decreases depending on the value of k. The result is a shift upward or downward. Vertical Shift Given a function f (x), a new function g(x) = f (x) + k, where k is a constant, is a vertical shift of the function f (x). All the output values change by k units. If k is positive, the graph will shift up. If k is negative, the graph will shift down. Example 3.49 Adding a Constant to a Function To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure 3.70 shows the area of open vents V (in square feet) throughout the day in hours after midnight, t. During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function. Figure 3.70 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 319 We can sketch a graph of this new function by adding 20 to each of the output values of the original function. This will have the effect of shifting the graph vertically up, as shown in Figure 3.71. Figure 3.71 Notice that in Figure 3.71, for each input value, the output value has increased by 20, so if we call the new function S(t), we could write S(t) = V(t) + 20 This notation tells us that, for any value of t, S(t) can be found by evaluating the function V at the same input and then adding 20 to the result. This defines S as a transformation of the function V, in this case a vertical shift up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change. See Table 3.25. 0 0 8 0 10 17 220 220 20 20 240 240 19 0 20 24 0 20 t V(t) S(t) Table 3.25 Given a tabular function, create a new row to represent a vertical shift. 1. Identify the output row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down. Example 3.50 Shifting a Tabular Function Vertically 320 Chapter 3 Functions A function f (x) is given in Table 3.26. Create a table for the function g(x) = f (x) − 3. x f(x) 2 1 4 3 6 7 8 11 Table 3.26 Solution The formula g(x) = f (x) − 3 tells us that we can find the output values of g by subtracting 3 from the output values of f . For example: Given f (2) = 1 g(x) = f (x) − 3 Given transformation g(2) = f (2) − 3 = 1 − 3 = −2 Subtracting 3 from each f (x) value, we can complete a table of values for g(x) as shown in Table 3.27. x f(x) 2 1 g(x) −2 4 3 0 6 7 4 8 11 8 Table 3.27 Analysis As with the earlier vertical shift, notice the input values stay the same and only the output values change. 3.32 The function h(t) = − 4.9t 2 + 30t gives the height h of a ball (in meters) thrown upward from the ground after t seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function b(t) to h(t), and then find a formula for b(t). Identifying Horizontal Shifts We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift, shown in Figure 3.72. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 321 Figure 3.72 Horizontal shift of the function f (x) = x3 . Note that h = + 1 shifts the graph to the left, that is, towards negative values of x. For example, if f (x) = x2, then g(x) = (x − 2)2 is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in f . Horizontal Shift Given a function f , a new function g(x) = f (x − h), where h is a constant, is a horizontal shift of the function f . If h is positive, the graph will shift right. If h is negative, the graph will shift left. Example 3.51 Adding a Constant to an Input Returning to our building airflow example from Figure 3.70, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function. Solution We can set V(t) to be the original program and F(t) to be the revised program. V(t) = the original venting plan F(t) = starting 2 hrs sooner In the new graph, at each time, the airflow is the same as the original
function V was 2 hours later. For example, in the original function V, the airflow starts to change at 8 a.m., whereas for the function F, the airflow starts to change at 6 a.m. The comparable function values are V(8) = F(6). See Figure 3.73. Notice also that the vents first opened to 220 ft2 at 10 a.m. under the original plan, while under the new plan the vents reach 220 ft2 at 8 a.m., so V(10) = F(8). In both cases, we see that, because F(t) starts 2 hours sooner, h = − 2. That means that the same output values are reached when F(t) = V(t − (−2)) = V(t + 2). 322 Chapter 3 Functions Figure 3.73 Analysis Note that V(t + 2) has the effect of shifting the graph to the left. Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function F(t) uses the same outputs as V(t), but matches those outputs to inputs 2 hours earlier than those of V(t). Said another way, we must add 2 hours to the input of V to find the corresponding output for F : F(t) = V(t + 2). Given a tabular function, create a new row to represent a horizontal shift. 1. Identify the input row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each input cell. Example 3.52 Shifting a Tabular Function Horizontally A function f (x) is given in Table 3.28. Create a table for the function g(x) = f (x − 3). x f(x) 2 1 4 3 6 7 8 11 Table 3.28 Solution The formula g(x) = f (x − 3) tells us that the output values of g are the same as the output value of f when the input value is 3 less than the original value. For example, we know that f (2) = 1. To get the same output from This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 323 the function g, we will need an input value that is 3 larger. We input a value that is 3 larger for g(x) because the function takes 3 away before evaluating the function f . g(5) = f (5 − 3) = f (2) = 1 We continue with the other values to create Table 3.29. 7 4 3 3 9 6 7 7 11 8 11 11 x x − 3 f(x) g(x) 5 2 1 1 Table 3.29 The result is that the function g(x) has been shifted to the right by 3. Notice the output values for g(x) remain the same as the output values for f (x), but the corresponding input values, x, have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11. Analysis Figure 3.74 represents both of the functions. We can see the horizontal shift in each point. 324 Chapter 3 Functions Example 3.53 Identifying a Horizontal Shift of a Toolkit Function Figure 3.75 represents a transformation of the toolkit function f (x) = x2. Relate this new function g(x) to f (x), and then find a formula for g(x). Figure 3.75 Solution Notice that the graph is identical in shape to the f (x) = x2 function, but the x-values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so g(x) = f (x − 2) Notice how we must input the value x = 2 to get the output value y = 0; the x-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the f (x) function to write a formula for g(x) by evaluating f (x − 2). f (x) = x2 g(x) = f (x − 2) g(x) = f (x − 2) = (x − 2)2 Analysis To determine whether the shift is + 2 or − 2 , consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, f (0) = 0. In our shifted function, g(2) = 0. To obtain the output value of 0 from the function f , we need to decide whether a plus or a minus sign will work to satisfy g(2) = f (x − 2) = f (0) = 0. For this to work, we will need to subtract 2 units from our input values. Example 3.54 Interpreting Horizontal versus Vertical Shifts This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 325 The function G(m) gives the number of gallons of gas required to drive m miles. Interpret G(m) + 10 and G(m + 10). Solution G(m) + 10 can be interpreted as adding 10 to the output, gallons. This is the gas required to drive m miles, plus another 10 gallons of gas. The graph would indicate a vertical shift. G(m + 10) can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than m miles. The graph would indicate a horizontal shift. 3.33 Given the function f (x) = x, graph the original function f (x) and the transformation g(x) = f (x + 2) on the same axes. Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units? Combining Vertical and Horizontal Shifts Now that we have two transformations, we can combine them. Vertical shifts are outside changes that affect the output (y-) values and shift the function up or down. Horizontal shifts are inside changes that affect the input (x-) values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and left or right. Given a function and both a vertical and a horizontal shift, sketch the graph. 1. Identify the vertical and horizontal shifts from the formula. 2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down for a negative constant. 3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right for a negative constant. 4. Apply the shifts to the graph in either order. Example 3.55 Graphing Combined Vertical and Horizontal Shifts Given f (x) = \|x\|, sketch a graph of h(x) = f (x + 1) − 3. Solution The function f is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h has transformed f in two ways: f (x + 1) is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in f (x + 1) − 3 is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 3.76. Let us follow one point of the graph of f (x) = \|x\|. • The point (0, 0) is transformed first by shifting left 1 unit: (0, 0) → (−1, 0) 326 Chapter 3 Functions • The point (−1, 0) is transformed next by shifting down 3 units: (−1, 0) → (−1, −3) Figure 3.76 Figure 3.77 shows the graph of h. Figure 3.77 3.34 Given f (x) = \|x\|, sketch a graph of h(x) = f (x − 2) + 4. Example 3.56 Identifying Combined Vertical and Horizontal Shifts Write a formula for the graph shown in Figure 3.78, which is a transformation of the toolkit square root function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 327 Figure 3.78 Solution The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as Using the formula for the square root function, we can write h(x) = f (x − 1) + 2 h(x) = x − 1 + 2 Analysis Note that this transformation has changed the domain and range of the function. This new graph has domain [1, ∞) and range [2, ∞). 3.35 Write a formula for a transformation of the toolkit reciprocal function f (x) = 1 x that shifts the function’s graph one unit to the right and one unit up. Graphing Functions Using Reflections about the Axes Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in Figure 3.79. 328 Chapter 3 Functions Figure 3.79 Vertical and horizontal reflections of a function. Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis. Reflections Given a function f (x), a new function g(x) = − f (x) is a vertical reflection of the function f (x), sometimes called a reflection about (or over, or through) the x-axis. Given a function f (x), a new function g(x) = f ( − x) is a horizontal reflection of the function f (x), sometimes called a reflection about the y-axis. Given a function, reflect the graph both vertically and horizontally. 1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis. 2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis. Example 3.57 Reflecting a Graph Horizontally and Vertically Reflect the graph of s(t) = t (a) vertically and (b) horizontally. Solution a. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as shown in Figure 3.80. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 329 Figure 3.80 Vertical reflection of the square root function Because each output value is the opposite of the original output value, we can write (3.3) Notice that this is an outside change, or vertical shift, that affects the output s(t) values, so the negative sign belongs outside of the function. V(t) = − s(t) or V(t) = − t b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in Figure 3.81. Figure 3.81 Horizontal reflection of the square root function Because each input value is the opposite of the original input value, we can write H(t) = s( − t) or H(t) = −t 330 Chapter 3 Functions Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function. Note that these transformations can affect the domain and range of the functions. While the
original square root function has domain [0, ∞) and range [0, ∞), the vertical reflection gives the V(t) function the range (−∞, 0] and the horizontal reflection gives the H(t) function the domain (−∞, 0]. 3.36 Reflect the graph of f (x) = \|x − 1\| (a) vertically and (b) horizontally. Example 3.58 Reflecting a Tabular Function Horizontally and Vertically A function f (x) is given as Table 3.30. Create a table for the functions below. a. b. g(x) = − f (x) h(x) = f (−x) x f(x) 2 1 4 3 6 7 8 11 Table 3.30 Solution a. For g(x), the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value. See Table 3.31. x 2 4 6 8 g(x) –1 –3 –7 –11 Table 3.31 b. For h(x), the negative sign inside the function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the h(x) values stay the same as the f (x) values. See Table 3.32. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 331 x −2 −4 −6 −8 h(x) 1 3 7 11 Table 3.32 3.37 A function f (x) is given as Table 3.33. Create a table for the functions below. a. g(x) = − f (x) b. h(x) = f (−x) x −2 0 2 4 f(x) 5 10 15 20 Table 3.33 Example 3.59 Applying a Learning Model Equation A common model for learning has an equation similar to k(t) = − 2−t mastery that can be achieved after t practice sessions. This is a transformation of the function f (t) = 2 in Figure 3.82. Sketch a graph of k(t). + 1, where k is the percentage of t shown 332 Chapter 3 Functions Figure 3.82 Solution This equation combines three transformations into one equation. • A horizontal reflection: f (−t) = 2−t • A vertical reflection: − f (−t) = − 2−t • A vertical shift: − f (−t) + 1 = − 2−t + 1 We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each of the three transformations. We will choose the points (0, 1) and (1, 2). 1. First, we apply a horizontal reflection: (0, 1) (–1, 2). 2. Then, we apply a vertical reflection: (0, −1) (1, –2). 3. Finally, we apply a vertical shift: (0, 0) (1, 1). This means that the original points, (0,1) and (1,2) become (0,0) and (1,1) after we apply the transformations. In Figure 3.83, the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 333 Figure 3.83 Analysis As a model for learning, this function would be limited to a domain of t ≥ 0, with corresponding range [0, 1). 3.38 Given the toolkit function f (x) = x2, graph g(x) = − f (x) and h(x) = f ( − x). Take note of any surprising behavior for these functions. Determining Even and Odd Functions Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions f (x) = x2 or f (x) = \|x\| will result in the original graph. We say that these types of graphs are symmetric about the y-axis. A function whose graph is symmetric about the y-axis is called an even function. If the graphs of f (x) = x3 or f (x) = 1 Figure 3.84. x were reflected over both axes, the result would be the original graph, as shown in 334 Chapter 3 Functions Figure 3.84 (a) The cubic toolkit function (b) Horizontal reflection of the cubic toolkit function (c) Horizontal and vertical reflections reproduce the original cubic function. We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function. Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f (x) = 2 nor odd. Also, the only function that is both even and odd is the constant function f (x) = 0. x is neither even Even and Odd Functions A function is called an even function if for every input x f (x) = f ( − x) The graph of an even function is symmetric about the y- axis. A function is called an odd function if for every input x The graph of an odd function is symmetric about the origin. f (x) = − f ( − x) Given the formula for a function, determine if the function is even, odd, or neither. 1. Determine whether the function satisfies f (x) = f ( − x). If it does, it is even. 2. Determine whether the function satisfies f (x) = − f ( − x). If it does, it is odd. 3. If the function does not satisfy either rule, it is neither even nor odd. Example 3.60 Determining whether a Function Is Even, Odd, or Neither Is the function f (x) = x3 + 2x even, odd, or neither? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 335 Solution Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Let’s begin with the rule for even functions. f ( − x) = ( − x)3 + 2( − x) = − x3 − 2x This does not return us to the original function, so this function is not even. We can now test the rule for odd functions. Because − f ( − x) = f (x), this is an odd function. − f ( − x) = − ⎝−x3 − 2x⎞ ⎛ ⎠ = x3 + 2x Analysis Consider the graph of f in Figure 3.85. Notice that the graph is symmetric about the origin. For every point (x, y) on the graph, the corresponding point (−x, − y) is also on the graph. For example, (1, 3) is on the graph of f , and the corresponding point (−1, −3) is also on the graph. Figure 3.85 3.39 Is the function f (s) = s4 + 3s2 + 7 even, odd, or neither? Graphing Functions Using Stretches and Compressions Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity. We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically. 336 Chapter 3 Functions Vertical Stretches and Compressions When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression. Figure 3.86 shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression. Figure 3.86 Vertical stretch and compression Vertical Stretches and Compressions Given a function f (x), a new function g(x) = a f (x), where a is a constant, is a vertical stretch or vertical compression of the function f (x). • • • If a > 1, then the graph will be stretched. If 0 < a < 1, then the graph will be compressed. If a < 0, then there will be combination of a vertical stretch or compression with a vertical reflection. Given a function, graph its vertical stretch. 1. Identify the value of a. 2. Multiply all range values by a. 3. If a > 1, the graph is stretched by a factor of a. If 0 < a < 1, the graph is compressed by a factor of a. If a < 0, the graph is either stretched or compressed and also reflected about the x-axis. Example 3.61 Graphing a Vertical Stretch A function P(t) models the population of fruit flies. The graph is shown in Figure 3.87. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 337 Figure 3.87 A scientist is comparing this population to another population, Q, whose growth follows the same pattern, but is twice as large. Sketch a graph of this population. Solution Because the population is always twice as large, the new population’s output values are always twice the original function’s output values. Graphically, this is shown in Figure 3.88. If we choose four reference points, (0, 1), (3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2. The following shows where the new points for the new graph will be located. (0, 1) → (0, 2) (3, 3) → (3, 6) (6, 2) → (6, 4) (7, 0) → (7, 0) Figure 3.88 Symbolically, the relationship is written as Q(t) = 2P(t) 338 Chapter 3 Functions This means that for any input t, the value of the function Q is twice the value of the function P. Notice that the effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal axis. The input values, t, stay the same while the output values are twice as large as before. Given a tabular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression. 1. Determine the value of a. 2. Multiply all of the output values by a. Example 3.62 Finding a Vertical Compression of a Tabular Function A function f is given as Table 3.34. Create a table for the function g(x) = 1 2 f (x). x f(x) 2 1 4 3 6 7 8 11 Table 3.34 Solution The formula g(x) = 1 2 f (x) tells us that the output values of g are half of the output values of f with the same inputs. For example, we know that f (4) = 3. Then We do the same for the other values to produce Table 3.35. g(4) = 1 2 f (4) = 1 2 (3) = 3 2 x g(x 11 2 Table 3.35 Analysis The result is that the function g(x) has been compressed vertically by 1 2 the graph is half the original height. . Each output value is divided in half, so This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 339 3.40 A function f is given as Table 3.36. Create a table for the function g(x) = 3 4 f (x). x 2 4 6 f (x) 12 16 20 8 0 Table 3.36 Example 3.63 Recognizing a Vertical Stretch The graph in Figure 3.89 is a transformation of the toolkit function f (x) = x3. Relate this new function g(x) to f (x), and then find a formula for g(x). Figure 3.89 Solution When trying to determine a vertical stretch or
shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g(2) = 2. With the basic cubic function at the same input, f (2) = 23 = 8. Based on that, it appears that the outputs of g are 1 4 f (x). the outputs of the function f because g(2) = 1 4 this we can fairly safely conclude that g(x) = 1 4 f (2). From We can write a formula for g by using the definition of the function f . g(x) = 1 4 f (x) = 1 4 x3 340 Chapter 3 Functions 3.41 Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units. Horizontal Stretches and Compressions Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function. Figure 3.90 Given a function y = f (x), the form y = f (bx) results in a horizontal stretch or compression. Consider the function y = x2. Observe Figure 3.90. The graph of y = (0.5x)2 is a horizontal stretch of the graph of the function y = x2 by a factor of 2. The graph of y = (2x)2 is a horizontal compression of the graph of the function y = x2 by a factor of 2. Horizontal Stretches and Compressions Given a function f (x), a new function g(x) = f (bx), where b is a constant, is a horizontal stretch or horizontal compression of the function f (x). • • • If b > 1, then the graph will be compressed by 1 b. If 0 < b < 1, then the graph will be stretched by 1 b. If b < 0, then there will be combination of a horizontal stretch or compression with a horizontal reflection. Given a description of a function, sketch a horizontal compression or stretch. 1. Write a formula to represent the function. 2. Set g(x) = f (bx) where b > 1 for a compression or 0 < b < 1 for a stretch. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 341 Example 3.64 Graphing a Horizontal Compression Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, R, will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population. Solution Symbolically, we could write R(1) = P(2), R(2) = P(4), and in general, R(t) = P(2t). See Figure 3.91 for a graphical comparison of the original population and the compressed population. Figure 3.91 (a) Original population graph (b) Compressed population graph Analysis Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis. Example 3.65 Finding a Horizontal Stretch for a Tabular Function A function f (x) is given as Table 3.37. Create a table for the function g(x) = f ⎛ ⎝ x⎞ ⎠. 1 2 342 Chapter 3 Functions x f(x) 2 1 4 3 6 7 8 11 Table 3.37 1 2 Solution x⎞ The formula g(x) = f ⎛ ⎠ tells us that the output values for g are the same as the output values for the ⎝ function f at an input half the size. Notice that we do not have enough information to determine g(2) because g(2) = f ⎛ ⎝ be twice as large to get inputs for f that we can evaluate. For example, we can determine g(4). ⎞ ⎠ = f (1), and we do not have a value for f (1) in our table. Our input values to g will need to ⋅ 2 1 2 We do the same for the other values to produce Table 3.38. g(42) = 1 x g(x) 4 1 8 3 12 16 7 11 Table 3.38 Figure 3.92 shows the graphs of both of these sets of points. Figure 3.92 Analysis Because each input value has been doubled, the result is that the function g(x) has been stretched horizontally by a factor of 2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 343 Example 3.66 Recognizing a Horizontal Compression on a Graph Relate the function g(x) to f (x) in Figure 3.93. Figure 3.93 Solution The graph of g(x) looks like the graph of f (x) horizontally compressed. Because f (x) ends at (6, 4) and g(x) ends at (2, 4), we can see that the x- values have been compressed by 1 3 ⎞ ⎠ = 2. We might also notice that g(2) = f (6) and g(1) = f (3). Either way, we can describe this relationship as g(x) = f (3x). This is a horizontal compression by 1 3 , because 6 1 3 ⎛ ⎝ . Analysis Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of 1 4 in our function: f ⎛ ⎝ 1 4 x⎞ ⎠. This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. 3.42 Write a formula for the toolkit square root function horizontally stretched by a factor of 3. Performing a Sequence of Transformations When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first. When we see an expression such as 2 f (x) + 3, which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of f (x), we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition. Horizontal transformations are a little trickier to think about. When we write g(x) = f (2x + 3), for example, we have to think about how the inputs to the function g relate to the inputs to the function f . Suppose we know f (7) = 12. What input to g would produce that output? In other words, what value of x will allow g(x) = f (2x + 3) = 12 ? We would need 344 Chapter 3 Functions 2x + 3 = 7. To solve for x, we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression. This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function. f (bx + p) = f ⎛ ⎝b⎛ ⎝x + ⎞ ⎞ ⎠ ⎠ p b Let’s work through an example. We can factor out a 2. f (x) = (2x + 4)2 f (x) = ⎛ ⎝2(x + 2)⎞ ⎠ 2 Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally. Combining Transformations When combining vertical transformations written in the form a f (x) + k, first vertically stretch by a and then vertically shift by k. When combining horizontal transformations written in the form f (bx + h), first horizontally shift by h and then horizontally stretch by 1 b. When combining horizontal transformations written in the form f (b(x + h)), first horizontally stretch by 1 b and then horizontally shift by h. Horizontal and vertical transformations are performed first. transformations are independent. It does not matter whether horizontal or vertical Example 3.67 Finding a Triple Transformation of a Tabular Function Given Table 3.39 for the function f (x), create a table of values for the function g(x) = 2 f (3x) + 1. x 6 12 18 24 f(x) 10 14 15 17 Table 3.39 Solution There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal transformations, f (3x) is a horizontal compression by 1 . See 3 , which means we multiply each x- value by 1 3 Table 3.40. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 345 x 2 4 6 8 f(3x) 10 14 15 17 Table 3.40 Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. See Table 3.41. x 2 4 6 8 2 f(3x) 20 28 30 34 Table 3.41 Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 3.42. x 2 4 6 8 g(x) = 2 f(3x) + 1 21 29 31 35 Table 3.42 Example 3.68 Finding a Triple Transformation of a Graph Use the graph of f (x) in Figure 3.94 to sketch a graph of k(x. 1 2 346 Chapter 3 Functions Figure 3.94 Solution To simplify, let’s start by factoring out the inside of the functionx + 2) ⎠ − 3 1 2 By factoring the inside, we can first horizontally stretch by 2, as indicated by the 1 2 on the inside of the function. Remember that twice the size of 0 is still 0, so the point (0,2) remains at (0,2) while the point (2,0) will stretch to (4,0). See Figure 3.95. Figure 3.95 Next, we horizontally shift left by 2 units, as indicated by x + 2. See Figure 3.96. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 347 Figure 3.96 Last, we vertically shift down by 3 to complete our sketch, as indicated by the − 3 on the outside of the function. See Figure 3.97. Figure 3.97 Access this online resource for additional instruction and practice with transformation of functions. • Function Transformations (http://openstaxcollege.org/l/functrans) 348 Chapter 3 Functions 3.5 EXERCISES Verbal 298. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal shift from a vertical shift? 299. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch? 300. When examining the formula of a function that is the result of multip
le transformations, how can you tell a horizontal compression from a vertical compression? 301. When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x-axis from a reflection with respect to the y-axis? How can you determine whether a function is odd or 302. even from the formula of the function? Algebraic For the following exercises, write a formula for the function obtained when the graph is shifted as described. 303. f (x) = x is shifted up 1 unit and to the left 2 units. 304. unit. 305. units. 306. f (x) = \|x\| is shifted down 3 units and to the right 1 f (x) = 1 x is shifted down 4 units and to the right 3 f (x) = 1 x2 is shifted up 2 units and to the left 4 units. For the following exercises, describe how the graph of the function is a transformation of the graph of the original function f . 307. y = f (x − 49) 308. y = f (x + 43) 309. y = f (x + 3) 310. y = f (x − 4) 311. y = f (x) + 5 312. y = f (x) + 8 313. y = f (x) − 2 This content is available for free at https://cnx.org/content/col11758/1.5 314. y = f (x) − 7 315. y = f (x − 2) + 3 316. y = f (x + 4) − 1 For the following exercises, determine the interval(s) on which the function is increasing and decreasing. 317. 318. f (x) = 4(x + 1)2 − 5 g(x) = 5(x + 3)2 − 2 319. a(x) = −x + 4 320. k(x) = − 3 x − 1 Graphical For the following exercises, use the graph of f (x) = 2 shown in Figure 3.98 to sketch a graph of each transformation of f (x). x Figure 3.98 321. g(x) = 2 322. h(x) = 2 x x + 1 − 3 323. w(x) = 2 x − 1 For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. f (t) = (t + 1)2 − 3 324. 325. Chapter 3 Functions 349 h(x) = \|x − 1\| + 4 326. k(x) = (x − 2)3 − 1 327. m(t) = 3 + t + 2 Numeric 328. Tabular representations for the functions f , g, and Write g(x) and h(x) as below. given h are transformations of f (x). x −2 −1 0 f(x) −2 −1 −3 x −1 0 1 g(x) −2 −1 −3 x −2 −1 0 h(x) −1 0 − 329. Tabular representations for the functions f , g, and Write g(x) and h(x) as below. given h are transformations of f (x). x −2 −1 f(x) −1 −3 0 4 x −3 −2 −1 g(x) −1 −2 −1 h(x) −2 −4 0 3 1 1 2 0 For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions. 330. 331. 332. 333. 350 Chapter 3 Functions 337. 334. 335. 336. This content is available for free at https://cnx.org/content/col11758/1.5 the following For of exercises, transformations of the square root function to find a formula for each of the functions. graphs use the 338. 339. Chapter 3 Functions 351 For the the following exercises, use the graphs of transformed toolkit functions to write a formula for each of the resulting functions. 340. 341. 342. 343. 347. f (x) = (x − 2)2 348. g(x) = 2x4 349. h(x) = 2x − x3 For the following exercises, describe how the graph of each function is a transformation of the graph of the original function f . 350. g(x) = − f (x) 351. g(x) = f ( − x) 352. g(x) = 4 f (x) 353. g(x) = 6 f (x) 354. g(x) = f (5x) 355. g(x) = f (2x) 356. 357. g(x) = f ⎛ ⎝ x⎞ ⎠ 1 3 g(x) = f ⎛ ⎝ x⎞ ⎠ 1 5 358. g(x) = 3 f (−x) 359. g(x) = − f (3x) For the following exercises, write a formula for the function g that results when the graph of a given toolkit function is transformed as described. 360. The graph of f (x) = \|x\| is reflected over the y -axis and horizontally compressed by a factor of 1 4 . 361. The graph of f (x) = x is reflected over the x -axis and horizontally stretched by a factor of 2. 362. The graph of f (x) = 1 x2 is vertically compressed by For the following exercises, determine whether the function is odd, even, or neither. a factor of 1 3 units. , then shifted to the left 2 units and down 3 344. f (x) = 3x4 345. g(x) = x 346. h(x) = 1 x + 3x 363. The graph of f (x) = 1 x is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units. 364. 352 Chapter 3 Functions Figure 3.99 375. g(x) = f (x) − 2 376. g(x) = − f (x) 377. g(x) = f (x + 1) 378. g(x) = f (x − 2) The graph of f (x) = x2 is vertically compressed by a factor of 1 2 , then shifted to the right 5 units and up 1 unit. 365. The graph of f (x) = x2 is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units. For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation. 366. 367. g(x) = 4(x + 1)2 − 5 g(x) = 5(x + 3)2 − 2 368. h(x) = − 2\|x − 4\| + 3 369. k(x) = − 3 x − 1 370. 371. 372. 373. m(x) = 1 2 x3 n(x) = 1 3\|x − 2\| x(x) = q(x) = 3 x⎞ ⎠ ⎛ ⎝ 1 4 + 1 374. a(x) = −x + 4 For the following exercises, use the graph in Figure 3.99 to sketch the given transformations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 353 3.6 \| Absolute Value Functions Learning Objectives In this section you will: 3.6.1 Graph an absolute value function. 3.6.2 Solve an absolute value equation. Figure 3.100 Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: "s58y"/Flickr) Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will continue our investigation of absolute value functions. Understanding Absolute Value Recall that in its basic form f (x) = \|x\|, the absolute value function is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems. Absolute Value Function The absolute value function can be defined as a piecewise function f (x) = \|x\| = x if x ≥ 0 ⎧ ⎨ −x if x < 0 ⎩ Example 3.69 Using Absolute Value to Determine Resistance Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters 354 Chapter 3 Functions vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ± 5%, or ± 10%. Suppose we have a resistor rated at 680 ohms, ± 5%. Use the absolute value function to express the range of possible values of the actual resistance. Solution We can find that 5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance R in ohms, \|R − 680\| ≤ 34 Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute 3.43 value notation. Graphing an Absolute Value Function The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure 3.101. Figure 3.101 Figure 3.102 shows the graph of y = 2\|x – 3\| + 4. The graph of y = \|x\| has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at (3, 4) for this transformed function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 355 Figure 3.102 Example 3.70 Writing an Equation for an Absolute Value Function Given a Graph Write an equation for the function graphed in Figure 3.103. Figure 3.103 Solution The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 3.104. 356 Chapter 3 Functions Figure 3.104 We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 3.105. Figure 3.105 From this information we can write the equation f (x) = 2\|x − 3\| − 2, f (x) = \|2(x − 3)\| − 2, treating the stretch as a vertical stretch,or treating the stretch as a horizontal compression. Analysis Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 357 If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it? Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for x and f (x). Now substituting in the point (1, 2) f (x) = a\|x − 3\| − 2 2 = a\|1 − 3\| − 2 4 = 2a a = 2 Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically 3.44 flipped, and vertically shifted up 3 units. Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis? Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero. No, they do not always intersect the
horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see Figure 3.106). Figure 3.106 (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points. Solving an Absolute Value Equation In Other Type of Equations, we touched on the concepts of absolute value equations. Now that we understand a little more about their graphs, we can take another look at these types of equations. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as 8 = \|2x − 6\|, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. This leads to two different equations we can solve independently. 2x − 6 = 8 2x = 14 x = 7 or 2x − 6 = −8 2x = −2 x = −1 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point. 358 Chapter 3 Functions An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example, \|x\| = 4, \|2x − 1\| = 3, or \|5x + 2\| − 4 = 9 Solutions to Absolute Value Equations For real numbers A and B , an equation of the form \|A\| = B, with B ≥ 0, will have solutions when A = B or A = − B. If B < 0, the equation \|A\| = B has no solution. Given the formula for an absolute value function, find the horizontal intercepts of its graph. 1. Isolate the absolute value term. 2. Use \|A\| = B to write A = B or −A = B, assuming B > 0. 3. Solve for x. Example 3.71 Finding the Zeros of an Absolute Value Function For the function f (x) = \|4x + 1\| − 7, find the values of x such that f (x) = 0. Solution 0 = \|4x + 1\| − 7 7 = \|4x + 1\| 7 = 4x + 1 6 = 4x x = 6 4 = 1.5 or −7 = 4x + 1 −8 = 4x x = −8 4 = − 2 Substitute 0 for f (x). Isolate the absolute value on one side of the equation. Break into two separate equations and solve. The function outputs 0 when x = 3 2 or x = − 2. See Figure 3.107. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 359 Figure 3.107 3.45 For the function f (x) = \|2x − 1\| − 3, find the values of x such that f (x) = 0. Should we always expect two answers when solving \|A\| = B? No. We may find one, two, or even no answers. For example, there is no solution to 2 + \|3x − 5\| = 1. Access these online resources for additional instruction and practice with absolute value. • Graphing Absolute Value Functions (http://openstaxcollege.org/l/graphabsvalue) • Graphing Absolute Value Functions 2 (http://openstaxcollege.org/l/graphabsvalue2) 360 Chapter 3 Functions 3.6 EXERCISES Verbal 379. How do you solve an absolute value equation? How can you tell whether an absolute value function 380. has two x-intercepts without graphing the function? 381. When solving an absolute value function, the isolated absolute value term is equal to a negative number. What does that tell you about the graph of the absolute value function? 382. How can you use the graph of an absolute value function to determine the x-values for which the function values are negative? Algebraic Describe all numbers x that are at a distance of 4 383. from the number 8. Express this set of numbers using absolute value notation. 384. Describe all numbers x that are at a distance of 1 2 from the number −4. Express this set of numbers using absolute value notation. Describe the situation in which the distance that point 385. x is from 10 is at least 15 units. Express this set of numbers using absolute value notation. Find all function values f (x) such that the distance 386. from f (x) to the value 8 is less than 0.03 units. Express this set of numbers using absolute value notation. For the following exercises, find the x- and y-intercepts of the graphs of each function. 387. f (x) = 4\|x − 3\| + 4 388. f (x) = − 3\|x − 2\| − 1 389. f (x) = − 2\|x + 1\| + 6 390. 391. 392. 393. f (x) = − 5\|x + 2\| + 15 f (x) = 2\|x − 1\| − 6 f (x) = \| − 2x + 1\| − 13 f (x) = − \|x − 9\| + 16 Graphical For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph. This content is available for free at https://cnx.org/content/col11758/1.5 394. y = \|x − 1\| 395. y = \|x + 1\| 396. y = \|x\| + 1 For the following exercises, graph the given functions by hand. 397. y = \|x\| − 2 398. y = − \|x\| 399. y = − \|x\| − 2 400. y = − \|x − 3\| − 2 401. 402. 403. f (x) = − \|x − 1\| − 2 f (x) = − \|x + 3\| + 4 f (x) = 2\|x + 3\| + 1 404. f (x) = 3\|x − 2\| + 3 405. f (x) = \|2x − 4\| − 3 406. f (x) = \|3x + 9\| + 2 407. f (x) = − \|x − 1\| − 3 408. f (x) = − \|x + 4\| − 3 409. f (x) = 1 2\|x + 4\| − 3 Technology 410. Use a graphing utility to graph f (x) = 10\|x − 2\| on Identify the corresponding the viewing window [0, 4]. range. Show the graph. a Use graphing 411. f (x) = − 100\|x\| + 100 on the viewing window ⎡ Identify the corresponding range. Show the graph. utility to graph ⎦. ⎣−5, 5⎤ For the following exercises, graph each function using a graphing utility. Specify the viewing window. 412. 413. f (x) = − 0.1\|0.1(0.2 − x)\| + 0.3 f (x) = 4×109\|x − ⎛ ⎝5×109⎞ ⎠\| + 2×109 361 Chapter 3 Functions Extensions For the following exercises, solve the inequality. If possible, find all values of a such that there are no 414. x- intercepts for f (x) = 2\|x + 1\| + a. If possible, find all values of a such that there are no 415. y -intercepts for f (x) = 2\|x + 1\| + a. Real-World Applications 416. line. Cities A and B are on the same east-west Assume that city A is located at the origin. If the distance from city A to city B is at least 100 miles and x represents the distance from city B to city A, express this using absolute value notation. 417. The true proportion p of people who give a favorable rating to Congress is 8% with a margin of error of 1.5%. Describe this statement using an absolute value equation. Students who score within 18 points of the number 82 418. will pass a particular test. Write this statement using absolute value notation and use the variable x for the score. 419. A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using x as the diameter of the bearing, write this statement using absolute value notation. 420. The tolerance for a ball bearing is 0.01. If the true diameter of the bearing is to be 2.0 inches and the measured value of the diameter is x inches, express the tolerance using absolute value notation. 362 Chapter 3 Functions 3.7 \| Inverse Functions Learning Objectives In this section, you will: 3.7.1 Verify inverse functions. 3.7.2 Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one. 3.7.3 Find or evaluate the inverse of a function. 3.7.4 Use the graph of a one-to-one function to graph its inverse function on the same axes. A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating. If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. Figure 3.108 provides a visual representation of this question. In this section, we will consider the reverse nature of functions. Figure 3.108 Can a function “machine” operate in reverse? Verifying That Two Functions Are Inverse Functions Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula and substitutes 75 for F to calculate C = 5 9 (F − 32) (75 − 32) ≈ 24°C 5 9 Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure 3.109 for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 363 Figure 3.109 At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for F after substituting a value for C. For example, to convert 26 degrees Celsius, she could write (F − 32) 26 = 5 9 = F − 32 26 ⋅ 9 5 F = 26 ⋅ 9 5 + 32 ≈ 79 After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature. The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function. Given a function f (x), we represent its inverse as f −1(x), read as “ f inverse of x.” The raised −1 is part of the notation. It is not an exponent; it does not imply a power of −1 . In other words, f −1(x) does not mean 1 f (x) because 1 f (x) is the reciprocal of f and not the inverse. The “exponent-like” notation comes from an analo
gy between function composition and multiplication: just as a−1 a = 1 (1 is the identity element for multiplication) for any nonzero number a, so f −1 ∘ f equals the identity function, that is, ⎛ ⎝ f −1 ∘ f ⎞ ⎠(x) = f −1 ⎛ ⎝ f (x)⎞ ⎠ = f −1 (y) = x This holds for all x in the domain of f . Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses. Given a function f (x), we can verify whether some other function g(x) is the inverse of f (x) by checking whether either g( f (x)) = x or f (g(x)) = x is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.) For example, y = 4x and y = 1 4 x are inverse functions. and ⎝ f −1 ∘ f ⎞ ⎛ ⎠(x) = f −1 (4x) = 1 4 (4x) = x ⎝ f ∘ f −1⎞ ⎛ ⎠(x) = f ⎛ ⎝ 1 4 x⎞ ⎠ = 4 ⎛ ⎝ x⎞ ⎠ = x 1 4 364 Chapter 3 Functions A few coordinate pairs from the graph of the function y = 4x are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function y = 1 4 x are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function. Inverse Function For any one-to-one function f (x) = y, a function f −1 (x) is an inverse function of f if f −1(y) = x. This can also be written as f −1( f (x)) = x for all x in the domain of f . It also follows that f ( f −1(x)) = x for all x in the domain of f −1 if f −1 is the inverse of f . The notation f −1 is read “ f so we will often write f −1(x), which we read as “ f inverse of x.” Keep in mind that inverse.” Like any other function, we can use any variable name as the input for f −1, f −1(x) ≠ 1 f (x) and not all functions have inverses. Example 3.72 Identifying an Inverse Function for a Given Input-Output Pair If for a particular one-to-one function f (2) = 4 and f (5) = 12, what are the corresponding input and output values for the inverse function? Solution The inverse function reverses the input and output quantities, so if f (2) = 4, then f −1(4) = 2; f (5) = 12, then f −1(12) = 5. Alternatively, if we want to name the inverse function g, then g(4) = 2 and g(12) = 5. Analysis Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 3.42. ⎛ ⎝x, f(x)⎞ ⎠ ⎛ ⎝x, g(x)⎞ ⎠ (2, 4) (4, 2) (5, 12) (12, 5) Table 3.42 3.46 Given that h−1(6) = 2, what are the corresponding input and output values of the original function h ? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 365 Given two functions f(x) and g(x), test whether the functions are inverses of each other. 1. Determine whether f (g(x)) = x or g( f (x)) = x. 2. If either statement is true, then both are true, and g = f −1 and f = g−1. If either statement is false, then both are false, and g ≠ f −1 and f ≠ g−1. Example 3.73 Testing Inverse Relationships Algebraically If f (x) = 1 x + 2 and g(x) = 1 x − 2, is g = f −1 ? Solution so g( f (x)) = − This is enough to answer yes to the question, but we can also verify the other formula. g = f −1 and f = g−1 f (g(x)) = Analysis Notice the inverse operations are in reverse order of the operations from the original function. 3.47 If f (x) = x3 − 4 and g(x) = x + 4 , is g = f −1 ? 3 Example 3.74 Determining Inverse Relationships for Power Functions If f (x) = x3 (the cube function) and g(x) = 1 3 x, is g = f −1 ? Solution No, the functions are not inverses. ⎝g(x)⎞ f ⎛ ⎠ = x3 27 ≠ x 366 Analysis Chapter 3 Functions The correct inverse to the cube is, of course, the cube root x3 = x multiplier. 1 3, that is, the one-third is an exponent, not a 3.48 If f (x) = (x − 1)3 and g(x) = x3 + 1, is g = f −1 ? Finding Domain and Range of Inverse Functions The outputs of the function f are the inputs to f −1, so the range of f is also the domain of f −1. Likewise, because the inputs to f are the outputs of f −1, the domain of f is the range of f −1. We can visualize the situation as in Figure 3.110. Figure 3.110 Domain and range of a function and its inverse When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of f (x) = x is f −1(x) = x2, because a square “undoes” a square root; but the square is only the inverse of the square root on the domain [0, ∞), since that is the range of f (x) = x. We can look at this problem from the other side, starting with the square (toolkit quadratic) function f (x) = x2. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function. In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is oneto-one. For example, we can make a restricted version of the square function f (x) = x2 with its range limited to [0, ∞), which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function). If f (x) = (x − 1)2 on [1, ∞), then the inverse function is f −1(x) = x + 1. • The domain of f = range of f −1 = [1, ∞). • The domain of f −1 = range of f = [0, ∞). Is it possible for a function to have more than one inverse? No. If two supposedly different functions, say, g and h, both meet the definition of being inverses of another function f , then you can prove that g = h. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 367 Domain and Range of Inverse Functions The range of a function f (x) is the domain of the inverse function f −1(x). The domain of f (x) is the range of f −1(x). Given a function, find the domain and range of its inverse. 1. 2. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function. Example 3.75 Finding the Inverses of Toolkit Functions Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 3.43. We restrict the domain in such a fashion that the function assumes all y-values exactly once. Constant Identity Quadratic Cubic Reciprocal f (x) = c f (x) = x f (x) = x2 f (x) = x3 f (x) = 1 x Reciprocal squared Cube root Square root Absolute value f (x) = x3 f (x) = x f (x) = \|x\| f (x) = 1 x2 Table 3.43 Solution The constant function is not one-to-one, and there is no domain (except a single point) on which it could be oneto-one, so the constant function has no inverse. The absolute value function can be restricted to the domain [0, ∞), where it is equal to the identity function. The reciprocal-squared function can be restricted to the domain (0, ∞). Analysis We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 3.111. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse. 368 Chapter 3 Functions Figure 3.111 (a) Absolute value (b) Reciprocal square 3.49 The domain of function f is (1, ∞) and the range of function f is (−∞, −2). Find the domain and range of the inverse function. Finding and Evaluating Inverse Functions Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. Inverting Tabular Functions Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range. Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Example 3.76 Interpreting the Inverse of a Tabular Function A function f (t) is given in Table 3.44, showing distance in miles that a car has traveled in t minutes. Find and interpret f −1(70). t (minutes) 30 50 70 90 f(t) (miles) 20 40 60 70 Table 3.44 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 369 Solution The inverse function takes an output of f and returns an input for f . So in the expression f −1(70), 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function f , 90 minutes, so f −1(70) = 90. The interpretation of this is that, to drive 70 miles, it took 90 minutes. Alter
natively, recall that the definition of the inverse was that if f (a) = b, then f −1(b) = a. By this definition, if we are given f −1(70) = a, then we are looking for a value a so that f (a) = 70. In this case, we are looking for a t so that f (t) = 70, which is when t = 90. 3.50 Using Table 3.45, find and interpret (a) f (60), and (b) f −1(60). t (minutes) 30 50 60 70 90 f(t) (miles) 20 40 50 60 70 Table 3.45 Evaluating the Inverse of a Function, Given a Graph of the Original Function We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. Given the graph of a function, evaluate its inverse at specific points. 1. Find the desired input on the y-axis of the given graph. 2. Read the inverse function’s output from the x-axis of the given graph. Example 3.77 Evaluating a Function and Its Inverse from a Graph at Specific Points A function g(x) is given in Figure 3.112. Find g(3) and g−1(3). 370 Chapter 3 Functions Figure 3.112 Solution To evaluate g(3), we find 3 on the x-axis and find the corresponding output value on the y-axis. The point (3, 1) tells us that g(3) = 1. To evaluate g−1(3), recall that by definition g−1(3) means the value of x for which g(x) = 3. By looking for the output value 3 on the vertical axis, we find the point (5, 3) on the graph, which means g(5) = 3, so by definition, g−1(3) = 5. See Figure 3.113. Figure 3.113 3.51 Using the graph in Figure 3.113, (a) find g−1(1), and (b) estimate g−1(4). Finding Inverses of Functions Represented by Formulas Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula—for example, y as a function of x— we can often find the inverse function by solving to obtain x as a function of y. Given a function represented by a formula, find the inverse. 1. Make sure f is a one-to-one function. 2. Solve for x. 3. Interchange x and y. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 371 Example 3.78 Inverting the Fahrenheit-to-Celsius Function Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. Solution C = 5 9 (F − 32) (F − 32) C = 5 9 = F − 32 C ⋅ 9 5 F = 9 5 C + 32 By solving in general, we have uncovered the inverse function. If then C = h(F) = 5 9 (F − 32), F = h−1(C) = 9 5 C + 32 In this case, we introduced a function h to represent the conversion because the input and output variables are descriptive, and writing C −1 could get confusing. 3.52 Solve for x in terms of y given y = 1 3 (x − 5). Example 3.79 Solving to Find an Inverse Function Find the inverse of the function f (x) = 2 x − 3 + 4. Solution or f −1 (x) = 2 x = + 3 + 3. x − 4 Set up an equation. Subtract 4 from both sides. Multiply both sides by x − 3 and divide by y − 4. Add 3 to both sides. So f −1 (y) = 2 y − 4 Analysis 372 Chapter 3 Functions The domain and range of f exclude the values 3 and 4, respectively. f and f −1 are equal at two points but are not the same function, as we can see by creating Table 3.45. x f(x) 1 3 2 2 5 5 f −1(y) y Table 3.45 Example 3.80 Solving to Find an Inverse with Radicals Find the inverse of the function f (x) = 2 + x − 4. Solution y = 2 + x − 4 (y − 2)2 = x − 4 x = (y − 2)2 + 4 So f −1 (x) = (x − 2)2 + 4. The domain of f is [4, ∞). Notice that the range of f is [2, ∞), so this means that the domain of the inverse function f −1 is also [2, ∞). Analysis The formula we found for f −1 (x) looks like it would be valid for all real x. However, f −1 itself must have an inverse (namely, f ) so we have to restrict the domain of f −1 to [2, ∞) in order to make f −1 a one-to-one function. This domain of f −1 is exactly the range of f . 3.53 What is the inverse of the function f (x) = 2 − x ? State the domains of both the function and the inverse function. Finding Inverse Functions and Their Graphs Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function f (x) = x2 restricted to the domain [0, ∞), on which this function is one-to-one, and graph it as in Figure 3.114. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 373 Figure 3.114 Quadratic function with domain restricted to [0, ∞). Restricting the domain to [0, ∞) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain. We already know that the inverse of the toolkit quadratic function is the square root function, that is, f −1(x) = x. What happens if we graph both f and f −1 on the same set of axes, using the x- axis for the input to both f and f −1 ? We notice a distinct relationship: The graph of f −1(x) is the graph of f (x) reflected about the diagonal line y = x, which we will call the identity line, shown in Figure 3.115. Figure 3.115 Square and square-root functions on the nonnegative domain This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes. Example 3.81 Finding the Inverse of a Function Using Reflection about the Identity Line Given the graph of f (x) in Figure 3.116, sketch a graph of f −1(x). 374 Chapter 3 Functions Figure 3.116 Solution This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of (0, ∞) and range of (−∞, ∞), so the inverse will have a domain of (−∞, ∞) and range of (0, ∞). If we reflect this graph over the line y = x, the point (1, 0) reflects to (0, 1) and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph gives Figure 3.117. Figure 3.117 The function and its inverse, showing reflection about the identity line 3.54 Draw graphs of the functions f and f −1 from Example 3.79. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 375 Is there any function that is equal to its own inverse? Yes. If f = f −1, then f ⎛ ⎝ f (x)⎞ function does, and so does the reciprocal function, because ⎠ = x, and we can think of several functions that have this property. The identity Any function f (x) = c − x, where c is a constant, is also equal to its own inverse. = x 1 1 x Access these online resources for additional instruction and practice with inverse functions. • Inverse Functions (http://openstaxcollege.org/l/inversefunction) • One-to-one Functions (http://openstaxcollege.org/l/onetoone) • Inverse Function Values Using Graph (http://openstaxcollege.org/l/inversfuncgraph) • Restricting the Domain and Finding the Inverse (http://openstaxcollege.org/l/ restrictdomain) this website (http://openstaxcollege.org/l/PreCalcLPC01) Visit Learningpod. for additional practice questions from 376 Chapter 3 Functions 3.7 EXERCISES Verbal Describe why the horizontal line test is an effective 421. way to determine whether a function is one-to-one? Why do we restrict 422. f (x) = x2 to find the function’s inverse? the domain of the function 423. Can a function be its own inverse? Explain. Are one-to-one functions either always increasing or 424. always decreasing? Why or why not? How do you find the inverse of a function 425. algebraically? Algebraic Show that 426. inverse for all real numbers a. the function f (x) = a − x is its own b. What does the answer tell us about the relationship between f (x) and g(x) ? For the following exercises, use function composition to verify that f (x) and g(x) are inverse functions. 437. 438. 3 f (x) = x − 1 and g(x) = x3 + 1 f (x) = − 3x + 5 and g(x) = x − 5 −3 Graphical For the following exercises, use a graphing utility to determine whether each function is one-to-one. 439. f (x) = x 440. 3 f (x) = 3x + 1 For the following exercises, find f −1(x) for each function. 441. f (x) = −5x + 1 442. f (x) = x3 − 27 For the following exercises, determine whether the graph represents a one-to-one function. 443. 427. f (x) = x + 3 428. f (x) = x + 5 429. f (x) = 2 − x 430. f (x) = 3 − x 431. f (x) = x x + 2 432. f (x) = 2x + 3 5x + 4 For the following exercises, find a domain on which each function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain. 444. 433. f (x) = (x + 7)2 434. f (x) = (x − 6)2 435. f (x) = x2 − 5 436. Given f (x) = x3 − 5 and g(x) = 2x 1 − x : a. Find f (g(x)) and g( f (x)). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 377 For the following exercises, use the graph of f shown in Figure 3.118. Numeric For the following exercises, evaluate or solve, assuming that the function f is one-to-one. 453. 454. 455. 456. If f (6) = 7, find f −1(7). If f (3) = 2, find f −1(2). If f −1 (−4) = − 8, find f ( − 8). If f −1 (−2) = − 1, find f ( − 1). For the following exercises, use the values listed in Table 3.46 to evaluate or solve. x f(x) Figure 3.118 445. Find f (0). 446. Solve f (x) = 0. 447. Find f −1 (0). 448. Solve f −1 (x) = 0. For the following exercises, use the graph of the one-to-one function shown in Figure 3.119. Figure 3.119 449. Sketch the graph of f −1. 450. Find f (6) and f −1(2). Table 3.46 If the complete graph of f is shown, find the do
main 451. of f . 457. Find f (1). 458. Solve f (x) = 3. 452. f . If the complete graph of f is shown, find the range of 459. Find f −1 (0). 460. Solve f −1 (x) = 7. 378 Chapter 3 Functions 461. Use the tabular representation of f in Table 3.47 to create a table for f −1 (x). x f(x) 3 1 6 4 9 7 13 14 12 16 Table 3.47 Technology For the following exercises, find the inverse function. Then, graph the function and its inverse. 462. f (x) = 3 x − 2 463. f (x) = x3 − 1 464. Find the inverse function of f (x) = 1 x − 1 . Use a graphing utility to find its domain and range. Write the domain and range in interval notation. Real-World Applications 465. To convert from x degrees Celsius to y degrees Fahrenheit, we use the formula f (x) = 9 5 x + 32. Find the inverse function, if it exists, and explain its meaning. The circumference C of a circle is a function of its 466. radius given by C(r) = 2πr. Express the radius of a circle as a function of its circumference. Call this function r(C). Find r(36π) and interpret its meaning. A car travels at a constant speed of 50 miles per hour. 467. The distance the car travels in miles is a function of time, t, in hours given by d(t) = 50t. Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function t(d). Find t(180) and interpret its meaning. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 379 CHAPTER 3 REVIEW KEY TERMS absolute maximum the greatest value of a function over an interval absolute minimum the lowest value of a function over an interval average rate of change the difference in the output values of a function found for two values of the input divided by the difference between the inputs composite function input of another the new function formed by function composition, when the output of one function is used as the decreasing function a function is decreasing in some open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a dependent variable an output variable domain the set of all possible input values for a relation even function the y- axis a function whose graph is unchanged by horizontal reflection, f (x) = f ( − x), and is symmetric about function a relation in which each input value yields a unique output value horizontal compression constant b > 1 a transformation that compresses a function’s graph horizontally, by multiplying the input by a horizontal line test a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once horizontal reflection a transformation that reflects a function’s graph across the y-axis by multiplying the input by −1 horizontal shift input a transformation that shifts a function’s graph left or right by adding a positive or negative constant to the horizontal stretch 0 < b < 1 a transformation that stretches a function’s graph horizontally by multiplying the input by a constant increasing function a function is increasing in some open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a independent variable an input variable input each object or value in a domain that relates to another object or value by a relationship known as a function interval notation a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion inverse function for any one-to-one function f (x), the inverse is a function f −1(x) such that f −1 ⎛ ⎝ f (x)⎞ ⎠ = x for all x in the domain of f ; this also implies that f ⎛ ⎞ ⎝ f −1 (x) ⎠ = x for all x in the domain of f −1 local extrema collectively, all of a function's local maxima and minima local maximum a value of the input where a function changes from increasing to decreasing as the input value increases. local minimum a value of the input where a function changes from decreasing to increasing as the input value increases. odd function a function whose graph is unchanged by combined horizontal and vertical reflection, f (x) = − f ( − x), and is symmetric about the origin one-to-one function a function for which each value of the output is associated with a unique input value 380 output each object or value in the range that is produced when an input value is entered into a function Chapter 3 Functions piecewise function a function in which more than one formula is used to define the output range the set of output values that result from the input values in a relation rate of change the change of an output quantity relative to the change of the input quantity relation a set of ordered pairs set-builder notation a method of describing a set by a rule that all of its members obey; it takes the form {x\| statement about x} vertical compression a function transformation that compresses the function’s graph vertically by multiplying the output by a constant 0 < a < 1 vertical line test a method of testing whether a graph represents a function by determining whether a vertical line intersects the graph no more than once vertical reflection a transformation that reflects a function’s graph across the x-axis by multiplying the output by −1 vertical shift a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the output vertical stretch a > 1 a transformation that stretches a function’s graph vertically by multiplying the output by a constant KEY EQUATIONS Constant function f (x) = c, where c is a constant Identity function f (x) = x Absolute value function f (x) = \|x\| Quadratic function Cubic function Reciprocal function Reciprocal squared function f (x) = x2 f (x) = x3 f (x) = 1 x f (x) = 1 x2 Square root function f (x) = x Cube root function f (x) = x3 Average rate of change Δy Δx = f (x2) − f (x1) x2 − x1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 381 Composite function ⎛ ⎝ f ∘ g⎞ ⎠(x) = f ⎛ ⎝g(x)⎞ ⎠ Vertical shift g(x) = f (x) + k (up for k > 0 ) Horizontal shift g(x) = f (x − h) (right for h > 0 ) Vertical reflection g(x) = − f (x) Horizontal reflection g(x) = f ( − x) Vertical stretch g(x) = a f (x) ( a > 0 ) Vertical compression g(x) = a f (x) (0 < a < 1) Horizontal stretch g(x) = f (bx) (0 < b < 1) Horizontal compression. g(x) = f (bx) ( b > 1 ) KEY CONCEPTS 3.1 Functions and Function Notation • A relation is a set of ordered pairs. A function is a specific type of relation in which each domain value, or input, leads to exactly one range value, or output. See Example 3.1 and Example 3.2. • Function notation is a shorthand method for relating the input to the output in the form y = f (x). See Example 3.3 and Example 3.4. • In tabular form, a function can be represented by rows or columns that relate to input and output values. See Example 3.5. • To evaluate a function, we determine an output value for a corresponding input value. Algebraic forms of a function can be evaluated by replacing the input variable with a given value. See Example 3.6 and Example 3.7. • To solve for a specific function value, we determine the input values that yield the specific output value. See Example 3.8. • An algebraic form of a function can be written from an equation. See Example 3.9 and Example 3.10. • Input and output values of a function can be identified from a table. See Example 3.11. • Relating input values to output values on a graph is another way to evaluate a function. See Example 3.12. • A function is one-to-one if each output value corresponds to only one input value. See Example 3.13. • A graph represents a function if any vertical line drawn on the graph intersects the graph at no more than one point. See Example 3.14. • The graph of a one-to-one function passes the horizontal line test. See Example 3.15. 382 Chapter 3 Functions 3.2 Domain and Range • The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number. • The domain of a function can be determined by listing the input values of a set of ordered pairs. See Example 3.16. • The domain of a function can also be determined by identifying the input values of a function written as an equation. See Example 3.17, Example 3.18, and Example 3.19. • Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation. See Example 3.20. • For many functions, the domain and range can be determined from a graph. See Example 3.21 and Example 3.22. • An understanding of toolkit functions can be used to find the domain and range of related functions. See Example 3.23, Example 3.24, and Example 3.25. • A piecewise function is described by more than one formula. See Example 3.26 and Example 3.27. • A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See Example 3.28. 3.3 Rates of Change and Behavior of Graphs • A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See Example 3.29. • Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example 3.30. • Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Example 3.31. • An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Example 3.32 and Example 3.33. • The average rate of change can sometimes be determined as an expression. See Example 3.34. • A function is increasing where its rate of change is positive and decreasing where its rate of change
is negative. See Example 3.35. • A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values. • A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values. • Minima and maxima are also called extrema. • We can find local extrema from a graph. See Example 3.36 and Example 3.37. • The highest and lowest points on a graph indicate the maxima and minima. See Example 3.38. 3.4 Composition of Functions • We can perform algebraic operations on functions. See Example 3.39. • When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function. • The function produced by combining two functions is a composite function. See Example 3.40 and Example 3.41. • The order of function composition must be considered when interpreting the meaning of composite functions. See Example 3.42. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 383 • A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function. • A composite function can be evaluated from a table. See Example 3.43. • A composite function can be evaluated from a graph. See Example 3.44. • A composite function can be evaluated from a formula. See Example 3.45. • The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Example 3.46 and Example 3.47. • Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions. • Functions can often be decomposed in more than one way. See Example 3.48. 3.5 Transformation of Functions • A function can be shifted vertically by adding a constant to the output. See Example 3.49 and Example 3.50. • A function can be shifted horizontally by adding a constant to the input. See Example 3.51, Example 3.52, and Example 3.53. • Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. See Example 3.54. • Vertical and horizontal shifts are often combined. See Example 3.55 and Example 3.56. • A vertical reflection reflects a graph about the x- axis. A graph can be reflected vertically by multiplying the output by –1. • A horizontal reflection reflects a graph about the y- axis. A graph can be reflected horizontally by multiplying the input by –1. • A graph can be reflected both vertically and horizontally. The order in which the reflections are applied does not affect the final graph. See Example 3.57. • A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or columns accordingly. See Example 3.58. • A function presented as an equation can be reflected by applying transformations one at a time. See Example 3.59. • Even functions are symmetric about the y- axis, whereas odd functions are symmetric about the origin. • Even functions satisfy the condition f (x) = f ( − x). • Odd functions satisfy the condition f (x) = − f ( − x). • A function can be odd, even, or neither. See Example 3.60. • A function can be compressed or stretched vertically by multiplying the output by a constant. See Example 3.61, Example 3.62, and Example 3.63. • A function can be compressed or stretched horizontally by multiplying the input by a constant. See Example 3.64, Example 3.65, and Example 3.66. • The order in which different transformations are applied does affect the final function. Both vertical and horizontal transformations must be applied in the order given. However, a vertical transformation may be combined with a horizontal transformation in any order. See Example 3.67 and Example 3.68. 3.6 Absolute Value Functions • Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See Example 3.69. • The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. See Example 3.70. 384 • • In an absolute value equation, an unknown variable is the input of an absolute value function. If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable. See Example 3.71. Chapter 3 Functions 3.7 Inverse Functions • If g(x) is the inverse of f (x), then g( f (x)) = f (g(x)) = x. See Example 3.72, Example 3.73, and Example 3.74. • Only some of the toolkit functions have an inverse. See Example 3.75. • For a function to have an inverse, it must be one-to-one (pass the horizontal line test). • A function that is not one-to-one over its entire domain may be one-to-one on part of its domain. • For a tabular function, exchange the input and output rows to obtain the inverse. See Example 3.76. • The inverse of a function can be determined at specific points on its graph. See Example 3.77. • To find the inverse of a formula, solve the equation y = f (x) for x as a function of y. Then exchange the labels x and y. See Example 3.78, Example 3.79, and Example 3.80. • The graph of an inverse function is the reflection of the graph of the original function across the line y = x. See Example 3.81. CHAPTER 3 REVIEW EXERCISES Functions and Function Notation 473. f (x) = 2\|3x − 1\| For the following exercises, determine whether the relation is a function. 468. {(a, b), (c, d), (e, d)} the following exercises, determine whether For functions are one-to-one. the 469. ⎧ ⎨(5, 2), (6, 1), (6, 2), (4, 8)⎫ ⎬ ⎭ ⎩ 474. f (x) = − 3x + 5 475. f (x) = \|x − 3\| 470. y2 + 4 = x, for x the independent variable and y the dependent variable 471. Is the graph in Figure 3.120 a function? For the following exercises, use the vertical line test to determine if the relation whose graph is provided is a function. 476. Figure 3.120 For the following exercises, evaluate the function at the indicated values: f ( − 3); f (2); f ( − a); − f (a); f (a + h). 472. f (x) = − 2x2 + 3x 477. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 385 482. f (−2) 483. If f (x) = −2, then solve for x. 484. If f (x) = 1, then solve for x. the following For exercises, function h(t) = − 16t 2 + 80t to find the values in simplest form. use the 478. 485. h(2) − h(1) 2 − 1 486. h(a) − h(1) a − 1 Domain and Range For the following exercises, find the domain of each function, expressing answers using interval notation. 487. f (x) = 2 3x + 2 488. f (x) = x − 3 x2 − 4x − 12 489. f (x) = x − 6 x − 4 490. f (x) = Graph this 2x − 3 x ≥ − 2 ⎩ piecewise function: For the following exercises, graph the functions. 479. f (x) = \|x + 1\| 480. f (x) = x2 − 2 Rates of Change and Behavior of Graphs For the following exercises, find the average rate of change of the functions from x = 1 to x = 2. 491. f (x) = 4x − 3 the following exercises, use Figure 3.121 to For approximate the values. 492. f (x) = 10x2 + x 493. f (x) = − 2 x2 For the following exercises, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant. 494. Figure 3.121 481. f (2) 386 Chapter 3 Functions 499. For the graph in Figure 3.122, the domain of the function is [−3, 3]. The range is [−10, 10]. Find the absolute minimum of the function on this interval. 500. Find the absolute maximum of the function graphed in Figure 3.122. 495. 496. Figure 3.122 Composition of Functions For the following exercises, find ( f ∘ g)(x) and (g ∘ f )(x) for each pair of functions. 501. f (x) = 4 − x, g(x) = − 4x 502. f (x) = 3x + 2, g(x) = 5 − 6x 503. f (x) = x2 + 2x, g(x) = 5x + 1 504. f (x) = x + 2, g(x) = 1 x 497. Find the local minimum of the function graphed in Exercise 3.494. 498. Find the local extrema for the function graphed in Exercise 3.495. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 387 505. f (x) = x + 3 2 , g(x) = 1 − x For the following exercises, sketch the graph of the function g if the graph of the function f is shown in Figure 3.123. For the following exercises, find ⎛ ⎝ f ∘ g⎞ ⎠ and the domain for ⎛ ⎝ f ∘ g⎞ ⎠(x) for each pair of functions. 506. f (x(x) = 1 x 507. f (x) = 1 x + 3 , g(x) = 1 x − 9 508. f (x) = 1 x, g(x) = x 509. f (x) = 1 x2 − 1 , g(x) = x + 1 For the following exercises, express each function H as functions f and g where a composition H(x) = ( f ∘ g)(x). two of 510. H(x) = 2x − 1 3x + 4 511. H(x) = 1 (3x2 − 4)−3 Transformation of Functions For the following exercises, sketch a graph of the given function. 512. f (x) = (x − 3)2 513. f (x) = (x + 4)3 514. f (x) = x + 5 515. f (x) = − x3 516. f (x) = −x3 517. f (x) = 5 −x − 4 518. f (x) = 4[\|x − 2\| − 6] 519. f (x) = − (x + 2)2 − 1 Figure 3.123 520. g(x) = f (x − 1) 521. g(x) = 3 f (x) For the following exercises, write the equation for the standard function represented by each of the graphs below. 522. 523. the following exercises, determine whether each For function below is even, odd, or neither. 524. f (x) = 3x4 525. g(x) = x 388 Chapter 3 Functions 526. h(x) = 1 x + 3x For the following exercises, analyze the graph and determine whether the graphed function is even, odd, or neither. 527. 528. 529. 531. 532. For the following exercises, graph the absolute value function. 533. f (x) = \|x − 5\| 534. f (x) = − \|x − 3\| 535. f (x) = \|2x − 4\| Inverse Functions For the following exercises, find f −1(x) for each function. 536. f (x) = 9 + 10x 537. f (x) = x x + 2 Absolute Value Functions For the following exercises, write an equation for the transformation of f (x) = \|x\|. 530. For the following exercise, find a domain on wh
ich the function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 538. f (x) = x2 + 1 389 539. Given f (x) = x3 − 5 and g(x) = x + 5 a. Find f (g(x)) and g( f (x)). b. What does the answer relationship between f (x) and g(x) ? : 3 tell us about the For the following exercises, use a graphing utility to determine whether each function is one-to-one. 540. f (x) = 1 x 541. f (x) = − 3x2 + x 542. If f (5) = 2, find f −1(2). If f (1) = 4, 543. CHAPTER 3 PRACTICE TEST find f −1(4). For the following exercises, determine whether each of the following relations is a function. 544. y = 2x + 8 545. {(2, 1), (3, 2), ( − 1, 1), (0, − 2)} the For following exercises, f (x) = − 3x2 + 2x at the given input. evaluate the function 546. f (−2) 547. f (a) 548. Show that the function f (x) = − 2(x − 1)2 + 3 is not one-to-one. 552. Find the average rate of change of the function f (x) = 3 − 2x2 + x by finding f (b) − f (a) form. in simplest b − a the following For functions exercises, f (x) = 3 − 2x2 + x and g(x) = x to find the composite functions. use the 553. 554. ⎛ ⎝g ∘ f ⎞ ⎠(x) ⎛ ⎝g ∘ f ⎞ ⎠(1) 555. Express H(x) = 5x2 − 3x functions, f and g, where ⎛ ⎝ f ∘ g⎞ 3 ⎠(x) = H(x). as a composition of two 549. Write the domain of the function f (x) = 3 − x in interval notation. the following exercises, graph the functions by stretching, and/or compressing a toolkit For translating, function. 550. Given f (x) = 2x2 − 5x, find f (a + 1) − f (1) in simplest form. 551. Graph the function f (x) = x + 1 if −2 < x < 3 ⎧ ⎨ − x if ⎩ x ≥ 3 556. f (x) = x + 6 − 1 557. f (x) = 1 x + 2 − 1 the following exercises, determine whether For functions are even, odd, or neither. the 558. f (x) = − 5 x2 + 9x6 390 Chapter 3 Functions 559. f (x) = − 5 x3 + 9x5 560. f (x) = 1 x 561. the Graph f (x) = − 2\|x − 1\| + 3. absolute value function For the following exercises, find the inverse of the function. 562. f (x) = 3x − 5 563. f (x) = 4 x + 7 Figure 3.125 568. Find f (2). 569. Find f (−2). For the following exercises, use the graph of g shown in Figure 3.124. 570. Write an equation for the piecewise function. For the following exercises, use the values listed in Table 3.48(x) 1 3 5 7 9 11 13 15 17 Table 3.48 Figure 3.124 564. On what intervals is the function increasing? 565. On what intervals is the function decreasing? 566. Approximate the local minimum of the function. Express the answer as an ordered pair. 567. Approximate the local maximum of the function. Express the answer as an ordered pair. For the following exercises, use the graph of the piecewise function shown in Figure 3.125. 571. Find F(6). 572. Solve the equation F(x) = 5. 573. Is the graph increasing or decreasing on its domain? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 391 574. Is the function represented by the graph one-to-one? 575. Find F −1(15). 576. Given f (x) = − 2x + 11, find f −1(x). 392 Chapter 3 Functions This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 393 4 \| LINEAR FUNCTIONS Chapter Outline 4.1 Linear Functions 4.2 Modeling with Linear Functions 4.3 Fitting Linear Models to Data Introduction Figure 4.1 A bamboo forest in China (credit: "JFXie"/Flickr) Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour. [1] In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function. Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data. 1. http://www.guinnessworldrecords.com/records-3000/fastest-growing-plant/ 394 Chapter 4 Linear Functions 4.1 \| Linear Functions Learning Objectives In this section you will: 4.1.1 Represent a linear function. 4.1.2 Determine whether a linear function is increasing, decreasing, or constant. 4.1.3 Interpret slope as a rate of change. 4.1.4 Write and interpret an equation for a linear function. 4.1.5 Graph linear functions. 4.1.6 Determine whether lines are parallel or perpendicular. 4.1.7 Write the equation of a line parallel or perpendicular to a given line. Figure 4.2 Shanghai MagLev Train (credit: "kanegen"/Flickr) Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train (Figure 4.2). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes[2]. Suppose a maglev train travels a long distance, and maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time. Representing Linear Functions The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change. This is a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method. Representing a Linear Function in Word Form Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship. • The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed. The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station. 2. http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 395 Representing a Linear Function in Function Notation Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where x is the input value, m is the rate of change, and b is the initial value of the dependent variable. Equation form Function notation y = mx + b f (x) = mx + b In the example of the train, we might use the notation D(t) where the total distance D is a function of the time t. The rate, m, is 83 meters per second. The initial value of the dependent variable b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train. Representing a Linear Function in Tabular Form D(t) = 83t + 250 A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 4.3. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time. Figure 4.3 Tabular representation of the function D showing selected input and output values Can the input in the previous example be any real number? No. The input represents time so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers. Representing a Linear Function in Graphical Form Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D(t) = 83t + 250, to draw a graph as represented in Figure 4.4. Notice the graph is a line. When we plot a linear function, the graph is always a line. The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph that the y-intercept in the train example we just saw is (0, 250) and represents the distance of the train from the station when it began moving at a constant speed. Figure 4.4 The graph of D(t) = 83t + 250 . Graphs of linear functions are lines because the rate of change is constant. 396 Chapter 4 Linear Functions Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f (x) = 2x + 1. Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product. Linear Function A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a
line f (x) = mx + b where b is the initial or starting value of the function (when input, x = 0 ), and m is the constant rate of change, or slope of the function. The y-intercept is at (0, b). Example 4.1 Using a Linear Function to Find the Pressure on a Diver The pressure, P, water surface, d, this function in words. in pounds per square inch (PSI) on the diver in Figure 4.5 depends upon her depth below the in feet. This relationship may be modeled by the equation, P(d) = 0.434d + 14.696. Restate Figure 4.5 (credit: Ilse Reijs and Jan-Noud Hutten) Solution To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths. Analysis The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 397 Determining Whether a Linear Function Is Increasing, Decreasing, or Constant The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 4.6(a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 4.6(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 4.6(c). Figure 4.6 Increasing and Decreasing Functions The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function. f (x) = mx + b is an increasing function if m > 0. f (x) = mx + b is a decreasing function if m < 0. f (x) = mx + b is a constant function if m = 0. Example 4.2 Deciding Whether a Function Is Increasing, Decreasing, or Constant Some recent studies suggest that a teenager sends an average of 60 texts per day[3]. For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant. a. The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent. b. A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month. c. A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month. 3. http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/ 398 Chapter 4 Linear Functions Solution Analyze each function. a. The function can be represented as f (x) = 60x where x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day. b. The function can be represented as f (x) = 500 − 60x where x is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after x days. c. The cost function can be represented as f (x) = 50 because the number of days does not affect the total cost. The slope is 0 so the function is constant. Interpreting Slope as a Rate of Change In the examples we have seen so far, the slope was provided to us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input, x1 and x2, and two corresponding values for the output, y1 and y2 —which can be represented by a set of points, (x1 , y1) and (x2 , y2) —we can calculate the slope m. m = change in output (rise) change in input (run) = Δy Δx = y2 − y1 x2 − x1 Note that in function notation we can obtain two corresponding values for the output y1 and y2 for the function f , y1 = f ⎛ ⎠, so we could equivalently write ⎠ and y2 = f ⎛ ⎝x2 ⎝x1 ⎞ ⎞ x2 x2 – x1 ⎝x1 ⎞ ⎠ Figure 4.7 indicates how the slope of the line between the points, (x1, y1) and (x2, y2), is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the slant is. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 399 Figure 4.7 The slope of a function is calculated by the change in y divided by the change in x. It does not matter which coordinate is used as the (x2, y2) and which is the (x1, y1), as long as each calculation is started with the elements from the same coordinate pair. Are the units for slope always units for the output units for the input ? Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input. Calculate Slope The slope, or rate of change, of a function m can be calculated according to the following: m = change in output (rise) change in input (run) = Δy Δx = y2 − y1 x2 − x1 where x1 and x2 are input values, y1 and y2 are output values. Given two points from a linear function, calculate and interpret the slope. 1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value. 400 Chapter 4 Linear Functions Example 4.3 Finding the Slope of a Linear Function If f (x) is a linear function, and (3, −2) and (8, 1) are points on the line, find the slope. Is this function increasing or decreasing? Solution The coordinate pairs are (3, −2) and (8, 1). To find the rate of change, we divide the change in output by the change in input. m = change in output change in input = 1 − (−2) 8 − 3 = 3 5 We could also write the slope as m = 0.6. The function is increasing because m > 0. Analysis As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used. Note that if we had reversed them, we would have obtained the same slope. m = (−2) − (1) 3 − 8 = −3 −5 = 3 5 4.1 If f (x) is a linear function, and (2, 3) and (0, 4) are points on the line, find the slope. Is this function increasing or decreasing? Example 4.4 Finding the Population Change from a Linear Function The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012. Solution The rate of change relates the change in population to the change in time. The population increased by 27, 800 − 23, 400 = 4400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years. So the population increased by 1,100 people per year. 4,400 people 4 years = 1,100 people year Analysis Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable. The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of 4.2 population per year if we assume the change was constant from 2009 to 2012. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 401 Writing and Interpreting an Equation for a Linear Function Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form. Now we can choose which method to use to write equations for linear functions based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 4.8. Figure 4.8 We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). m = y2 − y1 x2 − x1 = Now we can substitute the slope and the coordinates of one of the points into the point-slope form. If we want to rewrite the equation in the slope-intercept form, we would find y − y1 = m(x − x1) y − 4 = − 3 (x − 4x − 4) x + 3 x + 7 If we want to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line. So the function is f (x) = − 3 4 x + 7, and the linear equation would be y = − 3 4 x + 7. 402 Chapter 4 Linear Functions Given the graph of a linear function, write an equation to represent the function. 1. Identify two points on the line. 2. Use the two points to calculate the slope. 3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection. 4. Substitute the slope and y-intercept into the slope-intercept form of a line equation. Example 4.5 Writing an Equation for a Linear Function Write an equat
ion for a linear function given a graph of f shown in Figure 4.9. Figure 4.9 Solution Identify two points on the line, such as (0, 2) and (−2, −4). Use the points to calculate the slope. m = y2 − y1 x2 − x1 = −4 − 2 −2 − 0 = −6 −2 = 3 Substitute the slope and the coordinates of one of the points into the point-slope form. y − y1 = m(x − x1) y − (−4) = 3(x − (−2)) y + 4 = 3(x + 2) We can use algebra to rewrite the equation in the slope-intercept form. y + 4 = 3(x + 2) y + 4 = 3x + 6 y = 3x + 2 Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 403 This makes sense because we can see from Figure 4.10 that the line crosses the y-axis at the point (0, 2), which is the y-intercept, so b = 2. Figure 4.10 Example 4.6 Writing an Equation for a Linear Cost Function Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C where C(x) is the cost for x items produced in a given month. Solution The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by C(x) = 1250 + 37.5x. Analysis If Ben produces 100 items in a month, his monthly cost is found by substituting 100 for x. C(100) = 1250 + 37.5(100) = 5000 So his monthly cost would be $5,000. Example 4.7 Writing an Equation for a Linear Function Given Two Points If f is a linear function, with f (3) = −2, and f (8) = 1, find an equation for the function in slope-intercept form. 404 Chapter 4 Linear Functions Solution We can write the given points using coordinates. We can then use the points to calculate the slope. f (3) = −2 → (3, −2) f (8) = 1 → (8, 1) m = = y2 − y1 x2 − x1 1 − (−2) 8 − 3 = 3 5 Substitute the slope and the coordinates of one of the points into the point-slope form. y − y1 = m(x − x1) (x − 3) y − (−2) = 3 5 We can use algebra to rewrite the equation in the slope-intercept formx − 3) x − 9 5 x − 19 5 4.3 If f (x) is a linear function, with f (2) = –11, and f (4) = −25, write an equation for the function in slope-intercept form. Modeling Real-World Problems with Linear Functions In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems. Given a linear function f and the initial value and rate of change, evaluate f(c). 1. Determine the initial value and the rate of change (slope). 2. Substitute the values into f (x) = mx + b. 3. Evaluate the function at x = c. Example 4.8 Using a Linear Function to Determine the Number of Songs in a Music Collection This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 405 Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, the number of months. How many songs will he own at the end of one year? in his collection as a function of time, t, Solution The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200. The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line. We can write the formula N(t) = 15t + 200. With this formula, we can then predict how many songs Marcus will have at the end of one year (12 months). In other words, we can evaluate the function at t = 12. N(12) = 15(12) + 200 = 180 + 200 = 380 Marcus will have 380 songs in 12 months. Analysis Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well. Example 4.9 Using a Linear Function to Calculate Salary Based on Commission Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation. Solution The given information gives us two input-output pairs: (3, 760) and (5, 920). We start by finding the rate of change. m = 920 − 760 = 5 − 3 $160 2 policies = $80 per policy 406 Chapter 4 Linear Functions Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week. We can then solve for the initial value. I(n) = 80n + b 760 = 80(3) + b when n = 3, I(3) = 760 760 − 80(3) = b 520 = b The value of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold. We can now write the final equation. I(n) = 80n + 520 Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold. Example 4.10 Using Tabular Form to Write an Equation for a Linear Function Table 4.1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation. number of weeks, w 0 2 4 6 number of rats, P(w) 1000 1080 1160 1240 Table 4.1 Solution We can see from the table that the initial value for the number of rats is 1000, so b = 1000. Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week. P(w) = 40w + 1000 If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240) m = 1240 − 1080 6 − 2 = 160 4 = 40 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 407 Is the initial value always provided in a table of values like Table 4.1? No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f (x) = mx + b, and solve for b. 4.4 A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 4.2 shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment. x 0 2 4 8 12 H(x) 12.5 13.5 14.5 16.5 18.5 Table 4.2 Graphing Linear Functions Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third method is by using transformations of the identity function f (x) = x. Graphing a Function by Plotting Points To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f (x) = 2x, we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. Given a linear function, graph by plotting points. 1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points. Example 4.11 Graphing by Plotting Points Graph f (x) = − 2 3 x + 5 by plotting points. 408 Chapter 4 Linear Functions Solution Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6. Evaluate the function at each input value, and use the output value to identify coordinate pairs0) = − 2 3 f (3) = − 2 3 f (6) = − 2 3 (0) + 5 = 5 ⇒ (0, 5) (3) + 5 = 3 ⇒ (3, 3) (6) + 5 = 1 ⇒ (6, 1) Plot the coordinate pairs and draw a line through the points. Figure 4.11 represents the graph of the function f (x) = − 2 3 x + 5. Figure 4.11 The graph of the linear function f (x) = − 2 3 x + 5. Analysis The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function. 4.5 Graph f (x) = − 3 4 x + 6 by plottin
g points. Graphing a Function Using y-intercept and Slope Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set x = 0 in the equation. The other characteristic of the linear function is its slope. Let’s consider the following function. f (x) = 1 2 x + 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 409 The slope is 1 2 . Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is the point on the graph when x = 0. The graph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. We can begin graphing by plotting the point (0, 1). We know that the slope is the change in the y-coordinate over the change in the x-coordinate. This is commonly referred to as rise over run, m = rise , which means run. From our example, we have m = 1 2 that the rise is 1 and the run is 2. So starting from our y-intercept (0, 1), we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 4.12. Figure 4.12 Graphical Interpretation of a Linear Function In the equation f (x) = mx + b • b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis. • m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope: m = change in output (rise) change in input (run) = Δy Δx = y2 − y1 x2 − x1 Do all linear functions have y-intercepts? Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line is parallel to the y-axis does not have a y-intercept, but it is not a function.) Given the equation for a linear function, graph the function using the y-intercept and slope. 1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept. 4. Use rise run to determine at least two more points on the line. 5. Sketch the line that passes through the points. 410 Chapter 4 Linear Functions Example 4.12 Graphing by Using the y-intercept and Slope Graph f (x) = − 2 3 x + 5 using the y-intercept and slope. Solution Evaluate the function at x = 0 to find the y-intercept. The output value when x = 0 is 5, so the graph will cross the y-axis at (0, 5). According to the equation for the function, the slope of the line is − 2 3 . This tells us that for each vertical decrease in the “rise” of – 2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 4.13. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then drawing a line through the points. Figure 4.13 Graph of f (x) = −2/3x + 5 and shows how to calculate the rise over run for the slope. Analysis The graph slants downward from left to right, which means it has a negative slope as expected. 4.6 Find a point on the graph we drew in Example 4.14 that has a negative x-value. Graphing a Function Using Transformations Another option for graphing is to use a transformation of the identity function f (x) = x. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression. Vertical Stretch or Compression the m is acting as the vertical stretch or compression of the identity function. When m is In the equation f (x) = mx, negative, there is also a vertical reflection of the graph. Notice in Figure 4.14 that multiplying the equation of f (x) = x by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. This means the larger the absolute value of m, the steeper the slope. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 411 Figure 4.14 Vertical stretches and compressions and reflections on the function f (x) = x Vertical Shift the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. In f (x) = mx + b, Notice in Figure 4.15 that adding a value of b to the equation of f (x) = x shifts the graph of f a total of b units up if b is positive and \|b\| units down if b is negative. 412 Chapter 4 Linear Functions Figure 4.15 This graph illustrates vertical shifts of the function f (x) = x. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. Given the equation of a linear function, use transformations to graph the linear function in the form f(x) = mx + b. 1. Graph f (x) = x. 2. Vertically stretch or compress the graph by a factor m. 3. Shift the graph up or down b units. Example 4.13 Graphing by Using Transformations This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 413 Graph f (x) = 1 2 x − 3 using transformations. Solution The equation for the function shows that m = 1 2 so the identity function is vertically compressed by 1 2 . The equation for the function also shows that b = − 3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 4.16. Figure 4.16 The function, y = x, compressed by a factor of 1 2 Then show the vertical shift as in Figure 4.17. Figure 4.17 The function y = 1 2 x, shifted down 3 units 4.7 Graph f (x) = 4 + 2x using transformations. 414 Chapter 4 Linear Functions In Example 4.15, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2. (2) − 3 f (22 Writing the Equation for a Function from the Graph of a Line Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 4.18. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept. Figure 4.18 Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point ( – 2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be m = rise run = 4 2 = 2 Substituting the slope and y-intercept into the slope-intercept form of a line gives y = 2x + 4 Given a graph of linear function, find the equation to describe the function. 1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute the y-intercept and slope into the slope-intercept form of a line. Example 4.14 Matching Linear Functions to Their Graphs Match each equation of the linear functions with one of the lines in Figure 4.19. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 415 a. b. c. d. f (x) = 2x + 3 g(x) = 2x − 3 h(x) = −2x + 3 j(x) = 1 2 x + 3 Figure 4.19 Solution Analyze the information for each function. a. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I. b. This function also has a slope of 2, but a y-intercept of −3. It must pass through the point (0, −3) and slant upward from left to right. It must be represented by line III. c. This function has a slope of –2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right. d. This function has a slope of 1 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II. Now we can re-label the lines as in Figure 4.20. 416 Chapter 4 Linear Functions Figure 4.20 Finding the x-intercept of a Line So far we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis. Recall that a function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero. To find the x-intercept, set a function f (x) equal to zero and solve for the value of x. For example, consider the function shown. Set the function equal to 0 and solve for x. f (x) = 3x − 6 0 = 3x − 6 6 = 3x 2 = x x = 2 The graph of the function crosses the x-axis at the point (2, 0). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 417 Do all linear functions have x-intercepts? No. However, linear functions of the form y = c, where c is a nonzero real number are the only examples of linear func
tions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 4.21. Figure 4.21 x-intercept The x-intercept of the function is value of x when f (x) = 0. It can be solved by the equation 0 = mx + b. Example 4.15 Finding an x-intercept Find the x-intercept of f (x) = 1 2 x − 3. Solution Set the function equal to zero to solve for x The graph crosses the x-axis at the point (6, 0). Analysis A graph of the function is shown in Figure 4.22. We can see that the x-intercept is (6, 0) as we expected. 418 Chapter 4 Linear Functions Figure 4.22 4.8 Find the x-intercept of f (x) = 1 4 x − 4. Describing Horizontal and Vertical Lines There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 4.23, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation f (x) = mx + b, the equation simplifies to f (x) = b. In other words, the value of the function is a constant. This graph represents the function f (x) = 2. Figure 4.23 A horizontal line representing the function f (x) = 2 A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 419 function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined. Figure 4.24 Example of how a line has a vertical slope. 0 in the denominator of the slope. A vertical line, such as the one in Figure 4.25, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2. Figure 4.25 The vertical line, x = 2, which does not represent a function Horizontal and Vertical Lines Lines can be horizontal or vertical. A horizontal line is a line defined by an equation in the form f (x) = b. A vertical line is a line defined by an equation in the form x = a. Example 4.16 Writing the Equation of a Horizontal Line Write the equation of the line graphed in Figure 4.26. 420 Chapter 4 Linear Functions Figure 4.26 Solution For any x-value, the y-value is − 4, so the equation is y = − 4. Example 4.17 Writing the Equation of a Vertical Line Write the equation of the line graphed in Figure 4.27. Figure 4.27 Solution The constant x-value is 7, so the equation is x = 7. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 421 Determining Whether Lines are Parallel or Perpendicular The two lines in Figure 4.28 are parallel lines: they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the other, they would become coincident. Figure 4.28 Parallel lines We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. f (x) = − 2x + 6 ⎫ ⎬ parallel f (x) = − 2x − 4 ⎭ f (x) = 3x + 2 ⎫ ⎬ not parallel f (x) = 2x + 2 ⎭ Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 4.29 are perpendicular. Figure 4.29 Perpendicular lines Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if m1 and m2 are negative reciprocals of one another, they can be multiplied together to yield –1. 422 Chapter 4 Linear Functions m1 m2 = −1 To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is 1 8 , and the reciprocal of 1 8 is 8. To find the negative reciprocal, first find the reciprocal and then change the sign. As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular. f (x) = 1 4 x + 2 f (x) = −4x + 3 negative reciprocal of 1 4 is −4 negative reciprocal of −4 is 1 4 The product of the slopes is –1. − Parallel and Perpendicular Lines Two lines are parallel lines if they do not intersect. The slopes of the lines are the same. f (x) = m1 x + b1 and g(x) = m2 x + b2 are parallel if and only if m1 = m2 If and only if b1 = b2 and m1 = m2, we say the lines coincide. Coincident lines are the same line. Two lines are perpendicular lines if they intersect to form a right angle. f (x) = m1 x + b1 and g(x) = m2 x + b2 are perpendicular if and only if m1 m2 = − 1, so m2 = − 1 m1 Example 4.18 Identifying Parallel and Perpendicular Lines Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines. f (x) = 2x + 3 g(x) = 1 2 x − 4 h(x) = −2x + 2 j(x) = 2x − 6 Solution Parallel lines have the same slope. Because the functions f (x) = 2x + 3 and j(x) = 2x − 6 each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and 1 2 are negative reciprocals, the functions g(x) = 1 2 x − 4 and h(x) = −2x + 2 represent perpendicular lines. Analysis A graph of the lines is shown in Figure 4.30. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 423 Figure 4.30 The graph shows that the lines f (x) = 2x + 3 and j(x) = 2x – 6 are parallel, and the lines g(x) = 1 2 x – 4 and h(x) = − 2x + 2 are perpendicular. Writing the Equation of a Line Parallel or Perpendicular to a Given Line If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line. Writing Equations of Parallel Lines Suppose for example, we are given the equation shown. f (x) = 3x + 1 We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to f (x). So the lines formed by all of the following functions will be parallel to f (x). g(x) = 3x + 6 h(x) = 3x + 1 p(x) = 3x + 2 3 Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value of b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form. y − y1 = m(x − x1) y − 7 = 3(x − 1) y − 7 = 3x − 3 y = 3x + 4 So g(x) = 3x + 4 is parallel to f (x) = 3x + 1 and passes through the point (1, 7). 424 Chapter 4 Linear Functions Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point. 1. Find the slope of the function. 2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line. 3. Simplify. Example 4.19 Finding a Line Parallel to a Given Line Find a line parallel to the graph of f (x) = 3x + 6 that passes through the point (3, 0). Solution The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m = 3, x = 3, and f (x) = 0 into the slope-intercept form to find the y-intercept. g(x) = 3x + b 0 = 3(3) + b b = –9 The line parallel to f (x) that passes through (3, 0) is g(x) = 3x − 9. Analysis We can confirm that the two lines are parallel by graphing them. Figure 4.31 shows that the two lines will never intersect. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 425 Figure 4.31 Writing Equations of Perpendicular Lines We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the function shown. f (x) = 2x + 4 The slope of the line is 2, and its negative reciprocal is − 1 2 . Any function with a slope of − 1 2 f (x). So the lines formed by all of the following functions will be perpendicular to f (x). will be perpendicular to 426 Chapter 4 Linear Functions g(x) = − 1 2 h(x) = − 1 2 p(x As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to f (x) and passes through the point (4, 0). We already know that the slope is − 1 2 . Now we can use the point to find the y-intercept by substituting the given values into the slope-intercept form of a line and solving for b. (4) + b g(x) = mx + b 0 = − 1 2 0 = −2 + b 2 = b b = 2 The equation for the function with a slope of − 1 2 and a y-intercept of 2 is g(x) = − 1 2 x + 2 So g(x) = − 1 2 x + 2 is perpendicular to f (x) = 2x + 4 and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature. A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their sl
opes is not –1. Doesn’t this fact contradict the definition of perpendicular lines? No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted. Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line. 1. Find the slope of the function. 2. Determine the negative reciprocal of the slope. 3. Substitute the new slope and the values for x and y from the coordinate pair provided into g(x) = mx + b. 4. Solve for b. 5. Write the equation of the line. Example 4.20 Finding the Equation of a Perpendicular Line Find the equation of a line perpendicular to f (x) = 3x + 3 that passes through the point (3, 0). Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 427 The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or − 1 3 . Using this slope and the given point, we can find the equation of the line. x + b (3) + b g(x The line perpendicular to f (x) that passes through (3, 0) is g(x) = − 1 3 x + 1. Analysis A graph of the two lines is shown in Figure 4.32. Figure 4.32 Note that that if we graph perpendicular lines on a graphing calculator using standard zoom, the lines may not appear to be perpendicular. Adjusting the window will make it possible to zoom in further to see the intersection more closely. 428 Chapter 4 Linear Functions 4.9 Given the function h(x) = 2x − 4, write an equation for the line passing through (0, 0) that is a. parallel to h(x) b. perpendicular to h(x) Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point. 1. Determine the slope of the line passing through the points. 2. Find the negative reciprocal of the slope. 3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values. 4. Simplify. Example 4.21 Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point A line passes through the points (−2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5). Solution From the two points of the given line, we can calculate the slope of that line. Find the negative reciprocal of the slope. m1 = 5 − 6 4 − (−2) = −1 6 = − 1 6 m2 = −1 − 1 6 ⎛ ⎝− 6 = −1 1 ⎞ ⎠ = 6 We can then solve for the y-intercept of the line passing through the point (4, 5). g(x) = 6x + b 5 = 6(4) + b 5 = 24 + b −19 = b b = −19 The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is y = 6x − 19 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 429 4.10 A line passes through the points, (−2, −15) and (2, −3). Find the equation of a perpendicular line that passes through the point, (6, 4). Access this online resource for additional instruction and practice with linear functions. • Linear Functions (http://Openstaxcollege.org/l/linearfunctions) • Finding Input of Function from the Output and Graph (http://Openstaxcollege.org/l/ findinginput) • Graphing Functions using Tables (http://Openstaxcollege.org/l/graphwithtable) 430 Chapter 4 Linear Functions 4.1 EXERCISES Verbal in feet 1. Terry is skiing down a steep hill. Terry's elevation, after t seconds E(t), by E(t) = 3000 − 70t. Write a complete sentence describing Terry’s starting elevation and how it is changing over time. given is Jessica is walking home from a friend’s house. After 2 2. minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour? A boat is 100 miles away from the marina, sailing 3. directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours. If the graphs of two linear functions are perpendicular, 4. describe the relationship between the slopes and the yintercepts. If a horizontal line has the equation f (x) = a and a 5. vertical line has the equation x = a, what is the point of intersection? Explain why what you found is the point of intersection. Algebraic the following exercises, determine whether For equation of the curve can be written as a linear function. the 6. 7. 8. y = 1 4 x + 6 y = 3x − 5 y = 3x2 − 2 9. 3x + 5y = 15 10. 11. 12. 13. 3x2 + 5y = 15 3x + 5y2 = 15 −2x2 + 3y2 = 6 − x − 3 5 = 2y 16. a(x) = 5 − 2x 17. b(x) = 8 − 3x 18. h(x) = −2x + 4 19. k(x) = −4x + 1 20. 21. 22. 23. j(x) = 1 2 x − 3 p(x) = 1 4 x − 5 n(x) = − 1 3 x − 2 m(x) = − 3 8 x + 3 For the following exercises, find the slope of the line that passes through the two given points. 24. (2, 4) and (4, 10) 25. (1, 5) and (4, 11) 26. (–1, 4) and (5, 2) 27. (8, –2) and (4, 6) 28. (6, 11) and (–4, 3) For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. 29. f ( − 5) = −4, and f (5) = 2 30. f (−1) = 4, and f (5) = 1 31. Passes through (2, 4) and (4, 10) 32. Passes through (1, 5) and (4, 11) 33. Passes through (−1, 4) and (5, 2) 34. Passes through (−2, 8) and (4, 6) the following exercises, determine whether each For function is increasing or decreasing. 35. x intercept at (−2, 0) and y intercept at (0, −3) 14. f (x) = 4x + 3 15. g(x) = 5x + 6 36. x intercept at (−5, 0) and y intercept at (0, 4) For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 431 Write an equation for a line parallel to f (x) = − 5x − 3 and passing through the point (2, –12). Write an equation for a line parallel to g(x) = 3x − 1 53. and passing through the point (4, 9). Write an equation for a line perpendicular 54. h(t) = −2t + 4 and passing through the point (−4, –1). to Write an equation for a line perpendicular 55. p(t) = 3t + 4 and passing through the point (3, 1). to Graphical the following exercises, For line graphed. find the slope of the 56. 37. 38. 39. 40. 4x − 7y = 10 7x + 4y = 1 3y + x = 12 −y = 8x + 1 3y + 4x = 12 −6y = 8x + 1 6x − 9y = 10 3x + 2y = 1 For the following exercises, find the x- and y-intercepts of each equation. 41. f (x) = − x + 2 42. g(x) = 2x + 4 43. h(x) = 3x − 5 44. k(x) = −5x + 1 45. −2x + 5y = 20 46. 7x + 2y = 56 For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 47. Line 1: Passes through (0, 6) and (3, −24) Line 2: Passes through (−1, 19) and (8, −71) 48. Line 1: Passes through (−8, −55) and (10, 89) 57. Line 2: Passes through (9, − 44) and (4, − 14) 49. Line 1: Passes through (2, 3) and (4, −1) Line 2: Passes through (6, 3) and (8, 5) 50. Line 1: Passes through (1, 7) and (5, 5) Line 2: Passes through (−1, −3) and (1, 1) 51. Line 1: Passes through (2, 5) and (5, − 1) Line 2: Passes through (−3, 7) and (3, −5) For the following exercises, write an equation for the line described. 52. 432 Chapter 4 Linear Functions For the following exercises, write an equation for the line graphed. 58. 59. 60. 61. This content is available for free at https://cnx.org/content/col11758/1.5 62. 63. Chapter 4 Linear Functions 433 the following exercises, match the given linear For equation with its graph in Figure 4.33. 77. f (x) = −3x + 2 78. 79. f (x) = 1 3 x + 2 f (x) = 2 3 x − 3 80. f (t) = 3 + 2t 81. p(t) = −2 + 3t 82. x = 3 83. x = −2 84. r(x) = 4 For the following exercises, write the equation of the line shown in the graph. 85. Figure 4.33 64. f (x) = − x − 1 65. f (x) = −2x − 1 66. f (x) = − 1 2 x − 1 67. f (x) = 2 68. f (x) = 2 + x 69. f (x) = 3x + 2 For the following exercises, sketch a line with the given features. 70. An x-intercept of (–4, 0) and y-intercept of (0, –2) 71. An x-intercept (–2, 0) and y-intercept of (0, 4) 86. 72. 73. A y-intercept of (0, 7) and slope − 3 2 A y-intercept of (0, 3) and slope 2 5 74. Passing through the points (–6, –2) and (6, –6) 75. Passing through the points (–3, –4) and (3, 0) For the following exercises, sketch the graph of each equation. 76. f (x) = −2x − 1 434 Chapter 4 Linear Functions x g(x) x h(x) 0 5 0 5 5 10 15 –10 –25 –40 5 10 15 30 105 230 x 0 5 10 15 f (x) –5 20 45 70 x 5 10 20 25 k(x) 13 28 58 73 x g(x) 0 6 2 4 6 –19 –44 –69 x 2 4 8 10 h(x) 13 23 43 53 x 2 4 6 8 f (x) –4 16 36 56 89. 90. 91. 92. 93. 94. 95. 96. 87. 88. Numeric For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 435 x k(x) 0 6 2 6 8 31 106 231 Technology For the following exercises, use a calculator or graphing technology to complete the task. If f is 97. f (0.1) = 11.5, for the function. a function, and f (0.4) = –5.9, find an equation linear a the Graph domain function f on 98. of [–10, 10] : f (x) = 0.02x − 0.01. Enter the function in a graphing utility. For the viewing window, set the minimum value of x to be −10 and the maximum value of x to be 10. intercept is 31 16 −10 and 10. . Label the points for the input values of linear Graph the function f on a domain of 103. [−0.1, 0.1] for the function whose slope is 75 and yintercept is −22.5. Label the points for the input values of −0.1 and 0.1. Graph the linear function f where f (x) = ax + b on 104. the same set of axes on a domain of [−4, 4] for the following values of a and b. i. a = 2; b = 3 ii. a = 2; b = 4 iii. a = 2; b = –4 iv. a = 2; b = –5 Extensions Graph function f on 99. [–10, 10] : f x) = 2, 500x + 4, 000 the a domain of Find the value of x if a linear function goes through the following slope: 105. the following points and has (x, 2), (
−4, 6), m = 3 Table 4.3 shows the input, w, and output, k, for a 100. linear function k. a. Fill in the missing values of the table. b. Write the linear function k, round to 3 decimal places. Find the value of y if a linear function goes through the following slope: 106. the following points and has (10, y), (25, 100), m = −5 w k –10 5.5 67.5 b Find the equation of the line that passes through the 107. following points: 30 –26 a –44 (a, b) and (a, b + 1) Table 4.3 Table 4.4 shows the input, p, and output, q, for a 101. linear function q. a. Fill in the missing values of the table. b. Write the linear function k. p q 0.5 0.8 12 b 400 700 a 1,000,000 Table 4.4 102. Graph the linear function f on a domain of [−10, 10] for the function whose slope is 1 8 and y- Find the equation of the line that passes through the 108. following points: (2a, b) and (a, b + 1) Find the equation of the line that passes through the 109. following points: (a, 0) and (c, d) Find the equation of the line parallel to the line 110. g(x) = −0.01x+2.01 through the point (1, 2). Find the equation of the line perpendicular to the line 111. g(x) = −0.01x+2.01 through the point (1, 2). the For f (x) = −0.1x+200 and g(x) = 20x + 0.1. exercises, following use the functions 112. Find the point of intersection of the lines f and g. 113. 436 Chapter 4 Linear Functions c. Each year in the decade of the 1990s, average annual income increased by $1,054. d. Average annual income rose to a level of $23,286 by the end of 1999. When temperature temperature the 122. the Celsius 32. When Fahrenheit temperature corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C, is 0 degrees Celsius, is the the Celsius temperature, F(C). 100, is a. Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius. b. Find and interpret F(28). c. Find and interpret F(–40). Where is f (x) greater than g(x) ? Where is g(x) greater than f (x) ? Real-World Applications 114. At noon, a barista notices that she has $20 in her tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves n more customers during her shift? A gym membership with two personal 115. training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session? it can charge per shirt. In particular, historical data A clothing business finds there is a linear relationship it can sell and the price, 116. between the number of shirts, n, p, shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form p(n) = mn + b that gives the price p they can charge for n shirts. 117. A phone company charges for service according to the formula: C(n) = 24 + 0.1n, where n is the number of minutes talked, and C(n) is the monthly charge, in dollars. Find and interpret the rate of change and initial value. 118. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationships in the form y = mn + b that gives the yield when n stalks are planted. A city’s population in the year 1960 was 287,500. In 119. 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year. A town’s population has been growing linearly. In 120. 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, P(t), for the population t years after 2003. 121. Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: I(x) = 1054x + 23, 286, where x is the number of years after 1990. Which of the following interprets the slope in the context of the problem? a. As of 1990, average annual income was $23,286. b. In the ten-year period from 1990–1999, average annual income increased by a total of $1,054. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 437 4.2 \| Modeling with Linear Functions Learning Objectives In this section you will: 4.2.1 Build linear models from verbal descriptions. 4.2.2 Model a set of data with a linear function. Figure 4.34 (credit: EEK Photography/Flickr) Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models. Building Linear Models from Verbal Descriptions When building linear models to solve problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them: 1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system. 2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value. 3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret. 4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem. 5. When needed, write a formula for the function. 6. Solve or evaluate the function using the formula. 7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically. 8. Clearly convey your result using appropriate units, and answer in full sentences when necessary. Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units. Output: M, money remaining, in dollars Input: t, time, in weeks 438 Chapter 4 Linear Functions So, the amount of money remaining depends on the number of weeks: M(t) . We can also identify the initial value and the rate of change. Initial Value: She saved $3,500, so $3,500 is the initial value for M. Rate of Change: She anticipates spending $400 each week, so − $400 per week is the rate of change, or slope. Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week. The rate of change is constant, so we can start with the linear model M(t) = mt + b. Then we can substitute the intercept and slope provided. To find the x-intercept, we set the output to zero, and solve for the input. 0 = −400t + 3500 t = 3500 400 = 8.75 The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks. When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily uses a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75. In this example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model. Using a Given Intercept to Build a Model Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model. Now we can set the function equal to 0, and solve for x to find the x-intercept. f (x) = mx + b = −250x + 1000 0 = −250x + 1000 1000 = 250x 4 = x x = 4 The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 439 Using a Given Input and Output to Build a Model Many real-world applications are not a
s direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output. Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem. 1. Identify the input and output values. 2. Convert the data to two coordinate pairs. 3. Find the slope. 4. Write the linear model. 5. Use the model to make a prediction by evaluating the function at a given x-value. 6. Use the model to identify an x-value that results in a given y-value. 7. Answer the question posed. Example 4.22 Using a Linear Model to Investigate a Town’s Population A town’s population has been growing linearly. In 2004, the population was 6,200. By 2009, the population had grown to 8,100. Assume this trend continues. a. Predict the population in 2013. b. Identify the year in which the population will reach 15,000. Solution The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2000 years ago! To make computation a little nicer, we will define our input as the number of years since 2004. Input: t, years since 2004 Output: P(t), the town’s population To predict the population in 2013 ( t = 9 ), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope. To determine the rate of change, we will use the change in output per change in input. m = change in output change in input The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0, giving the point (0, 6200). Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to t = 5, giving the point (5, 8100). The two coordinate pairs are (0, 6200) and (5, 8100). Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope. 440 Chapter 4 Linear Functions m = 8100 − 6200 5 − 0 = 1900 5 = 380 people per year We already know the y-intercept of the line, so we can immediately write the equation: To predict the population in 2013, we evaluate our function at t = 9. P(t) = 380t + 6200 P(9) = 380(9) + 6,200 = 9,620 If the trend continues, our model predicts a population of 9,620 in 2013. To find when the population will reach 15,000, we can set P(t) = 15000 and solve for t. 15000 = 380t + 6200 8800 = 380t t ≈ 23.158 Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027. A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It 4.11 costs $0.25 to produce each doughnut. a. Write a linear model to represent the cost C of the company as a function of x, the number of doughnuts produced. b. Find and interpret the y-intercept. A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the 4.12 population was 36,800. Assume this trend continues. a. Predict the population in 2014. b. Identify the year in which the population will reach 54,000. Using a Diagram to Build a Model It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful. Example 4.23 Using a Diagram to Model Distance Walked This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 441 Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact? Solution In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: "How long will it take them to be 2 miles apart"? In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables. Input: t, time in hours. Output: A(t), distance in miles, and E(t), distance in miles Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 4.35. Figure 4.35 Initial Value: They both start at the same intersection so when t = 0, also be 0. Thus the initial value for each is 0. the distance traveled by each person should Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3. Using those values, we can write formulas for the distance each person has walked. A(t) = 4t E(t) = 3t For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure. We can then define a third variable, D, Showing the variables on the diagram is often helpful, as we can see from Figure 4.36. to be the measurement of the distance between Anna and Emanuel. 442 Chapter 4 Linear Functions Recall that we need to know how long it takes for D, for any given input t, the outputs A(t), E(t), and D(t) represent distances. the distance between them, to equal 2 miles. Notice that Figure 4.36 Figure 4.35 shows us that we can use the Pythagorean Theorem because we have drawn a right angle. Using the Pythagorean Theorem, we get: D(t)2 = A(t)2 + E(t)2 = (4t)2 + (3t)2 = 16t 2 + 9t 2 = 25t 2 D(t) = ± 25t 2 = ± 5\|t\| Solve for D(t) using the square root. In this scenario we are considering only positive values of t, so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t. D(t) = 2 5t = 2 t = 2 5 = 0.4 They will fall out of radio contact in 0.4 hour, or 24 minutes. Should I draw diagrams when given information based on a geometric shape? Yes. Sketch the figure and label the quantities and unknowns on the sketch. Example 4.24 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 443 Using a Diagram to Model Distance Between Cities There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough? Solution It might help here to draw a picture of the situation. See Figure 4.37. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0). Figure 4.37 Using this point along with the origin, we can find the slope of the line from Westborough to Agritown. Now we can write an equation to describe the road from Westborough to Agritown. m = 10 − 0 30 − 0 = 1 3 W(x) = 1 3 x From this, we can determine the perpendicular road to Eastborough will have slope m = – 3. Because the town of Eastborough is at the point (20, 0), we can find the equation. E(x) = −3x + b 0 = −3(20) + b b = 60 E(x) = −3x + 60 Substitute (20, 0)into the equation. We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal, 1 3 10 3 x = −3x + 60 x = 60 10x = 180 x = 18 y = W(18) = 1 3 = 6 (18) Substituting this back into W(x). The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction. 444 Chapter 4 Linear Functions distance = (x2 − x1)2 + (y2 − y1)2 = (18 − 0)2 + (6 − 0)2 ≈ 18.974 miles Analysis One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points. 4.13 There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located
22 miles directly east of the town of Timpson, how far is the road junction from Timpson? Modeling a Set of Data with Linear Functions Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 4.38. Figure 4.38 Given a situation that represents a system of linear equations, write the system of equations and identify the solution. 1. 2. Identify the input and output of each linear model. Identify the slope and y-intercept of each linear model. 3. Find the solution by setting the two linear functions equal to another and solving for x, or find the point of intersection on a graph. Example 4.25 Building a System of Linear Models to Choose a Truck Rental Company This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 445 Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile[4]. When will Keep on Trucking, Inc. be the better choice for Jamal? Solution The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions in Table 4.5. Input d, distance driven in miles Outputs K(d) : cost, in dollars, for renting from Keep on Trucking M(d) cost, in dollars, for renting from Move It Your Way Initial Value Up-front fee: K(0) = 20 and M(0) = 16 Rate of Change K(d) = $0.59 /mile and P(d) = $0.63 /mile Table 4.5 A linear function is of the form f (x) = mx + b. Using the rates of change and initial charges, we can write the equations K(d) = 0.59d + 20 M(d) = 0.63d + 16 Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K(d) < M(d). The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K(d) function is smaller. These graphs are sketched in Figure 4.39, with K(d) in blue. 4. Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ 446 Chapter 4 Linear Functions Figure 4.39 To find the intersection, we set the equations equal and solve: K(d) = M(d) 0.59d + 20 = 0.63d + 16 4 = 0.04d 100 = d d = 100 This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K(d) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100 . Access this online resource for additional instruction and practice with linear function models. • Interpreting a Linear Function (http://Openstaxcollege.org/l/interpretlinear) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 447 4.2 EXERCISES Verbal Explain how to find the input variable in a word 123. problem that uses a linear function. Find the linear function that models the town’s population P as a function of the year, t, where t is the number of years since the model began. Explain how to find the output variable in a word 124. problem that uses a linear function. 136. P. Find a reasonable domain and range for the function Explain how to interpret the initial value in a word 125. problem that uses a linear function. If the function P is graphed, find and interpret the x- 137. and y-intercepts. Explain how to determine the slope in a word problem 126. that uses a linear function. If the function P is graphed, find and interpret the 138. slope of the function. Algebraic 139. When will the population reach 100,000? Find the area of a parallelogram bounded by the ythe line f (x) = 1 + 2x, and the line 127. axis, the line x = 3, parallel to f (x) passing through (2, 7). 128. line f (x) = 12 – 1 3 Find the area of a triangle bounded by the x-axis, the x, and the line perpendicular to f (x) that passes through the origin. What is the population in the year 12 years from the 140. onset of the model? For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained onehalf pound a month for its first year. Find the linear function that models the baby’s weight 141. W as a function of the age of the baby, in months, t. 129. line f (x) = 9 – 6 7 Find the area of a triangle bounded by the y-axis, the x, and the line perpendicular to f (x) 142. W. Find a reasonable domain and range for the function that passes through the origin. Find the area of a parallelogram bounded by the xthe line f (x) = 3x, and the line 130. axis, the line g(x) = 2, parallel to f (x) passing through (6, 1). For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues. 131. Predict the population in 2016. 132. Identify the year in which the population will reach 0. For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues. 133. Predict the population in 2016. If the function W is graphed, find and interpret the x- 143. and y-intercepts. If the function W is graphed, find and interpret the 144. slope of the function. 145. When did the baby weight 10.4 pounds? 146. What is the output when the input is 6.2? For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were inflicted. Find the linear function that models the number of 147. people inflicted with the common cold C as a function of the year, t. Find a reasonable domain and range for the function 148. C. Identify the year in which the population will reach 134. 75,000. For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years. If the function C is graphed, find and interpret the x- 149. and y-intercepts. If the function C is graphed, find and interpret the 150. slope of the function. 135. 151. When will the output reach 0? 448 Chapter 4 Linear Functions 152. In what year will the number of people be 9,700? Numeric Graphical For the following exercises, use the graph in Figure 4.40, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980. Figure 4.40 153. Find the linear function y, where y depends on t, the number of years since 1980. 154. Find and interpret the y-intercept. 155. Find and interpret the x-intercept. 156. Find and interpret the slope. For the following exercises, use the graph in Figure 4.41, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980. Figure 4.41 For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 4.6. Assume that the house values are changing linearly. Year Mississippi Hawaii 1950 $25,200 $74,400 2000 $71,400 $272,700 Table 4.6 In which state have home values increased at a higher 161. rate? If these trends were to continue, what would be the 162. median home value in Mississippi in 2010? If we assume the linear trend existed before 1950 and 163. continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.) For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in Table 4.7. Assume that the house values are changing linearly. Year Indiana Alabama 1950 $37,700 $27,100 2000 $94,300 $85,100 Table 4.7 In which state have home values increased at a higher 164. rate? If these trends were to continue, what would be the 165. median home value in Indiana in 2010? If we assume the linear trend existed before 1950 and 166. continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.) 157. Find the linear function y, where y depends on t, the number of years since 1980. Real-World Applications 158. Find and interpret the y-intercept. 159. Find and interpret the x-intercept. 160. Find and interpret the slope. In 2004, a school population was 1001. By 2008 the 167. population had grown to 1697. Assume the population is changing linearly. a. How much did the population grow between the year 2004 and 2008? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 449 b. How long did it take the population to grow from a. Find a formula for the moose population, P since 1001 students to 1697 students? 1990. c. What is the average population growth per year? b. What does your model predict the moose d. What was the population in the year 2000? e. Find an equation for the population, P, of the school t years after 2000. f. Using your equation, predict the population of the school in 2011. In 2003, a town’s population was 1431. By 2007 the 168. population had grown to 2134. Assume the population is changing linearly. a. How much did the population grow between the year 2003 and 2007? b. How long did it take the population to grow from 1431 people to 2134 people? c. What is the average population growth per year? d. What was the population in the year 2000? e. Find an equation for the population, P, of the to
wn t years after 2000. f. Using your equation, predict the population of the town in 2014. A phone company has a monthly cellular plan where a 169. customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118. a. Find a linear equation for the monthly cost of the cell plan as a function of x, the number of monthly minutes used. b. Interpret the slope and y-intercept of the equation. c. Use your equation to find the total monthly cost if 687 minutes are used. A phone company has a monthly cellular data plan 170. where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80. a. Find a linear equation for the monthly cost of the the number of MB data plan as a function of x, used. b. Interpret the slope and y-intercept of the equation. c. Use your equation to find the total monthly cost if 250 MB are used. In 1991, the moose population in a park was 171. measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly. population to be in 2003? In 2003, the owl population in a park was measured to 172. be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 1990. a. Find a formula for the owl population, P. Let the input be years since 2003. b. What does your model predict the owl population to be in 2012? The Federal Helium Reserve held about 16 billion 173. cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year. a. Give a linear equation for the remaining federal the number of in terms of t, helium reserves, R, years since 2010. b. c. In 2015, what will the helium reserves be? If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted? Suppose the world’s oil reserves in 2014 are 1,820 reserves are 174. billion barrels. decreasing by 25 billion barrels of oil each year: If, on average, the total a. Give a linear equation for reserves, R, since now. in terms of t, the remaining oil the number of years b. Seven years from now, what will the oil reserves be? c. If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted? You are choosing between two different prepaid cell 175. phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable? You are choosing between two different window 176. washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable? When hired at a new job selling jewelry, you are given 177. two pay options: Option A: Base salary of $17,000 a year with a commission of 12% of your sales Option B: Base salary of $20,000 a year with a commission of 5% of your sales 450 Chapter 4 Linear Functions How much jewelry would you need to sell for option A to produce a larger income? When hired at a new job selling electronics, you are 178. given two pay options: Option A: Base salary of $14,000 a year with a commission of 10% of your sales Option B: Base salary of $19,000 a year with a commission of 4% of your sales How much electronics would you need to sell for option A to produce a larger income? When hired at a new job selling electronics, you are 179. given two pay options: Option A: Base salary of $20,000 a year with a commission of 12% of your sales Option B: Base salary of $26,000 a year with a commission of 3% of your sales How much electronics would you need to sell for option A to produce a larger income? When hired at a new job selling electronics, you are 180. given two pay options: Option A: Base salary of $10,000 a year with a commission of 9% of your sales Option B: Base salary of $20,000 a year with a commission of 4% of your sales How much electronics would you need to sell for option A to produce a larger income? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 451 4.3 \| Fitting Linear Models to Data Learning Objectives In this section you will: 4.3.1 Draw and interpret scatter diagrams. 4.3.2 Use a graphing utility to find the line of best fit. 4.3.3 Distinguish between linear and nonlinear relations. 4.3.4 Fit a regression line to a set of data and use the linear model to make predictions. A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot. Drawing and Interpreting Scatter Plots A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure 4.42 shows a sample scatter plot. Figure 4.42 A scatter plot of age and final exam score variables Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam. Example 4.26 Using a Scatter Plot to Investigate Cricket Chirps Table 4.8 shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[5]. Plot this data, and determine whether the data appears to be linearly related. 5. Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010 452 Chapter 4 Linear Functions Chirps 44 35 20.4 33 31 35 18.5 37 26 Temperature 80.5 70.5 57 66 68 72 52 73.5 53 Table 4.8 Cricket Chirps vs Air Temperature Solution Plotting this data, as depicted in Figure 4.43 suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so. Figure 4.43 Finding the Line of Best Fit Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can estimate the rise run. Example 4.27 Finding a Line of Best Fit Find a linear function that fits the data in Table 4.8 by “eyeballing” a line that seems to fit. Solution On a graph, we could try sketching a line. Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 453 m = 60 50 = 1.2 and a y-intercept at 30. This gives an equation of T(c) = 1.2c + 30 where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure 4.44. Figure 4.44 Analysis This linear equation can then be used to approximate answers to various questions we might ask about the trend. Recognizing Interpolation or Extrapolation While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of the data. Figure 4.45 compares the two processes for the cricket-chirp data addressed in Example 4.27. We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44. There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called model breakdown. For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when x = 50, that is in 50 years, the model would not apply because we could not account for factors fifty years in the future. 454 Chapter 4 Linear Functions Figure 4.45 Interpolation occurs within the domain and range of the provided data whereas extrapolation occurs outside. Interpolation and Extrapolation Different methods of making predictions are used to analyze data. The method of interpolation involves predicting a value inside the domain and/or range of the data. The method of extrapolation involves predicting a value outside the domain and/or range of the data. Model breakdown occurs at the point when the model no longer applies. Example 4.28 Understanding Interpolation and Extrapolation Use the cricket data from Table 4.8 to answer the following questions: a. Would predicting the temperat
ure when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. b. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. Solution a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model: T (30) = 30 + 1.2(30) = 66 degrees Based on the data we have, this value seems reasonable. b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model: 40 = 30 + 1.2c 10 = 1.2c c ≈ 8.33 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 455 We can compare the regions of interpolation and extrapolation using Figure 4.46. Figure 4.46 Analysis Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees. According to the data from Table 4.8, what temperature can we predict it is if we counted 20 chirps in 15 4.14 seconds? Finding the Line of Best Fit Using a Graphing Utility While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values[6]. One such technique is called least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[7]. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression. Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression. 1. Enter the input in List 1 (L1). 2. Enter the output in List 2 (L2). 3. On a graphing utility, select Linear Regression (LinReg). Example 4.29 Finding a Least Squares Regression Line 6. Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values. 7. For example, http://www.shodor.org/unchem/math/lls/leastsq.html 456 Chapter 4 Linear Functions Find the least squares regression line using the cricket-chirp data in Table 4.9. Solution 1. Enter the input (chirps) in List 1 (L1). 2. Enter the output (temperature) in List 2 (L2). See Table 4.9. L1 44 35 20.4 33 31 35 18.5 37 26 L2 80.5 70.5 57 66 68 72 52 73.5 53 Table 4.9 3. On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation: T(c) = 30.281 + 1.143c Analysis Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to: T(30) = 30.281 + 1.143(30) = 64.571 ≈ 64.6 degrees The graph of the scatter plot with the least squares regression line is shown in Figure 4.47. Figure 4.47 Will there ever be a case where two different lines will serve as the best fit for the data? No. There is only one best fit line. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 457 Distinguishing Between Linear and Nonlinear Models As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a ”diagnostic on” selection to find the correlation coefficient, which mathematicians label as r The correlation coefficient provides an easy way to get an idea of how close to a line the data falls. We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of r and the graph of the data, Figure 4.48 shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output. Figure 4.48 Plotted data and related correlation coefficients. (credit: “DenisBoigelot,” Wikimedia Commons) Correlation Coefficient The correlation coefficient is a value, r, between –1 and 1. • • r > 0 suggests a positive (increasing) relationship r < 0 suggests a negative (decreasing) relationship • The closer the value is to 0, the more scattered the data. • The closer the value is to 1 or –1, the less scattered the data is. Example 4.30 Finding a Correlation Coefficient Calculate the correlation coefficient for cricket-chirp data in Table 4.8. Solution 458 Chapter 4 Linear Functions Because the data appear to follow a linear pattern, we can use technology to calculate r Enter the inputs and corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation coefficient, r = 0.9509. This value is very close to 1, which suggests a strong increasing linear relationship. Note: For some calculators, the Diagnostics must be turned "on" in order to get the correlation coefficient when linear regression is performed: [2nd]>[0]>[alpha][x–1], then scroll to DIAGNOSTICSON. Fitting a Regression Line to a Set of Data Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data. Example 4.31 Using a Regression Line to Make Predictions Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in Table 4.10.[8] Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008. Year '94 '95 '96 '97 '98 '99 '00 '01 '02 '03 '04 Consumption (billions of gallons) Table 4.10 113 116 118 119 123 125 126 128 131 133 136 The scatter plot of the data, including the least squares regression line, is shown in Figure 4.49. Figure 4.49 8. http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 459 Solution We can introduce new input variable, t, representing years since 1994. The least squares regression equation is: C(t) = 113.318 + 2.209t Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend. Using this to predict consumption in 2008 (t = 14), The model predicts 144.244 billion gallons of gasoline consumption in 2008. C(14) = 113.318 + 2.209(14) = 144.244 Use the model we created using technology in Example 4.31 to predict the gas consumption in 2011. Is 4.15 this an interpolation or an extrapolation? Access these online resources for additional instruction and practice with fitting linear models to data. • Introduction to Regression Analysis (http://Openstaxcollege.org/l/introregress) • Linear Regression (http://Openstaxcollege.org/l/linearregress) 460 Chapter 4 Linear Functions 4.3 EXERCISES Verbal 190. Describe what it means if there is a model breakdown 181. when using a linear model. 100 250 300 450 600 750 182. What is interpolation when using a linear model? 12 12.6 13.1 14 14.5 15.2 183. What is extrapolation when using a linear model? Explain the difference between a positive and a 184. negative correlation coefficient. 191. Explain how to interpret 185. correlation coefficient. the absolute value of a 1 1 3 9 5 7 9 11 28 65 125 216 Algebraic 186. A regression was run to determine whether there is a relationship between hours of TV watched per day (x) and number of sit-ups a person can do (y). The results of the regression are given below. Use this to predict the number of sit-ups a person who watches 11 hours of TV can do. y = ax + b a = −1.341 b = 32.234 r = −0.896 A regression was run to determine whether there is a in inches) 187. relationship between the diameter of a tree ( x, and the tree’s age ( y, regression are given below. Use this to predict the age of a tree with diameter 10 inches. in years). The results of the y = ax + b a = 6.301 b = −1.044 r = −0.970 For the following exercises, draw a scatter plot for the data provided. Does the data appear to be linearly related? 0 2 4 6 8 10 –22 –19 –15 –11 –6 –2 188. 189. For the following data, draw a scatter plot. If we 192. wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Eyeball the line, and estimate the answer. Year Population 1990 11,500 1995 12,100 2000 12,700 2005 13,000 2010 13,750 For the following data, draw a scatter plot. If we 193. wanted to know when the temperature would reach 28°F, would the answer involve interpolation or extrapolation? Eyeball the line and estimate the answer. Temperature,°F 16 18 20 25 30 Time, seconds 46 50 54 55 62 1 2 3 4 5 6 Graphical 46 50 59 75 100 136 For the following exercises, match each scatterplot with one of the four specified correlations in Figure 4.50 and Figure 4.51. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 461 Figure 4.50 Figure 4.51 194. r = 0.95 195. r = −0.89 19
6. r = −0.26 197. r = −0.39 For the following exercises, draw a best-fit line for the plotted data. 198. 462 Chapter 4 Linear Functions Numeric 202. The U.S. Census tracks the percentage of persons 25 years or older who are college graduates. That data for several years is given in Table 4.11.[9] Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the percentage exceed 35%? Year Percent Graduates 199. 200. 201. 1990 21.3 1992 21.4 1994 22.2 1996 23.6 1998 24.4 2000 25.6 2002 26.7 2004 27.7 2006 28 2008 29.4 Table 4.11 The U.S. import of wine (in hectoliters) for several 203. years is given in Table 4.12. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will imports exceed 12,000 hectoliters? 9. Based on data from http://www.census.gov/hhes/socdemo/education/data/cps/historical/index.html. Accessed 5/1/ 2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 463 Year Imports Year Number Unemployed 1992 2665 1994 2688 1996 3565 1998 4129 2000 4584 2002 5655 2004 6549 2006 7950 2008 8487 2009 9462 1990 750 1992 670 1994 650 1996 605 1998 550 2000 510 2002 460 2004 420 2006 380 2008 320 Table 4.12 Table 4.13 204. Table 4.13 shows the year and the number of people unemployed in a particular city for several years. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the number of unemployed reach 5? Technology For the following exercises, use each set of data to calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to 3 decimal places of accuracy. 8 15 26 31 56 23 41 53 72 103 5 4 7 10 12 15 12 17 22 24 x y x y 205. 206. 207. 464 Chapter 4 Linear Functions x y x y 21.9 10 18.54 22.22 11 15.76 x y 21 25 30 31 40 50 17 11 2 –1 –18 –40 3 4 5 6 7 8 9 22.74 12 13.68 210. 22.26 13 14.1 20.78 14 14.02 17.6 15 11.94 16.52 16 12.76 x y 100 2000 80 60 55 40 20 1798 1589 1580 1390 1202 208. x y 4 5 6 7 8 9 44.8 43.1 38.8 39 38 32.7 10 30.1 11 29.3 12 27 13 25.8 209. This content is available for free at https://cnx.org/content/col11758/1.5 211. x y 900 988 1000 1010 1200 1205 70 80 82 84 105 108 Extensions Graph f (x) = 0.5x + 10. Pick a set of five ordered linear 212. pairs using inputs x = −2, 1, 5, 6, 9 and use regression to verify that the function is a good fit for the data. Graph f (x) = − 2x − 10. Pick a set of five ordered linear 213. pairs using inputs x = −2, 1, 5, 6, 9 and use regression to verify the function. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs shows dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years: (46, 1, 600), (48, 1, 550), (50, 1, 505), (52, 1, 540), (54, 1, 495). Chapter 4 Linear Functions 465 Use linear regression to determine a function P 214. where the profit in thousands of dollars depends on the number of units sold in hundreds. 215. Find to the nearest tenth and interpret the x-intercept. 216. Find to the nearest tenth and interpret the y-intercept. Real-World Applications For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population, year) for specific recorded years: (2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010) 217. Use linear regression to determine a function y, where the year depends on the population. Round to three decimal places of accuracy. 218. Predict when the population will hit 8,000. For the following exercises, consider this scenario: The profit of a company increased steadily over a ten-year span. The following ordered pairs show the number of units sold in hundreds and the profit in thousands of over the ten year span, (number of units sold, profit) for specific recorded years: (46, 250), (48, 305), (50, 350), (52, 390), (54, 410). 219. Use linear regression to determine a function y, where the profit in thousands of dollars depends on the number of units sold in hundreds. Predict when the profit will exceed one million 220. dollars. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs show dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span (number of units sold, profit) for specific recorded years: (46, 250), (48, 225), (50, 205), (52, 180), (54, 165). 221. Use linear regression to determine a function y, where the profit in thousands of dollars depends on the number of units sold in hundreds. Predict when the profit will dip below the $25,000 222. threshold. 466 Chapter 4 Linear Functions CHAPTER 4 REVIEW KEY TERMS correlation coefficient a value, r, between –1 and 1 that indicates the degree of linear correlation of variables, or how closely a regression line fits a data set. decreasing linear function a function with a negative slope: If f (x) = mx + b, then m < 0. extrapolation predicting a value outside the domain and range of the data horizontal line a line defined by f (x) = b, where b is a real number. The slope of a horizontal line is 0. increasing linear function a function with a positive slope: If f (x) = mx + b, then m > 0. interpolation predicting a value inside the domain and range of the data least squares regression the line and data values a statistical technique for fitting a line to data in a way that minimizes the differences between linear function line a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight model breakdown when a model no longer applies after a certain point parallel lines two or more lines with the same slope perpendicular lines two lines that intersect at right angles and have slopes that are negative reciprocals of each other point-slope form the equation for a line that represents a linear function of the form y − y1 = m(x − x1) slope the ratio of the change in output values to the change in input values; a measure of the steepness of a line slope-intercept form the equation for a line that represents a linear function in the form f (x) = mx + b vertical line a line defined by x = a, where a is a real number. The slope of a vertical line is undefined. KEY CONCEPTS 4.1 Linear Functions • Linear functions can be represented in words, function notation, tabular form, and graphical form. See Example 4.1. • An increasing linear function results in a graph that slants upward from left to right and has a positive slope. A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. A constant linear function results in a graph that is a horizontal line. See Example 4.2. • Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between yvalues by the difference in corresponding x-values of any two points on the line. See Example 4.3 and Example 4.4. • An equation for a linear function can be written from a graph. See Example 4.5. • The equation for a linear function can be written if the slope m and initial value b are known. See Example 4.6 and Example 4.7. • A linear function can be used to solve real-world problems given information in different forms. See Example 4.8, Example 4.9, and Example 4.10. • Linear functions can be graphed by plotting points or by using the y-intercept and slope. See Example 4.11 and Example 4.12. • Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See Example 4.13. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 467 • The equation for a linear function can be written by interpreting the graph. See Example 4.14. • The x-intercept is the point at which the graph of a linear function crosses the x-axis. See Example 4.15. • Horizontal lines are written in the form, f (x) = b. See Example 4.16. • Vertical lines are written in the form, x = b. See Example 4.17. • Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See Example 4.18. • A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x- and y-values of the given point into the equation, f (x) = mx + b, and using the b that results. Similarly, the point-slope form of an equation can also be used. See Example 4.19. • A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See Example 4.20 and Example 4.21. 4.2 Modeling with Linear Functions • We can use the same problem strategies that we would use for any type of function. • When modeling and solving a problem, identify the variables and look for key values, including the slope and y- intercept. See Example 4.22. • Draw a diagram, where appropriate. See Example 4.23 and Example 4.24. • Check for reasonableness of the answer. • Linear models may be built by identifying or calculating the slope and using the y-intercept. ◦ The x-intercept may be found by setting y = 0, which is setting the expression mx + b equal to 0. ◦ The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See Example 4.25. ◦ A graph of the system may be used to identify the points where one line falls below (or above) the other line. 4.3 Fitting Linear Models to Data • Scatter plots show the relationship between two sets of data. See Example 4.26. • Scatter plots m
ay represent linear or non-linear models. • The line of best fit may be estimated or calculated, using a calculator or statistical software. See Example 4.27. • Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data. See Example 4.28. • The correlation coefficient, r, indicates the degree of linear relationship between data. See Example 4.29. • A regression line best fits the data. See Example 4.30. • The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables. See Example 4.31. CHAPTER 4 REVIEW EXERCISES Linear Functions 223. Determine whether the algebraic equation is linear. 2x + 3y = 7 225. Determine whether the function is increasing or decreasing. f (x) = 7x − 2 224. Determine whether the algebraic equation is linear. 6x2 − y = 5 226. Determine whether the function is increasing or decreasing. g(x) = − x + 2 468 Chapter 4 Linear Functions 227. Given each set of information, find a linear equation that satisfies the given conditions, if possible. Passes through (7, 5) and (3, 17) 228. Given each set of information, find a linear equation that satisfies the given conditions, if possible. x-intercept at (6, 0) and y-intercept at (0, 10) 229. Find the slope of the line shown in the graph. 230. Find the slope of the line graphed. 232. Does the following table represent a linear function? If so, find the linear equation that models the data. x –4 0 2 10 g(x) 18 –2 –12 –52 233. Does the following table represent a linear function? If so, find the linear equation that models the data. x 6 8 12 26 g(x) –8 –12 –18 –46 234. On June 1st, a company has $4,000,000 profit. If the company then loses 150,000 dollars per day thereafter in the month of June, what is the company’s profit nth day after June 1st? For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 235. 2x − 6y = 12 −x + 3y = 1 231. Write an equation in slope-intercept form for the line shown. 236. x − 2 y = 1 3 3x + y = − 9 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 469 Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t. When will no one be afflicted? For the following exercises, use the graph in Figure 4.52 showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. For the following exercises, find the x- and y- intercepts of the given equation 237. 7x + 9y = − 63 238. f (x) = 2x − 1 For the following exercises, use the descriptions of the pairs of lines to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 239. Line 1: Passes through (5, 11) and (10, 1) Line 2: Passes through (−1, 3) and (−5, 11) 240. Line 1: Passes through (8, −10) and (0, −26) Line 2: Passes through (2, 5) and (4, 4) 241. Write an equation for a line perpendicular f (x) = 5x − 1 and passing through the point (5, 20). to Figure 4.52 242. Find the equation of a line with a y- intercept of (0, 2) and slope − 1 2 . 243. Sketch a graph of the linear function f (t) = 2t − 5. 244. Find the point of intersection for the 2 linear functions: x = y + 6 2x − y = 13 . 245. A car rental company offers two plans for renting a car. Plan A: 25 dollars per day and 10 cents per mile Plan B: 50 dollars per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? Modeling with Linear Functions 246. Find the area of a triangle bounded by the y axis, the line f (x) = 10 − 2x, and the line perpendicular to f that passes through the origin. 249. Find the linear function y, where y depends on x, number of years since 1980. the 250. Find and interpret the y-intercept. For the following exercise, consider this scenario: In 2004, a school population was 1,700. By 2012 the population had grown to 2,500. 251. Assume the population is changing linearly. a. How much did the population grow between the year 2004 and 2012? b. What year? c. Find an equation for the population, P, of the school t years after 2004. is the average population growth per For the following exercises, consider this scenario: In 2000, the moose population in a park was measured to be 6,500. the population was measured to be 12,500. By 2010, Assume the population continues to change linearly. 252. Find a formula for the moose population, P. 253. What does your model predict the moose population to be in 2020? 247. A town’s population increases at a constant rate. In 2010 the population was 55,000. By 2012 the population had increased to 76,000. If this trend continues, predict the population in 2016. For the following exercises, consider this scenario: The median home values in subdivisions Pima Central and East Valley (adjusted for inflation) are shown in Table 4.14. Assume that the house values are changing linearly. 248. The number of people afflicted with the common cold in the winter months dropped steadily by 50 each year since 2004 until 2010. In 2004, 875 people were inflicted. 470 Chapter 4 Linear Functions Year Pima Central East Valley Predicted Actual 1970 2010 Table 4.14 32,000 120,250 85,000 150,000 In which subdivision have home values increased at 254. a higher rate? If these trends were to continue, what would be the 255. median home value in Pima Central in 2015? Fitting Linear Models to Data 256. Draw a scatter plot for the data in Table 4.15. Then determine whether the data appears to be linearly related. 0 2 –105 –50 4 1 6 8 10 55 105 160 6 7 7 8 7 9 10 10 6 7 8 8 9 10 10 9 Table 4.15 Table 4.17 257. Draw a scatter plot for the data in Table 4.16. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? 259. Draw a best-fit line for the plotted data. Year Population 1990 5,600 1995 5,950 2000 6,300 2005 6,600 2010 6,900 Table 4.16 For the following exercises, consider the data in Table 4.18, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year. Year 2000 2002 2005 2007 2010 258. Eight students were asked to estimate their score on a 10-point quiz. Their estimated and actual scores are given in Table 4.17. Plot the points, then sketch a line that fits the data. Percent Graduates Table 4.18 6.5 7.0 7.4 8.2 9.0 260. Determine whether the trend appears to be linear. If so, and assuming the trend continues, find a linear This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 471 regression model to predict the percent of unemployed in a given year to three decimal places. 261. In what year will the percentage exceed 12%? 262. Based on the set of data given in Table 4.19, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. x y 17 20 23 26 29 15 25 31 37 40 Table 4.19 263. Based on the set of data given in Table 4.20, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. x y 10 12 15 18 20 36 34 30 28 22 Table 4.20 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the population and the year over the ten-year span (population, year) for specific recorded years: (3,600, 2000); (4,000, 2001); (4,700, 2003); (6,000, 2006) 264. Use linear regression to determine a function y, where the year depends on the population, to three decimal places of accuracy. 265. Predict when the population will hit 12,000. 266. What is the correlation coefficient for this model to three decimal places of accuracy? 267. According to the model, what is the population in 2014? CHAPTER 4 PRACTICE TEST 268. Determine whether the following algebraic equation can be written as a linear function. 2x + 3y = 7 269. Determine whether increasing or decreasing. f (x) = − 2x + 5 the following function is 472 Chapter 4 Linear Functions 270. Determine whether increasing or decreasing. f (x) = 7x + 9 the following function is x –6 0 2 4 271. Find a linear equation that passes through (5, 1) and (3, –9), if possible. g(x) 14 32 38 44 272. Find a linear equation, that has an x intercept at (–4, 0) and a y-intercept at (0, –6), if possible. Table 4.21 273. Find the slope of the line in Figure 4.53. 276. Does Table 4.22 represent a linear function? If so, find a linear equation that models the data. x g(x) 1 4 3 9 7 11 19 12 Table 4.22 277. At 6 am, an online company has sold 120 items that day. If the company sells an average of 30 items per hour for the remainder of the day, write an expression to represent the number of items that were sold n after 6 am. For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular. 278. x − 9 y = 3 4 −4x − 3y = 8 279. −2x + y = 3 3x + 3 2 y = 5 Find the x- and y-intercepts of 280. 2x + 7y = − 14. the equation 281. Given below are descriptions of two lines. Find the slopes of Line 1 and Line 2. Is the pair of lines parallel, perpendicular, or neither? Line 1: Passes through (−2, −6) and (3, 14) Line 2: Passes through (2, 6) and (4, 14) 282. Write an equation for a line perpendicular f (x) = 4x + 3 and passing through the point (8, 10). to 283. Sketch a line with a y-intercept of (0, 5) and slope − 5 2 . Figure 4.53 274. Write an equation for line in Figure 4.54. Figure 4.54 275. Does Table 4.21 represent a linear function? If so, find a linear equation that models the da
ta. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 4 Linear Functions 473 284. Graph of the linear function f (x) = − x + 6. 285. For the two linear functions, find the point of intersection: x = y + 2 . 2x − 3y = − 1 286. A car rental company offers two plans for renting a car. Plan A: $25 per day and $0.10 per mile Plan B: $40 per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? 287. Find the area of a triangle bounded by the y axis, the line f (x) = 12 − 4x, and the line perpendicular to f that passes through the origin. 288. A town’s population increases at a constant rate. In 2010 the population was 65,000. By 2012 the population had increased to 90,000. Assuming this trend continues, predict the population in 2018. 289. The number of people afflicted with the common cold in the winter months dropped steadily by 25 each year since 2002 until 2012. In 2002, 8,040 people were inflicted. Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t. When will less than 6,000 people be afflicted? For the following exercises, use the graph in Figure 4.55, showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. Figure 4.55 290. Find the linear function y, where y depends on x, the number of years since 1980. 291. Find and interpret the y-intercept. 292. In 2004, a school population was 1250. By 2012 the population had dropped to 875. Assume the population is changing linearly. a. How much did the population drop between the year 2004 and 2012? b. What year? c. Find an equation for the population, P, of the school t years after 2004. is the average population decline per 293. Draw a scatter plot for the data provided in Table 4.23. Then determine whether the data appears to be linearly related. 0 2 4 6 8 10 –450 –200 10 265 500 755 Table 4.23 294. Draw a best-fit line for the plotted data. For the following exercises, use Table 4.24, which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year. Year 2000 2002 2005 2007 2010 8.5 8.0 7.2 6.7 6.4 Percent Graduates Table 4.24 295. Determine whether the trend appears linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. 296. In what year will the percentage drop below 4%? 474 Chapter 4 Linear Functions 297. Based on the set of data given in Table 4.25, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracy. x y 16 18 20 24 26 106 110 115 120 125 Table 4.25 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population (in hundreds) and the year over the ten-year span, (population, year) for specific recorded years: (4, 500, 2000); (4, 700, 2001); (5, 200, 2003); (5, 800, 2006) 298. Use linear regression to determine a function y, where the year depends on the population. Round to three decimal places of accuracy. 299. Predict when the population will hit 20,000. 300. What is the correlation coefficient for this model? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 475 5 \| POLYNOMIAL AND RATIONAL FUNCTIONS Figure 5.1 35-mm film, once the standard for capturing photographic images, has been made largely obsolete by digital photography. (credit “film”: modification of work by Horia Varlan; credit “memory cards”: modification of work by Paul Hudson) Chapter Outline 5.1 Quadratic Functions 5.2 Power Functions and Polynomial Functions 5.3 Graphs of Polynomial Functions 5.4 Dividing Polynomials 5.5 Zeros of Polynomial Functions 5.6 Rational Functions 5.7 Inverses and Radical Functions 5.8 Modeling Using Variation Introduction Digital photography has dramatically changed the nature of photography. No longer is an image etched in the emulsion on a roll of film. Instead, nearly every aspect of recording and manipulating images is now governed by mathematics. An image becomes a series of numbers, representing the characteristics of light striking an image sensor. When we open an image file, software on a camera or computer interprets the numbers and converts them to a visual image. Photo editing software uses complex polynomials to transform images, allowing us to manipulate the image in order to crop details, change the color palette, and add special effects. Inverse functions make it possible to convert from one file format to another. In this chapter, we will learn about these concepts and discover how mathematics can be used in such applications. 476 Chapter 5 Polynomial and Rational Functions 5.1 \| Quadratic Functions Learning Objectives In this section, you will: 5.1.1 Recognize characteristics of parabolas. 5.1.2 Understand how the graph of a parabola is related to its quadratic function. 5.1.3 Determine a quadratic function’s minimum or maximum value. 5.1.4 Solve problems involving a quadratic function’s minimum or maximum value. Figure 5.2 An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr) Curved antennas, such as the ones shown in Figure 5.2, are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function. In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior. Recognizing Characteristics of Parabolas The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. These features are illustrated in Figure 5.3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 477 Figure 5.3 The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of x at which y = 0. 478 Chapter 5 Polynomial and Rational Functions Example 5.1 Identifying the Characteristics of a Parabola Determine the vertex, axis of symmetry, zeros, and y- intercept of the parabola shown in Figure 5.4. Figure 5.4 Solution The vertex is the turning point of the graph. We can see that the vertex is at (3, 1). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. This parabola does not cross the x- axis, so it has no zeros. It crosses the y- axis at (0, 7) so this is the y-intercept. Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions The general form of a quadratic function presents the function in the form where a, b, and c are real numbers and a ≠ 0. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry. f (x) = ax2 + bx + c (5.1) The axis of symmetry is defined by x = − 2a. If we use the quadratic formula, x = −b ± b2 − 4ac ax2 + bx + c = 0 for the x- intercepts, or zeros, we find the value of x halfway between them is always x = − 2a b , to solve b 2a, the equation for the axis of symmetry. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 479 Figure 5.5 represents the graph of the quadratic function written in general form as y = x2 + 4x + 3. In this form, a = 1, b = 4, and c = 3. Because a > 0, the parabola opens upward. The axis of symmetry is x = − 4 2(1) = −2. This also makes sense because we can see from the graph that the vertical line x = −2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, (−2, −1). The x- intercepts, those points where the parabola crosses the x- axis, occur at (−3, 0) and (−1, 0). Figure 5.5 The standard form of a quadratic function presents the function in the form f (x) = a(x − h)2 + k (5.2) where (h, k) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function. As with the general form, if a > 0, the parabola opens upward and the vertex is a minimum. If a < 0, the parabola opens downward, and the vertex is a maximum. Figure 5.6 represents the graph of the quadratic function written in standard form as y = −3(x + 2)2 + 4. Since x – h = x + 2 in this example, h = –2. In this form, a = −3, h = −2, and k = 4. Because a < 0, the parabola opens downward. The vertex is at (−2, 4). 480 Chapter 5 Polynomial and Rational Functions Figure 5.6 The standard form is useful for determining how the graph is transformed from the graph of y = x2. Figure 5.7 is the graph of this basic function. Figure 5.7 If k > 0, the graph shifts upward
, whereas if k < 0, the graph shifts downward. In Figure 5.6, k > 0, so the graph is shifted 4 units upward. If h > 0, the graph shifts toward the right and if h < 0, the graph shifts to the left. In Figure This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 481 5.6, h < 0, so the graph is shifted 2 units to the left. The magnitude of a indicates the stretch of the graph. If \|a\| > 1, the point associated with a particular x- value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if \|a\| < 1, the point associated with a particular x- value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5.6, \|a\| > 1, so the graph becomes narrower. The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form. a(x − h)2 + k = ax2 + bx + c ax2 − 2ahx + (ah2 + k) = ax2 + bx + c For the linear terms to be equal, the coefficients must be equal. –2ah = b, so h = − b 2a This is the axis of symmetry we defined earlier. Setting the constant terms equal: ah2 + k = c k = c − ah2 2 ⎛ ⎝ b 2a ⎞ ⎠ = c − a − = c − b2 4a In practice, though, it is usually easier to remember that k is the output value of the function when the input is h, so f (h) = k. Forms of Quadratic Functions A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is f (x) = ax2 + bx + c where a, b, and c are real numbers and a ≠ 0. The standard form of a quadratic function is f (x) = a(x − h)2 + k where a ≠ 0. The vertex (h, k) is located at h = – b 2a, k = f (h) = f ⎛ ⎝ −b 2a ⎞ ⎠ Given a graph of a quadratic function, write the equation of the function in general form. 1. Identify the horizontal shift of the parabola; this value is h. Identify the vertical shift of the parabola; this value is k. 2. Substitute the values f (x) = a(x – h)2 + k. of the horizontal and vertical shift for h and k. in the function 3. Substitute the values of any point, other than the vertex, on the graph of the parabola for x and f (x). 4. Solve for the stretch factor, \|a\|. 5. Expand and simplify to write in general form. 482 Chapter 5 Polynomial and Rational Functions Example 5.2 Writing the Equation of a Quadratic Function from the Graph Write an equation for the quadratic function g in Figure 5.8 as a transformation of f (x) = x2, and then expand the formula, and simplify terms to write the equation in general form. Figure 5.8 Solution We can see the graph of g is the graph of f (x) = x2 shifted to the left 2 and down 3, giving a formula in the form g(x) = a(x − (−2))2 − 3 = a(x + 2)2 – 3. Substituting the coordinates of a point on the curve, such as (0, −1), we can solve for the stretch factor. −1 = a(0 + 2)2 − 3 2 = 4a a = 1 2 In standard form, the algebraic model for this graph is (g)x = 1 2 (x + 2)2 – 3. To write this in general polynomial form, we can expand the formula and simplify terms. g(xx + 2)2 − 3 (x + 2)(x + 2) − 3 (x2 + 4x + 4) − 3 x2 + 2x + 2 − 3 x2 + 2x − 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 483 Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions. Analysis We can check our work using the table feature on a graphing utility. First enter Y1 = 1 2 (x + 2)2 − 3. Next, select TBLSET, then use TblStart = – 6 and ΔTbl = 2, and select TABLE. See Table 5.0. x y –6 –4 –2 0 5 –1 –3 –1 2 5 Table 5.0 The ordered pairs in the table correspond to points on the graph. A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 5.9. Find an 5.1 equation for the path of the ball. Does the shooter make the basket? Figure 5.9 (credit: modification of work by Dan Meyer) Given a quadratic function in general form, find the vertex of the parabola. 1. Identify a, b, and c. 2. Find h, the x-coordinate of the vertex, by substituting a and b into h = – b 2a. 3. Find k, the y-coordinate of the vertex, by evaluating k = f (h) = f ⎛ ⎝− ⎞ ⎠. b 2a 484 Chapter 5 Polynomial and Rational Functions Example 5.3 Finding the Vertex of a Quadratic Function Find the vertex of the quadratic function f (x) = 2x2 – 6x + 7. Rewrite the quadratic in standard form (vertex form). Solution The horizontal coordinate of the vertex will be at The vertical coordinate of the vertex will be at h = − b 2a = −6 2(2h Rewriting into standard form, the stretch factor will be the same as the a in the original quadratic. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the “a” from the general form. f (x) = ax2 + bx + c f (x) = 2x2 − 6x + 7 The standard form of a quadratic function prior to writing the function then becomes the following: f (x) = 2 2 ⎛ ⎝x – 3 2 ⎞ ⎠ + 5 2 Analysis One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, k, and where it occurs, x. 5.2 Given the equation g(x) = 13 + x2 − 6x, write the equation in general form and then in standard form. Finding the Domain and Range of a Quadratic Function Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 485 Domain and Range of a Quadratic Function The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions. The range of a quadratic function written in general form f (x) = ax2 + bx + c with a positive a value is f (x) ≥ f ⎛ ⎝− ⎞ ⎞ ⎠; the range of a quadratic function written in general form with a negative a ⎠, ∞ ⎝−∞, f ⎛ ⎝− value is f (x) ≤ f ⎛ ⎝− ⎠, or ⎡ ⎣ f ⎛ b ⎝− 2a ⎠, or ⎛ ⎞ b 2a b 2a b 2a ⎤ ⎞ ⎦. ⎠ ⎞ The range of a quadratic function written in standard form f (x) = a(x − h)2 + k with a positive a value is f (x) ≥ k; the range of a quadratic function written in standard form with a negative a value is f (x) ≤ k. Given a quadratic function, find the domain and range. 1. Identify the domain of any quadratic function as all real numbers. 2. Determine whether a is positive or negative. If a is positive, the parabola has a minimum. If a is negative, the parabola has a maximum. 3. Determine the maximum or minimum value of the parabola, k. 4. If the parabola has a minimum, the range is given by f (x) ≥ k, or ⎡ maximum, the range is given by f (x) ≤ k, or (−∞, k⎤ ⎦. ⎣k, ∞). If the parabola has a Example 5.4 Finding the Domain and Range of a Quadratic Function Find the domain and range of f (x) = − 5x2 + 9x − 1. Solution As with any quadratic function, the domain is all real numbers. Because a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x- value of the vertex. h = − b 2a = − 9 2(−5) The maximum value is given by f (h). = 9 10 2 ⎞ ⎠ + p⎛ ⎝ 9 10 ⎞ ⎠ − 1 f ⎛ ⎝ 9 10 9 10 ⎞ ⎛ ⎠ = 5 ⎝ = 61 20 The range is f (x) ≤ 61 20 , or ⎛ ⎝−∞, 61 20 ⎤ ⎦. 486 5.3 Find the domain and range of f (x) = 2 2 ⎛ ⎝x − 4 7 ⎞ ⎠ + 8 11 . Chapter 5 Polynomial and Rational Functions Determining the Maximum and Minimum Values of Quadratic Functions The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in Figure 5.10. Figure 5.10 There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. Example 5.5 Finding the Maximum Value of a Quadratic Function A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. a. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L. b. What dimensions should she make her garden to maximize the enclosed area? Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 487 Let’s use a diagram such as Figure 5.11 to record the given information. It is also helpful to introduce a temporary variable, W, to represent the width of the garden and the length of the fence section parallel to the backyard fence. Figure 5.11 a. We know we have only 80 feet of fence available, and L + W + L = 80, or more simply, 2L + W = 80. This allows us to represent the width, W, in terms of L. W = 80 − 2L Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so A = LW = L(80 − 2L) A(L) = 80L − 2L2 This formula represents the area of the fence in terms of the variable length L. The function, written in general form, is A(L) = −2L2 + 80L. b. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the e
quation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a is the coefficient of the squared term, a = −2, b = 80, and c = 0. To find the vertex: h = − b 2a = − 80 2(−2) = 20 k = A(20) and = 80(20) − 2(20)2 = 800 The maximum value of the function is an area of 800 square feet, which occurs when L = 20 feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet. Analysis This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 5.12. 488 Chapter 5 Polynomial and Rational Functions Figure 5.12 Given an application involving revenue, use a quadratic equation to find the maximum. 1. Write a quadratic equation for a revenue function. 2. Find the vertex of the quadratic equation. 3. Determine the y-value of the vertex. Example 5.6 Finding Maximum Revenue The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue? Solution Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p for price per subscription and Q for quantity, giving us the equation Revenue = pQ. Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently p = 30 and Q = 84,000. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, p = 32 and Q = 79,000. From this we can find a linear equation relating the two quantities. The slope will be This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 489 m = 79,000 − 84,000 32 − 30 = −5,000 2 = −2,500 This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept. Q = −2500p + b 84,000 = −2500(30) + b b = 159,000 Substitute in the pointQ = 84,000 and p = 30 Solve forb This gives us the linear equation Q = −2,500p + 159,000 relating cost and subscribers. We now return to our revenue equation. Revenue = pQ Revenue = p(−2,500p + 159,000) Revenue = −2,500p2 + 159,000p We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex. h = − 159,000 2(−2,500) = 31.8 The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function. maximum revenue = −2,500(31.8)2 + 159,000(31.8) = 2,528,100 Analysis This could also be solved by graphing the quadratic as in Figure 5.13. We can see the maximum revenue on a graph of the quadratic function. Figure 5.13 Finding the x- and y-Intercepts of a Quadratic Function Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y- intercept of a quadratic by evaluating the function at an input of zero, and we find 490 Chapter 5 Polynomial and Rational Functions the x- intercepts at locations where the output is zero. Notice in Figure 5.14 that the number of x- intercepts can vary depending upon the location of the graph. Figure 5.14 Number of x-intercepts of a parabola Given a quadratic function f(x), find the y- and x-intercepts. 1. Evaluate f (0) to find the y-intercept. 2. Solve the quadratic equation f (x) = 0 to find the x-intercepts. Example 5.7 Finding the y- and x-Intercepts of a Parabola Find the y- and x-intercepts of the quadratic f (x) = 3x2 + 5x − 2. Solution We find the y-intercept by evaluating f (0). f (0) = 3(0)2 + 5(0) − 2 = −2 So the y-intercept is at (0, −2). For the x-intercepts, we find all solutions of f (x) = 0. In this case, the quadratic can be factored easily, providing the simplest method for solution. 0 = 3x2 + 5x − 2 h = − b 2a = − 4 2(2) = −1 0 = (3x − 1)(x + 2) k = f (−1) = 2(−1)2 + 4(−1) − 4 = −6 So the x-intercepts are at ⎛ ⎝ ⎞ ⎠ and (−2, 0). , 0 1 3 Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 491 By graphing the function, we can confirm that the graph crosses the y-axis at (0, −2). We can also confirm that the graph crosses the x-axis at ⎛ ⎝ ⎞ ⎠ and (−2, 0). See Figure 5.15 , 0 1 3 Figure 5.15 Rewriting Quadratics in Standard Form In Example 5.7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form. Given a quadratic function, find the x- intercepts by rewriting in standard form. 1. Substitute a and b into h = − b 2a. 2. Substitute x = h into the general form of the quadratic function to find k. 3. Rewrite the quadratic in standard form using h and k. 4. Solve for when the output of the function will be zero to find the x- intercepts. Example 5.8 Finding the x-Intercepts of a Parabola Find the x- intercepts of the quadratic function f (x) = 2x2 + 4x − 4. Solution We begin by solving for when the output will be zero. 0 = 2x2 + 4x − 4 492 Chapter 5 Polynomial and Rational Functions Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form. We know that a = 2. Then we solve for h and k. f (x) = a(x − h)2 + k h = − b 2a = − 4 2(2) = −1 k = f (−1) = 2(−1)2 + 4(−1) − 4 = −6 So now we can rewrite in standard form. We can now solve for when the output will be zero. f (x) = 2(x + 1)2 − 6 0 = 2(x + 1)2 − 6 6 = 2(x + 1)2 3 = (x + 1) The graph has x-intercepts at (−1 − 3, 0) and (−1 + 3, 0). We can check our work by graphing the given function on a graphing utility and observing the x- intercepts. See Figure 5.16. Figure 5.16 Analysis We could have achieved the same results using the quadratic formula. Identify a = 2, b = 4 and c = −4. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 493 x = −b ± b2 − 4ac 2a = −4 ± 42 − 4(2)(−4) 2(2) = −4 ± 48 4 = −4 ± 3(16) 4 = −1 ± 3 So the x-intercepts occur at ⎛ ⎝−1 − 3, 0⎞ ⎠ and ⎛ ⎝−1 + 3, 0⎞ ⎠. 5.4 In a Try It, we found the standard and general form for the function g(x) = 13 + x2 − 6x. Now find the y- and x-intercepts (if any). Example 5.9 Applying the Vertex and x-Intercepts of a Parabola A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H(t) = − 16t 2 + 80t + 40. a. When does the ball reach the maximum height? b. What is the maximum height of the ball? c. When does the ball hit the ground? Solution a. The ball reaches the maximum height at the vertex of the parabola. h = − 80 2(−16) = 80 32 = 5 2 = 2.5 The ball reaches a maximum height after 2.5 seconds. b. To find the maximum height, find the y- coordinate of the vertex of the parabola. k = H⎛ ⎝− b 2a = H(2.5) = −16(2.5)2 + 80(2.5) + 40 = 140 The ball reaches a maximum height of 140 feet. ⎞ ⎠ c. To find when the ball hits the ground, we need to determine when the height is zero, H(t) = 0. We use the quadratic formula. 494 Chapter 5 Polynomial and Rational Functions t = −80 ± 802 − 4(−16)(40) 2(−16) = −80 ± 8960 −32 Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions. t = −80 − 8960 −32 ≈ 5.458 or t = −80 + 8960 −32 ≈ − 0.458 The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 5.17. Figure 5.17 Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph. A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet 5.5 per second. The rock’s height above ocean can be modeled by the equation H(t) = −16t 2 + 96t + 112. a. When does the rock reach the maximum height? b. What is the maximum height of the rock? c. When does the rock hit the ocean? Access these online resources for additional instruction and practice with quadratic equations. • Graphing Quadratic Functions in General Form (http://openstaxcollege.org/l/ graphquadgen) • Graphing Quadratic Functions in Standard Form (http://openstaxcollege.org/l/ graphquadstan) • Quadratic Function Review (http://openstaxcollege.org/l/quadfuncrev) • Characteristics of a Quadratic Function (http://openstaxcollege.org/l/characterquad) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 495 5.1 EXERCISES Verbal Explain the advantage of writing a quadratic function in 1. standard form. How can the vertex of a parabola be used in solving real- 2. world problems? 3. Explain why the condition of a ≠ 0 is imposed in the definition of the quadratic function. What 4. quadratic function? is another name for the standard form of a What two algebraic methods can be used to find the 5. horizontal intercepts of a quadratic function? Algebraic For the following exercises, rewrite the qu
adratic functions in standard form and give the vertex. 6. 7. 8. 9. 10. 11. 12. 13. f (x) = x2 − 12x + 32 g(x) = x2 + 2x − 3 f (x) = x2 − x f (x) = x2 + 5x − 2 h(x) = 2x2 + 8x − 10 k(x) = 3x2 − 6x − 9 f (x) = 2x2 − 6x f (x) = 3x2 − 5x − 1 For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry. y(x) = 2x2 + 10x + 12 f (x) = 2x2 − 10x + 4 f (x) = − x2 + 4x + 3 f (x) = 4x2 + x − 1 h(t) = −4t 2 + 6t − 1 14. 15. 16. 17. 18. 19. f (x) = 1 2 x2 + 3x + 1 20. f (x) = − 1 3 x2 − 2x + 3 For the following exercises, determine the domain and range of the quadratic function. 21. 22. 23. 24. 25. f (x) = (x − 3)2 + 2 f (x) = −2(x + 3)2 − 6 f (x) = x2 + 6x + 4 f (x) = 2x2 − 4x + 2 k(x) = 3x2 − 6x − 9 For the following exercises, use the vertex (h, k) and a point on the graph (x, y) to find the general form of the equation of the quadratic function. 26. (h, k) = (2, 0), (x, y) = (4, 4) 27. (h, k) = (−2, −1), (x, y) = (−4, 3) 28. (h, k) = (0, 1), (x, y) = (2, 5) 29. (h, k) = (2, 3), (x, y) = (5, 12) 30. (h, k) = ( − 5, 3), (x, y) = (2, 9) 31. (h, k) = (3, 2), (x, y) = (10, 1) 32. (h, k) = (0, 1), (x, y) = (1, 0) 33. (h, k) = (1, 0), (x, y) = (0, 1) Graphical For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. 34. 35. 36. 37. f (x) = x2 − 2x f (x) = x2 − 6x − 1 f (x) = x2 − 5x − 6 f (x) = x2 − 7x + 3 Chapter 5 Polynomial and Rational Functions 43. 44. 45. 496 38. 39. f (x) = −2x2 + 5x − 8 f (x) = 4x2 − 12x − 3 For the following exercises, write the equation for the graphed quadratic function. 40. 41. 42. Numeric For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function. 46. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 497 x y –2 –2 –1 1 0 –2 –1 –2 1 –2 –1 –8 –3 –2 –2 2 0 2 8 47. 48. 49. 50. Technology For the following exercises, use a calculator to find the answer. What appears to be the effect of adding or subtracting those numbers? The path of an object projected at a 45 degree angle 54. with initial velocity of 80 feet per second is given by the function h(x) = −32 (80)2 x2 + x where x is the horizontal distance traveled and h(x) is the height in feet. Use the TRACE feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally. A suspension bridge can be modeled by the quadratic 55. function h(x) = .0001x2 with −2000 ≤ x ≤ 2000 where \|x\| is the number of feet from the center and h(x) is height in feet. Use the TRACE feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet. Extensions For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function. 56. Vertex (1, −2), opens up. 57. Vertex (−1, 2) opens down. 58. Vertex (−5, 11), opens down. 59. Vertex (−100, 100), opens up. For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function. 60. Contains (1, 1) and has shape of f (x) = 2x2. Vertex is on the y- axis. 61. Contains (−1, 4) and has the shape of f (x) = 2x2. Vertex is on the y- axis. Graph on the same set of axes x2. 51. f (x) = x2, f (x) = 2x2, and f (x) = 1 3 the functions 62. Contains (2, 3) and has the shape of f (x) = 3x2. Vertex is on the y- axis. What appears to be the effect of changing the coefficient? on the same Graph 52. f (x) = x2, f (x) = x2 + 2 f (x) = x2, f (x) = x2 + 5 and f (x) = x2 − 3. appears to be the effect of adding a constant? set of axes and What Graph axes 53. f (x) = x2, f (x) = (x − 2)2, f (x − 3)2, and f (x) = (x + 4)2. same the set on of 63. Contains (1, −3) and has the shape of f (x) = − x2. Vertex is on the y- axis. 64. Contains (4, 3) and has the shape of f (x) = 5x2. Vertex is on the y- axis. 65. 498 Chapter 5 Polynomial and Rational Functions Contains (1, −6) has the shape of f (x) = 3x2. Vertex has x-coordinate of −1. Real-World Applications Find the dimensions of corral 66. producing the greatest enclosed area given 200 feet of fencing. rectangular the Find the dimensions of the rectangular corral split into 67. 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing. Find the dimensions of corral 68. producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing. rectangular the Among all of the pairs of numbers whose sum is 6, find 69. the pair with the largest product. What is the product? Among all of the pairs of numbers whose difference is 70. 12, find the pair with the smallest product. What is the product? Suppose that the price per unit in dollars of a cell phone 71. production is modeled by p = $45 − 0.0125x, where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x ⋅ p. Find the production level that will maximize revenue. A rocket is launched in the air. Its height, in meters 72. above sea level, as a function of time, in seconds, is given by h(t) = −4.9t 2 + 229t + 234. Find the maximum height the rocket attains. A ball is thrown in the air from the top of a building. Its 73. height, in meters above ground, as a function of time, in seconds, is given by h(t) = − 4.9t 2 + 24t + 8. How long does it take to reach maximum height? A soccer stadium holds 62,000 spectators. With a ticket 74. price of $11, the average attendance has been 26,000. When the price dropped to $9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue? A farmer finds that if she plants 75 trees per acre, each 75. tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 499 5.2 \| Power Functions and Polynomial Functions Learning Objectives In this section, you will: 5.2.1 Identify power functions. 5.2.2 Identify end behavior of power functions. 5.2.3 Identify polynomial functions. 5.2.4 Identify the degree and leading coefficient of polynomial functions. Figure 5.18 (credit: Jason Bay, Flickr) Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in Table 5.1. Year 2009 2010 2011 2012 2013 Bird Population 800 897 992 1, 083 1, 169 Table 5.1 The population can be estimated using the function P(t) = − 0.3t 3 + 97t + 800, where P(t) represents the bird population on the island t years after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we will examine functions that we can use to estimate and predict these types of changes. Identifying Power Functions Before we can understand the bird problem, it will be helpful to understand a different type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number. As an example, consider functions for area or volume. The function for the area of a circle with radius r is and the function for the volume of a sphere with radius r is A(r) = πr 2 V(r) = 4 3 πr 3 500 Chapter 5 Polynomial and Rational Functions Both of these are examples of power functions because they consist of a coefficient, π or 4 3 π, multiplied by a variable r raised to a power. Power Function A power function is a function that can be represented in the form f (x) = kx p where k and p are real numbers, and k is known as the coefficient. Is f(x) = 2 x a power function? No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function. Example 5.10 Identifying Power Functions Which of the following functions are power functions? f (x) = 1 f (x) = x f (x) = x2 f (x) = x3 f (x) = 1 x f (x) = 1 x2 f (x) = x f (x) = x3 Solution Constant function Identify function Quadratic function Cubic function Reciprocal function Reciprocal squared function Square root function Cube root function All of the listed functions are power functions. The constant and identity functions are power functions because they can be written as f (x) = x0 and f (x) = x1 respectively. The quadratic and cubic functions are power functions with whole number powers f (x) = x2 and f (x) = x3. The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as f (x) = x−1 and f (x) = x−2. The square and cube root functions are power functions with fractional powers because they can be written as f (x) = x 1 2 or f (x) = x 1 3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 501 5.6 Which functions are power functions? f (x) = 2x ⋅ 4x3 g(x) = −x5 + 5x3 h(x) = 2x5 − 1 3x2 + 4 Identifying End Behavior of Power Functions Figure 5.19 shows the graphs of f (x) = x2, g(x) = x4 and h(x) = x6, which are all power functions with even, wholenumber powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin. Figure 5.19 Even-power functions To describe the
behavior as numbers become larger and larger, we use the idea of infinity. We use the symbol ∞ for positive infinity and −∞ for negative infinity. When we say that “ x approaches infinity,” which can be symbolically written as x → ∞, we are describing a behavior; we are saying that x is increasing without bound. With the positive even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as x approaches positive or negative infinity, the f (x) values increase without bound. In symbolic form, we could write as x → ± ∞, f (x) → ∞ Figure 5.20 shows the graphs of f (x) = x3, g(x) = x5, and h(x) = x7, which are all power functions with odd, wholenumber powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin. 502 Chapter 5 Polynomial and Rational Functions Figure 5.20 Odd-power functions These examples illustrate that functions of the form f (x) = xn reveal symmetry of one kind or another. First, in Figure 5.19 we see that even functions of the form f (x) = xn , n even, are symmetric about the y- axis. In Figure 5.20 we see that odd functions of the form f (x) = xn , n odd, are symmetric about the origin. For these odd power functions, as x approaches negative infinity, f (x) decreases without bound. As x approaches positive infinity, f (x) increases without bound. In symbolic form we write as x → − ∞, as x → ∞, f (x) → ∞ f (x) → − ∞ The behavior of the graph of a function as the input values get very small ( x → −∞ ) and get very large ( x → ∞ ) is referred to as the end behavior of the function. We can use words or symbols to describe end behavior. Figure 5.21 shows the end behavior of power functions in the form f (x) = kxn where n is a non-negative integer depending on the power and the constant. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 503 Figure 5.21 Given a power function f(x) = kxn where n is a non-negative integer, identify the end behavior. 1. Determine whether the power is even or odd. 2. Determine whether the constant is positive or negative. 3. Use Figure 5.21 to identify the end behavior. Example 5.11 Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f (x) = x8. Solution The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As x approaches infinity, the output (value of f (x) ) increases without bound. We write as x → ∞, f (x) → ∞. As x approaches negative infinity, the output increases without bound. In symbolic form, as x → −∞, f (x) → ∞. We can graphically represent the function as shown in Figure 5.22. 504 Chapter 5 Polynomial and Rational Functions Figure 5.22 Example 5.12 Identifying the End Behavior of a Power Function. Describe the end behavior of the graph of f (x) = − x9. Solution The exponent of the power function is 9 (an odd number). Because the coefficient is –1 (negative), the graph is the reflection about the x- axis of the graph of f (x) = x9. Figure 5.23 shows that as x approaches infinity, the output decreases without bound. As x approaches negative infinity, the output increases without bound. In symbolic form, we would write as x → −∞, as x → ∞, f (x) → ∞ f (x) → −∞ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 505 Figure 5.23 Analysis We can check our work by using the table feature on a graphing utility. x f(x) –10 1,000,000,000 –5 1,953,125 0 5 0 –1,953,125 10 –1,000,000,000 Table 5.1 We can see from Table 5.1 that, when we substitute very small values for x, the output is very large, and when we substitute very large values for x, the output is very small (meaning that it is a very large negative value). 506 Chapter 5 Polynomial and Rational Functions 5.7 Describe in words and symbols the end behavior of f (x) = − 5x4. Identifying Polynomial Functions An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil slick by combining two functions. The radius r of the spill depends on the number of weeks w that have passed. This relationship is linear. We can combine this with the formula for the area A of a circle. r(w) = 24 + 8w A(r) = πr 2 Composing these functions gives a formula for the area in terms of weeks. Multiplying gives the formula. A(w) = A(r(w)) = A(24 + 8w) = π(24 + 8w)2 A(w) = 576π + 384πw + 64πw2 This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. Polynomial Functions Let n be a non-negative integer. A polynomial function is a function that can be written in the form f (x) = an xn + ...a1 x + a2 x2 + a1 x + a0 (5.3) This is called the general form of a polynomial function. Each ai is a coefficient and can be any real number other than zero. Each expression ai xi is a term of a polynomial function. Example 5.13 Identifying Polynomial Functions Which of the following are polynomial functions? f (x) = 2x3 ⋅ 3x + 4 g(x) = −x(x2 − 4) h(x) = 5 x + 2 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 507 The first two functions are examples of polynomial functions because they can be written in the form f (x) = an xn + ... + a2 x2 + a1 x + a0, where the powers are non-negative integers and the coefficients are real numbers. • • • f (x) can be written as f (x) = 6x4 + 4. g(x) can be written as g(x) = − x3 + 4x. h(x) cannot be written in this form and is therefore not a polynomial function. Identifying the Degree and Leading Coefficient of a Polynomial Function Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term. Terminology of Polynomial Functions We often rearrange polynomials so that the powers are descending. When a polynomial is written in this way, we say that it is in general form. Given a polynomial function, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree function. 2. 3. Identify the term containing the highest power of x to find the leading term. Identify the coefficient of the leading term. Example 5.14 Identifying the Degree and Leading Coefficient of a Polynomial Function Identify the degree, leading term, and leading coefficient of the following polynomial functions. f (x) = 3 + 2x2 − 4x3 g(t) = 5t 2 − 2t 3 + 7t h(p) = 6p − p3 − 2 508 Chapter 5 Polynomial and Rational Functions Solution For the function f (x), the highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4x3. The leading coefficient is the coefficient of that term, −4. For the function g(t), the highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t 5. The leading coefficient is the coefficient of that term, 5. For the function h(p), the highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, −p3. The leading coefficient is the coefficient of that term, −1. 5.8 Identify the degree, leading term, and leading coefficient of the polynomial f (x) = 4x2 − x6 + 2x − 6. Identifying End Behavior of Polynomial Functions Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the power function consisting of the leading term. See Table 5.2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 509 Polynomial Function Leading Term Graph of Polynomial Function f (x) = 5x4 + 2x3 − x − 4 5x4 −2x6 f (x) = − 2x6 − x5 + 3x4 + x3 Table 5.2 510 Chapter 5 Polynomial and Rational Functions Polynomial Function Leading Term Graph of Polynomial Function f (x) = 3x5 − 4x4 + 2x2 + 1 f (x) = − 6x3 + 7x2 + 3x + 1 3x5 −6x3 Table 5.2 Example 5.15 Identifying End Behavior and Degree of a Polynomial Function This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 511 Describe the end behavior and determine a possible degree of the polynomial function in Figure 5.24. Figure 5.24 Solution As the input values x get very large, the output values f (x) increase without bound. As the input values x get very small, the output values f (x) decrease without bound. We can describe the end behavior symbolically by writing as x → −∞, as x → ∞, f (x) → −∞ f (x) → ∞ In words, we could say that as x values approach infinity, the function values approach infinity, and as x va
lues approach negative infinity, the function values approach negative infinity. We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive. 512 Chapter 5 Polynomial and Rational Functions 5.9 Describe the end behavior, and determine a possible degree of the polynomial function in Figure 5.25. Figure 5.25 Example 5.16 Identifying End Behavior and Degree of a Polynomial Function Given the function f (x) = − 3x2(x − 1)(x + 4), express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function. Solution Obtain the general form by expanding the given expression for f (x). f (x) = −3x2(x − 1)(x + 4) ⎞ = −3x2 ⎛ ⎝x2 + 3x − 4 ⎠ = −3x4 − 9x3 + 12x2 The general form is f (x) = −3x4 − 9x3 + 12x2. The leading term is −3x4; polynomial is 4. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is therefore, the degree of the as x → − ∞, as x → ∞, f (x) → − ∞ f (x) → − ∞ 5.10 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. Identifying Local Behavior of Polynomial Functions In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 513 We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-intercept ⎛ ⎠. The x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept. See Figure 5.26. ⎝0, a0 ⎞ Figure 5.26 Intercepts and Turning Points of Polynomial Functions A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero. Given a polynomial function, determine the intercepts. 1. Determine the y-intercept by setting x = 0 and finding the corresponding output value. 2. Determine the x-intercepts by solving for the input values that yield an output value of zero. Example 5.17 Determining the Intercepts of a Polynomial Function Given the polynomial function f (x) = (x − 2)(x + 1)(x − 4), written in factored form for your convenience, determine the y- and x-intercepts. 514 Chapter 5 Polynomial and Rational Functions Solution The y-intercept occurs when the input is zero so substitute 0 for x. f (0) = (0)4 − 4(0)2 − 45 = −45 The y-intercept is (0, 8). The x-intercepts occur when the output is zero. x − 2 = 0 x = 2 or or 0 = (x − 2)(x + 1)(x − 4) x + 1 = 0 x = −1 or or x − 4 = 0 x = 4 The x-intercepts are (2, 0), ( – 1, 0), and (4, 0). We can see these intercepts on the graph of the function shown in Figure 5.27. Figure 5.27 Example 5.18 Determining the Intercepts of a Polynomial Function with Factoring Given the polynomial function f (x) = x4 − 4x2 − 45, determine the y- and x-intercepts. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 515 Solution The y-intercept occurs when the input is zero. f (0) = (0)4 − 4(0)2 − 45 = −45 The y-intercept is (0, −45). The x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial. = f (x) = x4 − 4x2 − 45 ⎞ ⎛ ⎞ ⎛ ⎝x2 + 5 ⎝x2 − 9 ⎠ ⎠ ⎞ ⎛ ⎝x2 + 5 = (x − 3)(x + 3) ⎠ ⎛ ⎞ ⎝x2 + 5 0 = (x − 3)(x + 3 or or x + 3 = 0 x = −3 or or x2 + 5 = 0 (no real solution) The x-intercepts are (3, 0) and (–3, 0). We can see these intercepts on the graph of the function shown in Figure 5.28. We can see that the function is even because f (x) = f (−x). Figure 5.28 5.11 Given the polynomial function f (x) = 2x3 − 6x2 − 20x, determine the y- and x-intercepts. Comparing Smooth and Continuous Graphs The degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function of nth degree is the product of n factors, so it will have at most n roots or zeros, or x-intercepts. The graph of the polynomial function of degree n must have at most n – 1 turning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors. 516 Chapter 5 Polynomial and Rational Functions A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth. Intercepts and Turning Points of Polynomials A polynomial of degree n will have, at most, n x-intercepts and n − 1 turning points. Example 5.19 Determining the Number of Intercepts and Turning Points of a Polynomial Without graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for f (x) = − 3x10 + 4x7 − x4 + 2x3. Solution The polynomial has a degree of 10, so there are at most 10 x-intercepts and at most 9 turning points. Without graphing the function, determine the maximum number of x-intercepts and turning points for 5.12 f (x) = 108 − 13x9 − 8x4 + 14x12 + 2x3. Example 5.20 Drawing Conclusions about a Polynomial Function from the Graph What can we conclude about the polynomial represented by the graph shown in Figure 5.29 based on its intercepts and turning points? Figure 5.29 Solution The end behavior of the graph tells us this is the graph of an even-degree polynomial. See Figure 5.30. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 517 Figure 5.30 The graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4. What can we conclude about the polynomial represented by the graph shown in Figure 5.31 based on its 5.13 intercepts and turning points? Figure 5.31 Example 5.21 Drawing Conclusions about a Polynomial Function from the Factors Given the function f (x) = − 4x(x + 3)(x − 4), determine the local behavior. 518 Chapter 5 Polynomial and Rational Functions Solution The y-intercept is found by evaluating f (0). f (0) = −4(0)(0 + 3)(0 − 4 = 0 The y-intercept is (0, 0). The x-intercepts are found by determining the zeros of the function. 0 = −4x(x + 3)(x − 4) x = 0 x = 0 or or x + 3 = 0 x = −3 or or x − 4 = 0 = 4 x The x-intercepts are (0, 0), (–3, 0), and (4, 0). The degree is 3 so the graph has at most 2 turning points. 5.14 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), determine the local behavior. Access these online resources for additional instruction and practice with power and polinomial functions. • Find Key Information about a Given Polynomial Function (http://openstaxcollege.org/l/ keyinfopoly) • End Behavior of a Polynomial Function (http://openstaxcollege.org/l/endbehavior) • Turning Points and x-intercepts of Polynomial Functions (http://openstaxcollege.org/l/ turningpoints) • Least Possible Degree of a Polynomial Function (http://openstaxcollege.org/l/ leastposdegree) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 519 5.2 EXERCISES Verbal Explain the difference between the coefficient of a 76. power function and its degree. 77. If a polynomial function is in factored form, what would be a good first step in order to determine the degree of the function? In general, explain the end behavior of a power is 78. function with odd degree if the leading coefficient positive. What 79. is the relationship between the degree of a polynomial function and the maximum number of turning points in its graph? 80. What can we conclude if, in general, the graph of a polynomial function exhibits the following end behavior? As x → − ∞, f (x) → − ∞ as x → ∞, f (x) → − ∞. and Algebraic For the following exercises, identify the function as a power function, a polynomial function, or neither. 81. 82. 83. 84. f (x) = x5 f (x) = 3 ⎛ ⎝x2⎞ ⎠ f (x) = x − x4 f (x) = x2 x2 − 1 85. f (x) = 2x(x + 2)(x − 1)2 86. f (x) = 3 x + 1 For the following exercises, determine the end behavior of the functions. 92. 93. 94. 95. 96. 97. 98. 99. f (x) = x4 f (x) = x3 f (x) = − x4 f (x) = − x9 f (x) = − 2x4 − 3x2 + x − 1 f (x) = 3x2 + x − 2 f (x) = x2 ⎛ ⎞ ⎝2x3 − x + 1 ⎠ f (x) = (2 − x)7 For the following exercises, find the intercepts of the functions. 100. f (t) = 2(t − 1)(t + 2)(t − 3) 101. g(n) = −2(3n − 1)(2n + 1) 102. f (x) = x4 − 16 103. f (x) = x3 + 27 104. f (x) = x⎛ ⎞ ⎝x2 − 2x − 8 ⎠ 105. ⎛ ⎝4x2 − 1 f (x) = (x + 3) ⎞ ⎠ Graphical For the following exercises, find the degree and leading coefficient for the given polynomial. For the following exercises, determine the least possible degree of the polynomial function shown. 87. −3x 88. 89. 90. 7 − 2x2 −2x2 − 3x5 + x − 6 ⎝4 − x2⎞ x⎛ ⎠(2x + 1) 91. x2 (2x − 3)2 106. 107. 520 Chapter 5 Polynomial and Rational Functions 108. 109. 110. 111. 112. 113. For the following exercises, determine whether th
e graph of the function provided is a graph of a polynomial function. If so, determine the number of turning points and the least possible degree for the function. 114. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 521 115. 119. 116. 120. 117. Numeric For the following exercises, make a table to confirm the end behavior of the function. 121. f (x) = − x3 122. f (x) = x4 − 5x2 123. f (x) = x2 (1 − x)2 124. f (x) = (x − 1)(x − 2)(3 − x) 125. f (x) = x5 10 − x4 118. 522 Technology For the following exercises, graph the polynomial functions using a calculator. Based on the graph, determine the intercepts and the end behavior. 126. f (x) = x3(x − 2) 127. f (x) = x(x − 3)(x + 3) 128. f (x) = x(14 − 2x)(10 − 2x) 129. f (x) = x(14 − 2x)(10 − 2x)2 130. f (x) = x3 − 16x 131. f (x) = x3 − 27 132. f (x) = x4 − 81 133. 134. 135. f (x) = − x3 + x2 + 2x f (x) = x3 − 2x2 − 15x f (x) = x3 − 0.01x Extensions For the following exercises, use the information about the graph of a polynomial function to determine the function. Assume the leading coefficient is 1 or –1. There may be more than one correct answer. The y- intercept 136. ( − 2, 0), (2, 0). Degree is 2. is (0, − 4). The x- intercepts are End as x → − ∞, f (x) → ∞, as x → ∞, f (x) → ∞. behavior: Chapter 5 Polynomial and Rational Functions End as x → − ∞, f (x) → ∞, as x → ∞, f (x) → − ∞. behavior: 140. The y- intercept is (0, 1). There is no x- intercept. Degree is 4. End as x → − ∞, f (x) → ∞, as x → ∞, f (x) → ∞. behavior: Real-World Applications For the following exercises, use the written statements to construct a polynomial function that represents the required information. An oil slick is expanding as a circle. The radius of the 141. circle is increasing at the rate of 20 meters per day. Express the area of the circle as a function of d, the number of days elapsed. A cube has an edge of 3 feet. The edge is increasing at 142. the rate of 2 feet per minute. Express the volume of the cube as a function of m, the number of minutes elapsed. A rectangle has a length of 10 inches and a width of 6 143. inches. If the length is increased by x inches and the width increased by twice that amount, express the area of the rectangle as a function of x. An open box is to be constructed by cutting out square 144. corners of x- inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of x. A rectangle is twice as long as it is wide. Squares of 145. side 2 feet are cut out from each corner. Then the sides are folded up to make an open box. Express the volume of the box as a function of the width ( x ). The y- intercept 137. ( − 3, 0), (3, 0). Degree is 2. is (0, 9). The x- intercepts are behavior: End as x → − ∞, f (x) → − ∞, as x → ∞, f (x) → − ∞. The y- intercept 138. (0, 0), (2, 0). Degree is 3. is (0, 0). The x- intercepts are End as x → − ∞, f (x) → − ∞, as x → ∞, f (x) → ∞. behavior: The y- intercept 139. (1, 0). Degree is 3. is (0, 1). The x- intercept is This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 523 5.3 \| Graphs of Polynomial Functions Learning Objectives In this section, you will: 5.3.1 Recognize characteristics of graphs of polynomial functions. 5.3.2 Use factoring to ﬁnd zeros of polynomial functions. 5.3.3 Identify zeros and their multiplicities. 5.3.4 Determine end behavior. 5.3.5 Understand the relationship between degree and turning points. 5.3.6 Graph polynomial functions. 5.3.7 Use the Intermediate Value Theorem. The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 5.3. Year 2006 2007 2008 2009 2010 2011 2012 2013 Revenues 52.4 52.8 51.2 49.5 48.6 48.6 48.7 47.1 Table 5.3 The revenue can be modeled by the polynomial function R(t) = − 0.037t 4 + 1.414t 3 − 19.777t 2 + 118.696t − 205.332 where R represents the revenue in millions of dollars and t represents the year, with t = 6 corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general. Recognizing Characteristics of Graphs of Polynomial Functions Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 5.32 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial. 524 Chapter 5 Polynomial and Rational Functions Figure 5.32 Example 5.22 Recognizing Polynomial Functions Which of the graphs in Figure 5.33 represents a polynomial function? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 525 Figure 5.33 Solution The graphs of f and h are graphs of polynomial functions. They are smooth and continuous. The graphs of g and k are graphs of functions that are not polynomials. The graph of function g has a sharp corner. The graph of function k is not continuous. Do all polynomial functions have as their domain all real numbers? Yes. Any real number is a valid input for a polynomial function. Using Factoring to Find Zeros of Polynomial Functions Recall that if f is a polynomial function, the values of x for which f (x) = 0 are called zeros of f . If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros. We can use this method to find x- intercepts because at the x- intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases: 526 Chapter 5 Polynomial and Rational Functions 1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring. 2. The polynomial is given in factored form. 3. Technology is used to determine the intercepts. Given a polynomial function f, find the x-intercepts by factoring. 1. Set f (x) = 0. 2. If the polynomial function is not given in factored form: a. Factor out any common monomial factors. b. Factor any factorable binomials or trinomials. 3. Set each factor equal to zero and solve to find the x- intercepts. Example 5.23 Finding the x-Intercepts of a Polynomial Function by Factoring Find the x-intercepts of f (x) = x6 − 3x4 + 2x2. Solution We can attempt to factor this polynomial to find solutions for f (x) = 0. x2 − 3x4 + 2x2 = 0 x2⎛ ⎞ ⎝x4 − 3x2 + 2 ⎠ = 0 ⎛ ⎞ ⎞ ⎝x2 − 1 ⎝x2 − 2 ⎠ = 0 ⎠ x2⎛ Factor out the greatest common factor. Factor the trinomial. Set each factor equal to zero. x2 = 0 x = 0 ⎛ ⎞ ⎝x2 − 1 ⎠ = 0 x2 = 1 x = ±1 or ⎛ ⎞ ⎝x2 − 2 ⎠ = 0 x2 = 2 x = ± 2 or This gives us five x-intercepts: (0, 0), (1, 0), (−1, 0), ( 2, 0), and ( − 2, 0). See Figure 5.34. We can see that this is an even function because it is symmetric about the y-axis. Figure 5.34 Example 5.24 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 527 Finding the x-Intercepts of a Polynomial Function by Factoring Find the x-intercepts of f (x) = x3 − 5x2 − x + 5. Solution Find solutions for f (x) = 0 by factoring. x3 − 5x2 − x + 5 = 0 x2(x − 5) − (x − 5) = 0 ⎞ ⎠(x − 5) = 0 (x + 1)(x − 1)(x − 5x2 − 1 or x = −1 Factor by grouping. Factor out the common factor. Factor the diffe ence of squares. Set each factor equal to zero. x − 1 = 0 x = 1 or x − 5 = 0 x = 5 There are three x-intercepts: (−1, 0), (1, 0), and (5, 0). See Figure 5.35. Figure 5.35 Example 5.25 Finding the y- and x-Intercepts of a Polynomial in Factored Form Find the y- and x-intercepts of g(x) = (x − 2)2(2x + 3). Solution The y-intercept can be found by evaluating g(0). g(0) = (0 − 2)2(2(0) + 3) = 12 528 Chapter 5 Polynomial and Rational Functions So the y-intercept is (0, 12). The x-intercepts can be found by solving g(x) = 0. (x − 2)2(2x + 3) = 0 (x − 2)2 = 0 x − 2 = 0 or (2x + 3) = 0 x = − 3 2 So the x-intercepts are (2, 0) and ⎛ ⎝− 3 2 x = 2 ⎞ , 0 ⎠. Analysis We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5.36. Figure 5.36 Example 5.26 Finding the x-Intercepts of a Polynomial Function Using a Graph This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 529 Find the x-intercepts of h(x) = x3 + 4x2 + x − 6. Solution This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities. Looking at the graph of this function, as shown in Figure 5.37, it appears that there are x-intercepts at x = −3, −2, and 1. Figure 5.37 We can check whether these are correct by substituting these values for x and verifying that h( − 3) = h( − 2) = h(1) = 0 Since h(x) = x3 + 4x2 + x − 6, we have: h(−3) = (−3)3 + 4(−3)2 + (−3) − 6 = −27 + 36 − 3 − 6 = 0 h(−2) = (−2)3 + 4(−2)2 + (−2) − 6 = −8 + 16 − 2 − 6 = 0 h(1) = (1)3 + 4(1)2 + (1 Each x-intercept corresponds to a zero of the polynomial function and each zero yields a
factor, so we can now write the polynomial in factored form. h(x) = x3 + 4x2 + x − 6 = (x + 3)(x + 2)(x − 1) 5.15 Find the y- and x-intercepts of the function f (x) = x4 − 19x2 + 30x. 530 Chapter 5 Polynomial and Rational Functions Identifying Zeros and Their Multiplicities Graphs behave differently at various x-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and "bounce" off. Suppose, for example, we graph the function shown. f (x) = (x + 3)(x − 2)2 (x + 1)3 Notice in Figure 5.38 that the behavior of the function at each of the x-intercepts is different. Figure 5.38 Identifying the behavior of the graph at an xintercept by examining the multiplicity of the zero. is the solution of equation (x + 3) = 0. The graph passes directly through the x-intercept at The x-intercept x = −3 x = −3. The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function. The x-intercept x = 2 is the repeated solution of equation (x − 2)2 = 0. The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. (x − 2)2 = (x − 2)(x − 2) The factor is repeated, that is, the factor (x − 2) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, x = 2, has multiplicity 2 because the factor (x − 2) occurs twice. The x-intercept x = − 1 is the repeated solution of factor (x + 1)3 = 0. The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function f (x) = x3. We call this a triple zero, or a zero with multiplicity 3. For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure 5.39 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 531 Figure 5.39 For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis. For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis. Graphical Behavior of Polynomials at x-Intercepts If a polynomial contains a factor of the form (x − h) power p. We say that x = h is a zero of multiplicity p. p , the behavior near the x- intercept h is determined by the The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities. The sum of the multiplicities is the degree of the polynomial function. Given a graph of a polynomial function of degree n, identify the zeros and their multiplicities. 1. 2. 3. If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. 4. The sum of the multiplicities is n. Example 5.27 Identifying Zeros and Their Multiplicities Use the graph of the function of degree 6 in Figure 5.40 to identify the zeros of the function and their possible multiplicities. 532 Chapter 5 Polynomial and Rational Functions Figure 5.40 Solution The polynomial function is of degree 6. The sum of the multiplicities must be 6. Starting from the left, the first zero occurs at x = −3. The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of −3 most likely has multiplicity 2. The next zero occurs at x = −1. The graph looks almost linear at this point. This is a single zero of multiplicity 1. The last zero occurs at x = 4. The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is 6. Use the graph of the function of degree 5 in Figure 5.41 to identify the zeros of the function and their 5.16 multiplicities. Figure 5.41 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 533 Determining End Behavior As we have already learned, the behavior of a graph of a polynomial function of the form f (x) = an xn + an − 1 xn − 1 + ... + a1 x + a0 will either ultimately rise or fall as x increases without bound and will either rise or fall as x decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000. Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, an xn , is an even power function, as x increases or decreases without bound, f (x) increases without bound. When the leading term is an odd power function, as x decreases without bound, f (x) also decreases without bound; as x increases without bound, f (x) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 5.42 summarizes all four cases. Figure 5.42 Understanding the Relationship between Degree and Turning Points In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). 534 Chapter 5 Polynomial and Rational Functions Look at the graph of the polynomial function f (x) = x4 − x3 − 4x2 + 4x in Figure 5.43. The graph has three turning points. Figure 5.43 This function f is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function. Interpreting Turning Points A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree n will have at most n − 1 turning points. Example 5.28 Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function Find the maximum number of turning points of each polynomial function. a. b. f (x) = − x3 + 4x5 − 3x2 + 1 f (x) = − (x − 1)2 ⎛ ⎝1 + 2x2⎞ ⎠ Solution a. First, rewrite the polynomial function in descending order: f (x) = 4x5 − x3 − 3x2 + 1 Identify the degree of the polynomial function. This polynomial function is of degree 5. The maximum number of turning points is 5 − 1 = 4. b. First, identify the leading term of the polynomial function if the function were expanded. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 535 Then, identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is 4 − 1 = 3. Graphing Polynomial Functions We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions. Given a polynomial function, sketch the graph. 1. Find the intercepts. 2. Check for symmetry. If the function is an even function, its graph is symmetrical about the y- axis, that is, f (−x) = f (x). If a function is an odd function, its graph is symmetrical about the origin, that is, f (−x) = − f (x). 3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the x- intercepts. 4. Determine the end behavior by examining the leading term. 5. Use the end behavior and the behavior at the intercepts to sketch a graph. 6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial. 7. Optionally, use technology to check the graph. Example 5.29 Sketching the Graph of a Polynomial Function Sketch a graph of f (x) = −2(x + 3)2(x − 5). Solution This graph has two x-intercepts. At x = −3, the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-intercept. At x = 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept. The y-intercept is found by evaluating f (0). f (0) = −2(0 + 3)2(0 − 5) = −2 ⋅ 9 ⋅ (−5) = 90 The y-intercept is (0, 90). Additionally, we can see the leading term, if this polynomial were multiplied out, would be − 2x3, so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 5.44. 536 Chapter 5 Polynomial and Rational Functions Figure 5.44 To sketch this, we consider that: • As x → − ∞ the function f (x) → ∞, so we know the graph starts in the second quadrant and is decreasing toward the x- axis. • Since f (−x) = −2(−x + 3)2 (−x – 5) is not equal to f (x), the graph does not display symmetry. • At (−3, 0), the graph bounces off of the x-axis, so the function must start increasing. At (0, 90), the graph crosses the y-axis at the y-intercept. See Figure 5.45. Figure 5.45 Somewhere after this
point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at (5, 0). See Figure 5.46. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 537 Figure 5.46 As x → ∞ the function f (x) → −∞, so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant. Using technology, we can create the graph for the polynomial function, shown in Figure 5.47, and verify that the resulting graph looks like our sketch in Figure 5.46. Figure 5.47 The complete graph of the polynomial function f (x) = − 2(x + 3)2(x − 5) 5.17 Sketch a graph of f (x) = 1 4 x(x − 1)4 (x + 3)3. Using the Intermediate Value Theorem In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial function f whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers a and b in the domain of f , if a < b then the function f takes on every value between f (a) and f (b). (While the theorem is intuitive, the and f (a) ≠ f (b), 538 Chapter 5 Polynomial and Rational Functions proof is actually quite complicated and requires higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function f at x = a lies above the x- axis and another point at x = b lies below the x- axis, there must exist a third point between x = a and x = b where the graph crosses the x- axis. Call this point ⎛ ⎠. This means that we are assured there is a solution c where f (c) = 0. ⎝c, f (c)⎞ In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x- axis. Figure 5.48 shows that there is a zero between a and b. Figure 5.48 Using the Intermediate Value Theorem to show there exists a zero. Intermediate Value Theorem Let f be a polynomial function. The Intermediate Value Theorem states that if f (a) and f (b) have opposite signs, then there exists at least one value c between a and b for which f (c) = 0. Example 5.30 Using the Intermediate Value Theorem Show that the function f (x) = x3 − 5x2 + 3x + 6 has at least two real zeros between x = 1 and x = 4. Solution As a start, evaluate f (x) at the integer values x = 1, 2, 3, and 4. See Table 5.4. x f(x) 1 5 2 0 3 –3 4 2 Table 5.4 We see that one zero occurs at x = 2. Also, since f (3) is negative and f (4) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 539 We have shown that there are at least two real zeros between x = 1 and x = 4. Analysis We can also see on the graph of the function in Figure 5.49 that there are two real zeros between x = 1 and x = 4. Figure 5.49 5.18 Show that the function f (x) = 7x5 − 9x4 − x2 has at least one real zero between x = 1 and x = 2. Writing Formulas for Polynomial Functions Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors. Factored Form of Polynomials If a polynomial of lowest degree p has horizontal intercepts at x = x1, x2, … , xn, then the polynomial can be pn where the powers pi on each factor can written in the factored form: f (x) = a(x − x1) be determined by the behavior of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the x-intercept. p2 ⋯ (x − xn) p1 (x − x2) 540 Chapter 5 Polynomial and Rational Functions Given a graph of a polynomial function, write a formula for the function. 1. Identify the x-intercepts of the graph to find the factors of the polynomial. 2. Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor. 3. Find the polynomial of least degree containing all the factors found in the previous step. 4. Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor. Example 5.31 Writing a Formula for a Polynomial Function from the Graph Write a formula for the polynomial function shown in Figure 5.50. Figure 5.50 Solution This graph has three x-intercepts: x = −3, 2, and 5. The y-intercept is located at (0, 2). At x = −3 and x = 5, the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At x = 2, the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us f (x) = a(x + 3)(x − 2)2(x − 5) To determine the stretch factor, we utilize another point on the graph. We will use the y- intercept (0, – 2), to solve for a. f (0) = a(0 + 3)(0 − 2)2(0 − 5) −2 = a(0 + 3)(0 − 2)2(0 − 5) −2 = −60a a = 1 30 The graphed polynomial appears to represent the function f (x) = 1 30 (x + 3)(x − 2)2(x − 5). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 541 5.19 Given the graph shown in Figure 5.51, write a formula for the function shown. Figure 5.51 Using Local and Global Extrema With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph. Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function. Local and Global Extrema A local maximum or local minimum at x = a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around x = a. If a function has a local maximum at a, then f (a) ≥ f (x) for all x in an open interval around x = a. If a function has a local minimum at a, then f (a) ≤ f (x) for all x in an open interval around x = a. A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at a, then f (a) ≥ f (x) for all x. If a function has a global minimum at a, then f (a) ≤ f (x) for all x. We can see the difference between local and global extrema in Figure 5.52. 542 Chapter 5 Polynomial and Rational Functions Figure 5.52 Do all polynomial functions have a global minimum or maximum? No. Only polynomial functions of even degree have a global minimum or maximum. For example, f (x) = x has neither a global maximum nor a global minimum. Example 5.32 Using Local Extrema to Solve Applications An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic and then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box. Solution We will start this problem by drawing a picture like that in Figure 5.53, labeling the width of the cut-out squares with a variable, w. Figure 5.53 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 543 Notice that after a square is cut out from each end, it leaves a (14 − 2w) cm by (20 − 2w) cm rectangle for the base of the box, and the box will be w cm tall. This gives the volume V(w) = (20 − 2w)(14 − 2w)w = 280w − 68w2 + 4w3 Notice, since the factors are w, 20 – 2w and 14 – 2w, the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values w may take on are greater than zero or less than 7. This means we will restrict the domain of this function to 0 < w < 7. Using technology to sketch the graph of V(w) on this reasonable domain, we get a graph like that in Figure 5.54. We can use this graph to estimate the maximum value for the volume, restricted to values for w that are reasonable for this problem—values from 0 to 7. Figure 5.54 From this graph, we turn our focus to only the portion on the reasonable domain, [0, 7]. We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 5.55. 544 Chapter 5 Polynomial and Rational Functions Figure 5.55 From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side. 5.20 Use technology to find the maximum and minimum values on the interval [−1, 4] of the function f (x) = − 0.2(x − 2)3 (x + 1)2(x − 4). Access the following online resource for additional instruction and practice with graphing polynomial functions. • Intermediate Value Theorem (http://openstaxcollege.org/l/ivt) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 545 5.3 EXERCISES Verbal What is the difference between an x- intercept and a 146. zero of a polynom
ial function f ? If a polynomial function of degree n has n distinct 147. zeros, what do you know about the graph of the function? Explain how the Intermediate Value Theorem can 148. assist us in finding a zero of a function. Explain how the factored form of the polynomial 149. helps us in graphing it. 150. If the graph of a polynomial just touches the x-axis and then changes direction, what can we conclude about the factored form of the polynomial? Algebraic For the following exercises, find the x- or t-intercepts of the polynomial functions. 151. C(t) = 2(t − 4)(t + 1)(t − 6) 152. C(t) = 3(t + 2)(t − 3)(t + 5) 153. 154. 155. 156. C(t) = 4t(t − 2)2(t + 1) C(t) = 2t(t − 3)(t + 1)2 C(t) = 2t 4 − 8t 3 + 6t 2 C(t) = 4t 4 + 12t 3 − 40t 2 157. f (x) = x4 − x2 f (x) = x3 + x2 − 20x f (x) = x3 + 6x2 − 7x f (x) = x3 + x2 − 4x − 4 f (x) = x3 + 2x2 − 9x − 18 f (x) = 2x3 − x2 − 8x + 4 f (x) = x6 − 7x3 − 8 f (x) = 2x4 + 6x2 − 8 158. 159. 160. 161. 162. 163. 164. 165. f (x) = x3 − 3x2 − x + 3 166. 167. 168. f (x) = x6 − 2x4 − 3x2 f (x) = x6 − 3x4 − 4x2 f (x) = x5 − 5x3 + 4x For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval. 169. 170. 171. 172. 173. 174. f (x) = x3 − 9x, between x = −4 and x = −2. f (x) = x3 − 9x, between x = 2 and x = 4. f (x) = x5 − 2x, between x = 1 and x = 2. f (x) = − x4 + 4, between x = 1 and x = 3 . f (x) = −2x3 − x, between x = –1 and x = 1. f (x) = x3 − 100x + 2, between x = 0.01 and x = 0.1 For the following exercises, find the zeros and give the multiplicity of each. 175. 176. 177. f (x) = (x + 2)3 (x − 3)2 f (x) = x2 (2x + 3)5 (x − 4)2 f (x) = x3 (x − 1)3 (x + 2) 178. f (x) = x2 ⎛ ⎞ ⎝x2 + 4x + 4 ⎠ 179. f (x) = (2x + 1)3 ⎛ ⎞ ⎝9x2 − 6x + 1 ⎠ 180. 181. 182. 183. 184. f (x) = (3x + 2)5 ⎛ ⎞ ⎝x2 − 10x + 25 ⎠ f (x) = x⎛ ⎞ ⎞ ⎛ ⎝x2 + 8x + 16 ⎝4x2 − 12x + 9 ⎠ ⎠ f (x) = x6 − x5 − 2x4 f (x) = 3x4 + 6x3 + 3x2 f (x) = 4x5 − 12x4 + 9x3 546 Chapter 5 Polynomial and Rational Functions 185. f (x) = 2x4 ⎛ ⎝x3 − 4x2 + 4x⎞ ⎠ 186. f (x) = 4x4 ⎛ ⎝9x4 − 12x3 + 4x2⎞ ⎠ Graphical For the following exercises, graph the polynomial functions. Note x- and y- intercepts, multiplicity, and end behavior. 187. 188. 189. 190. f (x) = (x + 3)2(x − 2) g(x) = (x + 4)(x − 1)2 h(x) = (x − 1)3 (x + 3)2 k(x) = (x − 3)3 (x − 2)2 191. m(x) = − 2x(x − 1)(x + 3) 192. n(x) = − 3x(x + 2)(x − 4) For the following exercises, use the graphs to write the formula for a polynomial function of least degree. 193. 194. 195. 196. 197. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 547 For the following exercises, use the graph to identify zeros and multiplicity. 201. 198. 199. 200. For the following exercises, use the given information about the polynomial graph to write the equation. Degree 3. Zeros at x = –2, 202. intercept at (0, –4). x = 1, and x = 3. y- Degree 3. Zeros at x = –5, 203. intercept at (0, 6) x = –2, and x = 1. y- Degree 5. Roots of multiplicity 2 at x = 3 and 204. x = 1 , and a root of multiplicity 1 at x = –3. y-intercept at (0, 9) Degree 4. Root of multiplicity 2 at x = 4, and a roots at 205. of multiplicity 1 at x = 1 and x = –2. y-intercept (0, –3). Degree 5. Double zero at x = 1, and triple zero at 206. x = 3. Passes through the point (2, 15). Degree 3. Zeros at x = 4, 207. intercept at (0, −24). x = 3, and x = 2. y- 548 Chapter 5 Polynomial and Rational Functions Degree 3. Zeros at x = −3, 208. y-intercept at (0, 12). x = −2 and x = 1. Degree 5. Roots of multiplicity 2 at x = −3 and 209. x = 2 and a root of multiplicity 1 at x = −2. y-intercept at (0, 4). 210. Degree 4. Roots of multiplicity 2 at x = 1 2 and roots of multiplicity 1 at x = 6 and x = −2. y-intercept at (0,18). Double zero at x = −3 and triple zero at x = 0. 211. Passes through the point (1, 32). Technology For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum. 212. 213. f (x) = x3 − x − 1 f (x) = 2x3 − 3x − 1 214. f (x) = x4 + x 215. 216. f (x) = − x4 + 3x − 2 f (x) = x4 − x3 + 1 Extensions For the following exercises, use the graphs to write a polynomial function of least degree. 219. 217. 218. This content is available for free at https://cnx.org/content/col11758/1.5 Real-World Applications For the following exercises, write the polynomial function that models the given situation. A rectangle has a length of 10 units and a width of 8 220. units. Squares of x by x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a polynomial function in terms of x. Consider the preceding the same rectangle of 221. problem. Squares of 2x by 2x units are cut out of each corner. Express the volume of the box as a polynomial in terms of x. 222. Chapter 5 Polynomial and Rational Functions 549 A square has sides of 12 units. Squares x + 1 by x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of x. A cylinder has a radius of x + 2 units and a height of 223. 3 units greater. Express the volume of the cylinder as a polynomial function. A right circular cone has a radius of 3x + 6 and a 224. height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is V = 1 πr 2 h 3 for radius r and height h. 550 Chapter 5 Polynomial and Rational Functions 5.4 \| Dividing Polynomials Learning Objectives In this section, you will: 5.4.1 Use long division to divide polynomials. 5.4.2 Use synthetic division to divide polynomials. Figure 5.56 Lincoln Memorial, Washington, D.C. (credit: Ron Cogswell, Flickr) The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters (m), width 40 m, and height 30 m.[1] We can easily find the volume using elementary geometry. V = l ⋅ w ⋅ h = 61.5 ⋅ 40 ⋅ 30 = 73,800 So the volume is 73,800 cubic meters (m ³ ). Suppose we knew the volume, length, and width. We could divide to find the height. h = V l ⋅ w = 73,800 61.5 ⋅ 40 = 30 As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any, or all, of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x; the width is given by x − 2. To find the height of the solid, we can use polynomial division, which is the focus of this section. Using Long Division to Divide Polynomials We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division. 1. National Park Service. "Lincoln Memorial Building Statistics." http://www.nps.gov/linc/historyculture/lincolnmemorial-building-statistics.htm. Accessed 4/3/2014 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 551 Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic. dividend = (divisor ⋅ quotient) + remainder 178 = (3 ⋅ 59) + 1 = 177 + 1 = 178 We call this the Division Algorithm and will discuss it more formally after looking at an example. Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide 2x3 − 3x2 + 4x + 5 by x + 2 using the long division algorithm, it would look like this: We have found or 2x3 − 3x2 + 4x + 5 x + 2 = 2x2 − 7x + 18 − 31 x + 2 2x3 − 3x2 + 4x + 5 x + 2 = (x + 2)(2x2 − 7x + 18) − 31 We can identify the dividend, the divisor, the quotient, and the remainder. 552 Chapter 5 Polynomial and Rational Functions Writing the result in this manner illustrates the Division Algorithm. The Division Algorithm The Division Algorithm states that, given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x) , there exist unique polynomials q(x) and r(x) such that f (x) = d(x)q(x) + r(x) (5.4) q(x) is the quotient and r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). If r(x) = 0, then d(x) divides evenly into f (x). This means that, in this case, both d(x) and q(x) are factors of f (x). Given a polynomial and a binomial, use long division to divide the polynomial by the binomial. 1. Set up the division problem. 2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor. 3. Multiply the answer by the divisor and write it below the like terms of the dividend. 4. Subtract the bottom binomial from the top binomial. 5. Bring down the next term of the dividend. 6. Repeat steps 2–5 until reaching the last term of the dividend. 7. If the remainder is non-zero, express as a fraction using the divisor as the denominator. Example 5.33 Using Long Division to Divide a Second-Degree Polynomial Divide 5x2 + 3x − 2 by x + 1. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 553 The quotient is 5x − 2. The remainder is 0. We write the result as or 5x2 + 3x − 2 x + 1 = 5x − 2 5x2 + 3x − 2 = (x + 1)(5x − 2) Analysis This division problem had a r
emainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend. Example 5.34 Using Long Division to Divide a Third-Degree Polynomial Divide 6x3 + 11x2 − 31x + 15 by 3x − 2. Solution There is a remainder of 1. We can express the result as: 6x3 + 11x2 − 31x + 15 3x − 2 = 2x2 + 5x − 7 + 1 3x − 2 Analysis We can check our work by using the Division Algorithm to rewrite the solution. Then multiply. 554 Chapter 5 Polynomial and Rational Functions (3x − 2)(2x2 + 5x − 7) + 1 = 6x3 + 11x2 − 31x + 15 Notice, as we write our result, • • • • the dividend is 6x3 + 11x2 − 31x + 15 the divisor is 3x − 2 the quotient is 2x2 + 5x − 7 the remainder is 1 5.21 Divide 16x3 − 12x2 + 20x − 3 by 4x + 5. Using Synthetic Division to Divide Polynomials As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1. To illustrate the process, recall the example at the beginning of the section. Divide 2x3 − 3x2 + 4x + 5 by x + 2 using the long division algorithm. The final form of the process looked like this: There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem. Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the leading coefficient. We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is 2x2 – 7x + 18 and the remainder is –31. The process will be made more clear in Example 5.35. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 555 Synthetic Division Synthetic division is a shortcut that can be used when the divisor is a binomial in the form x − k where k is a real number. In synthetic division, only the coefficients are used in the division process. Given two polynomials, use synthetic division to divide. 1. Write k for the divisor. 2. Write the coefficients of the dividend. 3. Bring the lead coefficient down. 4. Multiply the lead coefficient by k. Write the product in the next column. 5. Add the terms of the second column. 6. Multiply the result by k. Write the product in the next column. 7. Repeat steps 5 and 6 for the remaining columns. 8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on. Example 5.35 Using Synthetic Division to Divide a Second-Degree Polynomial Use synthetic division to divide 5x2 − 3x − 36 by x − 3. Solution Begin by setting up the synthetic division. Write k and the coefficients. Bring down the lead coefficient. Multiply the lead coefficient by k. Continue by adding the numbers in the second column. Multiply the resulting number by k. Write the result in the next column. Then add the numbers in the third column. The result is 5x + 12. The remainder is 0. So x − 3 is a factor of the original polynomial. Analysis Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the remainder. (x − 3)(5x + 12) + 0 = 5x2 − 3x − 36 556 Chapter 5 Polynomial and Rational Functions Example 5.36 Using Synthetic Division to Divide a Third-Degree Polynomial Use synthetic division to divide 4x3 + 10x2 − 6x − 20 by x + 2. Solution The binomial divisor is x + 2 so k = −2. Add each column, multiply the result by –2, and repeat until the last column is reached. The result is 4x2 + 2x − 10. The remainder is 0. Thus, x + 2 is a factor of 4x3 + 10x2 − 6x − 20. Analysis The graph of the polynomial function f (x) = 4x3 + 10x2 − 6x − 20 x = k = −2. This confirms that x + 2 is a factor of 4x3 + 10x2 − 6x − 20. in Figure 5.57 shows a zero at Figure 5.57 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 557 Example 5.37 Using Synthetic Division to Divide a Fourth-Degree Polynomial Use synthetic division to divide − 9x4 + 10x3 + 7x2 − 6 by x − 1. Solution Notice there is no x-term. We will use a zero as the coefficient for that term. The result is − 9x3 + x2 + 8x + 8 + 2 x − 1 . 5.22 Use synthetic division to divide 3x4 + 18x3 − 3x + 40 by x + 7. Using Polynomial Division to Solve Application Problems Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example. Example 5.38 Using Polynomial Division in an Application Problem The volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x and the width is given by x − 2. Find the height, t, of the solid. Solution There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in Figure 5.58. Figure 5.58 We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid. 3x4 − 3x3 − 33x2 + 54x = 3x ⋅ (x − 2 558 Chapter 5 Polynomial and Rational Functions To solve for h, first divide both sides by 3x. 3x ⋅ (x − 2) ⋅ h 3x (x − 2)h = x3 − x2 − 11x + 18 = 3x4 − 3x3 − 33x2 + 54x 3x Now solve for h using synthetic division. h = x3 − x2 − 11x + 18 x − 2 The quotient is x2 + x − 9 and the remainder is 0. The height of the solid is x2 + x − 9. 5.23 The area of a rectangle is given by 3x3 + 14x2 − 23x + 6. The width of the rectangle is given by x + 6. Find an expression for the length of the rectangle. Access these online resources for additional instruction and practice with polynomial division. • Dividing a Trinomial by a Binomial Using Long Division (http://openstaxcollege.org/l/ dividetribild) • Dividing a Polynomial by a Binomial Using Long Division (http://openstaxcollege.org/l/ dividepolybild) • Ex 2: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd2) • Ex 4: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 559 5.4 EXERCISES Verbal If division of a polynomial by a binomial results in a 225. remainder of zero, what can be conclude? 243. ⎞ ⎛ ⎝3x3 − 2x2 + x − 4 ⎠ ÷ (x + 3) 244. ⎞ ⎛ ⎝−6x3 + x2 − 4 ⎠ ÷ (2x − 3) If a polynomial of degree n is divided by a binomial 226. of degree 1, what is the degree of the quotient? 245. ⎞ ⎛ ⎝2x3 + 7x2 − 13x − 3 ⎠ ÷ (2x − 3) Algebraic For the following exercises, use long division to divide. Specify the quotient and the remainder. 227. ⎞ ⎛ ⎝x2 + 5x − 1 ⎠ ÷ (x − 1) 228. ⎞ ⎛ ⎝2x2 − 9x − 5 ⎠ ÷ (x − 5) 229. ⎛ ⎞ ⎝3x2 + 23x + 14 ⎠ ÷ (x + 7) 230. ⎛ ⎞ ⎝4x2 − 10x + 6 ⎠ ÷ (4x + 2) 231. ⎛ ⎞ ⎝6x2 − 25x − 25 ⎠ ÷ (6x + 5) 232. ⎛ ⎞ ⎝−x2 − 1 ⎠ ÷ (x + 1) 233. ⎛ ⎞ ⎝2x2 − 3x + 2 ⎠ ÷ (x + 2) 234. ⎞ ⎛ ⎝x3 − 126 ⎠ ÷ (x − 5) 235. ⎛ ⎞ ⎝3x2 − 5x + 4 ⎠ ÷ (3x + 1) 236. ⎞ ⎛ ⎝x3 − 3x2 + 5x − 6 ⎠ ÷ (x − 2) 237. ⎞ ⎛ ⎝2x3 + 3x2 − 4x + 15 ⎠ ÷ (x + 3) For the following exercises, use synthetic division to find the quotient. 238. ⎞ ⎛ ⎝3x3 − 2x2 + x − 4 ⎠ ÷ (x + 3) 239. ⎞ ⎛ ⎝2x3 − 6x2 − 7x + 6 ⎠ ÷ (x − 4) 240. ⎞ ⎛ ⎝6x3 − 10x2 − 7x − 15 ⎠ ÷ (x + 1) 241. ⎞ ⎛ ⎝4x3 − 12x2 − 5x − 1 ⎠ ÷ (2x + 1) 242. ⎛ ⎞ ⎝9x3 − 9x2 + 18x + 5 ⎠ ÷ (3x − 1) 246. ⎞ ⎛ ⎝3x3 − 5x2 + 2x + 3 ⎠ ÷ (x + 2) 247. ⎞ ⎛ ⎝4x3 − 5x2 + 13 ⎠ ÷ (x + 4) 248. ⎞ ⎛ ⎝x3 − 3x + 2 ⎠ ÷ (x + 2) 249. ⎞ ⎛ ⎝x3 − 21x2 + 147x − 343 ⎠ ÷ (x − 7) 250. ⎞ ⎛ ⎝x3 − 15x2 + 75x − 125 ⎠ ÷ (x − 5) 251. ⎞ ⎛ ⎝9x3 − x + 2 ⎠ ÷ (3x − 1) 252. ⎞ ⎛ ⎝6x3 − x2 + 5x + 2 ⎠ ÷ (3x + 1) 253. ⎞ ⎛ ⎝x4 + x3 − 3x2 − 2x + 1 ⎠ ÷ (x + 1) 254. ⎛ ⎞ ⎝x4 − 3x2 + 1 ⎠ ÷ (x − 1) 255. ⎞ ⎛ ⎝x4 + 2x3 − 3x2 + 2x + 6 ⎠ ÷ (x + 3) 256. ⎞ ⎛ ⎝x4 − 10x3 + 37x2 − 60x + 36 ⎠ ÷ (x − 2) 257. ⎞ ⎛ ⎝x4 − 8x3 + 24x2 − 32x + 16 ⎠ ÷ (x − 2) 258. ⎞ ⎛ ⎝x4 + 5x3 − 3x2 − 13x + 10 ⎠ ÷ (x + 5) 259. ⎞ ⎛ ⎝x4 − 12x3 + 54x2 − 108x + 81 ⎠ ÷ (x − 3) 260. ⎞ ⎛ ⎝4x4 − 2x3 − 4x + 2 ⎠ ÷ (2x − 1) 261. ⎞ ⎛ ⎝4x4 + 2x3 − 4x2 + 2x + 2 ⎠ ÷ (2x + 1) For the following exercises, use synthetic division to determine whether the first expression is a factor of the second. If it is, indicate the factorization. 262. x − 2, 4x3 − 3x2 − 8x + 4 Chapter 5 Polynomial and Rational Functions 271. Factor is x2 + x + 1 560 263. 264. 265. 266. 267. x − 2, 3x4 − 6x3 − 5x + 10 x + 3, − 4x3 + 5x2 + 8 x − 2, 4x4 − 15x2 − 4 x − 1 2 x + 1 3 , 2x4 − x3 + 2x − 1 , 3x4 + x3 − 3x + 1 Graphical For the following exercises, use the graph of the thirddegree polynomial and one factor to write the factored form of the polynomial suggested by the graph. The leading coefficient is one. 268. Factor is x2 − x + 3 269. Factor is (x2 + 2x + 4) 272. Factor is x2 + 2x + 2 270. Factor is x2 + 2x + 5 For the following exercises, use synthetic division to find the quotient and remainder. 273. 274. 275. 4x3 − 33 x − 2 2x3 + 25 x + 3 3x3 + 2x − 5 x − 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 276. 277. −4x3 − x2 − 12 x + 4 x4 − 22 x + 2 Technology For the following exercises, use a calculator with CAS to
answer the questions. 278. Consider xk − 1 x − 1 with k = 1, 2, 3. What do you expect the result to be if k = 4 ? Length is 2x + 5, area is 4x3 + 10x2 + 6x + 15 290. Length is 3x – 4, area 6x4 − 8x3 + 9x2 − 9x − 4 561 is For the following exercises, use the given volume of a box and its length and width to express the height of the box algebraically. 291. Volume is 12x3 + 20x2 − 21x − 36, length is 2x + 3, width is 3x − 4. 292. Volume is 18x3 − 21x2 − 40x + 48, length is 279. Consider xk + 1 x + 1 for k = 1, 3, 5. What do you 3x – 4, width is 3x – 4. expect the result to be if k = 7 ? 280. Consider x4 − k 4 x − k for k = 1, 2, 3. What do you expect the result to be if k = 4 ? 281. Consider xk x + 1 with k = 1, 2, 3. What do you expect the result to be if k = 4 ? 293. Volume is 10x3 + 27x2 + 2x − 24, length is 5x – 4, width is 2x + 3. 294. Volume is 10x3 + 30x2 − 8x − 24, length is 2, width is x + 3. For the following exercises, use the given volume and radius of a cylinder to express the height of the cylinder algebraically. 282. Consider xk x − 1 with k = 1, 2, 3. What do you 295. Volume is π⎛ ⎞ ⎝25x3 − 65x2 − 29x − 3 ⎠, radius is expect the result to be if k = 4 ? 5x + 1. 296. Volume is π⎛ ⎞ ⎝4x3 + 12x2 − 15x − 50 ⎠, radius is 2x + 5. 297. Volume radius is x + 4. is π⎛ ⎞ ⎝3x4 + 24x3 + 46x2 − 16x − 32 ⎠, Extensions For the following exercises, use synthetic division to determine the quotient involving a complex number. 283. x + 1 x − i 284. x2 + 1 x − i 285. x + 1 x + i 286. 287. x2 + 1 x + i x3 + 1 x − i Real-World Applications For the following exercises, use the given length and area of a rectangle to express the width algebraically. Length is x + 5, area is 2x2 + 9x − 5. 288. 289. 562 Chapter 5 Polynomial and Rational Functions 5.5 \| Zeros of Polynomial Functions Learning Objectives In this section, you will: 5.5.1 Evaluate a polynomial using the Remainder Theorem. 5.5.2 Use the Factor Theorem to solve a polynomial equation. 5.5.3 Use the Rational Zero Theorem to find rational zeros. 5.5.4 Find zeros of a polynomial function. 5.5.5 Use the Linear Factorization Theorem to find polynomials with given zeros. 5.5.6 Use Descartes’ Rule of Signs. 5.5.7 Solve real-world applications of polynomial equations A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be? This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations. Evaluating a Polynomial Using the Remainder Theorem In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. If the polynomial is divided by x – k, the remainder may be found quickly by evaluating the polynomial function at k, that is, f (k) Let’s walk through the proof of the theorem. Recall that the Division Algorithm states that, given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x) , there exist unique polynomials q(x) and r(x) such that If the divisor, d(x), is x − k, this takes the form f (x) = d(x)q(x) + r(x) f (x) = (x − k)q(x) + r Since the divisor x − k is linear, the remainder will be a constant, r. And, if we evaluate this for x = k, we have f (k) = (k − k)q(k) + r = 0 ⋅ q(k) + r = r In other words, f (k) is the remainder obtained by dividing f (x) by x − k. The Remainder Theorem If a polynomial f (x) is divided by x − k, then the remainder is the value f (k). Given a polynomial function f, evaluate f(x) at x = k using the Remainder Theorem. 1. Use synthetic division to divide the polynomial by x − k. 2. The remainder is the value f (k). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 563 Example 5.39 Using the Remainder Theorem to Evaluate a Polynomial Use the Remainder Theorem to evaluate f (x) = 6x4 − x3 − 15x2 + 2x − 7 at x = 2. Solution To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by x − 2. The remainder is 25. Therefore, f (2) = 25. Analysis We can check our answer by evaluating f (2). f (x) = 6x4 − x3 − 15x2 + 2x − 7 f (2) = 6(2)4 − (2)3 − 15(2)2 + 2(2) − 7 = 25 5.24 Use the Remainder Theorem to evaluate f (x) = 2x5 − 3x4 − 9x3 + 8x2 + 2 at x = − 3. Using the Factor Theorem to Solve a Polynomial Equation The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm. f (x) = (x − k)q(x) + r If k is a zero, then the remainder r is f (k) = 0 and f (x) = (x − k)q(x) + 0 or f (x) = (x − k)q(x). Notice, written in this form, x − k is a factor of f (x). We can conclude if k is a zero of f (x), then x − k is a factor of f (x). Similarly, if x − k is a factor of f (x), then the remainder of the Division Algorithm f (x) = (x − k)q(x) + r is 0. This tells us that k is a zero. This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree n in the complex number system will have n zeros. We can use the Factor Theorem to completely factor a polynomial into the product of n factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial. The Factor Theorem According to the Factor Theorem, k is a zero of f (x) if and only if (x − k) is a factor of f (x). 564 Chapter 5 Polynomial and Rational Functions Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial. 1. Use synthetic division to divide the polynomial by (x − k). 2. Confirm that the remainder is 0. 3. Write the polynomial as the product of (x − k) and the quadratic quotient. 4. If possible, factor the quadratic. 5. Write the polynomial as the product of factors. Example 5.40 Using the Factor Theorem to Solve a Polynomial Equation Show that (x + 2) is a factor of x3 − 6x2 − x + 30. Find the remaining factors. Use the factors to determine the zeros of the polynomial. Solution We can use synthetic division to show that (x + 2) is a factor of the polynomial. The remainder is zero, so (x + 2) is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient: ⎛ ⎝x2 − 8x + 15 (x + 2) ⎞ ⎠ We can factor the quadratic factor to write the polynomial as (x + 2)(x − 3)(x − 5) By the Factor Theorem, the zeros of x3 − 6x2 − x + 30 are –2, 3, and 5. 5.25 Use the Factor Theorem to find the zeros of f (x) = x3 + 4x2 − 4x − 16 given that (x − 2) is a factor of the polynomial. Using the Rational Zero Theorem to Find Rational Zeros Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial Consider a quadratic function with two zeros, x = 2 5 and x = 3 4 . By the Factor Theorem, these zeros have factors associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 565 Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5 and 4. We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros. The Rational Zero Theorem The Rational Zero Theorem states that, if the polynomial f (x) = an xn coefficients, then every rational zero of f (x) has the form p factor of the leading coefficient an. + an − 1 xn − 1 + ... + a1 x + a0 has integer q where p is a factor of the constant term a0 and q is a When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. Given a polynomial function f(x), use the Rational Zero Theorem to find rational zeros. 1. Determine all factors of the constant term and all factors of the leading coefficient. 2. Determine all possible values of p q , where p is a factor of the constant term and q is a factor of the leading coefficient. Be sure to include both positive and negative candidates. 3. Determine which possible zeros are actual zeros by evaluating each case of f ( p q). Example 5.41 Listing All Possible Rational Zeros List all possible rational zeros of f (x) = 2x4 − 5x3 + x2 − 4. Solution The only possible rational zeros of f (x) are the quotients of the factors of the last term, –4, and the factors of the leading coefficient, 2. The constant term is –4; the factors of –4 are p = ±1, ±2, ±4. The leading coefficient is 2; the factors of 2 are q = ±1, ±2. If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by one of the factors of 2. 566 Chapter 5 Polynomial and Rational Functions Note that 2 2 = 1 and 4 2 = 2, which have already been listed. So we can shorten our list. p q = Factors of the last Factors of the fir t = ±1, ±2, ±4, ± 1 2 Exam
ple 5.42 Using the Rational Zero Theorem to Find Rational Zeros Use the Rational Zero Theorem to find the rational zeros of f (x) = 2x3 + x2 − 4x + 1. Solution The Rational Zero Theorem tells us that if p q is a zero of f (x), then p is a factor of 1 and q is a factor of 2. p q = factor of constant term factor of leading coefficie = factor of 1 factor of 2 The factors of 1 are ±1 and the factors of 2 are ±1 and ±2. The possible values for p are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for x in f (x). q are ±1 and ± 1 2 . These f (−1) = 2(−1)3 + (−1)2 − 4(−1) + 1 = 4 f (1) = 1(1)3 + (1)2 − 4(1) + 1 = 0 f ⎛ ⎝− ⎛ ⎝− ⎞ ⎠ ⎛ ⎝− ⎛ ⎝− Of those, −1, − 1 2 , and 1 2 are not zeros of f (x). 1 is the only rational zero of f (x). 5.26 Use the Rational Zero Theorem to find the rational zeros of f (x) = x3 − 5x2 + 2x + 1. Finding the Zeros of Polynomial Functions The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 567 Given a polynomial function f, use synthetic division to find its zeros. 1. Use the Rational Zero Theorem to list all possible rational zeros of the function. 2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate. 3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic. 4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula. Example 5.43 Finding the Zeros of a Polynomial Function with Repeated Real Zeros Find the zeros of f (x) = 4x3 − 3x − 1. Solution The Rational Zero Theorem tells us that if p q is a zero of f (x), then p is a factor of –1 and q is a factor of 4. P Q = factor of constant term factor of leading coefficie = factor of –1 factor of 4 The factors of –1 are ±1 and the factors of 4 are ±1, ±2, and ±4. The possible values for p q are ±1, ± 1 2 , and ± 1 4 . These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1. Dividing by (x − 1) gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as The quadratic is a perfect square. f (x) can be written as (x − 1)(4x2 + 4x + 1) (x − 1)(2x + 1)2 We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0. 2x + 1 = 0 x = − 1 2 The zeros of the function are 1 and − 1 2 with multiplicity 2. Analysis 568 Chapter 5 Polynomial and Rational Functions Look at the graph of the function f in Figure 5.59. Notice, at x = − 0.5, the graph bounces off the x-axis, indicating the even multiplicity (2,4,6…) for the zero − 0.5. At x = 1, the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero x = 1. Figure 5.59 Using the Fundamental Theorem of Algebra Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations. Suppose f is a polynomial function of degree four, and f (x) = 0. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it c1. By the Factor Theorem, we can write f (x) as a product of x − c1 and a polynomial quotient. Since x − c1 is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it c2. So we can write the polynomial quotient as a product of x − c2 and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of f (x). The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra states that, if f (x) is a polynomial of degree n > 0, then f (x) has at least one complex zero. We can use this theorem to argue that, if f (x) is a polynomial of degree n > 0, and a is a non-zero real number, then f (x) has exactly n linear factors where c1, c2, ..., cn are complex numbers. Therefore, f (x) has n roots if we allow for multiplicities. f (x) = a(x − c1)(x − c2)...(x − cn) Does every polynomial have at least one imaginary zero? No. Real numbers are a subset of complex numbers, but not the other way around. A complex number is not necessarily imaginary. Real numbers are also complex numbers. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 569 Example 5.44 Finding the Zeros of a Polynomial Function with Complex Zeros Find the zeros of f (x) = 3x3 + 9x2 + x + 3. Solution The Rational Zero Theorem tells us that if p q is a zero of f (x), then p is a factor of 3 and q is a factor of 3. p q = factor of constant term factor of leading coefficie = factor of 3 factor of 3 The factors of 3 are ±1 and ±3. The possible values for p q , and therefore the possible rational zeros for the function, are ±3, ±1, and ± 1 3 . We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3. Dividing by (x + 3) gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as We can then set the quadratic equal to 0 and solve to find the other zeros of the function. ⎛ ⎝3x2 + 1 (x + 3) ⎞ ⎠ 3x2 + 1 = 0 x2 = − The zeros of f (x) are –3 and ± i 3 3 . Analysis Look at the graph of the function f in Figure 5.60. Notice that, at x = −3, the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero x = –3. Also note the presence of the two turning points. This means that, since there is a 3rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of x = −3 is 1 and there are two complex solutions, which is what we found, or the multiplicity at x = −3 is three. Either way, our result is correct. 570 Chapter 5 Polynomial and Rational Functions Figure 5.60 5.27 Find the zeros of f (x) = 2x3 + 5x2 − 11x + 4. Using the Linear Factorization Theorem to Find Polynomials with Given Zeros A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree n will have n zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into n factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (x − c), where c is a complex number. Let f be a polynomial function with real coefficients, and suppose a + bi, b ≠ 0, is a zero of f (x). Then, by the Factor Theorem, x − (a + bi) is a factor of f (x). For f to have real coefficients, x − (a − bi) must also be a factor of f (x). This is true because any factor other than x − (a − bi), when multiplied by x − (a + bi), will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function f with real coefficients has a complex zero a + bi, then the complex conjugate a − bi must also be a zero of f (x). This is called the Complex Conjugate Theorem. Complex Conjugate Theorem According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form (x − c) , where c is a complex number. If the polynomial function f has real coefficients and a complex zero in the form a + bi, conjugate of the zero, a − bi, is also a zero. then the complex Given the zeros of a polynomial function f and a point (c, f(c)) on the graph of f, use the Linear Factorization Theorem to find the polynomial function. 1. Use the zeros to construct the linear factors of the polynomial. 2. Multiply the linear factors to expand the polynomial. 3. Substitute ⎛ ⎝c, f (c)⎞ ⎠ into the function to determine the leading coefficient. 4. Simplify. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 571 Example 5.45 Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros Find a fourth degree polynomial with real coefficients that has zeros of –3, 2, i, such that f (−2) = 100. Solution Because x = i is a zero, by the Complex Conjugate Theorem x = – i is also a zero. The polynomial must have factors of (x + 3), (x − 2), (x − i), and (x + i). Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors. f (x) = a(x + 3)(x − 2)(x − i)(x + i) f (x) = a⎛ f (x) = a⎛ ⎞ ⎞ ⎛ ⎝x2 + 1 ⎝x2 + x − 6 ⎠ ⎠ ⎞ ⎝x4 + x3 − 5x2 + x − 6 ⎠ We need to find a to ensure f ( – 2) = 100. Substitute x = – 2 and f (2) = 100 into f (x). So the polynomial function is or 100 = a((−2)4 + (−2)3 − 5(−2)2 + (−2) − 6) 100 = a(−20) −5 = a ⎞ ⎛ ⎝x4 + x3 − 5x2 + x − 6 f (x) = −5 ⎠ f (x) = − 5x4 − 5x3 + 25x2 − 5x + 30 A
nalysis We found that both i and − i were zeros, but only one of these zeros needed to be given. If i is a zero of a polynomial with real coefficients, then − i must also be a zero of the polynomial because − i is the complex conjugate of i. If 2 + 3i were given as a zero of a polynomial with real coefficients, would 2 − 3i also need to be a zero? Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial. 5.28 Find a third degree polynomial with real coefficients that has zeros of 5 and − 2i such that f (1) = 10. Using Descartes’ Rule of Signs There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in f (x) and the number of positive real zeros. For example, the polynomial function below has one sign change. This tells us that the function must have 1 positive real zero. 572 Chapter 5 Polynomial and Rational Functions There is a similar relationship between the number of sign changes in f (−x) and the number of negative real zeros. In this case, f (−x) has 3 sign changes. This tells us that f (x) could have 3 or 1 negative real zeros. Descartes’ Rule of Signs According to Descartes’ Rule of Signs, if we let f (x) = an xn + an − 1 xn − 1 + ... + a1 x + a0 be a polynomial function with real coefficients: • The number of positive real zeros is either equal to the number of sign changes of f (x) or is less than the number of sign changes by an even integer. • The number of negative real zeros is either equal to the number of sign changes of f (−x) or is less than the number of sign changes by an even integer. Example 5.46 Using Descartes’ Rule of Signs Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for f (x) = − x4 − 3x3 + 6x2 − 4x − 12. Solution Begin by determining the number of sign changes. There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine f ( − x) to determine the number of negative real roots. f ( − x) = −( − x)4 − 3( − x)3 + 6( − x)2 − 4( − x) − 12 f ( − x) = −x4 + 3x3 + 6x2 + 4x − 12 Again, there are two sign changes, so there are either 2 or 0 negative real roots. There are four possibilities, as we can see in Table 5.5. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 573 Positive Real Zeros Negative Real Zeros Complex Zeros Total Zeros 2 2 0 0 Table 5. Analysis We can confirm the numbers of positive and negative real roots by examining a graph of the function. See Figure 5.61. We can see from the graph that the function has 0 positive real roots and 2 negative real roots. Figure 5.61 5.29 Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for f (x) = 2x4 − 10x3 + 11x2 − 15x + 12. Use a graph to verify the numbers of positive and negative real zeros for the function. Solving Real-World Applications We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section. Example 5.47 Solving Polynomial Equations 574 Chapter 5 Polynomial and Rational Functions A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be? Solution Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by V = lwh. We were given that the length must be four inches longer than the width, so we can express the length of the cake as l = w + 4. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as h = 1 3 w. Let’s write the volume of the cake in terms of width of the cake. ⎛ V = (w + 4)(w) ⎝ w⎞ ⎠ 1 3 V = 1 3 w3 + 4 3 w2 Substitute the given volume into this equation. w3 + 4 351 = 1 3 3 1053 = w3 + 4w2 w2 Substitute 351 for V. Multiply both sides by 3. 0 = w3 + 7w2 − 1053 Subtract 1053 from both sides. Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are ± 3, ± 9, ± 13, ± 27, ± 39, ± 81, ± 117, ± 351, and ± 1053. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check x = 1. Since 1 is not a solution, we will check x = 3. Since 3 is not a solution either, we will test x = 9. Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan = 13 and h = 1 3 w = 1 3 (9) = 3 The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches. 5.30 A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 575 Access these online resources for additional instruction and practice with zeros of polynomial functions. • Real Zeros, Factors, and Graphs of Polynomial Functions (http://openstaxcollege.org/l/ realzeros) • Complex Factorization Theorem (http://openstaxcollege.org/l/factortheorem) • Find the Zeros of a Polynomial Function (http://openstaxcollege.org/l/findthezeros) • Find the Zeros of a Polynomial Function 2 (http://openstaxcollege.org/l/findthezeros2) • Find the Zeros of a Polynomial Function 3 (http://openstaxcollege.org/l/findthezeros3) 576 Chapter 5 Polynomial and Rational Functions 5.5 EXERCISES Verbal 298. Describe a use for the Remainder Theorem. Explain why the Rational Zero Theorem does not 299. guarantee finding zeros of a polynomial function. What 300. zeros? is the difference between rational and real 301. If Descartes’ Rule of Signs reveals a no change of signs or one sign of changes, what specific conclusion can be drawn? If synthetic division reveals a zero, why should we try 302. that value again as a possible solution? Algebraic For the following exercises, use the Remainder Theorem to find the remainder. 303. ⎛ ⎞ ⎝x4 − 9x2 + 14 ⎠ ÷ (x − 2) 304. ⎞ ⎛ ⎝3x3 − 2x2 + x − 4 ⎠ ÷ (x + 3) 305. ⎞ ⎛ ⎝x4 + 5x3 − 4x − 17 ⎠ ÷ (x + 1) 306. ⎛ ⎞ ⎝−3x2 + 6x + 24 ⎠ ÷ (x − 4) 307. ⎛ ⎞ ⎝5x5 − 4x4 + 3x3 − 2x2 + x − 1 ⎠ ÷ (x + 6) 308. ⎛ ⎞ ⎝x4 − 1 ⎠ ÷ (x − 4) 309. ⎞ ⎛ ⎝3x3 + 4x2 − 8x + 2 ⎠ ÷ (x − 3) 310. ⎞ ⎛ ⎝4x3 + 5x2 − 2x + 7 ⎠ ÷ (x + 2) For the following exercises, use the Factor Theorem to find all real zeros for the given polynomial function and one factor. 311. 312. 313. 314. 315. f (x) = 2x3 − 9x2 + 13x − 6; x − 1 f (x) = 2x3 + x2 − 5x + 2; x + 2 f (x) = 3x3 + x2 − 20x + 12; x + 3 f (x) = 2x3 + 3x2 + x + 6; x + 2 f (x) = − 5x3 + 16x2 − 9; x − 3 This content is available for free at https://cnx.org/content/col11758/1.5 316. 317. 318. x3 + 3x2 + 4x + 12; x + 3 4x3 − 7x + 3; x − 1 2x3 + 5x2 − 12x − 30, 2x + 5 For the following exercises, use the Rational Zero Theorem to find all real zeros. 319. 320. 321. 322. 323. 324. 325. 326. 327. 328. 329. 330. 331. 332. 333. 334. 335. 336. x3 − 3x2 − 10x + 24 = 0 2x3 + 7x2 − 10x − 24 = 0 x3 + 2x2 − 9x − 18 = 0 x3 + 5x2 − 16x − 80 = 0 x3 − 3x2 − 25x + 75 = 0 2x3 − 3x2 − 32x − 15 = 0 2x3 + x2 − 7x − 6 = 0 2x3 − 3x2 − x + 1 = 0 3x3 − x2 − 11x − 6 = 0 2x3 − 5x2 + 9x − 9 = 0 2x3 − 3x2 + 4x + 3 = 0 x4 − 2x3 − 7x2 + 8x + 12 = 0 x4 + 2x3 − 9x2 − 2x + 8 = 0 4x4 + 4x3 − 25x2 − x + 6 = 0 2x4 − 3x3 − 15x2 + 32x − 12 = 0 x4 + 2x3 − 4x2 − 10x − 5 = 0 4x3 − 3x + 1 = 0 8x 4 + 26x3 + 39x2 + 26x + 6 For the following exercises, find all complex solutions (real and non-real). 337. x3 + x2 + x + 1 = 0 Chapter 5 Polynomial and Rational Functions 577 338. 339. 340. 341. 342. x3 − 8x2 + 25x − 26 = 0 x3 + 13x2 + 57x + 85 = 0 3x3 − 4x2 + 11x + 10 = 0 x4 + 2x3 + 22x2 + 50x − 75 = 0 2x3 − 3x2 + 32x + 17 = 0 Graphical For the following exercises, use Descartes’ Rule to determine the possible number of positive and negative solutions. Confirm with the given graph. 343. f (x) = x3 − 1 344. 345. 346. 347. 348. 349. 350. 351. 352. f (x) = x4 − x2 − 1 f (x) = x3 − 2x2 − 5x + 6 f (x) = x3 − 2x2 + x − 1 f (x) = x4 + 2x3 − 12x2 + 14x − 5 f (x) = 2x3 + 37x2 + 200x + 300 f (x) = x3 − 2x2 − 16x + 32 f (x) = 2x4 − 5x3 − 5x2 + 5x + 3 f (x) = 2x4 − 5x3 − 14x2 + 20x + 8 f (x) = 10x4 − 21x2 + 11 Numeric For the following exercises, list all possible rational zeros for the functions. 353. 354. 355. 356. 357. f (x) = x4 + 3x3 − 4x + 4 f (x) = 2x 3 + 3x2 − 8x + 5 f (x) = 3x 3 + 5x2 − 5x + 4 f (x) = 6x4 − 10x2 + 13x + 1 f (x) = 4x5 − 10x4 + 8x3 + x2 − 8 Technology For the following exercises, use your calculator to graph the polynomial function. Based on the graph, find the rational zeros. All real solutions are rational. 358. 359. 360. 361. 362. f (x) = 6x3 − 7x2 + 1 f (x) = 4x3 − 4x2 − 13x − 5 f (x) = 8x3 − 6x2 − 23x + 6 f (x) = 12x4 + 55x3 + 12x2 − 117x + 54 f (x) = 16x4 − 24x3 + x2 − 15x + 25 Extensions the following exercises, construct a polynomial least degree possible using the given For function of information. 363. Real
roots: –1, 1, 3 and ⎛ ⎝2, f (2)⎞ ⎠ = (2, 4) Real 364. ⎛ ⎝2, f (2)⎞ ⎠ = (2, 4) roots: –1 (with multiplicity 2 and 1) and 365. Real roots: –2, 1 2 ⎛ ⎝−3, f (−3)⎞ ⎠ = (−3, 5) (with multiplicity 2) and 366. 367. Real roots: − 1 2 , 0, 1 2 and ⎛ ⎝−2, f (−2)⎞ ⎠ = (−2, 6) Real roots: –4, –1, 1, 4 and ⎛ ⎝−2, f (−2)⎞ ⎠ = (−2, 10) Real-World Applications For the following exercises, find the dimensions of the box described. The length is twice as long as the width. The height is 368. 2 inches greater than the width. The volume is 192 cubic inches. The length, width, and height are consecutive whole 369. numbers. The volume is 120 cubic inches. The length is one inch more than the width, which is 370. one inch more than the height. The volume is 86.625 cubic inches. The length is three times the height and the height is 371. one inch less than the width. The volume is 108 cubic inches. The length is 3 inches more than the width. The width 372. is 2 inches more than the height. The volume is 120 cubic inches. 578 Chapter 5 Polynomial and Rational Functions For the following exercises, find the dimensions of the right circular cylinder described. The radius is 3 inches more than the height. The 373. volume is 16π cubic meters. The height is one less than one half the radius. The 374. volume is 72π cubic meters. The radius and height differ by one meter. The radius 375. is larger and the volume is 48π cubic meters. 376. The radius and height differ by two meters. The height is greater and the volume is 28.125π cubic meters. 377. The radius is 1 3 π cubic meters. volume is 98 9 meter greater than the height. The This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 579 5.6 \| Rational Functions Learning Objectives In this section, you will: 5.6.1 Use arrow notation. 5.6.2 Solve applied problems involving rational functions. 5.6.3 Find the domains of rational functions. 5.6.4 Identify vertical asymptotes. 5.6.5 Identify horizontal asymptotes. 5.6.6 Graph rational functions. Suppose we know that the cost of making a product is dependent on the number of items, x, produced. This is given by the equation C(x) = 15,000x − 0.1x2 + 1000. If we want to know the average cost for producing x items, we would divide the cost function by the number of items, x. The average cost function, which yields the average cost per item for x items produced, is f (x) = 15,000x − 0.1x2 + 1000 x Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power. In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator. Using Arrow Notation We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in Figure 5.62, and notice some of their features. Figure 5.62 Several things are apparent if we examine the graph of f (x) = 1 x. 1. On the left branch of the graph, the curve approaches the x-axis (y = 0) as x → – ∞. 580 Chapter 5 Polynomial and Rational Functions 2. As the graph approaches x = 0 from the left, the curve drops, but as we approach zero from the right, the curve rises. 3. Finally, on the right branch of the graph, the curves approaches the x-axis (y = 0) as x → ∞. To summarize, we use arrow notation to show that x or f (x) is approaching a particular value. See Table 5.6. Symbol Meaning x → a− x → a+ x → ∞ x approaches a from the left ( x < a but close to a ) x approaches a from the right ( x > a but close to a ) x approaches infinity ( x increases without bound) x → − ∞ x approaches negative infinity ( x decreases without bound) f (x) → ∞ the output approaches infinity (the output increases without bound) f (x) → − ∞ the output approaches negative infinity (the output decreases without bound) f (x) → a the output approaches a Table 5.6 Local Behavior of f(x) = 1 x Let’s begin by looking at the reciprocal function, f (x) = 1 x. We cannot divide by zero, which means the function is undefined at x = 0; so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in Table 5.7. x –0.1 –0.01 –0.001 –0.0001 –10 –100 –1000 –10,000 f(x) = 1 x Table 5.7 We write in arrow notation as x → 0− , f (x) → − ∞ As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in Table 5.8. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 581 x 0.1 0.01 0.001 0.0001 10 100 1000 10,000 f(x) = 1 x Table 5.8 We write in arrow notation See Figure 5.63. As x → 0+ , f (x) → ∞. Figure 5.63 This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line x = 0 as the input becomes close to zero. See Figure 5.64. Figure 5.64 582 Chapter 5 Polynomial and Rational Functions Vertical Asymptote A vertical asymptote of a graph is a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a. We write As x → a, f (x) → ∞, or as x → a, f (x) → − ∞. End Behavior of f(x) = 1 x As the values of x approach infinity, the function values approach 0. As the values of x approach negative infinity, the function values approach 0. See Figure 5.65. Symbolically, using arrow notation As x → ∞, f (x) → 0, and as x → − ∞, f (x) → 0. Figure 5.65 Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line y = 0. See Figure 5.66. Figure 5.66 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 583 Horizontal Asymptote A horizontal asymptote of a graph is a horizontal line y = b where the graph approaches the line as the inputs increase or decrease without bound. We write As x → ∞ or x → − ∞, f (x) → b. Example 5.48 Using Arrow Notation Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 5.67. Figure 5.67 Solution Notice that the graph is showing a vertical asymptote at x = 2, which tells us that the function is undefined at x = 2. As x → 2− , f (x) → − ∞, and as x → 2+ , f (x) → ∞. And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at y = 4. As the inputs increase without bound, the graph levels off at 4. As x → ∞, f (x) → 4 and as x → − ∞, f (x) → 4. 5.31 Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function. Example 5.49 584 Chapter 5 Polynomial and Rational Functions Using Transformations to Graph a Rational Function Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any. Solution Shifting the graph left 2 and up 3 would result in the function or equivalently, by giving the terms a common denominator, f (x) = 1 x + 2 + 3 The graph of the shifted function is displayed in Figure 5.68. f (x) = 3x + 7 x + 2 Figure 5.68 Notice that this function is undefined at x = −2, and the graph also is showing a vertical asymptote at x = −2. As x → − 2− , f (x) → − ∞, and as x → − 2+ , f (x) → ∞. As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at y = 3. As x → ± ∞, f (x) → 3. Analysis Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function. Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that 5.32 has been shifted right 3 units and down 4 units. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 585 Solving Applied Problems Involving Rational Functions In Example 5.49, we shifted a toolkit function in a way that resulted in the function f (x) = 3x + 7 x + 2 . This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions. Rational Function A rational function is a function that can be written as the quotient of two polynomial functions P(x) and Q(x). f (x) = P(x) Q(x) = a p x p bq xq + a p − 1 x + bq − 1 x p − 1 q − 1 + ... + a1 x + a0 + ... + b1 x + b0 , Q(x) ≠ 0 (5.5) Example 5.50 Solving an Applied Problem Involving a Rational Function A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning? Solution Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minut
e, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each: water: W(t) = 100 + 10t in gallons sugar: S(t) = 5 + 1t in pounds The concentration, C, will be the ratio of pounds of sugar to gallons of water The concentration after 12 minutes is given by evaluating C(t) at t = 12. C(t) = 5 + t 100 + 10t C(12) = 5 + 12 100 + 10(12) = 17 220 This means the concentration is 17 pounds of sugar to 220 gallons of water. At the beginning, the concentration is C(0) = 5 + 0 100 + 10(0) = 1 20 Since 17 220 ≈ 0.08 > 1 20 = 0.05, the concentration is greater after 12 minutes than at the beginning. 586 Chapter 5 Polynomial and Rational Functions 5.33 There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m. Finding the Domains of Rational Functions A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero. Domain of a Rational Function The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. Given a rational function, find the domain. 1. Set the denominator equal to zero. 2. Solve to find the x-values that cause the denominator to equal zero. 3. The domain is all real numbers except those found in Step 2. Example 5.51 Finding the Domain of a Rational Function Find the domain of f (x) = x + 3 x2 − 9 . Solution Begin by setting the denominator equal to zero and solving. x2 − 9 = 0 x2 = 9 x = ±3 The denominator is equal to zero when x = ± 3. The domain of the function is all real numbers except x = ± 3. Analysis A graph of this function, as shown in Figure 5.69, confirms that the function is not defined when x = ± 3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 587 Figure 5.69 There is a vertical asymptote at x = 3 and a hole in the graph at x = −3. We will discuss these types of holes in greater detail later in this section. 5.34 Find the domain of f (x) = 4x 5(x − 1)(x − 5) . Identifying Vertical Asymptotes of Rational Functions By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location. Vertical Asymptotes The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors. Given a rational function, identify any vertical asymptotes of its graph. 1. Factor the numerator and denominator. 2. Note any restrictions in the domain of the function. 3. Reduce the expression by canceling common factors in the numerator and the denominator. 4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur. 5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities, or “holes.” Example 5.52 Identifying Vertical Asymptotes Find the vertical asymptotes of the graph of k(x) = 5 + 2x2 2 − x − x2. 588 Chapter 5 Polynomial and Rational Functions Solution First, factor the numerator and denominator. k(x) = = 5 + 2x2 2 − x − x2 5 + 2x2 (2 + x)(1 − x) To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero: (2 + x)(1 − x) = 0 x = −2, 1 Neither x = –2 nor x = 1 are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in Figure 5.70 confirms the location of the two vertical asymptotes. Figure 5.70 Removable Discontinuities Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity. For example, the function f (x) = x2 − 1 x2 − 2x − 3 may be re-written by factoring the numerator and the denominator. f (x) = (x + 1)(x − 1) (x + 1)(x − 3) Notice that x + 1 is a common factor to the numerator and the denominator. The zero of this factor, x = −1, is the location of the removable discontinuity. Notice also that x – 3 is not a factor in both the numerator and denominator. The zero of this factor, x = 3, is the vertical asymptote. See Figure 5.71. [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.] This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 589 Figure 5.71 Removable Discontinuities of Rational Functions A removable discontinuity occurs in the graph of a rational function at x = a if a is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value. Example 5.53 Identifying Vertical Asymptotes and Removable Discontinuities for a Graph Find the vertical asymptotes and removable discontinuities of the graph of k(x) = x − 2 x2 − 4 . Solution Factor the numerator and the denominator. k(x) = x − 2 (x − 2)(x + 2) Notice that there is a common factor in the numerator and the denominator, x – 2. The zero for this factor is x = 2. This is the location of the removable discontinuity. Notice that there is a factor in the denominator that is not in the numerator, x + 2. The zero for this factor is x = −2. The vertical asymptote is x = −2. See Figure 5.72. 590 Chapter 5 Polynomial and Rational Functions Figure 5.72 The graph of this function will have the vertical asymptote at x = −2, but at x = 2 the graph will have a hole. 5.35 Find the vertical asymptotes and removable discontinuities of the graph of f (x) = x2 − 25 x3 − 6x2 + 5x . Identifying Horizontal Asymptotes of Rational Functions While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms. There are three distinct outcomes when checking for horizontal asymptotes: Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at y = 0. Example: f (x) = 4x + 2 x2 + 4x − 5 In this case, the end behavior is f (x) ≈ 4x x2 = 4 x. This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function g(x) = 4 asymptote at y = 0. See Figure 5.73. Note that this graph crosses the horizontal asymptote. x, and the outputs will approach zero, resulting in a horizontal This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 591 Figure 5.73 Horizontal asymptote y = 0 when f (x) = p(x) q(x) , q(x) ≠ 0 where degree of p < degree of q. Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote. Example: f (x) = 3x2 − 2x + 1 x − 1 In this case, the end behavior is f (x) ≈ 3x2 x = 3x. This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function g(x) = 3x. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of g(x) = 3x looks like a diagonal line, and since f will behave similarly to g, it will approach a line close to y = 3x. This line is a slant asymptote. To find the equation of the slant asymptote, divide 3x2 − 2x + 1 x − 1 . The quotient is 3x + 1, and the remainder is 2. The slant asymptote is the graph of the line g(x) = 3x + 1. See Figure 5.74. Figure 5.74 Slant asymptote when f (x) = where degree of p > degree of q by 1. p(x) q(x) , q(x) ≠ 0 592 Chapter 5 Polynomial and Rational Functions Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at y = an bn , where an and bn are the leading coefficients of p(x) and q(x) for f (x) = p(x) q(x) , q(x) ≠ 0. Example: f (x) = 3x2 + 2 x2 + 4x − 5 In this case, the end behavior is f (x) ≈ 3x2 x2 = 3. This tells us that as the inputs grow large, this function will behave like the function g(x) = 3, which is a horizontal line. As x → ± ∞, f (x) → 3, resulting in a horizontal asymptote at y = 3. See Figure 5.75. Note that this graph crosses the horizontal asymptote. Figure 5.75 Horizontal asymptote when f (x) = p(x) q(x) , q(x) ≠ 0 where degree of p = degree of q. Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote. It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one
, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function f (x) = 3x5 − x2 x + 3 with end behavior the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient. x → ± ∞, f (x) → ∞ f (x) ≈ 3x5 x = 3x4, Horizontal Asymptotes of Rational Functions The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator. • Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 593 • Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. • Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients. Example 5.54 Identifying Horizontal and Slant Asymptotes For the functions listed, identify the horizontal or slant asymptote. a. g(x) = 6x3 − 10x 2x3 + 5x2 b. h(x) = x2 − 4x + 1 x + 2 c. k(x) = x2 + 4x x3 − 8 Solution For these solutions, we will use f (x) = p(x) q(x) , q(x) ≠ 0. a. b. g(x) = 6x3 − 10x 2x3 + 5x2 : The degree of p = degree of q = 3, so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at y = 6 2 or y = 3. h(x) = x2 − 4x + 1 x + 2 asymptote found at x2 − 4x + 1 x + 2 . : The degree of p = 2 and degree of q = 1. Since p > q by 1, there is a slant The quotient is x – 2 and the remainder is 13. There is a slant asymptote at y = –x – 2. c. k(x) = x2 + 4x x3 − 8 : The degree of p = 2 < degree of q = 3, so there is a horizontal asymptote y = 0. Example 5.55 Identifying Horizontal Asymptotes In the sugar concentration problem earlier, we created the equation C(t) = 5 + t 100 + 10t. Find the horizontal asymptote and interpret it in context of the problem. Solution 594 Chapter 5 Polynomial and Rational Functions Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t, with coefficient 1. In the denominator, the leading term is 10t, with coefficient 10. The horizontal asymptote will be at the ratio of these values: This function will have a horizontal asymptote at y = 1 10 . t → ∞, C(t) → 1 10 This tells us that as the values of t increase, the values of C will approach 1 10 . In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or 1 10 pounds per gallon. Example 5.56 Identifying Horizontal and Vertical Asymptotes Find the horizontal and vertical asymptotes of the function f (x) = (x − 2)(x + 3) (x − 1)(x + 2)(x − 5) Solution First, note that this function has no common factors, so there are no potential removable discontinuities. The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at x = 1, – 2, and 5, indicating vertical asymptotes at these values. The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as x → ± ∞, f (x) → 0. This function will have a horizontal asymptote at y = 0. See Figure 5.76. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 595 Figure 5.76 5.36 Find the vertical and horizontal asymptotes of the function: f (x) = (2x − 1)(2x + 1) (x − 2)(x + 3) Intercepts of Rational Functions A rational function will have a y-intercept at f (0) , if the function is defined at zero. A rational function will not have a y-intercept if the function is not defined at zero. Likewise, a rational function will have x-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function is equal to zero. Example 5.57 Finding the Intercepts of a Rational Function Find the intercepts of f (x) = (x − 2)(x + 3) (x − 1)(x + 2)(x − 5) . Solution We can find the y-intercept by evaluating the function at zero 596 Chapter 5 Polynomial and Rational Functions f (0) = (0 − 2)(0 + 3) (0 − 1)(0 + 2)(0 − 5) = −6 10 = − 3 5 = −0.6 The x-intercepts will occur when the function is equal to zero: 0 = (x − 2)(x + 3) (x − 1)(x + 2)(x − 5) 0 = (x − 2)(x + 3) x = 2, −3 This is zero when the numerator is zero. The y-intercept is (0, –0.6), the x-intercepts are (2, 0) and (–3, 0). See Figure 5.77. Figure 5.77 Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational 5.37 function. Then, find the x- and y-intercepts and the horizontal and vertical asymptotes. Graphing Rational Functions In Example 5.56, we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials. The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See Figure 5.78. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 597 Figure 5.78 When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure 5.79. Figure 5.79 For example, the graph of f (x) = (x + 1)2(x − 3) (x + 3)2(x − 2) is shown in Figure 5.80. 598 Chapter 5 Polynomial and Rational Functions Figure 5.80 • At the x-intercept x = −1 corresponding to the (x + 1)2 factor of the numerator, the graph "bounces", consistent with the quadratic nature of the factor. • At the x-intercept x = 3 corresponding to the (x − 3) factor of the numerator, the graph passes through the axis as we would expect from a linear factor. • At the vertical asymptote x = −3 corresponding to the (x + 3)2 factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function f (x) = 1 x2. • At the vertical asymptote x = 2, corresponding to the (x − 2) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function f (x) = 1 x. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 599 Given a rational function, sketch a graph. 1. Evaluate the function at 0 to find the y-intercept. 2. Factor the numerator and denominator. 3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x-intercepts. 4. Find the multiplicities of the x-intercepts to determine the behavior of the graph at those points. 5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve. 6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve. 7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes. 8. Sketch the graph. Example 5.58 Graphing a Rational Function Sketch a graph of f (x) = (x + 2)(x − 3) (x + 1)2(x − 2) . Solution We can start by noting that the function is already factored, saving us a step. Next, we will find the intercepts. Evaluating the function at zero gives the y-intercept: f (0) = (0 + 2)(0 − 3) (0 + 1)2(0 − 2) = 3 To find the x-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find x-intercepts at x = –2 and x = 3. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept. We have a y-intercept at (0, 3) and x-intercepts at (–2, 0) and (3, 0). To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when x + 1 = 0 and when x – 2 = 0, giving us vertical asymptotes at x = –1 and x = 2. There are no common factors in the numerator and denominator. This means there are no removable discontinuities. Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at y = 0. To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x-intercepts between the vertical asymptotes, and the y-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in Figure 5.81. 600 Chapter 5 Polynomial and Rational Functions Figure 5.81 The factor associated with the vertical asymptote at x = −1 was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph wi
ll head toward positive infinity on the left as well. For the vertical asymptote at x = 2, the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See Figure 5.82. After passing through the x-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote. Figure 5.82 5.38 Given the function f (x) = (x + 2)2(x − 2) 2(x − 1)2(x − 3) , use the characteristics of polynomials and rational functions to describe its behavior and sketch the function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 601 Writing Rational Functions Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors. Writing Rational Functions from Intercepts and Asymptotes If a rational function has x-intercepts at x = x1, x2, ..., xn, vertical asymptotes at x = v1, v2, … , vm, and no xi = any v j, then the function can be written in the form: p1 (x − x2) q1 (x − v2) where the powers pi or qi on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor a can be determined given a value of the function other than the xintercept or by the horizontal asymptote if it is nonzero. f (x) = a(x − x1) (x − v1) p2 ⋯ (x − xn) q2 ⋯ (x − vm) pn qn Given a graph of a rational function, write the function. 1. Determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.) 2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers. 3. Use any clear point on the graph to find the stretch factor. Example 5.59 Writing a Rational Function from Intercepts and Asymptotes Write an equation for the rational function shown in Figure 5.83. 602 Chapter 5 Polynomial and Rational Functions Figure 5.83 Solution The graph appears to have x-intercepts at x = –2 and x = 3. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at x = –1 seems to exhibit the basic behavior similar to 1 x, with the graph heading toward positive infinity on one side and heading toward negative x2, with the graph heading infinity on the other. The asymptote at x = 2 is exhibiting a behavior similar to 1 toward negative infinity on both sides of the asymptote. See Figure 5.84. Figure 5.84 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 603 We can use this information to write a function of the form f (x) = a (x + 2)(x − 3) (x + 1)(x − 2)2 To find the stretch factor, we can use another clear point on the graph, such as the y-intercept (0, –2). −2 = a (0 + 2)(0 − 3) (0 + 1)(0 − 2)2 −2 = a−6 4 a = −8 −6 = 4 3 4(x + 2)(x − 3) 3(x + 1)(x − 2)2. This gives us a final function of f (x) = Access these online resources for additional instruction and practice with rational functions. • Graphing Rational Functions (http://openstaxcollege.org/l/graphrational) • Find the Equation of a Rational Function (http://openstaxcollege.org/l/equatrational) • Determining Vertical and Horizontal Asymptotes (http://openstaxcollege.org/l/asymptote) • Find the Intercepts, Asymptotes, and Hole of a Rational Function (http://openstaxcollege.org/l/interasymptote) 604 Chapter 5 Polynomial and Rational Functions 5.6 EXERCISES Verbal 378. What is the fundamental difference in the algebraic representation of a polynomial function and a rational function? What is the fundamental difference in the graphs of 379. polynomial functions and rational functions? If the graph of a rational function has a removable 380. discontinuity, what must be true of the functional rule? Can a graph of a rational function have no vertical 381. asymptote? If so, how? Can a graph of a rational function have no x- 382. intercepts? If so, how? Algebraic For the following exercises, find the domain of the rational functions. 383. f (x) = x − 1 x + 2 384. f (x) = x + 1 x2 − 1 385. 386. f (x) = x2 + 4 x2 − 2x − 8 f (x) = x2 + 4x − 3 x4 − 5x2 + 4 f (x) = x2 − 1 x3 + 9x2 + 14x 394. f (x) = x + 5 x2 − 25 395. 396. f (xx) = 4 − 2x 3x − 1 For the following exercises, find the x- and y-intercepts for the functions. 397. f (x) = 398. f (x) = x + 5 x2 + 4 x x2 − x 399. 400. 401. f (x) = x2 + 8x + 7 x2 + 11x + 30 f (x) = x2 + x + 6 x2 − 10x + 24 f (x) = 94 − 2x2 3x2 − 12 For the following exercises, describe the local and end behavior of the functions. For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. 402. f (x) = x 2x + 1 387. f (x) = 4 x − 1 388. f (x) = 389. f (x) = 2 5x + 2 x x2 − 9 390. f (x) = x x2 + 5x − 36 f (x) = 3 + x x3 − 27 f (x) = 3x − 4 x3 − 16x 391. 392. 393. 403. 404. 405. 406. f (x) = 2x x − 6 f (x) = −2x x − 6 f (x) = x2 − 4x + 3 x2 − 4x − 5 f (x) = 2x2 − 32 6x2 + 13x − 5 For the following exercises, find the slant asymptote of the functions. f (x) = 24x2 + 6x 2x + 1 407. 408. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 605 f (x) = 4x2 − 10 2x − 4 409. 410. 411. f (x) = 81x2 − 18 3x − 2 f (x) = 6x3 − 5x 3x2 + 4 f (x) = x2 + 5x + 4 x − 1 Graphical For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes. 412. The reciprocal function shifted up two units. The reciprocal function shifted down one unit and left 413. three units. The reciprocal squared function shifted to the right 2 414. units. The reciprocal squared function shifted down 2 units 415. and right 1 unit. For the following exercises, find the horizontal intercepts, the vertical the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph. intercept, 416. p(x) = 2x − 3 x + 4 417. q(x) = x − 5 3x − 1 418. s(x) = 419. r(x) = 4 (x − 2)2 5 (x + 1)2 420. 421. f (x) = 3x2 − 14x − 5 3x2 + 8x − 16 g(x) = 2x2 + 7x − 15 3x2 − 14 + 15 422. a(x) = x2 + 2x − 3 x2 − 1 423. b(x) = x2 − x − 6 x2 − 4 424. 425. h(x) = 2x2 + x − 1 x − 4 k(x) = 2x2 − 3x − 20 x − 5 426. w(x) = (x − 1)(x + 3)(x − 5) (x + 2)2(x − 4) 427. z(x) = (x + 2)2 (x − 5) (x − 3)(x + 1)(x + 4) For the following exercises, write an equation for a rational function with the given characteristics. Vertical 428. intercepts at (2, 0) and (−1, 0), y-intercept at (0, 4) at x = 5 and x = −5, x- asymptotes Vertical 429. x-intercepts at (1, 0) and (5, 0), y-intercept at (0, 7) at x = −4 and x = −1, asymptotes Vertical asymptotes at x = −4 and x = −5, x430. intercepts at (4, 0) and (−6, 0), Horizontal asymptote at y = 7 Vertical asymptotes at x = −3 and x = 6, x431. intercepts at (−2, 0) and (1, 0), Horizontal asymptote at y = −2 Vertical 432. x = 2, y-intercept at (0, 2) asymptote at x = −1, Double Vertical 433. x = 1, y-intercept at (0, 4) asymptote at x = 3, Double zero at zero at For the following exercises, use the graphs to write an equation for the function. 434. 435. 606 Chapter 5 Polynomial and Rational Functions 436. 440. 437. 441. 438. Numeric For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote 442. f (x) = 1 x − 2 443. f (x) = x x − 3 444. f (x) = 2x x + 4 445. f (x) = 2x (x − 3)2 439. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 607 459. bloodstream t hours C(t) = 2t drug as t increases? 460. bloodstream t hours C(t) = 100t 2t 2 + 75 of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t minutes. For the following exercises, use the given rational function to answer the question. The concentration C of a drug in a patient’s by injection given after in 3 + t 2. What happens to the concentration of the The concentration C of a drug in a patient’s by injection given after is . Use a calculator to approximate the time when the concentration is highest. For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question. An open box with a square base is to have a volume of 461. 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let x = length of the side of the base. A rectangular box with a square base is to have a 462. volume of 20 cubic feet. The material for the base costs 30 cents/ square foot. The material for the sides costs 10 cents/ square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let x = length of the side of the base. A right circular cylinder has volume of 100 cubic 463. inches. Find the radius and height that will yield minimum surface area. Let x = radius. A right circular cylinder with no top has a volume of 464. 50 cubic meters. Find the radius that will yield minimum surface area. Let x = radius. A right circular cylinder
is to have a volume of 40 465. cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost. Let x = radius. 446. f (x) = x2 x2 + 2x + 1 Technology For the following exercises, use a calculator to graph f (x). Use the graph to solve f (x) > 0. 447. 448. 449. 450. 451. f (x) = 2 x + 1 f (x) = 4 2x − 3 f (x) = 2 (x − 1)(x + 2) f (x) = x + 2 (x − 1)(x − 4) f (x) = (x + 3)2 (x − 1)2 (x + 1) Extensions the following exercises, For discontinuity. identify the removable 452. 453. 454. 455. 456. f (x) = f (x) = x2 − 4 x − 2 x3 + 1 x + 1 f (x) = x2 + x − 6 x − 2 f (x) = 2x2 + 5x − 3 x + 3 f (x) = x3 + x2 x + 1 Real-World Applications For the following exercises, express a rational function that describes the situation. A large mixing tank currently contains 200 gallons of 457. water, into which 10 pounds of sugar have been mixed. A tap will open, pouring 10 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 3 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t minutes. A large mixing tank currently contains 300 gallons of 458. water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate 608 Chapter 5 Polynomial and Rational Functions 5.7 \| Inverses and Radical Functions Learning Objectives In this section, you will: 5.7.1 Find the inverse of an invertible polynomial function. 5.7.2 Restrict the domain to find the inverse of a polynomial function. A mound of gravel is in the shape of a cone with the height equal to twice the radius. Figure 5.85 The volume is found using a formula from elementary geometry πr 2 h πr 2(2r) πr 3 We have written the volume V in terms of the radius r. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula 3 r = 3V 2π This function is the inverse of the formula for V in terms of r. In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process. Finding the Inverse of a Polynomial Function Two functions f and g are inverse functions if for every coordinate pair in f , (a, b), there exists a corresponding coordinate pair in the inverse function, g, (b, a). In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test. For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in Figure 5.86. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 609 Figure 5.86 Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x measured horizontally and y measured vertically, with the origin at the vertex of the parabola. See Figure 5.87. Figure 5.87 From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form y(x) = ax2. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a. Our parabolic cross section has the equation 18 = a62 a = 18 36 = 1 2 y(x) = 1 2 x2 We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y, the width will be given by 2x, so we need to solve the equation above for x and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative. 610 Chapter 5 Polynomial and Rational Functions To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x values. On this domain, we can find an inverse by solving for the input variable: x2 y = 1 2 2y = x2 x = ± 2y This is not a function as written. We are limiting ourselves to positive x values, so we eliminate the negative solution, giving us the inverse function we’re looking for. x2 2 Because x is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x. The trough is 3 feet (36 inches) long, so the surface area will then be: , x > 0 y = Area = l ⋅ w = 36 ⋅ 2x = 72x = 72 2y This example illustrates two important points: 1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one. 2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions. Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation f −1(x). Warning: f −1(x) is not the same as the reciprocal of the function f (x). This use of “–1” is reserved to denote inverse functions. To denote the reciprocal of a function f (x), we would need to write ⎛ ⎝ f (x)⎞ ⎠ −1 = 1 f (x) . An important relationship between inverse functions is that they “undo” each other. If f −1 is the inverse of a function f , then f is the inverse of the function f −1. In other words, whatever the function f does to x, versa. f −1 undoes it—and vice- f −1 ⎛ ⎝ f (x)⎞ ⎠ = x, for all x in the domain of f and Note that the inverse switches the domain and range of the original function. f ⎛ ⎞ ⎝ f −1 (x) ⎠ = x, for all x in the domain of f −1 Verifying Two Functions Are Inverses of One Another Two functions, f and g, are inverses of one another if for all x in the domain of f and g. g⎛ ⎝ f (x)⎞ ⎠ = f ⎛ ⎝g(x)⎞ ⎠ = x This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 611 Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one. 1. Replace f (x) with y. 2. Interchange x and y. 3. Solve for y, and rename the function f −1(x). Example 5.60 Verifying Inverse Functions Show that f (x) = 1 x + 1 and f −1 (x) = 1 x − 1 are inverses, for x ≠ 0, −1 . Solution We must show that f −1 ⎛ ⎝ f (x)⎞ ⎠ = x and f ⎛ ⎞ ⎝ f −1 (x) ⎠ = x. f −1( f (x)) = f −x + 1) − 1 = x f ( f −1(x)) = Therefore, f (x) = 1 x + 1 and f −1 (x) = 1 x − 1 are inverses. 5.39 Show that f (x) = x + 5 3 and f −1 (x) = 3x − 5 are inverses. Example 5.61 Finding the Inverse of a Cubic Function Find the inverse of the function f (x) = 5x3 + 1. Solution 612 Chapter 5 Polynomial and Rational Functions This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x. y = 5x3 + 1 x = 5y3 + 1 x − 1 = 5y3 x − 1 = y3 5 f −1(x) = 3 x − 1 5 Analysis Look at the graph of f and f –1. Notice that one graph is the reflection of the other about the line y = x. This is always the case when graphing a function and its inverse function. Also, since the method involved interchanging x and y, notice corresponding points. If (a, b) is on the graph of then (b, a) is on the graph of f –1. Since (0, 1) is on the graph of f , then (1, 0) is on the graph of f –1. f , Similarly, since (1, 6) is on the graph of f , then (6, 1) is on the graph of f –1. See Figure 5.88. Figure 5.88 5.40 Find the inverse function of f (x) = x + 4 . 3 Restricting the Domain to Find the Inverse of a Polynomial Function So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 613 restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in order to find their inverses. Restricting the Domain If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse. Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse. 1. Restrict the domain by determining a domain on which the original function is one-to-one. 2. Replace f (x) with y. 3. Interchange x and y. 4. Solve for y, and rename the function or pair of function f −1(x). 5. Revise the formula for f −1(x) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function. Example 5.62 Restricting the Domain to Find the Inverse of a Polynomial Function Find the inverse function of f : a. b. f (x) = (x − 4)2, x ≥ 4 f (x) = (x − 4)2, x ≤ 4 Solution The original function f (x) = (x − 4)2 is not one-to-one, but the function is restricted to a domain of x ≥ 4 or x ≤ 4 on which it is one-to-one. See Figure 5.89. Figure 5.89 614 Cha
pter 5 Polynomial and Rational Functions To find the inverse, start by replacing f (x) with the simple variable y. y = (x − 4)2 x = (y − 4) Interchangexand y. Take the square root. Add 4 to both sides. This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of x and y for the original f (x), we looked at the domain: the values x could assume. When we reversed the roles of x and y, this gave us the values y could assume. For this function, x ≥ 4, so for the inverse, we should have y ≥ 4, which is what our inverse function gives. a. The domain of the original function was restricted to x ≥ 4, so the outputs of the inverse need to be the same, f (x) ≥ 4, and we must use the + case: b. The domain of the original function was restricted to x ≤ 4, so the outputs of the inverse need to be the same, f (x) ≤ 4, and we must use the – case: f −1(x) = 4 + x f −1(x) = 4 − x Analysis On the graphs in Figure 5.90, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line y = x. The coordinate pair (4, 0) is on the graph 4) is on the graph of f −1. For any coordinate pair, if (a, b) is on the graph of f and the coordinate pair (0, of f , then (b, a) is on the graph of f −1. Finally, observe that the graph of f intersects the graph of f −1 on the line y = x. Points of intersection for the graphs of f and f −1 will always lie on the line y = x. Figure 5.90 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 615 Example 5.63 Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified Restrict the domain and then find the inverse of f (x) = (x − 2)2 − 3. Solution We can see this is a parabola with vertex at (2, –3) that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to x ≥ 2. To find the inverse, we will use the vertex form of the quadratic. We start by replacing f (x) with a simple variable, y, then solve for x. y = (x − 2)2 − 3 x = (y − 2)2 − 3 x + 3 = (y − 2)1(x) = 2 ± x + 3 Interchange x and y. Add 3 to both sides. Take the square root. Add 2 to both sides. Rename the function. Now we need to determine which case to use. Because we restricted our original function to a domain of x ≥ 2, the outputs of the inverse should be the same, telling us to utilize the + case f −1(x) = 2 + x + 3 If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain. Analysis Notice that we arbitrarily decided to restrict the domain on x ≥ 2. We could just have easily opted to restrict the domain on x ≤ 2, in which case f −1(x) = 2 − x + 3. Observe the original function graphed on the same set of axes as its inverse function in Figure 5.91. Notice that both graphs show symmetry about the line y = x. The coordinate pair (2, − 3) is on the graph of f and the coordinate pair (−3, 2) is on the graph of f −1. Observe from the graph of both functions on the same set of axes that and domain of f = range of f – 1 = [2, ∞) domain of f – 1 = range of f = [ – 3, ∞). Finally, observe that the graph of f intersects the graph of f −1 along the line y = x. 616 Chapter 5 Polynomial and Rational Functions Figure 5.91 5.41 Find the inverse of the function f (x) = x2 + 1, on the domain x ≥ 0. Solving Applications of Radical Functions Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the inverse of a radical function, we will need to restrict the domain of the answer because the range of the original function is limited. Given a radical function, find the inverse. 1. Determine the range of the original function. 2. Replace f (x) with y, then solve for x. 3. If necessary, restrict the domain of the inverse function to the range of the original function. Example 5.64 Finding the Inverse of a Radical Function Restrict the domain of the function f (x) = x − 4 and then find the inverse. Solution Note that the original function has range f (x) ≥ 0. Replace f (x) with y, then solve for x. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 617 x2 = y − 4 x2 + 4 = y f −1(x) = x2 + 4 Replace f (x) with y. Interchange x and y. Square each side. Add 4. Rename the function f −1(x). Recall that the domain of this function must be limited to the range of the original function. f −1(x) = x2 + 4, x ≥ 0 Analysis Notice in Figure 5.92 that the inverse is a reflection of the original function over the line y = x. Because the original function has only positive outputs, the inverse function has only positive inputs. Figure 5.92 5.42 Restrict the domain and then find the inverse of the function f (x) = 2x + 3. Solving Applications of Radical Functions Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section. Example 5.65 Solving an Application with a Cubic Function A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by 618 Chapter 5 Polynomial and Rational Functions V = 2 3 πr 3 Find the inverse of the function V = 2 3 πr 3 that determines the volume V of a cone and is a function of the radius r. Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use π = 3.14. Solution Start with the given function for V. Notice that the meaningful domain for the function is r > 0 since negative radii would not make sense in this context nor would a radius of 0. Also note the range of the function (hence, the domain of the inverse function) is V > 0. Solve for r in terms of V, using the method outlined previously. Note that in real-world applications, we do not swap the variables when finding inverses. Instead, we change which variable is considered to be the independent variable. πr 3 V = 2 3 r 3 = 3V 2π 3 r = 3V 2π Solve for r 3. Solve for r. This is the result stated in the section opener. Now evaluate this for V = 100 and π = 3.14. 3 r = 3V 2π 3 3 ⋅ 100 2 ⋅ 3.14 = 3 ≈ 47.7707 ≈ 3.63 Therefore, the radius is about 3.63 ft. Determining the Domain of a Radical Function Composed with Other Functions When radical functions are composed with other functions, determining domain can become more complicated. Example 5.66 Finding the Domain of a Radical Function Composed with a Rational Function Find the domain of the function f (x) = (x + 2)(x − 3) (x − 1) . Solution Because a square root is only defined when the quantity under the radical is non-negative, we need to determine where (x + 2)(x − 3) ≥ 0. The output of a rational function can change signs (change from positive to negative (x − 1) or vice versa) at x-intercepts and at vertical asymptotes. For this equation, the graph could change signs at x = – 2, and 3. 1, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 619 To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as shown in Figure 5.93. Figure 5.93 This function has two x-intercepts, both of which exhibit linear behavior near the x-intercepts. There is one vertical asymptote, corresponding to a linear factor; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a y-intercept at (0, 6). From the y-intercept and x-intercept at x = −2, we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph. From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function f (x) will be defined. f (x) has domain −2 ≤ x < 1 or x ≥ 3, or in interval notation, [−2, 1) ∪ [3, ∞). Finding Inverses of Rational Functions As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as in concentration applications. Example 5.67 Finding the Inverse of a Rational Function represents the concentration C of an acid solution after n mL of 40% solution has The function C = 20 + 0.4n 100 + n been added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for n in terms of C. Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution. 620 Chapter 5 Polynomial and Rational Functions Solution We first want the inverse of the function in order to determine how many mL we need for a given concentration. We will solve for n in terms of C. C = 20 + 0.4n 100 + n C(100 + n) = 20 + 0.4n 100C + Cn = 20 + 0.4n 100C − 20 = 0.4n − Cn 100C − 20 = (0.4n − C)n n = 100C − 20 0.4 − C Now evaluate this function at 35%, which is C = 0.35. n = 100(0.35) − 20 0.4 − 0.35 = 15 0.05 = 300 We can conclude that 300 mL of the 40% solution should be added. 5.43 Find the inverse of the function f (x) = x + 3 x − 2 . Access these online resources for additional instruction and practice with inverses and radical functions. • Graphing the Basic Square Root Function (http://openstaxcollege.org/l/graphsquareroot) • Find the Inverse of a Square Root Funct
ion (http://openstaxcollege.org/l/inversesquare) • Find the Inverse of a Rational Function (http://openstaxcollege.org/l/inverserational) • Find the Inverse of a Rational Function and an Inverse Function Value (http://openstaxcollege.org/l/rationalinverse) • Inverse Functions (http://openstaxcollege.org/l/inversefunction) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 621 5.7 EXERCISES Verbal Explain why we cannot find inverse functions for all 466. polynomial functions. 484. 485. f (x) = 6x − 8 + 5 f (x) = 9 + 2 x3 Why must we restrict the domain of a quadratic 467. function when finding its inverse? 486. f (x) = 3 − x3 When finding the inverse of a radical function, what 468. restriction will we need to make? The inverse of a quadratic function will always take 469. what form? Algebraic For the following exercises, find the inverse of the function on the given domain. 470. 471. 472. 473. 474. 475. 476. f (x) = (x − 4)2, [4, ∞) f (x) = (x + 2)2, [−2, ∞) f (x) = (x + 1)2 − 3, [−1, ∞) f (x) = 3x2 + 5, (∞, 0] f (x) = 12 − x2, [0, ∞) f (x) = 9 − x2, [0, ∞) f (x) = 2x2 + 4, [0, ∞) the following exercises, For functions. find the inverse of the 477. f (x) = x3 + 5 478. f (x) = 3x3 + 1 479. f (x) = 4 − x3 480. f (x) = 4 − 2x3 the following exercises, For functions. find the inverse of the 481. 482. 483. f (x) = 2x + 1 f (x) = 3 − 4x f (x) = 9 + 4x − 4 487. 488. f (x) = 2 x + 8 f (x) = 3 x − 4 489. f (x) = 490. f (x 491. 492. 493. 494. 495. f (x) = 3x + 4 5 − 4x f (x) = 5x + 1 2 − 5x f (x) = x2 + 2x, [−1, ∞) f (x) = x2 + 4x + 1, [−2, ∞) f (x) = x2 − 6x + 3, [3, ∞) Graphical For the following exercises, find the inverse of the function and graph both the function and its inverse. 496. 497. 498. 499. f (x) = x2 + 2, x ≥ 0 f (x) = 4 − x2, x ≥ 0 f (x) = (x + 3)2, x ≥ − 3 f (x) = (x − 4)2, x ≥ 4 500. f (x) = x3 + 3 501. f (x) = 1 − x3 f (x) = x2 + 4x, x ≥ − 2 f (x) = x2 − 6x + 1, x ≥ 3 502. 503. 504. 622 f (x) = 2 x 505. f (x) = 1 x2, x ≥ 0 For the following exercises, use a graph to help determine the domain of the functions. 506. 507. 508. 509. 510. f (x) = (x + 1)(x − 1) x f (x) = (x + 2)(x − 3) x − 1 f (x) = x(x + 3) x − 4 f (x) = x2 − x − 20 x − 2 f (x) = 9 − x2 x + 4 Technology For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with y-coordinates given. Chapter 5 Polynomial and Rational Functions Real-World Applications the following exercises, determine the function For described and then use it to answer the question. An object dropped from a height of 200 meters has a 521. height, h(t), in meters after t seconds have lapsed, such that h(t) = 200 − 4.9t 2. Express t as a function of height, h, and find the time to reach a height of 50 meters. An object dropped from a height of 600 feet has a 522. height, h(t), in feet after t seconds have elapsed, such that h(t) = 600 − 16t 2. Express t as a function of height h, and find the time to reach a height of 400 feet. 523. The volume, V, of a sphere in terms of its radius, πr 3. Express r as a function of r, is given by V(r) = 4 3 V, and find the radius of a sphere with volume of 200 cubic feet. 524. The surface area, A, of a sphere in terms of its radius, r, is a function of V, and find the radius of a sphere with a surface area of 1000 square inches. by A(r) = 4πr 2. Express r as given f (x) = x3 − x − 2, y = 1, 2, 3 f (x) = x3 + x − 2, y = 0, 1, 2 f (x) = x3 + 3x − 4, y = 0, 1, 2 A container holds 100 mL of a solution that is 25 mL 525. acid. If n mL of a solution that is 60% acid is added, the function C(n) = 25 + .6n 100 + n a function of the number of mL added, n. Express n as a function of C and determine the number of mL that need to be added to have a solution that is 50% acid. gives the concentration, C, as f (x) = x3 + 8x − 4, y = − 1, 0, 1 526. f (x) = x4 + 5x + 1, y = − 1, 0, 1 Extensions the following exercises, For functions with a, b, c positive real numbers. find the inverse of 516. f (x) = ax3 + b 517. f (x) = x2 + bx 518. 519. 520. f (x) = ax2 + b 3 f (x) = ax + b f (x) = ax + b x + c The period T, in seconds, of a simple pendulum as a by given feet, its length l, in of is function T(l) = 2π l 32.2 . Express l as a function of T and determine the length of a pendulum with period of 2 seconds. the 527. The volume of a cylinder , V, in terms of radius, r, and height, h, is given by V = πr 2 h. If a cylinder has a height of 6 meters, express the radius as a function of V and find the radius of a cylinder with volume of 300 cubic meters. 528. The surface area, A, of a cylinder in terms of its radius, r, and height, h, is given by A = 2πr 2 + 2πrh. If the height of the cylinder is 4 feet, express the radius as a function of V and find the radius if the surface area is 200 square feet. 511. 512. 513. 514. 515. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 623 529. The volume of a right circular cone, V, in terms of πr 2 h. its radius, r, and its height, h, is given by V = 1 3 Express r in terms of h if the height of the cone is 12 feet and find the radius of a cone with volume of 50 cubic inches. 530. Consider a cone with height of 30 feet. Express the radius, r, in terms of the volume, V, and find the radius of a cone with volume of 1000 cubic feet. 624 Chapter 5 Polynomial and Rational Functions 5.8 \| Modeling Using Variation Learning Objectives In this section, you will: 5.8.1 Solve direct variation problems. 5.8.2 Solve inverse variation problems. 5.8.3 Solve problems involving joint variation. A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate. Solving Direct Variation Problems In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula e = 0.16s tells us her earnings, e, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 5.9. s , sales price e = 0.16s Interpretation $4,600 $9,200 e = 0.16(4,600) = 736 A sale of a $4,600 vehicle results in $736 earnings. e = 0.16(9,200) = 1,472 A sale of a $9,200 vehicle results in $1472 earnings. $18,400 e = 0.16(18,400) = 2,944 A sale of a $18,400 vehicle results in $2944 earnings. Table 5.9 Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation. Each variable in this type of relationship varies directly with the other. Figure 5.94 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula y = kxn is used for direct variation. The value k is a nonzero constant greater than zero and is called the constant of variation. In this case, k = 0.16 and n = 1. We saw functions like this one when we discussed power functions. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 625 Figure 5.94 Direct Variation If x and y are related by an equation of the form y = kxn (5.6) then we say that the relationship is direct variation and y varies directly with, or is proportional to, the nth power y xn, where k is called the constant of of x. In direct variation relationships, there is a nonzero constant ratio k = variation, which help defines the relationship between the variables. Given a description of a direct variation problem, solve for an unknown. 1. Identify the input, x, and the output, y. 2. Determine the constant of variation. You may need to divide y by the specified power of x to determine the constant of variation. 3. Use the constant of variation to write an equation for the relationship. 4. Substitute known values into the equation to find the unknown. Example 5.68 Solving a Direct Variation Problem The quantity y varies directly with the cube of x. If y = 25 when x = 2, find y when x is 6. Solution The general formula for direct variation with a cube is y = kx3. The constant can be found by dividing y by the cube of x. 626 Chapter 5 Polynomial and Rational Functions k = y x3 = 25 23 = 25 8 Now use the constant to write an equation that represents this relationship. Substitute x = 6 and solve for y. y = 25 8 x3 y = 25 8 (6)3 = 675 Analysis The graph of this equation is a simple cubic, as shown in Figure 5.95. Figure 5.95 Do the graphs of all direct variation equations look like Example 5.68? No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0,0). 5.44 The quantity y varies directly with the square of x. If y = 24 when x = 3, find y when x is 4. Solving Inverse Variation Problems Water temperature in an ocean varies inversely to the water’s depth. The formula T = 14,000 d gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F. If we create Table 5.10, we observe that, as the depth increases, the water temperature decreases. This content is available for free at https://cnx.org/cont
ent/col11758/1.5 Chapter 5 Polynomial and Rational Functions 627 d, depth T = 14,000 d Interpretation 14,000 500 = 28 14,000 1000 = 14 14,000 2000 = 7 500 ft 1000 ft 2000 ft Table 5.10 At a depth of 500 ft, the water temperature is 28° F. At a depth of 1,000 ft, the water temperature is 14° F. At a depth of 2,000 ft, the water temperature is 7° F. We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations. For our example, Figure 5.96 depicts the inverse variation. We say the water temperature varies inversely with the depth k of the water because, as the depth increases, the temperature decreases. The formula y = x for inverse variation in this case uses k = 14,000. Figure 5.96 Inverse Variation If x and y are related by an equation of the form where k is a nonzero constant, proportional relationships, or inverse variations, there is a constant multiple k = xn y. then we say that y varies inversely with the nth power of x. In inversely y = k xn (5.7) Example 5.69 Writing a Formula for an Inversely Proportional Relationship 628 Chapter 5 Polynomial and Rational Functions A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives. Solution Recall that multiplying speed by time gives distance. If we let t represent the drive time in hours, and v represent the velocity (speed or rate) at which the tourist drives, then vt = distance. Because the distance is fixed at 100 miles, vt = 100 so t = 100/v. Because time is a function of velocity, we can write t(v). t(v) = 100 v = 100v−1 We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. We say that time varies inversely with velocity. Given a description of an indirect variation problem, solve for an unknown. 1. Identify the input, x, and the output, y. 2. Determine the constant of variation. You may need to multiply y by the specified power of x to determine the constant of variation. 3. Use the constant of variation to write an equation for the relationship. 4. Substitute known values into the equation to find the unknown. Example 5.70 Solving an Inverse Variation Problem A quantity y varies inversely with the cube of x. If y = 25 when x = 2, find y when x is 6. Solution The general formula for inverse variation with a cube is y = k x3. The constant can be found by multiplying y by the cube of x. k = x3 y = 23 ⋅ 25 = 200 Now we use the constant to write an equation that represents this relationship. y = k x3, k = 200 y = 200 x3 Substitute x = 6 and solve for y. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 629 y = 200 63 = 25 27 Analysis The graph of this equation is a rational function, as shown in Figure 5.97. Figure 5.97 5.45 A quantity y varies inversely with the square of x. If y = 8 when x = 3, find y when x is 4. Solving Problems Involving Joint Variation Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c, cost, varies jointly with the number of students, n, and the distance, d. Joint Variation Joint variation occurs when a variable varies directly or inversely with multiple variables. For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and inversely with z, we have x = ky z . Notice that we only use one constant in a joint variation equation. Example 5.71 Solving Problems Involving Joint Variation A quantity x varies directly with the square of y and inversely with the cube root of z. If x = 6 when y = 2 and z = 8, find x when y = 1 and z = 27. Solution 630 Chapter 5 Polynomial and Rational Functions Begin by writing an equation to show the relationship between the variables. Substitute x = 6, y = 2, and z = 8 to find the value of the constant k. x = ky2 z3 6 = k22 3 8 6 = 4k 2 3 = k Now we can substitute the value of the constant into the equation for the relationship. x = 3y2 z3 To find x when y = 1 and z = 27, we will substitute values for y and z into our equation. x = 3(1)2 3 27 = 1 5.46 A quantity x varies directly with the square of y and inversely with z. If x = 40 when y = 4 and z = 2, find x when y = 10 and z = 25. Access these online resources for additional instruction and practice with direct and inverse variation. • Direct Variation (http://openstaxcollege.org/l/directvariation) • Inverse Variation (http://openstaxcollege.org/l/inversevariatio) • Direct and Inverse Variation (http://openstaxcollege.org/l/directinverse) this website (http://openstaxcollege.org/l/PreCalcLPC03) Visit Learningpod. for additional practice questions from This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 631 5.8 EXERCISES Verbal What is true of the appearance of graphs that reflect a 531. direct variation between two variables? If two variables vary inversely, what will an equation 532. representing their relationship look like? jointly with x and z and when x = 2 and y varies z = 3, y = 36. y varies 547. jointly x = 1, z = 2, w = 5, then y = 100. as x, z, and w and when Is there a limit to the number of variables that can 533. vary jointly? Explain. y varies jointly as the square of x and the square of 548. z and when x = 3 and z = 4, then y = 72. Algebraic For the following exercises, write an equation describing the relationship of the given variables. 534. y varies directly as x and when x = 6, y = 12. y varies directly as the square of x and when 535. x = 4, y = 80. 536. x = 36, y varies directly as the square root of x and when y = 24. 552. y varies directly as 537. x = 36, y = 24. the cube of x and when y varies directly as the cube root of x and when 538. x = 27, y = 15. y varies jointly as x and the square root of z and 549. when x = 2 and z = 25, then y = 100. 550. y varies jointly as the square of x the cube of z and of W. When x = 1, z = 2, and square the w = 36, then y = 48. root 551. When x = 3, y varies jointly as x and z and inversely as w. z = 5, and w = 6, then y = 10. y varies jointly as the square of x and the square of w. When of z and root inversely x = 3, z = 4, and w = 3, then y = 6. cube the as 553. y varies jointly as x and z and inversely as the of t . When the square x = 3, z = 1, w = 25, and t = 2, then y = 6. of w and square root y varies directly as the fourth power of x and when 539. x = 1, y = 6. Numeric 540. y varies inversely as x and when x = 4, y = 2. For the following exercises, use the given information to find the unknown value. y varies inversely as the square of x and when 541. x = 3, y = 2. y varies directly as x. When x = 3, then y = 12. 554. Find y wneh x = 20. y varies inversely as the cube of x and when 542. x = 2, y = 5. y varies directly as the square of x. When x = 2, 555. then y = 16. Find y when x = 8. y varies inversely as the fourth power of x and 543. when x = 3, y = 1. y varies directly as the cube of x. When x = 3, 556. then y = 5. Find y when x = 4. y varies inversely as the square root of x and when y varies inversely as the cube root of x and when 544. x = 25, 545. x = 64, 546. y = 3. y = 5. y varies directly as the square root of x. When 557. x = 16, then y = 4. Find y when x = 36. y varies directly as the cube root of x. When 558. x = 125, then y = 15. Find y when x = 1,000. y varies 559. inversely y = 2. Find y when x = 1. with x. When x = 3, then 632 Chapter 5 Polynomial and Rational Functions y varies inversely with the square of x. When 560. x = 4, then y = 3. Find y when x = 2. y varies 561. x = 3, then y = 1. Find y when x = 1. inversely with the cube of x. When y varies inversely with the square root of x. When 562. x = 64, then y = 12. Find y when x = 36. y varies inversely with the cube root of x. When 563. x = 27, then y = 5. Find y when x = 125. y varies 564. z = 2, then y = 16. Find y when x = 3 and z = 3. as x and z. When x = 4 and jointly y varies jointly as x, 565. z = 1, and w = 12, then y = 72. Find y when x = 1, z = 2, and w = 3. z, and w. When x = 2, y varies jointly as x and the square of z. When 566. x = 2 and z = 4, then y = 144. Find y when x = 4 and z = 5. y varies jointly as the square of x and the square 567. root of z. When x = 2 and z = 9, then y = 24. Find y when x = 3 and z = 25. y varies jointly as x and z and inversely as w. 568. When x = 5, z = 2, Find y when x = 3 and z = 8, and w = 20, then y = 4. and w = 48. y varies jointly as the square of x and the cube of z 569. and inversely as the square root of w. When x = 2, z = 2, and w = 64, then y = 12. Find y when x = 1, z = 3, and w = 4. y varies jointly as the square of x and of z and 570. inversely as the square root of w and of t . When x = 2, z = 3, w = 16, and t = 3, then y = 1. Find y when w = 36, and t = 5. x = 3, z = 2, Technology For the following exercises, use a calculator to graph the equation implied by the given variation. y varies directly with the square of x and when 571. x = 2, y = 3. 572. This content is available for free at https://cnx.org/content/col11758/1.5 directly as the cube of x and when y varies x = 2, y = 4. y varies directly as the square root of x and when 573. x = 36, y = 2. y varies inversely with x and when y varies inversely as the square of x and when 574. x = 6, 575. x = 1, y = 2. y = 4. Extensions For the following exercises, use Kep
ler’s Law, which states that the square of the time, T, required for a planet to orbit the Sun varies directly with the cube of the mean distance, a, that the planet is from the Sun. Using Earth’s time of 1 year and mean distance of 93 576. million miles, find the equation relating T and a. 577. Use the result from the previous exercise to determine the time required for Mars to orbit the Sun if its mean distance is 142 million miles. Using Earth’s distance of 150 million kilometers, find 578. the equation relating T and a. 579. Use the result from the previous exercise to determine the time required for Venus to orbit the Sun if its mean distance is 108 million kilometers. 580. Using Earth’s distance of 1 astronomical unit (A.U.), determine the time for Saturn to orbit the Sun if its mean distance is 9.54 A.U. Real-World Applications For the following exercises, use the given information to answer the questions. The distance s that an object falls varies directly with 581. the square of the time, t, of the fall. If an object falls 16 feet in one second, how long for it to fall 144 feet? The velocity v of a falling object varies directly to 582. the time, t, of the fall. If after 2 seconds, the velocity of the object is 64 feet per second, what is the velocity after 5 seconds? 583. The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 24 inches long and vibrates 128 times per second, what is the length of a string that vibrates 64 times per second? 584. The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of Chapter 5 Polynomial and Rational Functions 633 a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what is the volume when the pressure is 300 millimeters of mercury? 585. The weight of an object above the surface of Earth varies inversely with the square of the distance from the center of Earth. If a body weighs 50 pounds when it is 3960 miles from Earth’s center, what would it weigh it were 3970 miles from Earth’s center? The intensity of light measured in foot-candles varies 586. inversely with the square of the distance from the light source. Suppose the intensity of a light bulb is 0.08 footcandles at a distance of 3 meters. Find the intensity level at 8 meters. The current in a circuit varies inversely with its 587. resistance measured in ohms. When the current in a circuit is 40 amperes, the resistance is 10 ohms. Find the current if the resistance is 12 ohms. The force exerted by the wind on a plane surface 588. varies jointly with the square of the velocity of the wind and with the area of the plane surface. If the area of the surface is 40 square feet surface and the wind velocity is 20 miles per hour, the resulting force is 15 pounds. Find the force on a surface of 65 square feet with a velocity of 30 miles per hour. The horsepower (hp) that a shaft can safely transmit 589. varies jointly with its speed (in revolutions per minute (rpm) and the cube of the diameter. If the shaft of a certain material 3 inches in diameter can transmit 45 hp at 100 rpm, what must the diameter be in order to transmit 60 hp at 150 rpm? The kinetic energy K of a moving object varies 590. jointly with its mass m and the square of its velocity v. If an object weighing 40 kilograms with a velocity of 15 meters per second has a kinetic energy of 1000 joules, find the kinetic energy if the velocity is increased to 20 meters per second. 634 Chapter 5 Polynomial and Rational Functions CHAPTER 5 REVIEW KEY TERMS arrow notation a way to represent symbolically the local and end behavior of a function by using arrows to indicate that an input or output approaches a value axis of symmetry a vertical line drawn through the vertex of a parabola, that opens up or down, around which the parabola is symmetric; it is defined by x = − b 2a. coefficient a nonzero real number multiplied by a variable raised to an exponent constant of variation the non-zero value k that helps define the relationship between variables in direct or inverse variation continuous function breaks in the graph a function whose graph can be drawn without lifting the pen from the paper because there are no degree the highest power of the variable that occurs in a polynomial Descartes’ Rule of Signs a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of f (x) and f ( − x) direct variation the relationship between two variables that are a constant multiple of each other; as one quantity increases, so does the other Division Algorithm given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x) , there exist unique polynomials q(x) and r(x) such that f (x) = d(x)q(x) + r(x) where q(x) is the quotient and r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). end behavior the behavior of the graph of a function as the input decreases without bound and increases without bound Factor Theorem k is a zero of polynomial function f (x) if and only if (x − k) is a factor of f (x) Fundamental Theorem of Algebra a polynomial function with degree greater than 0 has at least one complex zero general form of a quadratic function the function that describes a parabola, written in the form f (x) = ax2 + bx + c , where a, b, and c are real numbers and a ≠ 0. global maximum highest turning point on a graph; f (a) where f (a) ≥ f (x) for all x. global minimum lowest turning point on a graph; f (a) where f (a) ≤ f (x) for all x. horizontal asymptote a horizontal line y = b where the graph approaches the line as the inputs increase or decrease without bound. Intermediate Value Theorem for two numbers a and b in the domain of f , if a < b and f (a) ≠ f (b), then the function f takes on every value between f (a) and f (b) ; specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the x- axis inverse variation the relationship between two variables in which the product of the variables is a constant inversely proportional a relationship where one quantity is a constant divided by the other quantity; as one quantity increases, the other decreases invertible function any function that has an inverse function joint variation a relationship where a variable varies directly or inversely with multiple variables leading coefficient the coefficient of the leading term This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 635 leading term the term containing the highest power of the variable Linear Factorization Theorem allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form (x − c) , where c is a complex number multiplicity the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form (x − h) p , x = h is a zero of multiplicity p. polynomial function a function that consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. power function a function that can be represented in the form f (x) = kx p where k is a constant, the base is a variable, and the exponent, p , is a constant rational function a function that can be written as the ratio of two polynomials Rational Zero Theorem the possible rational zeros of a polynomial function have the form p q where p is a factor of the constant term and q is a factor of the leading coefficient. Remainder Theorem if a polynomial f (x) is divided by x − k , then the remainder is equal to the value f (k) removable discontinuity a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function roots in a given function, the values of x at which y = 0 , also called zeros smooth curve a graph with no sharp corners standard form of a quadratic function the function that describes a parabola, written in the form f (x) = a(x − h)2 + k , where (h, k) is the vertex synthetic division a shortcut method that can be used to divide a polynomial by a binomial of the form x − k term of a polynomial function any ai xi of f (x) = an xn + ... + a2 x2 + a1 x + a0 a polynomial function in the form turning point the location at which the graph of a function changes direction varies directly a relationship where one quantity is a constant multiplied by the other quantity varies inversely a relationship where one quantity is a constant divided by the other quantity vertex the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function vertex form of a quadratic function another name for the standard form of a quadratic function vertical asymptote approach a a vertical line x = a where the graph tends toward positive or negative infinity as the inputs zeros in a given function, the values of x at which y = 0 , also called roots KEY EQUATIONS general form of a quadratic function f (x) = ax2 + bx + c standard form of a quadratic function f (x) = a(x − h)2 + k 636 Chapter 5 Polynomial and Rational Functions general form of a polynomial function f (x) = an xn + ... + a2 x2 + a1 x + a0 Division Algorithm f (x) = d(x)q(x) + r(x) where q(x) ≠ 0 Rational Function f (x) = P(x) Q(x) = a p x p bq xq + a p − 1 x + bq − 1 x p − 1 q − 1 + ... + a1 x + a0 + ... + b1 x + b0 , Q(x) ≠ 0 Direct variation y = kxn , k is a nonzero constant. Inverse variation y = k xn, k is a nonzero constant. KEY CONCEPTS 5.1 Quadratic Functions • A polynomial function of degree two is called a quadratic functi
on. • The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down. • The axis of symmetry is the vertical line passing through the vertex. The zeros, or x- intercepts, are the points at which the parabola crosses the x- axis. The y- intercept is the point at which the parabola crosses the y- axis. See Example 5.1, Example 5.7, and Example 5.8. • Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph. See Example 5.2. • The vertex can be found from an equation representing a quadratic function. See Example 5.3. • The domain of a quadratic function is all real numbers. The range varies with the function. See Example 5.4. • A quadratic function’s minimum or maximum value is given by the y- value of the vertex. • The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. See Example 5.5 and Example 5.6. • The vertex and the intercepts can be identified and interpreted to solve real-world problems. See Example 5.9. 5.2 Power Functions and Polynomial Functions • A power function is a variable base raised to a number power. See Example 5.10. • The behavior of a graph as the input decreases beyond bound and increases beyond bound is called the end behavior. • The end behavior depends on whether the power is even or odd. See Example 5.11 and Example 5.12. • A polynomial function is the sum of terms, each of which consists of a transformed power function with positive whole number power. See Example 5.13. • The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term containing the highest power of the variable is called the leading term. The coefficient of the leading term is called the leading coefficient. See Example 5.14. • The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. See Example 5.15 and Example 5.16. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 637 • A polynomial of degree n will have at most n x-intercepts and at most n − 1 turning points. See Example 5.17, Example 5.18, Example 5.19, Example 5.20, and Example 5.21. 5.3 Graphs of Polynomial Functions • Polynomial functions of degree 2 or more are smooth, continuous functions. See Example 5.22. • To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero. See Example 5.23, Example 5.24, and Example 5.25. • Another way to find the x- intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the x- axis. See Example 5.26. • The multiplicity of a zero determines how the graph behaves at the x- intercepts. See Example 5.27. • The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity. • The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity. • The end behavior of a polynomial function depends on the leading term. • The graph of a polynomial function changes direction at its turning points. • A polynomial function of degree n has at most n − 1 turning points. See Example 5.28. • To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most n − 1 turning points. See Example 5.29 and Example 5.31. • Graphing a polynomial function helps to estimate local and global extremas. See Example 5.32. • The Intermediate Value Theorem tells us that if f (a) and f (b) have opposite signs, then there exists at least one value c between a and b for which f (c) = 0. See Example 5.30. 5.4 Dividing Polynomials • Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See Example 5.33 and Example 5.34. • The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder. • Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form x − k. See Example 5.35, Example 5.36, and Example 5.37. • Polynomial division can be used to solve application problems, including area and volume. See Example 5.38. 5.5 Zeros of Polynomial Functions • To find f (k), determine the remainder of the polynomial f (x) when it is divided by x − k. This is known as the Remainder Theorem. See Example 5.39. • According to the Factor Theorem, k is a zero of f (x) if and only if (x − k) is a factor of f (x). See Example 5.40. • According to the Rational Zero Theorem, each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of the leading coefficient. See Example 5.41 and Example 5.42. • When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. • Synthetic division can be used to find the zeros of a polynomial function. See Example 5.43. • According to the Fundamental Theorem, every polynomial function has at least one complex zero. See Example 5.44. • Every polynomial function with degree greater than 0 has at least one complex zero. 638 Chapter 5 Polynomial and Rational Functions • Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form (x − c), where c is a complex number. See Example 5.45. • The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer. • The number of negative real zeros of a polynomial function is either the number of sign changes of f ( − x) or less than the number of sign changes by an even integer. See Example 5.46. • Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division. See Example 5.47. 5.6 Rational Functions • We can use arrow notation to describe local behavior and end behavior of the toolkit functions f (x) = 1 x and f (x) = 1 x2. See Example 5.48. • A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote. See Example 5.49. • Application problems involving rates and concentrations often involve rational functions. See Example 5.50. • The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. See Example 5.51. • The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. See Example 5.52. • A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. See Example 5.53. • A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. See Example 5.54, Example 5.55, Example 5.56, and Example 5.57. • Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. See • Example 5.58. If a rational function has x-intercepts at x = x1, x2, … , xn, vertical asymptotes at x = v1, v2, … , vm, and no xi = any v j, then the function can be written in the form f (x) = a(x − x1) (x − v1) p1 (x − x2) q1 (x − v2) p2 ⋯ (x − xn) q2 ⋯ (x − vm) pn qn See Example 5.59. 5.7 Inverses and Radical Functions • The inverse of a quadratic function is a square root function. • If f −1 is the inverse of a function f , then f is the inverse of the function f −1. See Example 5.60. • While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See Example 5.61. • To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one. See Example 5.62 and Example 5.63. • When finding the inverse of a radical function, we need a restriction on the domain of the answer. See Example 5.64 and Example 5.66. • Inverse and radical and functions can be used to solve application problems. See Example 5.65 and Example 5.67. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 639 5.8 Modeling Using Variation • A relationship where one quantity is a constant multiplied by another quantity is called direct variation. See Example 5.68. • Two variables that are directly proportional to one another will have a constant ratio. • A relationship where one quantity is a constant divided by another quantity is called inverse variation. See Example 5.69. • Two variables that are inversely proportional to one another will have a constant multiple. See Example 5.70. • In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation. See Example 5.71. CHAPTER 5 REVIEW EXERCISES Quadratic Functions For the following exercises, write the quadratic function in standard form. Then give the vertex and axes intercepts. Finally, graph the function. 598. f (x) = 5 x + 1 − x2 599. f (x) = x2 ⎛ ⎝3 − 6x + x2⎞ ⎠ 591. f (x) = x2 − 4x − 5 592. f (x) = − 2x2 − 4x For the following exercises, determine end behavior of the polynomial function. 600. f (x) = 2x4 + 3x3 − 5x2 + 7 For the following exercises, find the equation of the quadratic function using the given information. 593. The vertex is ( – 2, 3) and a point on the graph is (3, 6). 601. f (x) = 4x3 − 6x2 + 2 602. f (x) = 2x2(1 + 3x − x2) 594. The vertex is ( – 3, 6.5) and a point on the graph is (2, 6). Graphs of Polynomial Functions For the following exercises, find all ze
ros of the polynomial function, noting multiplicities. For the following exercises, complete the task. 603. f (x) = (x + 3)2(2x − 1)(x + 1)3 604. f (x) = x5 + 4x4 + 4x3 605. f (x) = x3 − 4x2 + x − 4 For the following exercises, based on the given graph, determine the zeros of the function and note multiplicity. 606. 595. A rectangular plot of land is to be enclosed by fencing. One side is along a river and so needs no fence. If the total fencing available is 600 meters, find the dimensions of the plot to have maximum area. 596. An object projected from the ground at a 45 degree angle with initial velocity of 120 feet per second has height, h, in terms of horizontal distance traveled, x, given by h(x) = −32 (120)2 x2 + x. Find the maximum height the object attains. Power Functions and Polynomial Functions For the following exercises, determine if the function is a polynomial function and, if so, give the degree and leading coefficient. 597. f (x) = 4x5 − 3x3 + 2x − 1 640 607. 608. Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the function f (x) = x3 − 5x + 1 Chapter 5 Polynomial and Rational Functions 616. 3x3 + 11x2 + 8x − 4 = 0 617. 2x4 − 17x3 + 46x2 − 43x + 12 = 0 618. 4x4 + 8x3 + 19x2 + 32x + 12 = 0 For the following exercises, use Descartes’ Rule of Signs to find the possible number of positive and negative solutions. 619. x3 − 3x2 − 2x + 4 = 0 620. 2x4 − x3 + 4x2 − 5x + 1 = 0 Rational Functions For the following exercises, find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph of the function. Dividing Polynomials For the following exercises, use long division to find the quotient and remainder. 622. f (x) = 621. f (x) = x + 2 x − 5 x2 + 1 x2 − 4 609. x3 − 2x2 + 4x + 4 x − 2 610. 3x4 − 4x2 + 4x + 8 x + 1 For the following exercises, use synthetic division to find the quotient. If the divisor is a factor, then write the factored form. 611. x3 − 2x2 + 5x − 1 x + 3 612. x3 + 4x + 10 x − 3 613. 2x3 + 6x2 − 11x − 12 x + 4 614. 3x4 + 3x3 + 2x + 2 x + 1 Zeros of Polynomial Functions 623. f (x) = 3x2 − 27 x2 + x − 2 624. f (x) = x + 2 x2 − 9 For the following exercises, find the slant asymptote. 625. f (x) = x2 − 1 x + 2 626. f (x) = 2x3 − x2 + 4 x2 + 1 Inverses and Radical Functions For the following exercises, find the inverse of the function with the domain given. 627. f (x) = (x − 2)2, x ≥ 2 628. f (x) = (x + 4)2 − 3, x ≥ − 4 For the following exercises, use the Rational Zero Theorem to help you solve the polynomial equation. 629. f (x) = x2 + 6x − 2, x ≥ − 3 615. 2x3 − 3x2 − 18x − 8 = 0 630. f (x) = 2x3 − 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 5 Polynomial and Rational Functions 641 631. f (x) = 4x + 5 − 3 632. f (x) = x − 3 2x + 1 Modeling Using Variation For the following exercises, find the unknown value. y varies directly as the square of x. 633. x = 3, y = 36, find y if x = 4. If when y varies inversely as the square root of x If when 634. x = 25, y = 2, find y if x = 4. y varies jointly as the cube of x and as z. If when 635. x = 1 and z = 2, y = 6, find y if x = 2 and z = 3. y varies jointly as x and the square of z and 636. inversely as the cube of w. If when x = 3, z = 4, and w = 2, y = 48, find y if x = 4, z = 5, and w = 3. For the following exercises, solve the application problem. 637. The weight of an object above the surface of the earth varies inversely with the distance from the center of the earth. If a person weighs 150 pounds when he is on the surface of the earth (3,960 miles from center), find the weight of the person if he is 20 miles above the surface. 638. The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K. CHAPTER 5 PRACTICE TEST Give the degree and leading coefficient of the following polynomial function. Given information about the graph of a quadratic function, find its equation. 639. f (x) = x3 ⎛ ⎝3 − 6x2 − 2x2⎞ ⎠ 643. Vertex (2, 0) and point on graph (4, 12). Determine the end behavior of the polynomial function. 640. f (x) = 8x3 − 3x2 + 2x − 4 641. f (x) = − 2x2(4 − 3x − 5x2) Solve the following application problem. 644. A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed. Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function. Find all zeros of the following polynomial functions, noting multiplicities. 642. f (x) = x2 + 2x − 8 645. f (x) = (x − 3)3(3x − 1)(x − 1)2 642 Chapter 5 Polynomial and Rational Functions 646. f (x) = 2x6 − 6x5 + 18x4 657. 8x3 − 21x2 + 6 = 0 Based on the graph, determine the zeros of the function and multiplicities. For the following rational functions, find the intercepts and horizontal and vertical asymptotes, and sketch a graph. 647. Use long division to find the quotient. 648. 2x3 + 3x − 4 x + 2 Use synthetic division to find the quotient. If the divisor is a factor, write the factored form. 649. x4 + 3x2 − 4 x − 2 650. 2x3 + 5x2 − 7x − 12 x + 3 658. f (x) = x + 4 x2 − 2x − 3 659. f (x) = x2 + 2x − 3 x2 − 4 Find the slant asymptote of the rational function. 660. f (x) = x2 + 3x − 3 x − 1 Find the inverse of the function. 661. f (x) = x − 2 + 4 662. f (x) = 3x3 − 4 663. f (x) = 2x + 3 3x − 1 Find the unknown value. y varies inversely as the square of x and when 664. x = 3, y = 2. Find y if x = 1. Use the Rational Zero Theorem to help you find the zeros of the polynomial functions. 651. f (x) = 2x3 + 5x2 − 6x − 9 y varies jointly with x and the cube root of z. If 665. when x = 2 and z = 27, y = 12, find y if x = 5 and z = 8. 652. f (x) = 4x4 + 8x3 + 21x2 + 17x + 4 Solve the following application problem. 666. The distance a body falls varies directly as the square of the time it falls. If an object falls 64 feet in 2 seconds, how long will it take to fall 256 feet? 653. f (x) = 4x4 + 16x3 + 13x2 − 15x − 18 654. f (x) = x5 + 6x4 + 13x3 + 14x2 + 12x + 8 Given the following information about a polynomial function, find the function. It has a double zero at x = 3 and zeros at x = 1 and 655. x = − 2 . Its y-intercept is (0, 12). 656. It has a zero of multiplicity 3 at x = 1 2 zero at x = − 3 . It contains the point (1, 8). and another Use Descartes’ Rule of Signs to determine the possible number of positive and negative solutions. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 643 6 \| EXPONENTIAL AND LOGARITHMIC FUNCTIONS Figure 6.1 Electron micrograph of E.Coli bacteria (credit: “Mattosaurus,” Wikimedia Commons) Chapter Outline 6.1 Exponential Functions 6.2 Graphs of Exponential Functions 6.3 Logarithmic Functions 6.4 Graphs of Logarithmic Functions 6.5 Logarithmic Properties 6.6 Exponential and Logarithmic Equations 6.7 Exponential and Logarithmic Models 6.8 Fitting Exponential Models to Data Introduction Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see hundreds of thousands of microscopic organisms. They are bacteria, and they are not only on your skin, but in your mouth, 644 Chapter 6 Exponential and Logarithmic Functions nose, and even your intestines. In fact, the bacterial cells in your body at any given moment outnumber your own cells. But that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to the body. Bacteria commonly reproduce through a process called binary fission, during which one bacterial cell splits into two. When conditions are right, bacteria can reproduce very quickly. Unlike humans and other complex organisms, the time required to form a new generation of bacteria is often a matter of minutes or hours, as opposed to days or years.[1] For simplicity’s sake, suppose we begin with a culture of one bacterial cell that can divide every hour. Table 6.1 shows the number of bacterial cells at the end of each subsequent hour. We see that the single bacterial cell leads to over one thousand bacterial cells in just ten hours! And if we were to extrapolate the table to twenty-four hours, we would have over 16 million! Hour Bacteria 0 1 1 2 2 4 3 8 Table 6.1 4 5 6 7 8 9 10 16 32 64 128 256 512 1024 In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data. 6.1 \| Exponential Functions Learning Objectives In this section, you will: 6.1.1 Evaluate exponential functions. 6.1.2 Find the equation of an exponential function. 6.1.3 Use compound interest formulas. 6.1.4 Evaluate exponential functions with base e. India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year[2]. If this rate continues, the population of India will exceed China’s population by the year 2031. When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth. Identifying Exponential Functions When exploring linear growth, we observed a constant rate o
f change—a constant number by which the output increased for each unit increase in input. For example, in the equation f (x) = 3x + 4, the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people. Defining an Exponential Function A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2050. 1. Todar, PhD, Kenneth. Todar's Online Textbook of Bacteriology. http://textbookofbacteriology.net/growth_3.html. 2. http://www.worldometers.info/world-population/. Accessed February 24, 2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 645 What exactly does it mean to grow exponentially? What does the word double have in common with percent increase? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media. • Percent change refers to a change based on a percent of the original amount. • Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time. • Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time. For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth. We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See Table 6.2. f(x) = 2 x g(x) = 2x 1 2 4 8 16 32 64 0 2 4 6 8 10 12 x 0 1 2 3 4 5 6 Table 6.2 From Table 6.2 we can infer that for these two functions, exponential growth dwarfs linear growth. • Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain. • Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain. Apparently, the difference between “the same percentage” and “the same amount” is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one. The general form of the exponential function is f (x) = ab x not equal to 1. , where a is any nonzero number, b is a positive real number • • If b > 1, the function grows at a rate proportional to its size. If 0 < b < 1, the function decays at a rate proportional to its size. Let’s look at the function f (x) = 2 outputs over an interval in the domain from −3 to 3. x from our example. We will create a table (Table 6.3) to determine the corresponding 646 Chapter 6 Exponential and Logarithmic Functions x −3 −2 −1 0 1 2 3 f(x) = 2 x 2−3 = 1 8 2−2 = 1 4 2−1 = 1 2 20 = 1 21 = 2 22 = 4 23 = 8 Table 6.3 Let us examine the graph of f by plotting the ordered pairs we observe on the table in Figure 6.2, and then make a few observations. Figure 6.2 Let’s define the behavior of the graph of the exponential function f (x) = 2 x and highlight some its key characteristics. • • the domain is (−∞, ∞), the range is (0, ∞), • as x → ∞, f (x) → ∞, • as x → − ∞, f (x) → 0, • • • • f (x) is always increasing, the graph of f (x) will never touch the x-axis because base two raised to any exponent never has the result of zero. y = 0 is the horizontal asymptote. the y-intercept is 1. Exponential Function For any real number x, an exponential function is a function with the form f (x) = ab x (6.1) where This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 647 • • a is the a non-zero real number called the initial value and b is any positive real number such that b ≠ 1. • The domain of f is all real numbers. • The range of f is all positive real numbers if a > 0. • The range of f is all negative real numbers if a < 0. • The y-intercept is (0, a), and the horizontal asymptote is y = 0. Example 6.1 Identifying Exponential Functions Which of the following equations are not exponential functions? • • • • f (x) = 4 3(x − 2) g(x) = x3 h(x(x) = (−2) x Solution By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, g(x) = x3 does not represent an exponential function because the base is an independent variable. In fact, g(x) = x3 is a power function. Recall that the base b of an exponential function is always a positive constant, and b ≠ 1. Thus, j(x) = (−2) does not represent an exponential function because the base, −2, is less than 0. x 6.1 Which of the following equations represent exponential functions? • • • • f (x) = 2x2 − 3x + 1 g(x) = 0.875 x h(x) = 1.75x + 2 j(x) = 1095.6−2x Evaluating Exponential Functions Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive: • Let b = − 9 and x = 1 2 . Then f (x) = f ⎛ ⎝ ⎞ ⎠ = (−9) 1 2 1 2 = −9, which is not a real number. 648 Chapter 6 Exponential and Logarithmic Functions Why do we limit the base to positive values other than 1 ? Because base 1 results in the constant function. Observe what happens if the base is 1 : • Let b = 1. Then f (x) = 1 = 1 for any value of x. To evaluate an exponential function with the form f (x) = b x the resulting power. For example: x , we simply substitute x with the given value, and calculate Let f (x) = 2 x . What is f (3) ? x f (x) = 2 f (3) = 23 Substitute x = 3. = 8 Evaluate the power. To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example: Let f (x) = 30(2) x . What is f (3) ? x f (x) = 30(2) f (3) = 30(2)3 Substitute x = 3. = 30(8) Simplify the power fir t. = 240 Multiply. Note that if the order of operations were not followed, the result would be incorrect: f (3) = 30(2)3 ≠ 603 = 216,000 Example 6.2 Evaluating Exponential Functions Let f (x) = 5(3) x + 1. Evaluate f (2) without using a calculator. Solution Follow the order of operations. Be sure to pay attention to the parentheses. x + 1 f (x) = 5(3) f (2) = 5(3)2 + 1 Substitute x = 2. = 5(3)3 = 5(27) = 135 Add the exponents. Simplify the power. Multiply. 6.2 Let f (x) = 8(1.2) x − 5. Evaluate f (3) using a calculator. Round to four decimal places. Defining Exponential Growth Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 649 Exponential Growth A function that models exponential growth grows by a rate proportional to the amount present. For any real number x and any positive real numbers a and b such that b ≠ 1, an exponential growth function has the form f (x) = ab x where • a is the initial or starting value of the function. • b is the growth factor or growth multiplier per unit x . In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function A(x) = 100 + 50x. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function B(x) = 100(1 + 0.5) x . A few years of growth for these companies are illustrated in Table 6.4. Year, x Stores, Company A Stores, Company B 100 + 50(0) = 100 100(1 + 0.5)0 = 100 100 + 50(1) = 150 100(1 + 0.5)1 = 150 100 + 50(2) = 200 100(1 + 0.5)2 = 225 100 + 50(3) = 250 100(1 + 0.5)3 = 337.5 A(x) = 100 + 50x B(x) = 100(1 + 0.5) x 0 1 2 3 x Table 6.4 The graphs comparing the number of stores for each company over a five-year period are shown in Figure 6.3. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth. 650 Chapter 6 Exponential and Logarithmic Functions Figure 6.3 The graph shows the numbers of stores Companies A and B opened over a five-year period. Notice that the domain for both functions is [0, ∞), and the range for both functions is [100, ∞). After year 1, Company B always has more stores than Company A. . Now we will turn our attention to the function representing the number of stores for Company B, B(x) = 100(1 + 0.5) In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and 1 + 0.5 = 1.5 represents the growth factor. Generalizing further, we can write this function as B(x) = 100(1.5) , where 100 is the initial value, 1.5 is called the base,
and x is called the exponent. x x Example 6.3 Evaluating a Real-World Exponential Model At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 651 P(t) = 1.25(1.012) population of India be in 2031? t , where t is the number of years since 2013. To the nearest thousandth, what will the Solution To estimate the population in 2031, we evaluate the models for t = 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth, There will be about 1.549 billion people in India in the year 2031. P(18) = 1.25(1.012)18 ≈ 1.549 6.3 The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function P(t) = 1.39(1.006) , where t is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 6.3? t Finding Equations of Exponential Functions In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a and b, and evaluate the function. Given two data points, write an exponential model. 1. 2. If one of the data points has the form (0, a), point into the equation f (x) = a(b) x , and solve for b. then a is the initial value. Using a, substitute the second If neither of the data points have the form (0, a), f (x) = a(b) x . Solve the resulting system of two equations in two unknowns to find a and b. substitute both points into two equations with the form 3. Using the a and b found in the steps above, write the exponential function in the form f (x) = a(b) x . Example 6.4 Writing an Exponential Model When the Initial Value Is Known In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population (N) of deer over time t. Solution We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation N(t) = 80bt to find b : 652 Chapter 6 Exponential and Logarithmic Functions N(t) = 80bt 180 = 80b6 9 4 = b6 .1447 Substitute using point (6, 180). Divide and write in lowest terms. Isolate b using properties of exponents. Round to 4 decimal places. NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section. . (Note that this exponential function The exponential model for the population of deer is N(t) = 80(1.1447) models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.) t We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure 6.4 passes through the initial points given in the problem, (0, 80) and (6, 180). We can also see that the domain for the function is [0, ∞), and the range for the function is [80, ∞). Figure 6.4 Graph showing the population of deer over time, N(t) = 80(1.1447) t years after 2006 , t This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 653 6.4 A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013, the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N of wolves over time t. Example 6.5 Writing an Exponential Model When the Initial Value is Not Known Find an exponential function that passes through the points (−2, 6) and (2, 1). Solution Because we don’t have the initial value, we substitute both points into an equation of the form f (x) = ab x then solve the system for a and b. , and • Substituting (−2, 6) gives 6 = ab−2 • Substituting (2, 1) gives 1 = ab2 Use the first equation to solve for a in terms of b : Substitute a in the second equation, and solve for b : Use the value of b in the first equation to solve for the value of a : Thus, the equation is f (x) = 2.4492(0.6389) x . We can graph our model to check our work. Notice that the graph in Figure 6.5 passes through the initial points given in the problem, (−2, 6) and (2, 1). The graph is an example of an exponential decay function. 654 Chapter 6 Exponential and Logarithmic Functions Figure 6.5 The graph of f (x) = 2.4492(0.6389) exponential decay. x models 6.5 Given the two points (1, 3) and (2, 4.5), find the equation of the exponential function that passes through these two points. Do two points always determine a unique exponential function? Yes, provided the two points are either both above the x-axis or both below the x-axis and have different xcoordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x, which in many real world cases involves time. Given the graph of an exponential function, write its equation. 1. First, identify two points on the graph. Choose the y-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error. 2. 3. If one of the data points is the y-intercept (0, a) , then a is the initial value. Using a, x second point into the equation f (x) = a(b) , and solve for b. substitute the If neither of the data points have the form (0, a), f (x) = a(b) x . Solve the resulting system of two equations in two unknowns to find a and b. substitute both points into two equations with the form 4. Write the exponential function, f (x) = a(b) x . Example 6.6 Writing an Exponential Function Given Its Graph This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 655 Find an equation for the exponential function graphed in Figure 6.6. Figure 6.6 Solution We can choose the y-intercept of the graph, (0, 3), as our first point. This gives us the initial value, a = 3. Next, choose a point on the curve some distance away from (0, 3) that has integer coordinates. One such point is (2, 12). y = ab x y = 3b x 12 = 3b2 4 = b2 b = ± 2 Write the general form of an exponential equation. Substitute the initial value 3 for a. Substitute in 12 for y and 2 for x. Divide by 3. Take the square root. Because we restrict ourselves to positive values of b, we will use b = 2. Substitute a and b into the standard form to yield the equation f (x) = 3(2) x . 6.6 Find an equation for the exponential function graphed in Figure 6.7. Figure 6.7 656 Chapter 6 Exponential and Logarithmic Functions Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 1. Press [STAT]. 2. Clear any existing entries in columns L1 or L2. 3. 4. In L1, enter the x-coordinates given. In L2, enter the corresponding y-coordinates. 5. Press [STAT] again. Cursor right to CALC, scroll down to ExpReg (Exponential Regression), and press [ENTER]. 6. The screen displays the values of a and b in the exponential equation y = a ⋅ b x . Example 6.7 Using a Graphing Calculator to Find an Exponential Function Use a graphing calculator to find the exponential equation that includes the points (2, 24.8) and (5, 198.4). Solution Follow the guidelines above. First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4. Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values a = 6.2 and b = 2 will be displayed. The exponential equation is y = 6.2 ⋅ 2 x . Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 6.7 481.07). Applying the Compound-Interest Formula Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account. The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing. We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t, principal P, APR r, and number of compounding periods in a year n : A(t) = P⎛ ⎝1 + r n nt ⎞ ⎠ For example, observe Table 6.5, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 6 Exponential and Logarithmic Functions 657 Frequency Value aft

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