Split (1)

train · 15k rows

text stringlengths 235 3.08k
culus involved, followed by several pages of algebra. 1.6 Graphs of Functions 105 Example 1.6.5. Let f (x) =. Use a graphing calculator to approximate the intervals on which f is increasing and those on which it is decreasing. Approximate all extrema. 15x x2 + 3 Solution. Entering this function into the calculator gives Using the Minimum and Maximum features, we get To two decimal places, f appears to have its only local minimum at (−1.73, −4.33) and its only local maximum at (1.73, 4.33). Given the symmetry about the origin suggested by the graph, the relation between these points shouldn’t be too surprising. The function appears to be increasing on [−1.73, 1.73] and decreasing on (−∞, −1.73]∪[1.73, ∞). This makes −4.33 the (absolute) minimum and 4.33 the (absolute) maximum. Example 1.6.6. Find the points on the graph of y = (x − 3)2 which are closest to the origin. Round your answers to two decimal places. Solution. Suppose a point (x, y) is on the graph of y = (x − 3)2. Its distance to the origin (0, 0) is given by d = (x − 0)2 + (y − 0)2 = = = x2 + y2 x2 + [(x − 3)2]2 x2 + (x − 3)4 Since y = (x − 3)2 x2 + (x − 3)4 is the distance from (0, 0) to the point (x, y) Given a value for x, the formula d = on the curve y = (x − 3)2. What we have deﬁned, then, is a function d(x) which we wish to 106 Relations and Functions minimize over all values of x. To accomplish this task analytically would require Calculus so as we’ve mentioned before, we can use a graphing calculator to ﬁnd an approximate solution. Using the calculator, we enter the function d(x) as shown below and graph. Using the Minimum feature, we see above on the right that the (absolute) minimum occurs near x = 2. Rounding to two decimal places, we get that the minimum distance occurs when x =
2.00. To ﬁnd the y value on the parabola associated with x = 2.00, we substitute 2.00 into the equation to get y = (x − 3)2 = (2.00 − 3)2 = 1.00. So, our ﬁnal answer is (2.00, 1.00).16 (What does the y value listed on the calculator screen mean in this problem?) 16It seems silly to list a ﬁnal answer as (2.00, 1.00). Indeed, Calculus conﬁrms that the exact answer to this problem is, in fact, (2, 1). As you are well aware by now, the authors are overly pedantic, and as such, use the decimal places to remind the reader that any result garnered from a calculator in this fashion is an approximation, and should be treated as such. 1.6 Graphs of Functions 107 1.6.2 Exercises In Exercises 1 - 12, sketch the graph of the given function. State the domain of the function, identify any intercepts and test for symmetry. 1. f (x) = 2 − x 4. f (x) = 4 − x2 7. f (x) = x(x − 1)(x + 2) 2. f (x) = x − 2 3 5. f (x) = 2 √ 8. f (x) = x − 2 10. f (x) = 3 − 2 √ x + 2 √ 11. f (x) = 3 x 3. f (x) = x2 + 1 6. f (x) = x3 √ 9. f (x) = 5 − x 12. f (x) = 1 x2 + 1 In Exercises 13 - 20, sketch the graph of the given piecewise-deﬁned function. 13. f (x) = 15. f (x) = 17. f (x) = 19. f (x) =       4 − x 2 if if x ≤ 3 x > 3 −3 2x − 3 3 if if if 2x − 4 3x if if x < 0 x ≥ 0 x2 3 − x 4 x ≤ −2 if if −2
< x < 2 x ≥ 2 if 14. f (x) = 16. f (x) =    x2 2x if if x ≤ 0 x > 0 x2 − 4 4 − x2 x2 − 4 x ≤ −2 if if −2 < x < 2 x ≥ 2 if 18. f (x if −4 ≤ x < 5 x ≥ 5 if 20. f (x) =    1 x x x √ if −6 < x < −1 if − if In Exercises 21 - 41, determine analytically if the following functions are even, odd or neither. 21. f (x) = 7x 22. f (x) = 7x + 2 23. f (x) = 7 24. f (x) = 3x2 − 4 25. f (x) = 4 − x2 26. f (x) = x2 − x − 6 27. f (x) = 2x3 − x 30. f (x) = x3 + x2 + x + 1 28. f (x) = −x5 + 2x3 − x √ 31. f (x) = 1 − x 33. f (x) = 0 √ 34. f (x) = 3 x 29. f (x) = x6 − x4 + x2 + 9 √ 1 − x2 32. f (x) = 35. f (x) = 3√ x2 108 Relations and Functions 36. f (x) = 3 x2 39. f (x) = x2 − 3 x − 4x3 37. f (x) = 2x − 1 x + 1 40. f (x) = √ 9 4 − x2 38. f (x) = 41. f (x) = 3x x2 + 1 3√ x3 + x 5x In Exercises 42 - 57, use the graph of y = f (x) given below to answer the question. y 5 4 3 2 1 −5 −4 −3 −2 −1 −1 1 2 3 4 5 x −2 −3 −4 −5 42. Find the domain of f. 43. Find the range of f. 44. Determine f (−2). 45
. Solve f (x) = 4. 46. List the x-intercepts, if any exist. 47. List the y-intercepts, if any exist. 48. Find the zeros of f. 49. Solve f (x) ≥ 0. 50. Find the number of solutions to f (x) = 1. 51. Does f appear to be even, odd, or neither? 52. List the intervals where f is increasing. 53. List the intervals where f is decreasing. 54. List the local maximums, if any exist. 55. List the local minimums, if any exist. 56. Find the maximum, if it exists. 57. Find the minimum, if it exists. 1.6 Graphs of Functions 109 In Exercises 58 - 73, use the graph of y = f (x) given below to answer the question. y 5 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −5 58. Find the domain of f. 59. Find the range of f. 60. Determine f (2). 61. Solve f (x) = −5. 62. List the x-intercepts, if any exist. 63. List the y-intercepts, if any exist. 64. Find the zeros of f. 65. Solve f (x) ≤ 0. 66. Find the number of solutions to f (x) = 3. 67. Does f appear to be even, odd, or neither? 68. List the intervals where f is increasing. 69. List the intervals where f is decreasing. 70. List the local maximums, if any exist. 71. List the local minimums, if any exist. 72. Find the maximum, if it exists. 73. Find the minimum, if it exists. In Exercises 74 - 77, use your graphing calculator to approximate the local and absolute extrema of the given function. Approximate the intervals on which the function is increasing and those on which it is decreasing. Round your answers to two decimal places. 74. f (x) = x4 − 3x3 − 24x2 + 28x + 48 75. f (x) = x2/3(x − 4) 76. f (x) = √ 9 − x2 77. f (x) = x √ 9 − x2 110 Relations
and Functions In Exercises 78 - 85, use the graphs of y = f (x) and y = g(x) below to ﬁnd the function valuex) y = g(x) 78. (f + g)(0) 79. (f + g)(1) 80. (f − g)(1) 81. (g − f )(2) 82. (f g)(2) 83. (f g)(1) 84. f g (4) 85. g f (2) The graph below represents the height h of a Sasquatch (in feet) as a function of its age N in years. Use it to answer the questions in Exercises 86 - 90. y 8 6 4 2 15 30 45 60 N y = h(N ) 86. Find and interpret h(0). 87. How tall is the Sasquatch when she is 15 years old? 88. Solve h(N ) = 6 and interpret. 89. List the interval over which h is constant and interpret your answer. 90. List the interval over which h is decreasing and interpret your answer. 1.6 Graphs of Functions 111 For Exercises 91 - 93, let f (x) = x be the greatest integer function as deﬁned in Exercise 75 in Section 1.4. 91. Graph y = f (x). Be careful to correctly describe the behavior of the graph near the integers. 92. Is f even, odd, or neither? Explain. 93. Discuss with your classmates which points on the graph are local minimums, local maximums or both. Is f ever increasing? Decreasing? Constant? In Exercises 94 - 95, use your graphing calculator to show that the given function does not have any extrema, neither local nor absolute. 94. f (x) = x3 + x − 12 95. f (x) = −5x + 2 96. In Exercise 71 in Section 1.4, we saw that the population of Sasquatch in Portage County could be modeled by the function P (t) = 150t t + 15 your graphing calculator to analyze the general function behavior of P. Will there ever be a time when 200 Sasquatch roam Portage County?, where t = 0 represents the year 1803. Use 97. Suppose f and g are both even functions. What can be said about the functions f + g, f − g, f
g and f g? What if f and g are both odd? What if f is even but g is odd? 98. One of the most important aspects of the Cartesian Coordinate Plane is its ability to put Algebra into geometric terms and Geometry into algebraic terms. We’ve spent most of this chapter looking at this very phenomenon and now you should spend some time with your classmates reviewing what we’ve done. What major results do we have that tie Algebra and Geometry together? What concepts from Geometry have we not yet described algebraically? What topics from Intermediate Algebra have we not yet discussed geometrically? It’s now time to “thoroughly vet the pathologies induced” by the precise deﬁnitions of local maximum and local minimum. We’ll do this by providing you and your classmates a series of Exercises to discuss. You will need to refer back to Deﬁnition 1.10 (Increasing, Decreasing and Constant) and Deﬁnition 1.11 (Maximum and Minimum) during the discussion. 99. Consider the graph of the function f given below. y 3 2 1 −2 −1 −1 1 2 x −2 −3 112 Relations and Functions (a) Show that f has a local maximum but not a local minimum at the point (−1, 1). (b) Show that f has a local minimum but not a local maximum at the point (1, 1). (c) Show that f has a local maximum AND a local minimum at the point (0, 1). (d) Show that f is constant on the interval [−1, 1] and thus has both a local maximum AND a local minimum at every point (x, f (x)) where −1 < x < 1. 100. Using Example 1.6.4 as a guide, show that the function g whose graph is given below does not have a local maximum at (−3, 5) nor does it have a local minimum at (3, −3). Find its extrema, both local and absolute. What’s unique about the point (0, −4) on this graph? Also ﬁnd the intervals on which g is increasing and those on which g is decreasing. y 5 4 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 −4 101. We said earlier in the section that
it is not good enough to say local extrema exist where a function changes from increasing to decreasing or vice versa. As a previous exercise showed, we could have local extrema when a function is constant so now we need to examine some functions whose graphs do indeed change direction. Consider the functions graphed below. Notice that all four of them change direction at an open circle on the graph. Examine each for local extrema. What is the eﬀect of placing the “dot” on the y-axis above or below the open circle? What could you say if no function value were assigned to x = 02 −1 −1 1 2 x −2 −1 −1 1 2 x (a) Function I (b) Function II 1.6 Graphs of Functions 113 y 4 3 2 1 −2 −1 −2 −1 1 2 x (c) Function III (d) Function IV 114 1.6.3 Answers 1. f (x) = 2 − x Domain: (−∞, ∞) x-intercept: (2, 0) y-intercept: (0, 2) No symmetry 2. f (x) = x − 2 3 Domain: (−∞, ∞) x-intercept: (2, 0) y-intercept: 0, − 2 3 No symmetry 3. f (x) = x2 + 1 Domain: (−∞, ∞) x-intercept: None y-intercept: (0, 1) Even 4. f (x) = 4 − x2 Domain: (−∞, ∞) x-intercepts: (−2, 0), (2, 0) y-intercept: (0, 4) Even 5. f (x) = 2 Domain: (−∞, ∞) x-intercept: None y-intercept: (0, 2) Even Relations and Functions y 3 2 1 −2 −1 1 2 3 x −1 y 1 −1 1 2 3 4 x −2 −1 −2 −1 1 2 x 1 2 x −2 −1 1 2 x 1.6 Graphs of Functions 115 6. f (x) = x3 Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Odd 7. f (x) = x(x − 1)(x + 2) Domain
: (−∞, ∞) x-intercepts: (−2, 0), (0, 0), (1, 0) y-intercept: (0, 0) No symmetry √ 8. f (x) = x − 2 Domain: [2, ∞) x-intercept: (2, 0) y-intercept: None No symmetry 9. f (x) = √ 5 − x Domain: (−∞, 5] x-intercept: (5, 0) √ y-intercept: (0, 5) No symmetry 2−1 −1 1 2 x −2 −3 −4 −5 −6 −7 −8 y 4 3 2 1 −2 −4 −3 −2 −1 1 2 3 4 5 x 116 Relations and Functions 10. f (x) = 3 − 2 √ x + 2 Domain: [−2, ∞) 4, 0 x-intercept: 1 y-intercept: (0, 3 − 2 √ 2) No symmetry √ 11. f (x) = 3 x Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Odd 12. f (x) = 1 x2 + 1 Domain: (−∞, ∞) x-intercept: None y-intercept: (0, 1) Even 13. 15. y 5 4 3 2 1 −4−3−2−1 −1 −2 −1 1 2 x y 2 1 −8−7−6−5−4−3−2−1 −2 y 1 −2 −3−2−1 1 2 3 x 14. 16. y 5 4 3 2 1 −2−1 1 2 x 1.6 Graphs of Functions 117 17. 19. y 3 2 1 −2 −1 1 x −1 −2 −3 −4 6 5 4 3 2 1 y −2 −1 1 2 3 x 21. odd 24. even 27. odd 22. neither 25. even 28. odd 30. neither 31. neither 33. even and odd 34. odd 36. even 39. odd 42. [−5, 3] 45. x = −3 37. neither 40. even 43. [−5, 4] 18. 20. y 3 2 1 −4 −3 −2 −6−5−4−3
−2−1 23. even 26. neither 29. even 32. even 35. even 38. odd 41. even 44. f (−2) = 2 48. −4, −1, 1 49. [−4, −1] ∪ [1, 3] 50. 4 46. (−4, 0), (−1, 0), (1, 0) 47. (0, −1) 118 Relations and Functions 51. neither 52. [−5, −3], [0, 2] 53. [−3, 0], [2, 3] 54. f (−3) = 4, f (2) = 3 56. f (−3) = 4 55. f (0) = −1 57. f (−5) = −5 58. [−4, 4] 61. x = −2 64. −4, 0, 4 67. neither 70. none 72. none 74. No absolute maximum Absolute minimum f (4.55) ≈ −175.46 Local minimum at (−2.84, −91.32) Local maximum at (0.54, 55.73) Local minimum at (4.55, −175.46) Increasing on [−2.84, 0.54], [4.55, ∞) Decreasing on (−∞, −2.84], [0.54, 4.55] 59. [−5, 5) 60. f (2) = 3 62. (−4, 0), (0, 0), (4, 0) 63. (0, 0) 65. [−4, 0] ∪ {4} 66. 3 68. [−2, 2) 69. [−4, −2], (2, 4] 71. f (−2) = −5, f (2) = 3 73. f (−2) = −5 75. No absolute maximum No absolute minimum Local maximum at (0, 0) Local minimum at (1.60, −3.28) Increasing on (−∞, 0], [1.60, ∞) Decreasing on [0, 1.60] 76. Absolute maximum f (0) = 3 77. Absolute maximum f (2.12) ≈ 4.50 Absolute minimum f (±3) = 0 Local maximum at (0, 3) No local minimum Increasing on [−3, 0] Decreasing on [0, 3]
Absolute minimum f (−2.12) ≈ −4.50 Local maximum (2.12, 4.50) Local minimum (−2.12, −4.50) Increasing on [−2.12, 2.12] Decreasing on [−3, −2.12], [2.12, 3] 78. (f + g)(0) = 4 79. (f + g)(1) = 5 80. (f − g)(1) = −1 81. (g − f )(2) = 0 82. (f g)(2) = 9 83. (f g)(1) = 6 84. f g (4) = 0 85. g f (2) = 1 86. h(0) = 2, so the Sasquatch is 2 feet tall at birth. 87. h(15) = 6, so the Saquatch is 6 feet tall when she is 15 years old. 88. h(N ) = 6 when N = 15 and N = 60. This means the Sasquatch is 6 feet tall when she is 15 and 60 years old. 89. h is constant on [30, 45]. This means the Sasquatch’s height is constant (at 8 feet) for these years. 1.6 Graphs of Functions 119 90. h is decreasing on [45, 60]. This means the Sasquatch is getting shorter from the age of 45 to the age of 60. (Sasquatchteoporosis, perhaps?) 91. 92. Note that f (1.1) = 1, but f (−1.1) = −2, so f is neither even nor odd.... y 6 5 4 3 2 1 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6... The graph of f (x) = x. 120 Relations and Functions 1.7 Transformations In this section, we study how the graphs of functions change, or transform, when certain specialized modiﬁcations are made to their formulas. The transformations we will study fall into three broad categories: shifts, reﬂections and scalings, and we will present them in that order. Suppose the graph below is the complete graph of a function f. y 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3x)
The Fundamental Graphing Principle for Functions says that for a point (a, b) to be on the graph, f (a) = b. In particular, we know f (0) = 1, f (2) = 3, f (4) = 3 and f (5) = 5. Suppose we wanted to graph the function deﬁned by the formula g(x) = f (x) + 2. Let’s take a minute to remind ourselves of what g is doing. We start with an input x to the function f and we obtain the output f (x). The function g takes the output f (x) and adds 2 to it. In order to graph g, we need to graph the points (x, g(x)). How are we to ﬁnd the values for g(x) without a formula for f (x)? The answer is that we don’t need a formula for f (x), we just need the values of f (x). The values of f (x) are the y values on the graph of y = f (x). For example, using the points indicated on the graph of f, we can make the following table. x 0 2 4 5 (x, f (x)) f (x) g(x) = f (x) + 2 (x, g(x)) (0, 1) (2, 3) (4, 3) (5, 50, 3) (2, 5) (4, 5) (5, 7) In general, if (a, b) is on the graph of y = f (x), then f (a) = b, so g(a) = f (a) + 2 = b + 2. Hence, (a, b+2) is on the graph of g. In other words, to obtain the graph of g, we add 2 to the y-coordinate of each point on the graph of f. Geometrically, adding 2 to the y-coordinate of a point moves the point 2 units above its previous location. Adding 2 to every y-coordinate on a graph en masse is usually described as ‘shifting the graph up 2 units’. Notice that the graph retains the same basic shape as before, it is just 2 units above its original location. In other words, we connect the four points we moved in the same manner in which they were connected before
. We have the results side-by-side at the top of the next page. 1.7 Transformations 121 y 7 6 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3) (5, 7) (2, 5) (4, 5) y 7 6 5 4 (0, 3x) shift up 2 units −−−−−−−−−−−−→ add 2 to each y-coordinate 1 2 3 4 5 x y = g(x) = f (x) + 2 You’ll note that the domain of f and the domain of g are the same, namely [0, 5], but that the range of f is [1, 5] while the range of g is [3, 7]. In general, shifting a function vertically like this will leave the domain unchanged, but could very well aﬀect the range. You can easily imagine what would happen if we wanted to graph the function j(x) = f (x) − 2. Instead of adding 2 to each of the y-coordinates on the graph of f, we’d be subtracting 2. Geometrically, we would be moving the graph down 2 units. We leave it to the reader to verify that the domain of j is the same as f, but the range of j is [−1, 3]. What we have discussed is generalized in the following theorem. Theorem 1.2. Vertical Shifts. Suppose f is a function and k is a positive number. To graph y = f (x) + k, shift the graph of y = f (x) up k units by adding k to the y-coordinates of the points on the graph of f. To graph y = f (x) − k, shift the graph of y = f (x) down k units by subtracting k from the y-coordinates of the points on the graph of f. The key to understanding Theorem 1.2 and, indeed, all of the theorems in this section comes from an understanding of the Fundamental Graphing Principle for Functions. If (a, b) is on the graph of f, then f (a) = b. Substituting x = a into the equation y = f (x) + k gives y = f (a) + k = b + k. Hence, (a, b + k) is
on the graph of y = f (x) + k, and we have the result. In the language of ‘inputs’ and ‘outputs’, Theorem 1.2 can be paraphrased as “Adding to, or subtracting from, the output of a function causes the graph to shift up or down, respectively.” So what happens if we add to or subtract from the input of the function? Keeping with the graph of y = f (x) above, suppose we wanted to graph g(x) = f (x + 2). In other words, we are looking to see what happens when we add 2 to the input of the function.1 Let’s try to generate a table of values of g based on those we know for f. We quickly ﬁnd that we run into some diﬃculties. 1We have spent a lot of time in this text showing you that f (x + 2) and f (x) + 2 are, in general, wildly diﬀerent algebraic animals. We will see momentarily that their geometry is also dramatically diﬀerent. 122 Relations and Functions x 0 2 4 5 (x, f (x)) f (x) (0, 1) (2, 3) (4, 3) (5, 5) 1 3 3 5 g(x) = f (x + 2) f (0 + 2) = f (2) = 3 f (2 + 2) = f (4) = 3 f (4 + 2) = f (6) =? f (5 + 2) = f (7) =? (x, g(x)) (0, 3) (2, 3) When we substitute x = 4 into the formula g(x) = f (x + 2), we are asked to ﬁnd f (4 + 2) = f (6) which doesn’t exist because the domain of f is only [0, 5]. The same thing happens when we attempt to ﬁnd g(5). What we need here is a new strategy. We know, for instance, f (0) = 1. To determine the corresponding point on the graph of g, we need to ﬁgure out what value of x we must substitute into g(x) = f (x + 2) so that the quantity x + 2, works out to be
0. Solving x + 2 = 0 gives x = −2, and g(−2) = f ((−2) + 2) = f (0) = 1 so (−2, 1) is on the graph of g. To use the fact f (2) = 3, we set x + 2 = 2 to get x = 0. Substituting gives g(0) = f (0 + 2) = f (2) = 3. Continuing in this fashion, we get x −(x) = f (x + 2) g(−2) = f (0) = 1 g(0) = f (2) = 3 g(2) = f (4) = 3 g(3) = f (5) = 5 (x, g(x)) (−2, 1) (0, 3) (2, 3) (3, 5) In summary, the points (0, 1), (2, 3), (4, 3) and (5, 5) on the graph of y = f (x) give rise to the points (−2, 1), (0, 3), (2, 3) and (3, 5) on the graph of y = g(x), respectively. In general, if (a, b) is on the graph of y = f (x), then f (a) = b. Solving x + 2 = a gives x = a − 2 so that g(a − 2) = f ((a − 2) + 2) = f (a) = b. As such, (a − 2, b) is on the graph of y = g(x). The point (a − 2, b) is exactly 2 units to the left of the point (a, b) so the graph of y = g(x) is obtained by shifting the graph y = f (x) to the left 2 units, as pictured below. y 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3) y (3, 5) (2, 3) 5 4 (0, 3) 2 1 (−2, 1) −2 −x) shift left 2 units −−−−−−−−−−−−→ subtract 2 from each x-coordinate −2 −(x) = f (x + 2) Note that while the ranges of f and g are the same
, the domain of g is [−2, 3] whereas the domain of f is [0, 5]. In general, when we shift the graph horizontally, the range will remain the same, but the domain could change. If we set out to graph j(x) = f (x − 2), we would ﬁnd ourselves adding 1.7 Transformations 123 2 to all of the x values of the points on the graph of y = f (x) to eﬀect a shift to the right 2 units. Generalizing these notions produces the following result. Theorem 1.3. Horizontal Shifts. Suppose f is a function and h is a positive number. To graph y = f (x + h), shift the graph of y = f (x) left h units by subtracting h from the x-coordinates of the points on the graph of f. To graph y = f (x − h), shift the graph of y = f (x) right h units by adding h to the x-coordinates of the points on the graph of f. In other words, Theorem 1.3 says that adding to or subtracting from the input to a function amounts to shifting the graph left or right, respectively. Theorems 1.2 and 1.3 present a theme which will run common throughout the section: changes to the outputs from a function aﬀect the y-coordinates of the graph, resulting in some kind of vertical change; changes to the inputs to a function aﬀect the x-coordinates of the graph, resulting in some kind of horizontal change. Example 1.7.1. 1. Graph f (x) = √ x. Plot at least three points. 2. Use your graph in 1 to graph g(x) = √ x − 1. 3. Use your graph in 1 to graph j(x) = 4. Use your graph in 1 to graph m(x) = Solution. √ x − 1. √ x + 3 − 2. 1. Owing to the square root, the domain of f is x ≥ 0, or [0, ∞). We choose perfect squares to build our table and graph below. From the graph we verify the domain of f is [0, ∞) and the range of f is also [0, ∞). x 0 1 4 f (x) 0 1 2 (x, f (x))
(0, 0) (1, 1) (4, 2) (4, 2) y 2 (1, 1) 1 (0, 0x) = √ x 2. The domain of g is the same as the domain of f, since the only condition on both functions is that x ≥ 0. If we compare the formula for g(x) with f (x), we see that g(x) = f (x) − 1. In other words, we have subtracted 1 from the output of the function f. By Theorem 1.2, we know that in order to graph g, we shift the graph of f down one unit by subtracting 1 from each of the y-coordinates of the points on the graph of f. Applying this to the three points we have speciﬁed on the graph, we move (0, 0) to (0, −1), (1, 1) to (1, 0), and (4, 2) to (4, 1). 124 Relations and Functions The rest of the points follow suit, and we connect them with the same basic shape as before. We conﬁrm the domain of g is [0, ∞) and ﬁnd the range of g to be [−1, ∞). (4, 2) y 2 (1, 1) 1 (0, 01, 0) (4, 1) 1 2 3 4 x (0, −1) y = f (x) = √ x shift down 1 unit −−−−−−−−−−−−→ subtract 1 from each y-coordinate y = g(x) = √ x − 1 3. Solving x − 1 ≥ 0 gives x ≥ 1, so the domain of j is [1, ∞). To graph j, we note that j(x) = f (x − 1). In other words, we are subtracting 1 from the input of f. According to Theorem 1.3, this induces a shift to the right of the graph of f. We add 1 to the x-coordinates of the points on the graph of f and get the result below. The graph reaﬃrms that the domain of j is [1, ∞) and tells us that the range of j is [0, ∞). (4, 2) y 2 (1, 1) 1 (
0, 0) y 2 1 (2, 1) (5, 2x) = √ x shift right 1 unit −−−−−−−−−−−−→ add 1 to each x-coordinate (1, 0) 2 3 4 5 x y = j(x) = √ x − 1 4. To ﬁnd the domain of m, we solve x + 3 ≥ 0 and get [−3, ∞). Comparing the formulas of f (x) and m(x), we have m(x) = f (x + 3) − 2. We have 3 being added to an input, indicating a horizontal shift, and 2 being subtracted from an output, indicating a vertical shift. We leave it to the reader to verify that, in this particular case, the order in which we perform these transformations is immaterial; we will arrive at the same graph regardless as to which transformation we apply ﬁrst.2 We follow the convention ‘inputs ﬁrst’,3 and to that end we ﬁrst tackle the horizontal shift. Letting m1(x) = f (x + 3) denote this intermediate step, Theorem 1.3 tells us that the graph of y = m1(x) is the graph of f shifted to the left 3 units. Hence, we subtract 3 from each of the x-coordinates of the points on the graph of f. 2We shall see in the next example that order is generally important when applying more than one transformation to a graph. 3We could equally have chosen the convention ‘outputs ﬁrst’. 1.7 Transformations 125 (4, 2) y 2 (1, 1) 1 (0, 0) −3 −2 −1 1 2 3 4 x −1 −2 y = f (x) = √ x y (1, 2) (−2, 1) 2 1 (−3, 0) −3 −2 −1 1 2 3 4 x −1 −2 shift left 3 units −−−−−−−−−−−−→ subtract 3 from each x-coordinate y = m1(x) = f (x + 3) = √ x + 3 Since m(x) = f (x + 3) − 2 and f (x + 3) = m1(x), we have m(x) = m
1(x) − 2. We can apply Theorem 1.2 and obtain the graph of m by subtracting 2 from the y-coordinates of each of the points on the graph of m1(x). The graph veriﬁes that the domain of m is [−3, ∞) and we ﬁnd the range of m to be [−2, ∞). y (1, 2) (−2, 1) 2 1 (−3, 0) −3 −2 −1 1 2 3 4 x −1 −2 y = m1(x) = f (x + 3) = √ x + 3 shift down 2 units −−−−−−−−−−−−→ subtract 2 from each y-coordinate y 2 1 (1, 0) −3 −2 −1 1 2 3 4 (−2, −1) (−3, −2) −1 −2 x y = m(x) = m1(x Keep in mind that we can check our answer to any of these kinds of problems by showing that any of the points we’ve moved lie on the graph of our ﬁnal answer. For example, we can check that (−3, −2) is on the graph of m by computing m(−3) = 0 − 2 = −2 (−3) + 3 − 2 = √ We now turn our attention to reﬂections. We know from Section 1.1 that to reﬂect a point (x, y) across the x-axis, we replace y with −y. If (x, y) is on the graph of f, then y = f (x), so replacing y with −y is the same as replacing f (x) with −f (x). Hence, the graph of y = −f (x) is the graph of f reﬂected across the x-axis. Similarly, the graph of y = f (−x) is the graph of f reﬂected across the y-axis. Returning to the language of inputs and outputs, multiplying the output from a function by −1 reﬂects its graph across the x-axis, while multiplying the input to a function by −1 reﬂects the graph across the y-axis.4 4The expressions −f (x) and f (−x) should look familiar - they
are the quantities we used in Section 1.6 to test if a function was even, odd or neither. The interested reader is invited to explore the role of reﬂections and symmetry of functions. What happens if you reﬂect an even function across the y-axis? What happens if you reﬂect an odd function across the y-axis? What about the x-axis? 126 Relations and Functions Theorem 1.4. Reﬂections. Suppose f is a function. To graph y = −f (x), reﬂect the graph of y = f (x) across the x-axis by multiplying the y-coordinates of the points on the graph of f by −1. To graph y = f (−x), reﬂect the graph of y = f (x) across the y-axis by multiplying the x-coordinates of the points on the graph of f by −1. Applying Theorem 1.4 to the graph of y = f (x) given at the beginning of the section, we can graph y = −f (x) by reﬂecting the graph of f about the x-axis y 5 4 3 2 (5, 5) (2, 3) (4, 3) 1 2 3 4 5 x (0, 1) −1 −2 −3 −4 −5 y = f (x) reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −0, −1) −2 −3 −4 −5 (4, −3) (2, −3) (5, −5) y = −f (x) By reﬂecting the graph of f across the y-axis, we obtain the graph of y = f (−x). y 5 4 3 2 (0, 1) (5, 5) (−5, 5) (2, 3) (4, 3) (−2, 3) (−4, 3) y 5 4 3 2 (0, 1) −5 −4 −3 −2 −x) reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −5 −4 −3 −2 − (−x) With the addition of reﬂections,
it is now more important than ever to consider the order of transformations, as the next example illustrates. √ Example 1.7.2. Let f (x) = functions. Also, state their domains and ranges. √ √ 1. g(x) = −x 2. j(x) = 3 − x 3. m(x) = 3 − √ x x. Use the graph of f from Example 1.7.1 to graph the following 1.7 Transformations 127 Solution. √ 1. The mere sight of √ −x usually causes alarm, if not panic. When we discussed domains in Section 1.4, we clearly banished negatives from the radicands of even roots. However, we must remember that x is a variable, and as such, the quantity −x isn’t always negative. For −(−4) = 2 is perfectly well-deﬁned. To ﬁnd the example, if x = −4, −x = 4, thus domain analytically, we set −x ≥ 0 which gives x ≤ 0, so that the domain of g is (−∞, 0]. Since g(x) = f (−x), Theorem 1.4 tells us that the graph of g is the reﬂection of the graph of f across the y-axis. We accomplish this by multiplying each x-coordinate on the graph of f by −1, so that the points (0, 0), (1, 1), and (4, 2) move to (0, 0), (−1, 1), and (−4, 2), respectively. Graphically, we see that the domain of g is (−∞, 0] and the range of g is the same as the range of f, namely [0, ∞). −x = y (1, 1) 2 1 (0, 0) (4, 2) (−4, 2) y 2 1 (−1, 1) (0, 0) −4 −3 −2 −x) = √ x reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −4 −3 −2 −1 1 2 3 4 x y = g(x) = f (−x) = √ −x √ 2. To determine the domain of j(x) = √ √ −
x + 3 = f (−x + 3). Comparing this formula with f (x) = 3 − x, we solve 3 − x ≥ 0 and get x ≤ 3, or (−∞, 3]. To determine which transformations we need to apply to the graph of f to obtain the graph of j, we rewrite j(x) = x, we see that not only are we multiplying the input x by −1, which results in a reﬂection across the y-axis, but also we are adding 3, which indicates a horizontal shift to the left. Does it matter in which order we do the transformations? If so, which order is the correct order? Let’s consider the point (4, 2) on the graph of f. We refer to the discussion leading up to Theorem 1.3. We know f (4) = 2 and wish to ﬁnd the point on y = j(x) = f (−x + 3) which corresponds to (4, 2). We set −x + 3 = 4 and solve. Our ﬁrst step is to subtract 3 from both sides to get −x = 1. Subtracting 3 from the x-coordinate 4 is shifting the point (4, 2) to the left. From −x = 1, we then multiply5 both sides by −1 to get x = −1. Multiplying the x-coordinate by −1 corresponds to reﬂecting the point about the y-axis. Hence, we perform the horizontal shift ﬁrst, then follow it with the reﬂection about the y-axis. Starting with f (x) = x, we let j1(x) be the intermediate function which shifts the graph of f 3 units to the left, j1(x) = f (x + 3). √ y (1, 1) 2 1 (0, 0) (4, 2) y (1, 2) (−2, 1) 2 1 (−3, 0) −4 −3 −2 −x) = √ x shift left 3 units −−−−−−−−−−−−→ subtract 3 from each x-coordinate −4 −3 −2 −1 1 2 3 4 x y = j1(x) = f (x + 3) = √ x + 3 5Or divide - it amounts to the same thing. 128
Relations and Functions To obtain the function j, we reﬂect the graph of j1 about y-axis. Theorem 1.4 tells us we −x + 3, have j(x) = j1(−x). Putting it all together, we have j(x) = j1(−x) = f (−x + 3) = which is what we want.6 From the graph, we conﬁrm the domain of j is (−∞, 3] and we get that the range is [0, ∞). √ y (1, 2) (−2, 1) 2 1 (−3, 0) y 2 (−1, 2) (2, 1) (3, 0) −4 −3 −2 −1 1 2 3 4 x y = j1(x) = √ x + 3 reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −4 −3 −2 −1 1 2 3 4 x y = j(x) = j1(−x) = √ −x + 3 √ 3. The domain of m works out to be the domain of f, [0, ∞). Rewriting m(x) = − x + 3, we see m(x) = −f (x) + 3. Since we are multiplying the output of f by −1 and then adding 3, we once again have two transformations to deal with: a reﬂection across the x-axis and a vertical shift. To determine the correct order in which to apply the transformations, we imagine trying to determine the point on the graph of m which corresponds to (4, 2) on the graph of f. Since in the formula for m(x), the input to f is just x, we substitute to ﬁnd m(4) = −f (4) + 3 = −2 + 3 = 1. Hence, (4, 1) is the corresponding point on the graph of m. If we closely examine the arithmetic, we see that we ﬁrst multiply f (4) by −1, which corresponds to the reﬂection across the x-axis, and then we add 3, which corresponds to the vertical shift. If we deﬁne an intermediate function m1(x) = −f (x) to
take care of the reﬂection, we get y (1, 1) (4, 20, 0) −1 −2 y 3 2 1 (0, 0) −1 1 2 3 4 x (1, −1) −2 y = f (x) = √ x reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −1 y = m1(x) = −f (x) = − (4, −2) √ x To shift the graph of m1 up 3 units, we set m(x) = m1(x) + 3. Since m1(x) = −f (x), when we put it all together, we get m(x) = m1(x) + 3 = −f (x) + 3 = − x + 3. We see from the graph that the range of m is (−∞, 3]. √ 6If we had done the reﬂection ﬁrst, then j1(x) = f (−x). Following this by a shift left would give us j(x) = j1(x + 3) = f (−(x + 3)) = f (−x − 3) = −x − 3 which isn’t what we want. However, if we did the reﬂection ﬁrst and followed it by a shift to the right 3 units, we would have arrived at the function j(x). We leave it to the reader to verify the details. √ 1.7 Transformations 129 y 3 2 1 (0, 0) −1 1 2 3 4 x (1, −1) −2 (4, −2) shift up 3 units −−−−−−−−−−−−→ add 3 to each y-coordinate (0, 3) y (1, 2) (4, 1) 1 2 3 4 x 2 1 −1 −2 y = m1(x) = − √ x y = m(x) = m1(x) + 3 = − √ x + 3 We now turn our attention to our last class of transformations known as scalings. A thorough discussion of scalings can get complicated because they are not as straight-forward as the previous transformations. A quick review of what we’ve covered so far
, namely vertical shifts, horizontal shifts and reﬂections, will show you why those transformations are known as rigid transformations. Simply put, they do not change the shape of the graph, only its position and orientation in the plane. If, however, we wanted to make a new graph twice as tall as a given graph, or one-third as wide, we would be changing the shape of the graph. This type of transformation is called nonrigid for obvious reasons. Not only will it be important for us to diﬀerentiate between modifying inputs versus outputs, we must also pay close attention to the magnitude of the changes we make. As you will see shortly, the Mathematics turns out to be easier than the associated grammar. Suppose we wish to graph the function g(x) = 2f (x) where f (x) is the function whose graph is given at the beginning of the section. From its graph, we can build a table of values for g as before. y 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3x) x 0 2 4 5 (x, f (x)) f (x) g(x) = 2f (x) (0, 1) (2, 3) (4, 3) (5, 5) 1 3 3 5 2 6 6 10 (x, g(x)) (0, 2) (2, 6) (4, 6) (5, 10) In general, if (a, b) is on the graph of f, then f (a) = b so that g(a) = 2f (a) = 2b puts (a, 2b) on the graph of g. In other words, to obtain the graph of g, we multiply all of the y-coordinates of the points on the graph of f by 2. Multiplying all of the y-coordinates of all of the points on the graph of f by 2 causes what is known as a ‘vertical scaling7 by a factor of 2’, and the results are given on the next page. 7Also called a ‘vertical stretching’, ‘vertical expansion’ or ‘vertical dilation’ by a factor of 2. 130 Relations and Functions y 10 9 8 7 6 5 4 3 2 (0, 1) (5, 5) (2, 3) (4,
3) y 10 (5, 10) (2, 6) (4, 6) 9 8 7 6 5 4 3 (0, 2x) vertical scaling by a factor of 2 −−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 2 1 2 3 4 5 x y = 2f (x) If we wish to graph y = 1 of f by 1 2. This creates a ‘vertical scaling8 by a factor of 1 2 ’ as seen below. 2 f (x), we multiply the all of the y-coordinates of the points on the graph y 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3x) vertical scaling by a factor of 1 2 −−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 1 2 These results are generalized in the following theorem. y 5 4 3 2 1 0, 1 2 5, 5 2 2, 3 2 4x) Theorem 1.5. Vertical Scalings. Suppose f is a function and a > 0. To graph y = af (x), multiply all of the y-coordinates of the points on the graph of f by a. We say the graph of f has been vertically scaled by a factor of a. If a > 1, we say the graph of f has undergone a vertical stretching (expansion, dilation) by a factor of a. If 0 < a < 1, we say the graph of f has undergone a vertical shrinking (compression, contraction) by a factor of 1 a. 8Also called ‘vertical shrinking’, ‘vertical compression’ or ‘vertical contraction’ by a factor of 2. 1.7 Transformations 131 A few remarks about Theorem 1.5 are in order. First, a note about the verbiage. To the authors, the words ‘stretching’, ‘expansion’, and ‘dilation’ all indicate something getting bigger. Hence, ‘stretched by a factor of 2’ makes sense if we are scaling something by multiplying it by 2. Similarly, we believe words like ‘shrinking’, ‘compression’ and ‘contraction’ all indicate something getting smaller, so if we scale something
by a factor of 1 2, we would say it ‘shrinks by a factor of 2’ - not ‘shrinks by a factor of 1 2 ’. This is why we have written the descriptions ‘stretching by a factor of a’ and ‘shrinking by a factor of 1 a ’ in the statement of the theorem. Second, in terms of inputs and outputs, Theorem 1.5 says multiplying the outputs from a function by positive number a causes the graph to be vertically scaled by a factor of a. It is natural to ask what would happen if we multiply the inputs of a function by a positive number. This leads us to our last transformation of the section. Referring to the graph of f given at the beginning of this section, suppose we want to graph g(x) = f (2x). In other words, we are looking to see what eﬀect multiplying the inputs to f by 2 has on its graph. If we attempt to build a table directly, we quickly run into the same problem we had in our discussion leading up to Theorem 1.3, as seen in the table on the left below. We solve this problem in the same way we solved this problem before. For example, if we want to determine the point on g which corresponds to the point (2, 3) on the graph of f, we set 2x = 2 so that x = 1. Substituting x = 1 into g(x), we obtain g(1) = f (2 · 1) = f (2) = 3, so that (1, 3) is on the graph of g. Continuing in this fashion, we obtain the table on the lower rightx, f (x)) f (x) (0, 1) (2, 3) (4, 3) (5, 5) (x, g(x)) (0, 1) (2, 3) g(x) = f (2x) f (2 · 0) = f (0) = 1 f (2 · 2) = f (4) = 3 f (2 · 4) = f (8) =? f (2 · 5) = f (10) =? In general, if (a, b) is on the graph of f, then f (a) = b. Hence g a = f (a) = b so that 2 2, b is on the graph of
g. In other words, to graph g we divide the x-coordinates of the points on a the graph of f by 2. This results in a horizontal scaling9 by a factor of 1 2. g(x) = f (2x) g(0) = f (0) = 1 g(1) = f (2) = 3 g(2) = f (4) = 3 g 5 = f (5x, g(x)) (0, 0) (1, 3) (2, 3) 2, 5 5 2x 0, 1) (5, 5) (2, 3) (4, 3) y 2, 5 5 (1, 3) (2, 3) 5 4 3 2 (0, 1x) horizontal scaling by a factor of 1 2 −−−−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by (x) = f (2x) 9Also called ‘horizontal shrinking’, ‘horizontal compression’ or ‘horizontal contraction’ by a factor of 2. 132 Relations and Functions If, on the other hand, we wish to graph y = f 1 2 x, we end up multiplying the x-coordinates of the points on the graph of f by 2 which results in a horizontal scaling10 by a factor of 2, as demonstrated below. y (5, 5) (2, 3) (4, 3) 5 4 3 2 (0, 1) y 5 4 3 2 (0, 1) (10, 5) (4, 3) (8, 3 10 x y = f (x) horizontal scaling by a factor of 2 −−−−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by 10 x y = g(x) = f 1 2 x We have the following theorem. Theorem 1.6. Horizontal Scalings. Suppose f is a function and b > 0. To graph y = f (bx), divide all of the x-coordinates of the points on the graph of f by b. We say the graph of f has been horizontally scaled by a factor of 1 b. If 0 < b < 1, we say the graph of f has undergone a horizontal stretching (expansion, dilation) by a factor of 1 b. If b > 1,
we say the graph of f has undergone a horizontal shrinking (compression, con- traction) by a factor of b. Theorem 1.6 tells us that if we multiply the input to a function by b, the resulting graph is scaled horizontally by a factor of 1 b since the x-values are divided by b to produce corresponding points on the graph of y = f (bx). The next example explores how vertical and horizontal scalings sometimes interact with each other and with the other transformations introduced in this section. Example 1.7.3. Let f (x) = functions. Also, state their domains and ranges. √ x. Use the graph of f from Example 1.7.1 to graph the following 1. g(x) = 3 √ x 2. j(x) = √ 9x 3. m(x) = 1 − x+3 2 Solution. 1. First we note that the domain of g is [0, ∞) for the usual reason. Next, we have g(x) = 3f (x) so by Theorem 1.5, we obtain the graph of g by multiplying all of the y-coordinates of the points on the graph of f by 3. The result is a vertical scaling of the graph of f by a factor of 3. We ﬁnd the range of g is also [0, ∞). 10Also called ‘horizontal stretching’, ‘horizontal expansion’ or ‘horizontal dilation’ by a factor of 2. 1.7 Transformations 133 y (4, 2) (1, 1) 6 5 4 3 2 1 (0, 0) y (4, 6) 6 5 4 3 2 1 (0, 0) (1, 3x) = √ x vertical scale by a factor of 3 −−−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 3 1 2 3 4 x y = g(x) = 3f (x) = 3 √ x 2. To determine the domain of j, we solve 9x ≥ 0 to ﬁnd x ≥ 0. Our domain is once again [0, ∞). We recognize j(x) = f (9x) and by Theorem 1.6, we obtain the graph of j by dividing the x-coordinates of the points on the graph of
f by 9. From the graph, we see the range of j is also [0, ∞). (4, 2) y (1, 1) 2 1 (0, 00, 0x) = √ x horizontal scale by a factor of 1 9 −−−−−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by (x) = f (9x) = √ 9x 3. Solving x+3 2 x + 3 2 + 1, or m(x) = −f 1 2 corresponds to a shift to the left by 3 2 ≥ 0 gives x ≥ −3, so the domain of m is [−3, ∞). To take advantage of what 1 + 1. 2 x + 3 we know of transformations, we rewrite m(x) = − 2 2 x + 3 Focusing on the inputs ﬁrst, we note that the input to f in the formula for m(x) is 1 2. Multiplying the x by 1 2 corresponds to a horizontal stretching by a factor of 2, and adding the 3 2. As before, we resolve which to perform ﬁrst by thinking about how we would ﬁnd the point on m corresponding to a point on f, in this case, (4, 2). To use f (4) = 2, we solve 1 2 = 4. Our ﬁrst step is to subtract the 3 2 x + 3 2 (the horizontal shift) to obtain 1 2. Next, we multiply by 2 (the horizontal stretching) and obtain x = 5. We deﬁne two intermediate functions to handle ﬁrst the shift, then the 2 will shift the In accordance with Theorem 1.3, m1(x stretching. graph of f to the left 3 2 units. 134 Relations and Functions 1, 1) (4, 2) −3 −2 1 (0, 0) −1 −1 −2 1 2 3 4 5 x −3 −2 − 3 −x) = √ x shift left 3 2 units −−−−−−−−−−−−→ subtract 3 2 from each x-coordinate −2 y = m1(x Next, m2(x) = m1 graph of m1 by a factor of 2 will, according to Theorem 1.6, horizontally stretch the 3
−2 − 3 −2 y = m1(x) = x + 3 2 y 2 (−1, 1) (5, 2) −2 (−3, 0) −1 1 2 3 4 5 x −1 −2 horizontal scale by a factor of 2 −−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by 2 y = m2(x) = m1 We now examine what’s happening to the outputs. From m(x) = −f 1 + 1, we see that the output from f is being multiplied by −1 (a reﬂection about the x-axis) and then a 1 is added (a vertical shift up 1). As before, we can determine the correct order by looking at how the point (4, 2) is moved. We already know that to make use of the equation f (4) = 2, we need to substitute x = 5. We get m(5) = −f 1 + 1 = −f (4) + 1 = −2 + 1 = −1. We see that f (4) (the output from f ) is ﬁrst multiplied by −1 then the 1 is added meaning we ﬁrst reﬂect the graph about the x-axis then shift up 1. Theorem 1.4 tells us m3(x) = −m2(x) will handle the reﬂection. 2 (5 (−1, 1) (5, 2) y 2 1 (−3, 0) −2 (−3, 0) −1 1 2 3 4 5 x −2 −1 1 2 3 4 5 x −1 −2 y = m2(x) = 1 2 x + 3 2 (−1, −1) −2 (5, −2) reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −1 y = m3(x) = −m2(x.7 Transformations 135 Finally, to handle the vertical shift, Theorem 1.2 gives m(x) = m3(x) + 1, and we see that the range of m is (−∞, 1]. y 2 1 (−3, 0) y (−3, 1) 2 (−1, 0) −2 −1 1 2 3 4 5
x −2 −1 1 2 3 4 5 x (−1, −1) −2 (5, −2) −2 (5, −1) y = m3(x) = −m2(x) = − 1 2 x + 3 2 shift up 1 unit −−−−−−−−−−−−→ add 1 to each y-coordinate y = m(x) = m3(x √ √ √ √ 9x = x = 3 Some comments about Example 1.7.3 are in order. First, recalling the properties of radicals from Intermediate Algebra, we know that the functions g and j are the same, since j and g have the same domains and j(x) = x = g(x). (We invite the reader to verify that all of the points we plotted on the graph of g lie on the graph of j and vice-versa.) Hence, for f (x) = x, a vertical stretch by a factor of 3 and a horizontal shrinking by a factor of 9 result in the same transformation. While this kind of phenomenon is not universal, it happens commonly enough with some of the families of functions studied in College Algebra that it is worthy of note. Secondly, to graph the function m, we applied a series of four transformations. While it would have been easier on the authors to simply inform the reader of which steps to take, we have strived to explain why the order in which the transformations were applied made sense. We generalize the procedure in the theorem below. √ Theorem 1.7. Transformations. Suppose f is a function. If A = 0 and B = 0, then to graph g(x) = Af (Bx + H) + K 1. Subtract H from each of the x-coordinates of the points on the graph of f. This results in a horizontal shift to the left if H > 0 or right if H < 0. 2. Divide the x-coordinates of the points on the graph obtained in Step 1 by B. This results in a horizontal scaling, but may also include a reﬂection about the y-axis if B < 0. 3. Multiply the y-coordinates of the points on the graph obtained in Step 2 by A. This results in a vertical scaling, but may also include a reﬂection about the x-axis if A < 0. 4. Add K to
each of the y-coordinates of the points on the graph obtained in Step 3. This results in a vertical shift up if K > 0 or down if K < 0. Theorem 1.7 can be established by generalizing the techniques developed in this section. Suppose (a, b) is on the graph of f. Then f (a) = b, and to make good use of this fact, we set Bx + H = a If B and solve. We ﬁrst subtract the H (causing the horizontal shift) and then divide by B. 136 Relations and Functions = Af B · a−H is a positive number, this induces only a horizontal scaling by a factor of 1 If B < 0, then B. we have a factor of −1 in play, and dividing by it induces a reﬂection about the y-axis. So we have x = a−H B as the input to g which corresponds to the input x = a to f. We now evaluate g a−H B + H + K = Af (a) + K = Ab + K. We notice that the output from f is B ﬁrst multiplied by A. As with the constant B, if A > 0, this induces only a vertical scaling. If A < 0, then the −1 induces a reﬂection across the x-axis. Finally, we add K to the result, which is our vertical shift. A less precise, but more intuitive way to paraphrase Theorem 1.7 is to think of the quantity Bx + H is the ‘inside’ of the function f. What’s happening inside f aﬀects the inputs or x-coordinates of the points on the graph of f. To ﬁnd the x-coordinates of the corresponding points on g, we undo what has been done to x in the same way we would solve an equation. What’s happening to the output can be thought of as things happening ‘outside’ the function, f. Things happening outside aﬀect the outputs or y-coordinates of the points on the graph of f. Here, we follow the usual order of operations agreement: we ﬁrst multiply by A then add K to ﬁnd the corresponding y-coordinates on the graph of g. Example 1.7.4. Below is the complete graph of y = f (
x). Use it to graph g(x) = 4−3f (1−2x) 2. y (0, 3) 3 2 1 (−2, 0) (2, 0) −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 (−4, −3) (4, −3) Solution. We use Theorem 1.7 to track the ﬁve ‘key points’ (−4, −3), (−2, 0), (0, 3), (2, 0) and (4, −3) indicated on the graph of f to their new locations. We ﬁrst rewrite g(x) in the form presented in Theorem 1.7, g(x) = − 3 2 f (−2x + 1) + 2. We set −2x + 1 equal to the x-coordinates of the key points and solve. For example, solving −2x + 1 = −4, we ﬁrst subtract 1 to get −2x = −5 then divide by −2 to get x = 5 2. Subtracting the 1 is a horizontal shift to the left 1 unit. Dividing by −2 can be thought of as a two step process: dividing by 2 which compresses the graph horizontally by a factor of 2 followed by dividing (multiplying) by −1 which causes a reﬂection across the y-axis. We summarize the results in the table on the next page. 1.7 Transformations 137 a −2x + 1 = a x (a, f (a)) x = 5 (−4, −3) −4 −2x + 1 = −4 2 x = 3 (−2, 0) −2 −2x + 1 = −2 2 x = 1 −2x + 1 = 0 2 −2x + 1 = 2 x = − 1 2 −2x + 1 = 4 x = − 3 2 (2, 0) (4, −3) (0, 3) 4 2 0 Next, we take each of the x values and substitute them into g(x) = − 3 corresponding y-values. Substituting x = 5 2, and using the fact that f (−4) = −3, we get 2 f (−2x + 1) + 2 to get the 5 2 g = − 3 2 f − (−4) + 2 = −
3 2 (−3) + 2 = 9 2 + 2 = 13 2 We see that the output from f is ﬁrst multiplied by − 3 2. Thinking of this as a two step process, multiplying by 3 2 followed by a reﬂection across the x-axis. Adding 2 results in a vertical shift up 2 units. Continuing in this manner, we get the table below. 2 then by −1, we have a vertical stretching by a factor of (x) 13 2 2 − 5 2 2 13 2 2 (x, g(x)) 5 2, 13, 13 2 2 To graph g, we plot each of the points in the table above and connect them in the same order and fashion as the points to which they correspond. Plotting f and g side-by-side gives y (0, 3) 6 5 4 3 2 1 (2, 0) (−2, 0) − 3 2, 13 2 y 5 2, 13 4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x (−4, −3) −2 −3 −4 (4, −3) −2 −3 −4 1 2, − 5 2 138 Relations and Functions The reader is strongly encouraged11 to graph the series of functions which shows the gradual transformation of the graph of f into the graph of g. We have outlined the sequence of transformations in the above exposition; all that remains is to plot the ﬁve intermediate stages. Our last example turns the tables and asks for the formula of a function given a desired sequence of transformations. If nothing else, it is a good review of function notation. Example 1.7.5. Let f (x) = x2. Find and simplify the formula of the function g(x) whose graph is the result of f undergoing the following sequence of transformations. Check your answer using a graphing calculator. 1. Vertical shift up 2 units 2. Reﬂection across the x-axis 3. Horizontal shift right 1 unit 4. Horizontal stretching by a factor of 2 Solution. We build up to a formula for g(x) using intermediate functions as we’ve seen in previous examples. We let g1 take care of our ﬁrst step. Theorem 1.2 tells us g1(x) = f (x)+2 = x2 +2. Next, we re�
�ect the graph of g1 about the x-axis using Theorem 1.4: g2(x) = −g1(x) = − x2 + 2 = −x2 − 2. We shift the graph to the right 1 unit, according to Theorem 1.3, by setting g3(x) = g2(x − 1) = −(x − 1)2 − 2 = −x2 + 2x − 3. Finally, we induce a horizontal stretch by a factor of 2 2 x2 4 x2 + x − 3. using Theorem 1.6 to get g(x) = g3 We use the calculator to graph the stages below to conﬁrm our result. 2 x − 3 which yields g(x shift up 2 units −−−−−−−−−−−−→ add 2 to each y-coordinate y = f (x) = x2 y = g1(x) = f (x) + 2 = x2 + 2 reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −1 y = g1(x) = x2 + 2 y = g2(x) = −g1(x) = −x2 − 2 11You really should do this once in your life. 1.7 Transformations 139 y = g2(x) = −x2 − 2 y = g3(x) = g2(x − 1) = −x2 + 2x − 3 shift right 1 unit −−−−−−−−−−−−→ add 1 to each x-coordinate y = g3(x) = −x2 + 2x − 3 y = g(x) = g3 2 x = − 1 1 4 x2 + x − 3 horizontal stretch by a factor of 2 −−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by 2 We have kept the viewing window the same in all of the graphs above. This had the undesirable consequence of making the last graph look ‘incomplete’ in that we cannot see the original shape of f (x) = x2. Altering the viewing window results in a more complete graph of the transformed function as seen below. This example brings our ﬁrst chapter to a close.
In the chapters which lie ahead, be on the lookout for the concepts developed here to resurface as we study diﬀerent families of functions. y = g(x) 140 1.7.1 Exercises Relations and Functions Suppose (2, −3) is on the graph of y = f (x). In Exercises 1 - 18, use Theorem 1.7 to ﬁnd a point on the graph of the given transformed function. 1. y = f (x) + 3 2. y = f (x + 3) 3. y = f (x) − 1 4. y = f (x − 1) 5. y = 3f (x) 6. y = f (3x) 7. y = −f (x) 8. y = f (−x) 9. y = f (x − 3) + 1 10. y = 2f (x + 1) 11. y = 10 − f (x) 12. y = 3f (2x) − 1 13. y = 1 2 f (4 − x) 16. y = f 7 − 2x 4 14. y = 5f (2x + 1) + 3 15. y = 2f (1 − x) − 1 17. y = f (3x) − 1 2 18. y = 4 − f (3x − 1) 7 The complete graph of y = f (x) is given below. In Exercises 19 - 27, use it and Theorem 1.7 to graph the given transformed function. y 4 3 2 1 (2, 2) (−2, 2) −4 −3 −2 −1 (0, 0) 2 3 4 x The graph for Ex. 19 - 27 19. y = f (x) + 1 20. y = f (x) − 2 21. y = f (x + 1) 22. y = f (x − 2) 23. y = 2f (x) 24. y = f (2x) 25. y = 2 − f (x) 26. y = f (2 − x) 27. y = 2 − f (2 − x) 28. Some of the answers to Exercises 19 - 27 above should be the same. Which ones match up? What properties of the graph of y = f (x) contribute to the duplication? 1.7 Transformations 141 The complete graph of y
= f (x) is given below. In Exercises 29 - 37, use it and Theorem 1.7 to graph the given transformed function. y (0, 42, 0) (4, −2) −4 −3 −1 (−2, 0) −1 −2 −3 −4 The graph for Ex. 29 - 37 29. y = f (x) − 1 30. y = f (x + 1) 32. y = f (2x) 33. y = −f (x) 31. y = 1 2 f (x) 34. y = f (−x) 35. y = f (x + 1) − 1 36. y = 1 − f (x) 37. y = 1 2 f (x + 1) − 1 The complete graph of y = f (x) is given below. In Exercises 38 - 49, use it and Theorem 1.7 to graph the given transformed function. y (0, 3) 3 2 1 −3 −2 −1 1 2 (−3, 0) −1 x 3 (3, 0) The graph for Ex. 38 - 49 38. g(x) = f (x) + 3 39. h(x) = f (x) − 1 2 41. a(x) = f (x + 4) 42. b(x) = f (x + 1) − 1 44. d(x) = −2f (x) 45. k(x) = f 2 3 x 47. n(x) = 4f (x − 3) − 6 48. p(x) = 4 + f (1 − 2x) 40. j(x) = f x − 2 3 43. c(x) = 3 5 f (x) 46. m(x) = − 1 49. q(x) = − 1 4 f (3x) 2 f x+4 2 − 3 142 Relations and Functions The complete graph of y = S(x) is given below. y (1, 3) (0, 0) 1 (2, 0) x 3 2 1 −1 −2 (−2, 0) −2 −1 −3 (−1, −3) The graph of y = S(x) The purpose of Exercises 50 - 53 is to graph y = 1 one step at a time. 2 S(−x + 1) + 1 by grap
hing each transformation, 50. y = S1(x) = S(x + 1) 51. y = S2(x) = S1(−x) = S(−x + 1) 52. y = S3(x) = 1 2 S2(x) = 1 2 S(−x + 1) 53. y = S4(x) = S3(x) + 1 = 1 2 S(−x + 1) + 1 √ Let f (x) = sequence of transformations. x. Find a formula for a function g whose graph is obtained from f from the given 54. (1) shift right 2 units; (2) shift down 3 units 55. (1) shift down 3 units; (2) shift right 2 units 56. (1) reﬂect across the x-axis; (2) shift up 1 unit 57. (1) shift up 1 unit; (2) reﬂect across the x-axis 58. (1) shift left 1 unit; (2) reﬂect across the y-axis; (3) shift up 2 units 59. (1) reﬂect across the y-axis; (2) shift left 1 unit; (3) shift up 2 units 60. (1) shift left 3 units; (2) vertical stretch by a factor of 2; (3) shift down 4 units 61. (1) shift left 3 units; (2) shift down 4 units; (3) vertical stretch by a factor of 2 62. (1) shift right 3 units; (2) horizontal shrink by a factor of 2; (3) shift up 1 unit 63. (1) horizontal shrink by a factor of 2; (2) shift right 3 units; (3) shift up 1 unit 1.7 Transformations 143 √ 64. The graph of y = f (x) = 3 x is given below on the left and the graph of y = g(x) is given on the right. Find a formula for g based on transformations of the graph of f. Check your answer by conﬁrming that the points shown on the graph of g satisfy the equation y = g(x). 11−10−9−8−7−6−5−4−3−2−1 −11−10−9−8−7−6−5−4−3−2
−1 −2 −3 −4 −5 y = 3√ x −2 −3 −4 −5 y = g(x) √ 65. For many common functions, the properties of Algebra make a horizontal scaling the same as a vertical scaling by (possibly) a diﬀerent factor. For example, we stated earlier that √ x. With the help of your classmates, ﬁnd the equivalent vertical scaling produced. What about √ by the horizontal scalings y = (2x)3, y = \|5x\|, y = 3 √ 2 x2 y = (−2x)3, y = \| − 5x\|, y = 3? −27x and y = − 1 27x and y = 1 2 x2 9x = 3 66. We mentioned earlier in the section that, in general, the order in which transformations are applied matters, yet in our ﬁrst example with two transformations the order did not matter. (You could perform the shift to the left followed by the shift down or you could shift down and then left to achieve the same result.) With the help of your classmates, determine the situations in which order does matter and those in which it does not. 67. What happens if you reﬂect an even function across the y-axis? 68. What happens if you reﬂect an odd function across the y-axis? 69. What happens if you reﬂect an even function across the x-axis? 70. What happens if you reﬂect an odd function across the x-axis? 71. How would you describe symmetry about the origin in terms of reﬂections? 72. As we saw in Example 1.7.5, the viewing window on the graphing calculator aﬀects how we see the transformations done to a graph. Using two diﬀerent calculators, ﬁnd viewing windows so that f (x) = x2 on the one calculator looks like g(x) = 3x2 on the other. 144 1.7.2 Answers 1. (2, 0) 4. (3, −3) 7. (2, 3) 10. (1, −6) 13. 2, − 3 2 16. − 1 2, −3 Relations and Functions 2. (−1, −3) 5. (2, −9) 8.
(−2, −3) 3. (2, −4) 6. 2 3, −3 9. (5, −2) 11. (2, 13) 12. y = (1, −10) 14. 1 17. 2 2, −12 3, −2 19. y = f (x) + 1 y (−2, 3) 4 3 2 1 (2, 3) (0, 1) −4 −3 −2 −1 1 2 3 4 21. y = f (x + 1) y 4 3 2 1 (1, 2) (−3, 2) −5 −4 −3 (−1, 0) 1 2 3 23. y = 2f (x) (−2, 4) (2, 4) y 4 3 2 1 −4 −3 −2 −1 (0, 0) 2 3 4 x x x 15. (−1, −7) 18. (1, 1) 20. y = f (x) − 2 y 2 1 −4 (−2, 0) −1 −1 1 (2, 0) x 4 −2 (0, −2) 22. y = f (x − 2) y 4 2 1 (0, 2) (4, 2) −2 −1 (2, 0) 3 4 5 6 24. y = f (2x) y 4 3 2 1 (1, 2) (−1, 2) −4 −3 −2 −1 (0, 0) 2 3 4 x x 1.7 Transformations 25. y = 2 − f (x) y (0, 2) 2 1 −4 −3 (−2, 0) (2, 0) 3 4 x 145 26. y = f (2 − x) y 4 2 1 (0, 2) (4, 2) −2 −1 (2, 0) 3 4 5 6 x 27. y = 2 − f (2 − x) 29. y = f (x) − 1 y 4 3 2 1 (2, 2) y (0, 3) 4 3 2 1 −2 −1 (0, 0) 2 (4, 0) 5 6 x −4 −3 −2 −1 (−2, −1) 1 2 3 4 x (2, −1) (4, −3) −1 −2 −3 −4 30. y = f (x + 1) (−1, 4)
y 4 3 2 1 −4 −3 −2 (−3, 0) −1 −1 1 (1, 0) 2 3 4 x −2 −3 −4 (3, −2) 31. y = 1 2 f (x) −4 −3 y (0, 2) 4 3 2 1 −1 −2 −3 −4 −1 (−2, 0) 1 3 4 x (2, 0) (4, −1) 146 Relations and Functions 32. y = f (2x) 33. y = −f (x) y (0, 4) (1, 0) 2 3 4 x (2, −2) y (0, 4) x 4 1 3 (2, 0) 4 3 2 1 −4 −3 −2 (−1, 0) 34. y = f (−x) −2 −3 −4 4 3 2 1 −4 −3 (−4, −2) −1 (−2, 0) −1 −2 −3 −4 36. y = 1 − f (x) y 4 3 2 1 (−2, 1) (4, 3) (2, 1) −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 (0, −3) y 4 3 2 1 (−2, 0) −4 −3 −2 −1 −1 (4, 2) (2, 0) 1 2 3 4 x −2 −3 −4 (0, −4) 35. y = f (x + 1) − 1 y (−1, 3) 4 3 2 1 −4 −3 −2 −1 −1 1 3 2 (1, −1) 4 x (−3, −1) −2 −3 −4 (3, −3) 37. y = 1 2 f (x + 1) − 1 y 4 3 2 1 (−1, 1) −4 −3 −2 (−3, −1) −1 −1 1 3 2 (1, −1) 4 x −2 −3 −4 (3, −2) 1.7 Transformations 147 38. g(x) = f (x) + 3 y 39. h(x) = f (x) − 1 2 y (0, 6) 6 5 4 3 2 1 (3, 3) (−3, 3) −3 −2 −1 1 2 3 x −1 0, 5
2 3 2 1 1 −1 2 3 3, − 1 2 x −3 −2 −1 −3, − 1 2 40. j(x 41. a(x) = f (x + 4) (−4, 3) y 3 2 1 3 2 1 −3 −2 −1 3, 0 − 7 −1 1 2 3 11 3, 0 x −7 −6 −5 −4 −3 −2 −1 (−7, 0) (−1, 0) x 42. b(x) = f (x + 1) − 1 y (−1, 2) 2 1 43. c(x) = 3 5 f (x) 2 1 y 0, 9 5 −4 −3 −2 −1 1 2 x −1 (−4, −1) (2, −1) −3 −2 −1 1 2 (−3, 0) −1 x 3 (3, 0) 44. d(x) = −2f (x) y (−3, 0) (3, 0) −3 −2 −1 1 2 3 x −1 −2 −3 −4 −5 −6 (0, −6) 45. k(x) = f 2 3 x y (0, 3) 3 2 1 −4 −3 −2 −1 2, 0 − 9 148 Relations and Functions 46. m(x) = − 1 4 f (3x) y 47. n(x) = 4f (x − 3) − 6 y (3, 6) (−1, 0) −1 (1, 0) 1 x −1 01 −2 −3 −4 −5 −6 1 2 3 4 5 6 x (0, −6) (6, −6) 48. p(x) = 4 + f (1 − 2x) = f (−2x + 1) + 4 y 1 2, 7 (2, 4) 7 6 5 4 (−1, 4) 3 2 1 −1 −1 1 2 x 49. q(x) = − 1 2 f x+10−9−8−7−6−5−4−3−2−1 −1 (−10, −3) −2 −3 −4 −4, − 9 2 1 2 x (2, −3) 1.7 Transformations 149 50. y = S1(x) = S(x + 1) 51. y = S2(
x) = S1(−x) = S(−x + 1) y (0, 3) 3 2 1 (−3, 0) (−1, 0) −3 −2 −1 x (1, 0) −1 −2 −3 y (0, 3) 3 2 1 (1, 0) (3, 0) 1 2 3 x (−1, 0) −1 −2 −3 (−2, −3) (2, −3) 52. y = S3(x) = 1 2 S2(x) = 1 2 S(−x + 1) y 2 1 0, 3 2 (1, 0) (3, 0) 1 2 3 x 2, − 3 2 (−1, 0) −1 −2 54. g(x 56. g(x) = − √ 58. g(x) = 60. g(x) = 2 √ 53. y = S4(x) = S3(x) + 1 = 1 2 S(−x + 1) + 1 y 3 2 0, 5 2 (1, 1) (3, 1) 1 (−1, 1) −1 1 −1 2, − 1 2 x 3 55. g(x) = √ x − 2 − 3 57. g(x) = −( √ x + 1) = − √ −(x + 1) + 2 = 59. g(x) = 61. g(x) = 2 √ x − 1 √ −(x − 3) + 1 = 2x − 6 + 1 62. g(x) = 2x − 3 + 1 √ 64. g(x) = −2 3 √ x + 3 − 1 or g(x) = 2 3 63. g(x) = −x − 3 − 1 150 Relations and Functions Chapter 2 Linear and Quadratic Functions 2.1 Linear Functions We now begin the study of families of functions. Our ﬁrst family, linear functions, are old friends as we shall soon see. Recall from Geometry that two distinct points in the plane determine a unique line containing those points, as indicated below. P (x0, y0) Q (x1, y1) To give a sense of the ‘steepness’ of the line, we recall that we can compute the slope of the line using the formula below. Equation
2.1. The slope m of the line containing the points P (x0, y0) and Q (x1, y1) is: provided x1 = x0. m = y1 − y0 x1 − x0, A couple of notes about Equation 2.1 are in order. First, don’t ask why we use the letter ‘m’ to represent slope. There are many explanations out there, but apparently no one really knows for sure.1 Secondly, the stipulation x1 = x0 ensures that we aren’t trying to divide by zero. The reader is invited to pause to think about what is happening geometrically; the anxious reader can skip along to the next example. Example 2.1.1. Find the slope of the line containing the following pairs of points, if it exists. Plot each pair of points and the line containing them. 1See www.mathforum.org or www.mathworld.wolfram.com for discussions on this topic. 152 Linear and Quadratic Functions 1. P (0, 0), Q(2, 4) 2. P (−1, 2), Q(3, 4) 3. P (−2, 3), Q(2, −3) 4. P (−3, 2), Q(4, 2) 5. P (2, 3), Q(2, −1) 6. P (2, 3), Q(2.1, −1) Solution. In each of these examples, we apply the slope formula, Equation 2.1. 1. m = 4 − 2 3 − (−1) = 2 4 = 1 2 3. m = −3 − 3 2 − (−2) = −6 4 = − 3 2 4. m = 2 − 2 4 − (−33 −2 −1 1 2 3 x −1 −2 −3 −4 Q y 3 2 1 P Q −4 −3 −2 −1 1 2 3 4 x 2.1 Linear Functions 153 5. m = −1 − 3 2 − 2 = −4 0, which is undeﬁned 6. m = −1 − 3 2.1 − 2 = −4 0.1 = −40 1 −2 −3 3 2 1 −1 −2 −3 A few comments about Example 2.1.1 are in order. First, for reasons which will be made clear soon, if the slope is positive
then the resulting line is said to be increasing. If it is negative, we say the line is decreasing. A slope of 0 results in a horizontal line which we say is constant, and an undeﬁned slope results in a vertical line.2 Second, the larger the slope is in absolute value, the steeper the line. You may recall from Intermediate Algebra that slope can be described as the ratio ‘ rise run ’. For example, in the second part of Example 2.1.1, we found the slope to be 1 2. We can interpret this as a rise of 1 unit upward for every 2 units to the right we travel along the line, as shown below. y 4 3 2 1 ‘up 1’ ‘over 2’ −1 1 2 3 x 2Some authors use the unfortunate moniker ‘no slope’ when a slope is undeﬁned. It’s easy to confuse the notions of ‘no slope’ with ‘slope of 0’. For this reason, we will describe slopes of vertical lines as ‘undeﬁned’. 154 Linear and Quadratic Functions Using more formal notation, given points (x0, y0) and (x1, y1), we use the Greek letter delta ‘∆’ to write ∆y = y1 − y0 and ∆x = x1 − x0. In most scientiﬁc circles, the symbol ∆ means ‘change in’. Hence, we may write m = ∆y ∆x, which describes the slope as the rate of change of y with respect to x. Rates of change abound in the ‘real world’, as the next example illustrates. Example 2.1.2. Suppose that two separate temperature readings were taken at the ranger station on the top of Mt. Sasquatch: at 6 AM the temperature was 24◦F and at 10 AM it was 32◦F. 1. Find the slope of the line containing the points (6, 24) and (10, 32). 2. Interpret your answer to the ﬁrst part in terms of temperature and time. 3. Predict the temperature at noon. Solution. 1. For the slope, we have m = 32−24 10−6 = 8 4 = 2. 2. Since the values in the numerator correspond to the temperatures in �
�F, and the values in 2◦ F the denominator correspond to time in hours, we can interpret the slope as 2 =, 1 hour or 2◦F per hour. Since the slope is positive, we know this corresponds to an increasing line. Hence, the temperature is increasing at a rate of 2◦F per hour. 2 1 = 3. Noon is two hours after 10 AM. Assuming a temperature increase of 2◦F per hour, in two hours the temperature should rise 4◦F. Since the temperature at 10 AM is 32◦F, we would expect the temperature at noon to be 32 + 4 = 36◦F. Now it may well happen that in the previous scenario, at noon the temperature is only 33◦F. This doesn’t mean our calculations are incorrect, rather, it means that the temperature change throughout the day isn’t a constant 2◦F per hour. As discussed in Section 1.4.1, mathematical models are just that: models. The predictions we get out of the models may be mathematically accurate, but may not resemble what happens in the real world. In Section 1.2, we discussed the equations of vertical and horizontal lines. Using the concept of slope, we can develop equations for the other varieties of lines. Suppose a line has a slope of m and contains the point (x0, y0). Suppose (x, y) is another point on the line, as indicated below. (x, y) (x0, y0) 2.1 Linear Functions Equation 2.1 yields 155 y − y0 x − x0 m (x − x0) = y − y0 m = y − y0 = m (x − x0) We have just derived the point-slope form of a line.3 Equation 2.2. The point-slope form of the line with slope m containing the point (x0, y0) is the equation y − y0 = m (x − x0). Example 2.1.3. Write the equation of the line containing the points (−1, 3) and (2, 1). Solution. In order to use Equation 2.2 we need to ﬁnd the slope of the line in question so we use Equation 2.1 to get m = ∆y 3. We are spoiled for choice for a point (x0, y0). We’ll
use (−1, 3) and leave it to the reader to check that using (2, 1) results in the same equation. Substituting into the point-slope form of the line, we get 2−(−1) = − 2 ∆x = 1−3 y − y0 = m (x − x0x − (−1)) (x + 1) x − x + 2 3 7 3. We can check our answer by showing that both (−1, 3) and (2, 1) are on the graph of y = − 2 algebraically, as we did in Section 1.2.1. 3 x + 7 3 In simplifying the equation of the line in the previous example, we produced another form of a line, the slope-intercept form. This is the familiar y = mx + b form you have probably seen in Intermediate Algebra. The ‘intercept’ in ‘slope-intercept’ comes from the fact that if we set x = 0, we get y = b. In other words, the y-intercept of the line y = mx + b is (0, b). Equation 2.3. The slope-intercept form of the line with slope m and y-intercept (0, b) is the equation y = mx + b. Note that if we have slope m = 0, we get the equation y = b which matches our formula for a horizontal line given in Section 1.2. The formula given in Equation 2.3 can be used to describe all lines except vertical lines. All lines except vertical lines are functions (Why is this?) so we have ﬁnally reached a good point to introduce linear functions. 3We can also understand this equation in terms of applying transformations to the function I(x) = x. See the Exercises. 156 Linear and Quadratic Functions Deﬁnition 2.1. A linear function is a function of the form f (x) = mx + b, where m and b are real numbers with m = 0. The domain of a linear function is (−∞, ∞). For the case m = 0, we get f (x) = b. These are given their own classiﬁcation. Deﬁnition 2.2. A constant function is a function of the form f (x) = b, where b is real
number. The domain of a constant function is (−∞, ∞). Recall that to graph a function, f, we graph the equation y = f (x). Hence, the graph of a linear function is a line with slope m and y-intercept (0, b); the graph of a constant function is a horizontal line (a line with slope m = 0) and a y-intercept of (0, b). Now think back to Section 1.6.1, speciﬁcally Deﬁnition 1.10 concerning increasing, decreasing and constant functions. A line with positive slope was called an increasing line because a linear function with m > 0 is an increasing function. Similarly, a line with a negative slope was called a decreasing line because a linear function with m < 0 is a decreasing function. And horizontal lines were called constant because, well, we hope you’ve already made the connection. Example 2.1.4. Graph the following functions. Identify the slope and y-intercept. 1. f (x) = 3 2. f (x) = 3x − 1 Solution. 3. f (x) = 4. f (x) = 3 − 2x 4 x2 − 4 x − 2 1. To graph f (x) = 3, we graph y = 3. This is a horizontal line (m = 0) through (0, 3). 2. The graph of f (x) = 3x − 1 is the graph of the line y = 3x − 1. Comparison of this equation with Equation 2.3 yields m = 3 and b = −1. Hence, our slope is 3 and our y-intercept is (0, −1). To get another point on the line, we can plot (1, f (1)) = (1, 2). 2.1 Linear Functions 157 y 4 3 2 1 −3 −2 −1 1 2 3 x f (x) = 3 y 4 3 2 1 −2−1 −1 1 2 x f (x) = 3x − 1 3. At ﬁrst glance, the function f (x) = 3−2x 4 − 2x rearranging we get f (x) = 3−2x our graph is a line with a slope of − 1 point, we can choose (1, f (1)) to get 1, 1 4 4 = 3. 4 does not
ﬁt the form in Deﬁnition 2.1 but after some 2 and b = 3 4 = − 1 4. Hence,. Plotting an additional 2 and a y-intercept of 0, 3 4. We identify. If we simplify the expression for f, we get f (x) = x2 − 4 x − 2 = (x − 2)(x + 2) (x − 2) = x + 2. If we were to state f (x) = x + 2, we would be committing a sin of omission. Remember, to ﬁnd the domain of a function, we do so before we simplify! In this case, f has big problems when x = 2, and as such, the domain of f is (−∞, 2) ∪ (2, ∞). To indicate this, we write f (x) = x + 2, x = 2. So, except at x = 2, we graph the line y = x + 2. The slope m = 1 and the y-intercept is (0, 2). A second point on the graph is (1, f (1)) = (1, 3). Since our function f is not deﬁned at x = 2, we put an open circle at the point that would be on the line y = x + 2 when x = 2, namely (2, 4). y 2 1 −3 −2 −1 1 2 3 x f (x) = 3 − 2x 4 y 4 3 2 1 −1 1 2 3 x f (x) = x2 − 4 x − 2 158 Linear and Quadratic Functions The last two functions in the previous example showcase some of the diﬃculty in deﬁning a linear function using the phrase ‘of the form’ as in Deﬁnition 2.1, since some algebraic manipulations may be needed to rewrite a given function to match ‘the form’. Keep in mind that the domains of linear and constant functions are all real numbers (−∞, ∞), so while f (x) = x2−4 x−2 simpliﬁed to a formula f (x) = x + 2, f is not considered a linear function since its domain excludes x = 2. However, we would consider f (x) = 2x2 + 2 x2 + 1 to be a constant
function since its domain is all real numbers (Can you tell us why?) and f (x) = 2x2 + 2 x2 + 1 = 2 x2 + 1 x2 + 1 = 2 The following example uses linear functions to model some basic economic relationships. Example 2.1.5. The cost C, in dollars, to produce x PortaBoy4 game systems for a local retailer is given by C(x) = 80x + 150 for x ≥ 0. 1. Find and interpret C(10). 2. How many PortaBoys can be produced for $15,000? 3. Explain the signiﬁcance of the restriction on the domain, x ≥ 0. 4. Find and interpret C(0). 5. Find and interpret the slope of the graph of y = C(x). Solution. 1. To ﬁnd C(10), we replace every occurrence of x with 10 in the formula for C(x) to get C(10) = 80(10) + 150 = 950. Since x represents the number of PortaBoys produced, and C(x) represents the cost, in dollars, C(10) = 950 means it costs $950 to produce 10 PortaBoys for the local retailer. 2. To ﬁnd how many PortaBoys can be produced for $15,000, we solve C(x) = 15000, or 80x + 80 = 185.625. Since we can only produce a whole 150 = 15000. Solving, we get x = 14850 number amount of PortaBoys, we can produce 185 PortaBoys for $15,000. 3. The restriction x ≥ 0 is the applied domain, as discussed in Section 1.4.1. In this context, x represents the number of PortaBoys produced. It makes no sense to produce a negative quantity of game systems.5 4The similarity of this name to PortaJohn is deliberate. 5Actually, it makes no sense to produce a fractional part of a game system, either, as we saw in the previous part of this example. This absurdity, however, seems quite forgivable in some textbooks but not to us. 2.1 Linear Functions 159 4. We ﬁnd C(0) = 80(0) + 150 = 150. This means it costs $150 to produce 0 PortaBoys. As mentioned on page 82, this
is the ﬁxed, or start-up cost of this venture. 5. If we were to graph y = C(x), we would be graphing the portion of the line y = 80x + 150 for x ≥ 0. We recognize the slope, m = 80. Like any slope, we can interpret this as a rate of change. Here, C(x) is the cost in dollars, while x measures the number of PortaBoys so m = ∆y ∆x = ∆C ∆x = 80 = 80 1 = $80 1 PortaBoy. In other words, the cost is increasing at a rate of $80 per PortaBoy produced. This is often called the variable cost for this venture. The next example asks us to ﬁnd a linear function to model a related economic problem. Example 2.1.6. The local retailer in Example 2.1.5 has determined that the number x of PortaBoy game systems sold in a week is related to the price p in dollars of each system. When the price was $220, 20 game systems were sold in a week. When the systems went on sale the following week, 40 systems were sold at $190 a piece. 1. Find a linear function which ﬁts this data. Use the weekly sales x as the independent variable and the price p as the dependent variable. 2. Find a suitable applied domain. 3. Interpret the slope. 4. If the retailer wants to sell 150 PortaBoys next week, what should the price be? 5. What would the weekly sales be if the price were set at $150 per system? Solution. 1. We recall from Section 1.4 the meaning of ‘independent’ and ‘dependent’ variable. Since x is to be the independent variable, and p the dependent variable, we treat x as the input variable and p as the output variable. Hence, we are looking for a function of the form p(x) = mx + b. To determine m and b, we use the fact that 20 PortaBoys were sold during the week when the price was 220 dollars and 40 units were sold when the price was 190 dollars. Using function notation, these two facts can be translated as p(20) = 220 and p(40) = 190. Since m represents the rate of change of p with respect to x, we have m = ∆p ∆x = 190
− 220 40 − 20 = −30 20 = −1.5. We now have determined p(x) = −1.5x + b. To determine b, we can use our given data again. Using p(20) = 220, we substitute x = 20 into p(x) = 1.5x + b and set the result equal to 220: −1.5(20) + b = 220. Solving, we get b = 250. Hence, we get p(x) = −1.5x + 250. We can check our formula by computing p(20) and p(40) to see if we get 220 and 190, respectively. You may recall from page 82 that the function p(x) is called the price-demand (or simply demand) function for this venture. 160 Linear and Quadratic Functions 2. To determine the applied domain, we look at the physical constraints of the problem. Certainly, we can’t sell a negative number of PortaBoys, so x ≥ 0. However, we also note that the slope of this linear function is negative, and as such, the price is decreasing as more units are sold. Thus another constraint on the price is p(x) ≥ 0. Solving −1.5x + 250 ≥ 0 results in −1.5x ≥ −250 or x ≤ week, we round down to 166. As a result, a reasonable applied domain for p is [0, 166]. = 166.6. Since x represents the number of PortaBoys sold in a 500 3 3. The slope m = −1.5, once again, represents the rate of change of the price of a system with respect to weekly sales of PortaBoys. Since the slope is negative, we have that the price is decreasing at a rate of $1.50 per PortaBoy sold. (Said diﬀerently, you can sell one more PortaBoy for every $1.50 drop in price.) 4. To determine the price which will move 150 PortaBoys, we ﬁnd p(150) = −1.5(150)+250 = 25. That is, the price would have to be $25. 5. If the price of a PortaBoy were set at $150, we have p(x) = 150, or, −1.5x+250 = 150. Solving, we get
−1.5x = −100 or x = 66.6. This means you would be able to sell 66 PortaBoys a week if the price were $150 per system. Not all real-world phenomena can be modeled using linear functions. Nevertheless, it is possible to use the concept of slope to help analyze non-linear functions using the following. Deﬁnition 2.3. Let f be a function deﬁned on the interval [a, b]. The average rate of change of f over [a, b] is deﬁned as: ∆f ∆x = f (b) − f (a) b − a Geometrically, if we have the graph of y = f (x), the average rate of change over [a, b] is the slope of the line which connects (a, f (a)) and (b, f (b)). This is called the secant line through these points. For that reason, some textbooks use the notation msec for the average rate of change of a function. Note that for a linear function m = msec, or in other words, its rate of change over an interval is the same as its average rate of change. y = f (x) (b, f (b)) (a, f (a)) The graph of y = f (x) and its secant line through (a, f (a)) and (b, f (b)) The interested reader may question the adjective ‘average’ in the phrase ‘average rate of change’. In the ﬁgure above, we can see that the function changes wildly on [a, b], yet the slope of the secant line only captures a snapshot of the action at a and b. This situation is entirely analogous to the 2.1 Linear Functions 161 average speed on a trip. Suppose it takes you 2 hours to travel 100 miles. Your average speed is 100 miles 2 hours = 50 miles per hour. However, it is entirely possible that at the start of your journey, you traveled 25 miles per hour, then sped up to 65 miles per hour, and so forth. The average rate of change is akin to your average speed on the trip. Your speedometer measures your speed at any one instant along the trip, your instantaneous rate of change, and this is one of the central themes of Calculus.6 When interpreting rates of change, we interpret them the same way
we did slopes. In the context of functions, it may be helpful to think of the average rate of change as: change in outputs change in inputs Example 2.1.7. Recall from page 82, the revenue from selling x units at a price p per unit is given by the formula R = xp. Suppose we are in the scenario of Examples 2.1.5 and 2.1.6. 1. Find and simplify an expression for the weekly revenue R(x) as a function of weekly sales x. 2. Find and interpret the average rate of change of R(x) over the interval [0, 50]. 3. Find and interpret the average rate of change of R(x) as x changes from 50 to 100 and compare that to your result in part 2. 4. Find and interpret the average rate of change of weekly revenue as weekly sales increase from 100 PortaBoys to 150 PortaBoys. Solution. 1. Since R = xp, we substitute p(x) = −1.5x + 250 from Example 2.1.6 to get R(x) = x(−1.5x + 250) = −1.5x2 + 250x. Since we determined the price-demand function p(x) is restricted to 0 ≤ x ≤ 166, R(x) is restricted to these values of x as well. 2. Using Deﬁnition 2.3, we get that the average rate of change is ∆R ∆x = R(50) − R(0) 50 − 0 = 8750 − 0 50 − 0 = 175. Interpreting this slope as we have in similar situations, we conclude that for every additional PortaBoy sold during a given week, the weekly revenue increases $175. 3. The wording of this part is slightly diﬀerent than that in Deﬁnition 2.3, but its meaning is to ﬁnd the average rate of change of R over the interval [50, 100]. To ﬁnd this rate of change, we compute ∆R ∆x = R(100) − R(50) 100 − 50 = 10000 − 8750 50 = 25. 6Here we go again... 162 Linear and Quadratic Functions In other words, for each additional PortaBoy sold, the revenue increases by $25. Note that while the revenue is still increasing by selling more game systems, we aren’
t getting as much of an increase as we did in part 2 of this example. (Can you think of why this would happen?) 4. Translating the English to the mathematics, we are being asked to ﬁnd the average rate of change of R over the interval [100, 150]. We ﬁnd ∆R ∆x = R(150) − R(100) 150 − 100 = 3750 − 10000 50 = −125. This means that we are losing $125 dollars of weekly revenue for each additional PortaBoy sold. (Can you think why this is possible?) We close this section with a new look at diﬀerence quotients which were ﬁrst introduced in Section 1.4. If we wish to compute the average rate of change of a function f over the interval [x, x + h], then we would have ∆f ∆x = f (x + h) − f (x) (x + h) − x = f (x + h) − f (x) h As we have indicated, the rate of change of a function (average or otherwise) is of great importance in Calculus.7 Also, we have the geometric interpretation of diﬀerence quotients which was promised to you back on page 81 – a diﬀerence quotient yields the slope of a secant line. 7So we are not torturing you with these for nothing. 2.1 Linear Functions 2.1.1 Exercises 163 In Exercises 1 - 10, ﬁnd both the point-slope form and the slope-intercept form of the line with the given slope which passes through the given point. 1. m = 3, P (3, −1) 3. m = −1, P (−7, −1) 5. m = − 1 5, P (10, 4) 7. m = 0, P (3, 117) 9. m = −5, P ( √ √ 3) 3, 2 2. m = −2, P (−5, 8) 4. m = 2 3, P (−2, 1) 6. m = 1 7, P (−1, 4) √ 8. m = − 2, P (0, −3) 10. m = 678, P (−1, −12) In Exercises 11 - 20, ﬁnd the slope-
intercept form of the line which passes through the given points. 11. P (0, 0), Q(−3, 5) 13. P (5, 0), Q(0, −8) 15. P (−1, 5), Q(7, 5) 12. P (−1, −2), Q(3, −2) 14. P (3, −5), Q(7, 4) 16. P (4, −8), Q(5, −8) 17. P 1 2, 3 19, − √ √ 2 2, 18. P 2 20, −1, Q √ 2 3, 1 In Exercises 21 - 26, graph the function. Find the slope, y-intercept and x-intercept, if any exist. 21. f (x) = 2x − 1 23. f (x) = 3 25. f (x) = 2 3 x + 1 3 22. f (x) = 3 − x 24. f (x) = 0 26. f (x) = 1 − x 2 27. Find all of the points on the line y = 2x + 1 which are 4 units from the point (−1, 3). 28. Jeﬀ can walk comfortably at 3 miles per hour. Find a linear function d that represents the total distance Jeﬀ can walk in t hours, assuming he doesn’t take any breaks. 29. Carl can stuﬀ 6 envelopes per minute. Find a linear function E that represents the total number of envelopes Carl can stuﬀ after t hours, assuming he doesn’t take any breaks. 30. A landscaping company charges $45 per cubic yard of mulch plus a delivery charge of $20. Find a linear function which computes the total cost C (in dollars) to deliver x cubic yards of mulch. 164 Linear and Quadratic Functions 31. A plumber charges $50 for a service call plus $80 per hour. If she spends no longer than 8 hours a day at any one site, ﬁnd a linear function that represents her total daily charges C (in dollars) as a function of time t (in hours) spent at any one given location. 32. A salesperson is paid $200 per week plus 5% commission on her weekly sales of x dollars. Find a linear function that represents her total weekly pay, W (in dollars)
in terms of x. What must her weekly sales be in order for her to earn $475.00 for the week? 33. An on-demand publisher charges $22.50 to print a 600 page book and $15.50 to print a 400 page book. Find a linear function which models the cost of a book C as a function of the number of pages p. Interpret the slope of the linear function and ﬁnd and interpret C(0). 34. The Topology Taxi Company charges $2.50 for the ﬁrst ﬁfth of a mile and $0.45 for each additional ﬁfth of a mile. Find a linear function which models the taxi fare F as a function of the number of miles driven, m. Interpret the slope of the linear function and ﬁnd and interpret F (0). 35. Water freezes at 0◦ Celsius and 32◦ Fahrenheit and it boils at 100◦C and 212◦F. (a) Find a linear function F that expresses temperature in the Fahrenheit scale in terms of degrees Celsius. Use this function to convert 20◦C into Fahrenheit. (b) Find a linear function C that expresses temperature in the Celsius scale in terms of degrees Fahrenheit. Use this function to convert 110◦F into Celsius. (c) Is there a temperature n such that F (n) = C(n)? 36. Legend has it that a bull Sasquatch in rut will howl approximately 9 times per hour when it is 40◦F outside and only 5 times per hour if it’s 70◦F. Assuming that the number of howls per hour, N, can be represented by a linear function of temperature Fahrenheit, ﬁnd the number of howls per hour he’ll make when it’s only 20◦F outside. What is the applied domain of this function? Why? 37. Economic forces beyond anyone’s control have changed the cost function for PortaBoys to C(x) = 105x + 175. Rework Example 2.1.5 with this new cost function. 38. In response to the economic forces in Exercise 37 above, the local retailer sets the selling price of a PortaBoy at $250. Remarkably, 30 units were sold each week. When the systems went on sale for $220, 40 units per week were sold. Rework Examples 2.1.6 and
2.1.7 with this new data. What diﬃculties do you encounter? 39. A local pizza store oﬀers medium two-topping pizzas delivered for $6.00 per pizza plus a $1.50 delivery charge per order. On weekends, the store runs a ‘game day’ special: if six or more medium two-topping pizzas are ordered, they are $5.50 each with no delivery charge. Write a piecewise-deﬁned linear function which calculates the cost C (in dollars) of p medium two-topping pizzas delivered during a weekend. 2.1 Linear Functions 165 40. A restaurant oﬀers a buﬀet which costs $15 per person. For parties of 10 or more people, a group discount applies, and the cost is $12.50 per person. Write a piecewise-deﬁned linear function which calculates the total bill T of a party of n people who all choose the buﬀet. 41. A mobile plan charges a base monthly rate of $10 for the ﬁrst 500 minutes of air time plus a charge of 15¢ for each additional minute. Write a piecewise-deﬁned linear function which calculates the monthly cost C (in dollars) for using m minutes of air time. HINT: You may want to revisit Exercise 74 in Section 1.4 42. The local pet shop charges 12¢ per cricket up to 100 crickets, and 10¢ per cricket thereafter. Write a piecewise-deﬁned linear function which calculates the price P, in dollars, of purchasing c crickets. 43. The cross-section of a swimming pool is below. Write a piecewise-deﬁned linear function which describes the depth of the pool, D (in feet) as a function of: (a) the distance (in feet) from the edge of the shallow end of the pool, d. (b) the distance (in feet) from the edge of the deep end of the pool, s. (c) Graph each of the functions in (a) and (b). Discuss with your classmates how to trans- form one into the other and how they relate to the diagram of the pool. d ft. 37 ft. 8 ft. 15 ft. s ft. 10 ft. 2 ft. In Exercises 44 - 49, compute the average
rate of change of the function over the speciﬁed interval. 44. f (x) = x3, [−1, 2] 46. f (x) = √ x, [0, 16] 48. f (x5, 7] 45. f (x) = 1 x, [1, 5] 47. f (x) = x2, [−3, 3] 49. f (x) = 3x2 + 2x − 7, [−4, 2] 166 Linear and Quadratic Functions In Exercises 50 - 53, compute the average rate of change of the given function over the interval [x, x + h]. Here we assume [x, x + h] is in the domain of the function. 50. f (x) = x3 52. f (x) = x + 4 x − 3 51. f (x) = 1 x 53. f (x) = 3x2 + 2x − 7 54. The height of an object dropped from the roof of an eight story building is modeled by: h(t) = −16t2 + 64, 0 ≤ t ≤ 2. Here, h is the height of the object oﬀ the ground in feet, t seconds after the object is dropped. Find and interpret the average rate of change of h over the interval [0, 2]. 55. Using data from Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = −0.0076t2 + 0.45t + 16, 0 ≤ t ≤ 28, where t is the number of years since 1980. Find and interpret the average rate of change of F over the interval [0, 28]. 56. The temperature T in degrees Fahrenheit t hours after 6 AM is given by: T (t) = − 1 2 t2 + 8t + 32, 0 ≤ t ≤ 12 (a) Find and interpret T (4), T (8) and T (12). (b) Find and interpret the average rate of change of T over the interval [4, 8]. (c) Find and interpret the average rate of change of T from t = 8 to t = 12. (d) Find and interpret the average rate of temperature change between 10 AM and 6 PM. 57. Suppose C(x) = x2 − 10x + 27 represents the costs
, in hundreds, to produce x thousand pens. Find and interpret the average rate of change as production is increased from making 3000 to 5000 pens. 58. With the help of your classmates ﬁnd several other “real-world” examples of rates of change that are used to describe non-linear phenomena. (Parallel Lines) Recall from Intermediate Algebra that parallel lines have the same slope. (Please note that two vertical lines are also parallel to one another even though they have an undeﬁned slope.) In Exercises 59 - 64, you are given a line and a point which is not on that line. Find the line parallel to the given line which passes through the given point. 59. y = 3x + 2, P (0, 0) 60. y = −6x + 5, P (3, 2) 2.1 Linear Functions 61. y = 2 3 x − 7, P (6, 0) 63. y = 6, P (3, −2) 167 621, −1) 64. x = 1, P (−5, 0) (Perpendicular Lines) Recall from Intermediate Algebra that two non-vertical lines are perpendicular if and only if they have negative reciprocal slopes. That is to say, if one line has slope m1 and the other has slope m2 then m1 · m2 = −1. (You will be guided through a proof of this result in Exercise 71.) Please note that a horizontal line is perpendicular to a vertical line and vice versa, so we assume m1 = 0 and m2 = 0. In Exercises 65 - 70, you are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point. 65. y = 1 3 x + 2, P (0, 0) 67. y = 2 3 x − 7, P (6, 0) 69. y = 6, P (3, −2) 66. y = −6x + 5, P (3, 2) 681, −1) 70. x = 1, P (−5, 0) 71. We shall now prove that y = m1x + b1 is perpendicular to y = m2x + b2 if and only if m1 · m2 = −1. To make our lives easier we shall assume that m1 > 0 and m2 < 0. We can also
“move” the lines so that their point of intersection is the origin without messing things up, so we’ll assume b1 = b2 = 0. (Take a moment with your classmates to discuss why this is okay.) Graphing the lines and plotting the points O(0, 0), P (1, m1) and Q(1, m2) gives us the following set up. y O P Q x The line y = m1x will be perpendicular to the line y = m2x if and only if OP Q is a right triangle. Let d1 be the distance from O to P, let d2 be the distance from O to Q and let d3 be the distance from P to Q. Use the Pythagorean Theorem to show that OP Q is a right triangle if and only if m1 · m2 = −1 by showing d2 3 if and only if m1 · m2 = −1. 2 = d2 1 + d2 168 Linear and Quadratic Functions 72. Show that if a = b, the line containing the points (a, b) and (b, a) is perpendicular to the line y = x. (Coupled with the result from Example 1.1.7 on page 13, we have now shown that the line y = x is a perpendicular bisector of the line segment connecting (a, b) and (b, a). This means the points (a, b) and (b, a) are symmetric about the line y = x. We will revisit this symmetry in section 5.2.) 73. The function deﬁned by I(x) = x is called the Identity Function. (a) Discuss with your classmates why this name makes sense. (b) Show that the point-slope form of a line (Equation 2.2) can be obtained from I using a sequence of the transformations deﬁned in Section 1.7. 2.1 Linear Functions 2.1.2 Answers 1. y + 1 = 3(x − 3) y = 3x − 10 3. y + 1 = −(x + 7) y = −x − 8 169 2. y − 8 = −2(x + 5) y = −2x − 2 4x + 2) 5x − 10) 7. y − 117 = 0 y = 117 √ 9. y − 2 3 = −5
(x − √ y = −5x + 7 3 11. y = − 5 3 x 13. y = 8 5 x − 8 15. y = 5 17. y = − 5 4 x + 11 8 19. y = −x 21. f (x) = 2x − 1 slope: m = 2 y-intercept: (0, −1) 2, 0 x-intercept: 1 22. f (x) = 3 − x slope: m = −1 y-intercept: (0, 3) x-intercept: (3, 0) √ 3) 10. y + 12 = 678(x + 1) y = 678x + 666 12. y = −2 6x + 1) 7 x + 29 7 √ 8. y + 3 = − y = − √ 2x − 3 2(x − 0) 14. y = 9 4 x − 47 4 16. y = −8 18. y = 2x + 13 6 202 −1 1 2 x −1 −2 −3 y 4 3 2 1 −1 1 2 3 4 x −1 170 Linear and Quadratic Functions 23. f (x) = 3 slope: m = 0 y-intercept: (0, 3) x-intercept: none 24. f (x) = 0 slope: m = 0 y-intercept: (0, 0) x-intercept: {(x, 0) \| x is a real number} 25. f (x) = 2 3 3 x + 1 slope: m = 2 3 y-intercept: 0, 1 3 x-intercept: − 1 2, 0 26. f (x) = 1 − x 2 slope: m = − 1 2 y-intercept: 0, 1 2 x-intercept: (1, 0) y y 4 3 2 1 1 −2 −1 1 2 x −2 −1 1 2 x −1 2 1 −1 2 1 y y −2 1 2 x −2 −1 1 2 x −1 27. (−1, −1) and 11 5, 27 5 28. d(t) = 3t, t ≥ 0. 29. E(t) = 360t, t ≥ 0. 30. C(x) = 45x + 20, x ≥ 0. 31. C(t) = 80t + 50, 0 ≤ t ≤ 8
. 32. W (x) = 200 +.05x, x ≥ 0 She must make $5500 in weekly sales. 33. C(p) = 0.035p + 1.5 The slope 0.035 means it costs 3.5¢ per page. C(0) = 1.5 means there is a ﬁxed, or start-up, cost of $1.50 to make each book. 34. F (m) = 2.25m + 2.05 The slope 2.25 means it costs an additional $2.25 for each mile beyond the ﬁrst 0.2 miles. F (0) = 2.05, so according to the model, it would cost $2.05 for a trip of 0 miles. Would this ever really happen? Depends on the driver and the passenger, we suppose. 2.1 Linear Functions 171 35. (a) F (C) = 9 5 C + 32 (c) F (−40) = −40 = C(−40). (b) C(F ) = 5 9 (F − 32) = 5 9 F − 160 9 36. N (T ) = − 2 15 T + 43 3 and N (20) = 35 3 ≈ 12 howls per hour. Having a negative number of howls makes no sense and since N (107.5) = 0 we can put an upper bound of 107.5◦F on the domain. The lower bound is trickier because there’s nothing other than common sense to go on. As it gets colder, he howls more often. At some point it will either be so cold that he freezes to death or he’s howling non-stop. So we’re going to say that he can withstand temperatures no lower than −60◦F so that the applied domain is [−60, 107.5]. 39. C(p) = 40. T (n) = 41. C(m) = 42. P (c) = 43. (a) (b) (c) 6p + 1.5 5.5p if if 1 ≤ p ≤ 5 p ≥ 6 15n 12.5n if if 1 ≤ n ≤ 9 n ≥ 10 10 10 + 0.15(m − 500) 0 ≤ m ≤ 500 if if m > 500 0.12c 12 + 0.1(c − 100) if if
1 ≤ c ≤ 100 c > 100 D(d) =    − 1 8 2 d + 31 2 2 if if if 0 ≤ d ≤ 15 15 ≤ d ≤ 27 27 ≤ d ≤ 37 D(s) =    2 1 2 s − 3 8 if if if 0 ≤ s ≤ 10 10 ≤ s ≤ 22 22 ≤ s ≤ 37 8 2 8 2 15 27 37 10 22 37 y = D(d) y = D(s) 172 44. 46. 48 Linear and Quadratic Functions 45. 47 32 − (−3)2 3 − (−3) = 0 49. (3(2)2 + 2(2) − 7) − (3(−4)2 + 2(−4) − 7) 2 − (−4) = −4 23 − (−1)3 2 − (−1) √ √ 0 16 − 16 − 0 7−3 − 5+4 7 − 5 7+4 5−3 50. 3x2 + 3xh + h2 52. −7 (x − 3)(x + h − 3) 51. −1 x(x + h) 53. 6x + 3h + 2 54. The average rate of change is h(2)−h(0) = −32. During the ﬁrst two seconds after it is dropped, the object has fallen at an average rate of 32 feet per second. (This is called the average velocity of the object.) 2−0 55. The average rate of change is F (28)−F (0) = 0.2372. During the years from 1980 to 2008, the average fuel economy of passenger cars in the US increased, on average, at a rate of 0.2372 miles per gallon per year. 28−0 56. (a) T (4) = 56, so at 10 AM (4 hours after 6 AM), it is 56◦F. T (8) = 64, so at 2 PM (8 hours after 6 AM), it is 64◦F. T (12) = 56, so at 6 PM (12 hours after 6 AM), it is 56◦F. (b) The average rate of change is T (8)−T (4) 8−4 = 2. Between 10 AM and 2 PM,
the temperature increases, on average, at a rate of 2◦F per hour. (c) The average rate of change is T (12)−T (8) 12−8 = −2. Between 2 PM and 6 PM, the temperature decreases, on average, at a rate of 2◦F per hour. (d) The average rate of change is T (12)−T (4) ture, on average, remains constant. 12−4 = 0. Between 10 AM and 6 PM, the tempera- 57. The average rate of change is C(5)−C(3) 5−3 = −2. As production is increased from 3000 to 5000 pens, the cost decreases at an average rate of $200 per 1000 pens produced (20¢ per pen.) 59. y = 3x 60. y = −6x + 20 62 65. y = −3x 63. y = −2 66. y = 1 6 x + 3 2 68. y = 3x − 4 69. x = 3 61. y = 2 3 x − 4 64. x = −5 67. y = − 3 2 x + 9 70. y = 0 2.2 Absolute Value Functions 173 2.2 Absolute Value Functions There are a few ways to describe what is meant by the absolute value \|x\| of a real number x. You may have been taught that \|x\| is the distance from the real number x to 0 on the number line. So, for example, \|5\| = 5 and \| − 5\| = 5, since each is 5 units from 0 on the number line. distance is 5 units distance is 5 units −5 −4 −3 −2 −5)2 = x2. Using this deﬁnition, we have Another way to deﬁne absolute value is by the equation \|x\| = \|5\| = 25 = 5. The long and short of both of these procedures is that \|x\| takes negative real numbers and assigns them to their positive counterparts while it leaves positive numbers alone. This last description is the one we shall adopt, and is summarized in the following deﬁnition. 25 = 5 and \| − 5\| = (−5)2 = √ Deﬁnition 2.4. The absolute value of a real number x, denoted \|x\|, is given by \|x\| = −x, x, if x < 0
if x ≥ 0 In Deﬁnition 2.4, we deﬁne \|x\| using a piecewise-deﬁned function. (See page 62 in Section 1.4.) To check that this deﬁnition agrees with what we previously understood as absolute value, note that since 5 ≥ 0, to ﬁnd \|5\| we use the rule \|x\| = x, so \|5\| = 5. Similarly, since −5 < 0, we use the rule \|x\| = −x, so that \| − 5\| = −(−5) = 5. This is one of the times when it’s best to interpret the expression ‘−x’ as ‘the opposite of x’ as opposed to ‘negative x’. Before we begin studying absolute value functions, we remind ourselves of the properties of absolute value. Theorem 2.1. Properties of Absolute Value: Let a, b and x be real numbers and let n be an integer.a Then Product Rule: \|ab\| = \|a\|\|b\| Power Rule: \|an\| = \|a\|n whenever an is deﬁned Quotient Rule: a b = \|a\| \|b\|, provided b = 0 Equality Properties: \|x\| = 0 if and only if x = 0. For c > 0, \|x\| = c if and only if x = c or −x = c. For c < 0, \|x\| = c has no solution. aSee page 2 if you don’t remember what an integer is. 174 Linear and Quadratic Functions The proofs of the Product and Quotient Rules in Theorem 2.1 boil down to checking four cases: when both a and b are positive; when they are both negative; when one is positive and the other is negative; and when one or both are zero. For example, suppose we wish to show that \|ab\| = \|a\|\|b\|. We need to show that this equation is true for all real numbers a and b. If a and b are both positive, then so is ab. Hence, \|a\| = a, \|b\| = b and \|ab\| = ab. Hence, the equation \|ab\| = \|a\|\|b\| is the same as ab = ab which is true. If both a and b are negative, then ab is positive. Hence, \|a
\| = −a, \|b\| = −b and \|ab\| = ab. The equation \|ab\| = \|a\|\|b\| becomes ab = (−a)(−b), which is true. Suppose a is positive and b is negative. Then ab is negative, and we have \|ab\| = −ab, \|a\| = a and \|b\| = −b. The equation \|ab\| = \|a\|\|b\| reduces to −ab = a(−b) which is true. A symmetric argument shows the equation \|ab\| = \|a\|\|b\| holds when a is negative and b is positive. Finally, if either a or b (or both) are zero, then both sides of \|ab\| = \|a\|\|b\| are zero, so the equation holds in this case, too. All of this rhetoric has shown that the equation \|ab\| = \|a\|\|b\| holds true in all cases. The proof of the Quotient Rule is very similar, with the exception that b = 0. The Power Rule can be shown by repeated application of the Product Rule. The ‘Equality Properties’ can be proved using Deﬁnition 2.4 and by looking at the cases when x ≥ 0, in which case \|x\| = x, or when x < 0, in which case \|x\| = −x. For example, if c > 0, and \|x\| = c, then if x ≥ 0, we have x = \|x\| = c. If, on the other hand, x < 0, then −x = \|x\| = c, so x = −c. The remaining properties are proved similarly and are left for the Exercises. Our ﬁrst example reviews how to solve basic equations involving absolute value using the properties listed in Theorem 2.1. Example 2.2.1. Solve each of the following equations. 1. \|3x − 1\| = 6 2. 3 − \|x + 5\| = 1 3. 3\|2x + 1\| − 5 = 0 4. 4 − \|5x + 3\| = 5 5. \|x\| = x2 − 6 6. \|x − 2\| + 1 = x Solution. 1. The equation \|3x − 1\| = 6 is of the form \|x\| = c for c > 0, so by the Equality Properties, \|3x − 1\| = 6 is equivalent
to 3x − 1 = 6 or 3x − 1 = −6. Solving the former, we arrive at x = 7 3, and solving the latter, we get x = − 5 3. We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out. 2. To use the Equality Properties to solve 3 − \|x + 5\| = 1, we ﬁrst isolate the absolute value. 3 − \|x + 5\| = 1 −\|x + 5\| = −2 \|x + 5\| = 2 subtract 3 divide by −1 From the Equality Properties, we have x + 5 = 2 or x + 5 = −2, and get our solutions to be x = −3 or x = −7. We leave it to the reader to check both answers in the original equation. 2.2 Absolute Value Functions 175 3. As in the previous example, we ﬁrst isolate the absolute value in the equation 3\|2x+1\|−5 = 0 3 or 2x + 1 = − 5 3. 3. As usual, we may and get \|2x + 1\| = 5 Solving the former gives x = 1 substitute both answers in the original equation to check. 3. Using the Equality Properties, we have 2x + 1 = 5 3 and solving the latter gives x = − 4 4. Upon isolating the absolute value in the equation 4 − \|5x + 3\| = 5, we get \|5x + 3\| = −1. At this point, we know there cannot be any real solution, since, by deﬁnition, the absolute value of anything is never negative. We are done. 5. The equation \|x\| = x2 − 6 presents us with some diﬃculty, since x appears both inside and outside of the absolute value. Moreover, there are values of x for which x2 − 6 is positive, negative and zero, so we cannot use the Equality Properties without the risk of introducing extraneous solutions, or worse, losing solutions. For this reason, we break equations like this into cases by rewriting the term in absolute values, \|x\|, using Deﬁnition 2.4. For x < 0, \|x\| = −x, so for x < 0, the equation \|x\| = x2 − 6 is equivalent to −x = x2 − 6. Rearranging this gives
us x2 + x − 6 = 0, or (x + 3)(x − 2) = 0. We get x = −3 or x = 2. Since only x = −3 satisﬁes x < 0, this is the answer we keep. For x ≥ 0, \|x\| = x, so the equation \|x\| = x2 − 6 becomes x = x2 − 6. From this, we get x2 − x − 6 = 0 or (x − 3)(x + 2) = 0. Our solutions are x = 3 or x = −2, and since only x = 3 satisﬁes x ≥ 0, this is the one we keep. Hence, our two solutions to \|x\| = x2 − 6 are x = −3 and x = 3. 6. To solve \|x − 2\| + 1 = x, we ﬁrst isolate the absolute value and get \|x − 2\| = x − 1. Since we see x both inside and outside of the absolute value, we break the equation into cases. The term with absolute values here is \|x − 2\|, so we replace ‘x’ with the quantity ‘(x − 2)’ in Deﬁnition 2.4 to get \|x − 2\| = −(x − 2), (x − 2), if if (x − 2) < 0 (x − 2) ≥ 0 Simplifying yields \|x − 2\| = −x + 2, x − 2, if x < 2 if x ≥ 2 So, for x < 2, \|x − 2\| = −x + 2 and our equation \|x − 2\| = x − 1 becomes −x + 2 = x − 1, which gives x = 3 2. Since this solution satisﬁes x < 2, we keep it. Next, for x ≥ 2, \|x − 2\| = x − 2, so the equation \|x − 2\| = x − 1 becomes x − 2 = x − 1. Here, the equation reduces to −2 = −1, which signiﬁes we have no solutions here. Hence, our only solution is x = 3 2. Next, we turn our attention to graphing absolute value functions. Our strategy in the next example is to make liberal use of Deﬁnition 2.4 along with what we know about grap
hing linear functions (from Section 2.1) and piecewise-deﬁned functions (from Section 1.4). Example 2.2.2. Graph each of the following functions. 1. f (x) = \|x\| 2. g(x) = \|x − 3\| 3. h(x) = \|x\| − 3 4. i(x) = 4 − 2\|3x + 1\| 176 Linear and Quadratic Functions Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and ﬁnd the relative and absolute extrema, if they exist. Solution. 1. To ﬁnd the zeros of f, we set f (x) = 0. We get \|x\| = 0, which, by Theorem 2.1 gives us x = 0. Since the zeros of f are the x-coordinates of the x-intercepts of the graph of y = f (x), we get (0, 0) as our only x-intercept. To ﬁnd the y-intercept, we set x = 0, and ﬁnd y = f (0) = 0, so that (0, 0) is our y-intercept as well.1 Using Deﬁnition 2.4, we get f (x) = \|x\| = −x, x, if x < 0 if x ≥ 0 Hence, for x < 0, we are graphing the line y = −x; for x ≥ 0, we have the line y = x. Proceeding as we did in Section 1.6, we get 3 −2 −1 1 2 3 x f (x) = \|x\|, x < 0 −3 −2 −1 1 2 3 x f (x) = \|x\|, x ≥ 0 Notice that we have an ‘open circle’ at (0, 0) in the graph when x < 0. As we have seen before, this is due to the fact that the points on y = −x approach (0, 0) as the x-values approach 0. Since x is required to be strictly less than zero on this stretch, the open circle is drawn at the origin. However,
notice that when x ≥ 0, we get to ﬁll in the point at (0, 0), which eﬀectively ‘plugs’ the hole indicated by the open circle. Thus we get, y 4 3 2 1 −3 −2 −1 1 2 3 x f (x) = \|x\| 1Actually, since functions can have at most one y-intercept (Do you know why?), as soon as we found (0, 0) as the x-intercept, we knew this was also the y-intercept. 2.2 Absolute Value Functions 177 By projecting the graph to the x-axis, we see that the domain is (−∞, ∞). Projecting to the y-axis gives us the range [0, ∞). The function is increasing on [0, ∞) and decreasing on (−∞, 0]. The relative minimum value of f is the same as the absolute minimum, namely 0 which occurs at (0, 0). There is no relative maximum value of f. There is also no absolute maximum value of f, since the y values on the graph extend inﬁnitely upwards. 2. To ﬁnd the zeros of g, we set g(x) = \|x − 3\| = 0. By Theorem 2.1, we get x − 3 = 0 so that x = 3. Hence, the x-intercept is (3, 0). To ﬁnd our y-intercept, we set x = 0 so that y = g(0) = \|0 − 3\| = 3, which yields (0, 3) as our y-intercept. To graph g(x) = \|x − 3\|, we use Deﬁnition 2.4 to rewrite g as g(x) = \|x − 3\| = −(x − 3), (x − 3), if if (x − 3) < 0 (x − 3) ≥ 0 Simplifying, we get g(x) = −x + 3, x − 3, if x < 3 if x ≥ 3 As before, the open circle we introduce at (3, 0) from the graph of y = −x + 3 is ﬁlled by the point (3, 0) from the line y = x − 3. We determine the domain as (−∞, ∞) and the range as [0,
∞). The function g is increasing on [3, ∞) and decreasing on (−∞, 3]. The relative and absolute minimum value of g is 0 which occurs at (3, 0). As before, there is no relative or absolute maximum value of g. 3. Setting h(x) = 0 to look for zeros gives \|x\| − 3 = 0. As in Example 2.2.1, we isolate the absolute value to get \|x\| = 3 so that x = 3 or x = −3. As a result, we have a pair of xintercepts: (−3, 0) and (3, 0). Setting x = 0 gives y = h(0) = \|0\| − 3 = −3, so our y-intercept is (0, −3). As before, we rewrite the absolute value in h to get h(x) = −x − 3, x − 3, if x < 0 if x ≥ 0 Once again, the open circle at (0, −3) from one piece of the graph of h is ﬁlled by the point (0, −3) from the other piece of h. From the graph, we determine the domain of h is (−∞, ∞) and the range is [−3, ∞). On [0, ∞), h is increasing; on (−∞, 0] it is decreasing. The relative minimum occurs at the point (0, −3) on the graph, and we see −3 is both the relative and absolute minimum value of h. Also, h has no relative or absolute maximum value. 178 Linear and Quadratic Functions (x) = \|x − 3\| −3 −2 −1 y 1 −1 −2 −3 −4 1 2 3 x h(x) = \|x\| − 3 x 4. As before, we set i(x) = 0 to ﬁnd the zeros of i and get 4−2\|3x+1\| = 0. Isolating the absolute 3 or x = −1, so 3, 0 and (−1, 0). Substituting x = 0 gives y = i(0) = 4−2\|3(0)+1\| = 2, value term gives \|3x + 1\| = 2, so either 3x + 1 = 2 or 3x + 1 = −2. We get x = 1 our x-
intercepts are 1 for a y-intercept of (0, 2). Rewriting the formula for i(x) without absolute values gives i(x) = 4 − 2(−(3x + 1)), 4 − 2(3x + 1), if if (3x + 1) < 0 (3x + 1) ≥ 0 = 6x + 6, −6x + 2, if x < − 1 3 if x ≥ − 1 3 The usual analysis near the trouble spot x = − 1 and we get the distinctive ‘∨’ shape: 3 gives the ‘corner’ of this graph is − 1 3, 4, y 5 3 2 1 −1 −1 1 x i(x) = 4 − 2\|3x + 1\| and decreasing on − 1 The domain of i is (−∞, ∞) while the range is (−∞, 4]. The function i is increasing on 3, 4 −∞, − 1 3 and the relative and absolute maximum value of i is 4. Since the graph of i extends downwards forever more, there is no absolute minimum value. As we can see from the graph, there is no relative minimum, either. 3, ∞. The relative maximum occurs at the point − 1 Note that all of the functions in the previous example bear the characteristic ‘∨’ shape of the graph of y = \|x\|. We could have graphed the functions g, h and i in Example 2.2.2 starting with the graph of f (x) = \|x\| and applying transformations as in Section 1.7 as our next example illustrates. 2.2 Absolute Value Functions 179 Example 2.2.3. Graph the following functions starting with the graph of f (x) = \|x\| and using transformations. 1. g(x) = \|x − 3\| 2. h(x) = \|x\| − 3 3. i(x) = 4 − 2\|3x + 1\| Solution. We begin by graphing f (x) = \|x\| and labeling three points, (−1, 1), (0, 0) and (1, 1). y 4 3 2 1 (1, 1) (−1, 1) −3 −2 −1 (0, 0) 1 2 3 x f (x) = \|x\| 1. Since g(x) = \|x − 3\| = f (x
− 3), Theorem 1.7 tells us to add 3 to each of the x-values of the points on the graph of y = f (x) to obtain the graph of y = g(x). This shifts the graph of y = f (x) to the right 3 units and moves the point (−1, 1) to (2, 1), (0, 0) to (3, 0) and (1, 1) to (4, 1). Connecting these points in the classic ‘∨’ fashion produces the graph of y = g(x). y 4 3 2 1 (1, 1) (−1, 1) y 4 3 2 1 (2, 1) (4, 1) −3 −2 −1 (0, 0) 1 2 3 x f (x) = \|x\| shift right 3 units −−−−−−−−−−−−→ add 3 to each x-coordinate 1 2 (3, 0) 4 5 6 x g(x) = f (x − 3) = \|x − 3\| 2. For h(x) = \|x\| − 3 = f (x) − 3, Theorem 1.7 tells us to subtract 3 from each of the y-values of the points on the graph of y = f (x) to obtain the graph of y = h(x). This shifts the graph of y = f (x) down 3 units and moves (−1, 1) to (−1, −2), (0, 0) to (0, −3) and (1, 1) to (1, −2). Connecting these points with the ‘∨’ shape produces our graph of y = h(x). y 4 3 2 1 (1, 1) (−1, 1) −3 −2 −1 (0, 0) 1 2 3 x f (x) = \|x\| shift down 3 units −−−−−−−−−−−−→ subtract 3 from each y-coordinate y 1 −1 −3 −2 −1 1 2 3 x (−1, −2) −2 (1, −2) (0, −3) −3 −4 h(x) = f (x) − 3 = \|x\| − 3 180 Linear and Quadratic Functions 3. We re-write i(x) = 4 − 2\|3x + 1\|
= 4 − 2f (3x + 1) = −2f (3x + 1) + 4 and apply Theorem 1.7. First, we take care of the changes on the ‘inside’ of the absolute value. Instead of \|x\|, we have \|3x + 1\|, so, in accordance with Theorem 1.7, we ﬁrst subtract 1 from each of the x-values of points on the graph of y = f (x), then divide each of those new values by 3. This eﬀects a horizontal shift left 1 unit followed by a horizontal shrink by a factor of 3. These transformations move (−1, 1) to − 2 3, 0 and (1, 1) to (0, 1). Next, we take 3, 1, (0, 0) to − 1 care of what’s happening ‘outside of’ the absolute value. Theorem 1.7 instructs us to ﬁrst multiply each y-value of these new points by −2 then add 4. Geometrically, this corresponds to a vertical stretch by a factor of 2, a reﬂection across the x-axis and ﬁnally, a vertical shift up 4 units. These transformations move − 2 3, 4, and (0, 1) to (0, 2). Connecting these points with the usual ‘∨’ shape produces our graph of y = i(x). 3, 0 to − 1 3, 1 to − 2 3, 2, − 1 y 4 3 2 1 (1, 1) (−1, 10, 2) −1 −1 1 x −3 −2 −1 (0, 0) 1 2 3 x −−−−−−−−−−−−→ f (x) = \|x\| i(x) = −2f (3x + 1) + 4 = −2\|3x + 1\| + 4 While the methods in Section 1.7 can be used to graph an entire family of absolute value functions, not all functions involving absolute values posses the characteristic ‘∨’ shape. As the next example illustrates, often there is no substitute for appealing directly to the deﬁnition. Example 2.2.4. Graph each of the following functions. Find the zeros of each function and the x- and y-intercepts
of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and ﬁnd the relative and absolute extrema, if they exist. 1. f (x) = \|x\| x Solution. 2. g(x) = \|x + 2\| − \|x − 3\| + 1 1. We ﬁrst note that, due to the fraction in the formula of f (x), x = 0. Thus the domain is (−∞, 0) ∪ (0, ∞). To ﬁnd the zeros of f, we set f (x) = \|x\| x = 0. This last equation implies \|x\| = 0, which, from Theorem 2.1, implies x = 0. However, x = 0 is not in the domain of f, 2.2 Absolute Value Functions 181 which means we have, in fact, no x-intercepts. We have no y-intercepts either, since f (0) is undeﬁned. Re-writing the absolute value in the function gives f (x) =    −x x x x,, if x < 0 if x > 0 = −1, 1, if x < 0 if x > 0 To graph this function, we graph two horizontal lines: y = −1 for x < 0 and y = 1 for x > 0. We have open circles at (0, −1) and (0, 1) (Can you explain why?) so we get y −3 −2 −1 1 2 3 x f (x) = \|x\| x As we found earlier, the domain is (−∞, 0) ∪ (0, ∞). The range consists of just two y-values: {−1, 1}. The function f is constant on (−∞, 0) and (0, ∞). The local minimum value of f is the absolute minimum value of f, namely −1; the local maximum and absolute maximum values for f also coincide − they both are 1. Every point on the graph of f is simultaneously a relative maximum and a relative minimum. (Can you remember why in light of Deﬁnition 1.11? This was explored in the Exercises in Section 1.6.2
.) 2. To ﬁnd the zeros of g, we set g(x) = 0. The result is \|x + 2\| − \|x − 3\| + 1 = 0. Attempting to isolate the absolute value term is complicated by the fact that there are two terms with absolute values. In this case, it easier to proceed using cases by re-writing the function g with two separate applications of Deﬁnition 2.4 to remove each instance of the absolute values, one at a time. In the ﬁrst round we get −(x + 2) − \|x − 3\| + 1, (x + 2) − \|x − 3\| + 1, if if (x + 2) < 0 (x + 2) ≥ 0 = −x − 1 − \|x − 3\|, x + 3 − \|x − 3\|, if x < −2 if x ≥ −2 g(x) = Given that \|x − 3\| = −(x − 3), x − 3, if if (x − 3) < 0 (x − 3) ≥ 0 = −x + 3, x − 3, if x < 3 if x ≥ 3, we need to break up the domain again at x = 3. Note that if x < −2, then x < 3, so we replace \|x − 3\| with −x + 3 for that part of the domain, too. Our completed revision of the form of g yields 182 Linear and Quadratic Functions g(x) =    −x − 1 − (−x + 3), x + 3 − (−x + 3), x + 3 − (x − 3), if x < −2 if x ≥ −2 and x < 3 if x ≥ 3 =    −4, 2x, 6, if x < −2 if −2 ≤ x < 3 if x ≥ 3 To solve g(x) = 0, we see that the only piece which contains a variable is g(x) = 2x for −2 ≤ x < 3. Solving 2x = 0 gives x = 0. Since x = 0 is in the interval [−2, 3), we keep this solution and have (0, 0) as our only x-intercept. Accordingly, the y-intercept is
also (0, 0). To graph g, we start with x < −2 and graph the horizontal line y = −4 with an open circle at (−2, −4). For −2 ≤ x < 3, we graph the line y = 2x and the point (−2, −4) patches the hole left by the previous piece. An open circle at (3, 6) completes the graph of this part. Finally, we graph the horizontal line y = 6 for x ≥ 3, and the point (3, 6) ﬁlls in the open circle left by the previous part of the graph. The ﬁnished graph is y 6 5 4 3 2 1 −1 −2 −3 −4 −4 −3 −2 −1 1 2 3 4 x g(x) = \|x + 2\| − \|x − 3\| + 1 The domain of g is all real numbers, (−∞, ∞), and the range of g is all real numbers between −4 and 6 inclusive, [−4, 6]. The function is increasing on [−2, 3] and constant on (−∞, −2] and [3, ∞). The relative minimum value of f is −4 which matches the absolute minimum. The relative and absolute maximum values also coincide at 6. Every point on the graph of y = g(x) for x < −2 and x > 3 yields both a relative minimum and relative maximum. The point (−2, −4), however, gives only a relative minimum and the point (3, 6) yields only a relative maximum. (Recall the Exercises in Section 1.6.2 which dealt with constant functions.) Many of the applications that the authors are aware of involving absolute values also involve absolute value inequalities. For that reason, we save our discussion of applications for Section 2.4. 2.2 Absolute Value Functions 183 2.2.1 Exercises In Exercises 1 - 15, solve the equation. 1. \|x\| = 6 4. 4 − \|x\| = 3 7. 5 − \|x\| 2 = 1 2. \|3x − 1\| = 10 3. \|4 − x\| = 7 5. 2\|5x + 1\| − 3 = 0 6. \|7x − 1\| + 2 = 0 8. 2 3 \|5 − 2x\| − 1 2 = 5 9. \|x\| = x + 3 10.
\|2x − 1\| = x + 1 11. 4 − \|x\| = 2x + 1 12. \|x − 4\| = x − 5 13. \|x\| = x2 14. \|x\| = 12 − x2 15. \|x2 − 1\| = 3 Prove that if \|f (x)\| = \|g(x)\| then either f (x) = g(x) or f (x) = −g(x). Use that result to solve the equations in Exercises 16 - 21. 16. \|3x − 2\| = \|2x + 7\| 17. \|3x + 1\| = \|4x\| 18. \|1 − 2x\| = \|x + 1\| 19. \|4 − x\| − \|x + 2\| = 0 20. \|2 − 5x\| = 5\|x + 1\| 21. 3\|x − 1\| = 2\|x + 1\| In Exercises 22 - 33, graph the function. Find the zeros of each function and the x- and y-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and ﬁnd the relative and absolute extrema, if they exist. 22. f (x) = \|x + 4\| 23. f (x) = \|x\| + 4 24. f (x) = \|4x\| 25. f (x) = −3\|x\| 26. f (x) = 3\|x + 4\| − 4 27. f (x) = 1 3 \|2x − 1\| 28. f (x) = \|x + 4\| x + 4 29. f (x) = \|2 − x\| 2 − x 30. f (x) = x + \|x\| − 3 31. f (x) = \|x + 2\| − x 32. f (x) = \|x + 2\| − \|x\| 33. f (x) = \|x + 4\| + \|x − 2\| 34. With the help of your classmates, ﬁnd an absolute value function whose graph is given below. y 4 3 2 1 −8−7−6−5−4−3−2− 35. With help from your classmates, prove the second, third and ﬁf
th parts of Theorem 2.1. 36. Prove The Triangle Inequality: For all real numbers a and b, \|a + b\| ≤ \|a\| + \|b\|. 184 Linear and Quadratic Functions 2.2.2 Answers 1. x = −6 or x = 6 4. x = −1 or x = 1 7. x = −3 or x = 3 2. x = −3 or x = 11 3 3. x = −3 or x = 11 5. x = − 1 2 or x = 1 10 8. x = − 13 8 or x = 53 8 6. no solution 9. x = − 3 2 12. no solution 10. x = 0 or x = 2 11. x = 1 13. x = −1, x = 0 or x = 1 14. x = −3 or x = 3 15. x = −2 or x = 2 16. x = −1 or x = 9 17. x = − 1 7 or x = 1 19. x = 1 20. x = − 3 10 18. x = 0 or x = 2 21. x = 1 5 or x = 5 22. f (x) = \|x + 4\| f (−4) = 0 x-intercept (−4, 0) y-intercept (0, 4) Domain (−∞, ∞) Range [0, ∞) Decreasing on (−∞, −4] Increasing on [−4, ∞) Relative and absolute min. at (−4, 0) No relative or absolute maximum 23. f (x) = \|x\| + 4 No zeros No x-intercepts y-intercept (0, 4) Domain (−∞, ∞) Range [4, ∞) Decreasing on (−∞, 0] Increasing on [0, ∞) Relative and absolute minimum at (0, 4) No relative or absolute maximum y 4 3 2 1 −8 −7 −6 −5 −4 −3 −2 −4 −3 −2 −1 1 2 3 4 x 2.2 Absolute Value Functions 185 24. f (x) = \|4x\| f (0) = 0 x-intercept (0, 0) y-intercept (0, 0) Domain (−∞, ∞) Range [0, ∞) Decreasing on (−∞, 0] Increasing on [0, ∞
) Relative and absolute minimum at (0, 0) No relative or absolute maximum 25. f (x) = −3\|x\| f (0) = 0 x-intercept (0, 0) y-intercept (0, 0) Domain (−∞, ∞) Range (−∞, 0] Increasing on (−∞, 0] Decreasing on [0, ∞) Relative and absolute maximum at (0, 0) No relative or absolute minimum 26. f (x) = 3\|x + 4\| − 4 = 0, − 8 f − 16 3 x-intercepts − 16 y-intercept (0, 8) Domain (−∞, ∞) Range [−4, ∞) Decreasing on (−∞, −4] Increasing on [−4, ∞) Relative and absolute min. at (−4, −4) No relative or absolute maximum 3 \|2x − 1\| 2, 0 27. f (x) = 1 f 1 = 0 2 x-intercepts 1 y-intercept 0, 1 3 Domain (−∞, ∞) Range [0, ∞) Decreasing on −∞, 1 2 Increasing on 2 −1 1 2 x y 1 2 x −2 −1 −1 −2 −3 −4 −5 −8−7−6−5−4−3−2−1 −1 x 1 −2 −3 −4 Relative and absolute min. at 1 No relative or absolute maximum 2, 0 y 2 1 −3 −2 −1 1 2 3 4 x 186 28. f (x) = \|x + 4\| x + 4 No zeros No x-intercept y-intercept (0, 1) Domain (−∞, −4) ∪ (−4, ∞) Range {−1, 1} Constant on (−∞, −4) Constant on (−4, ∞) Absolute minimum at every point (x, −1) 29. f (x) = \|2 − x\| 2 − x No zeros No x-intercept y-intercept (0, 1) Domain (−∞, 2) ∪ (2, ∞) Range {−1, 1} Constant on (−∞, 2) Constant on (2, ∞) Absolute minimum at every point (x, −1) 30. Re-write f (x) = x + \|
x\| − 3 as x < 0 x ≥ 0 −3 2x − 3 f (x) = if if = 0 f 3 2 x-intercept 3 2, 0 y-intercept (0, −3) Domain (−∞, ∞) Range [−3, ∞) Increasing on [0, ∞) Constant on (−∞, 0] Absolute minimum at every point (x, −3) where x ≤ 0 No absolute maximum Linear and Quadratic Functions where x < −4 Absolute maximum at every point (x, 1) where x > −4 Relative maximum AND minimum at every point on the graph y 1 −8 −7 −6 −5 −4 −3 −2 −1 −1 1 x where x > 2 Absolute maximum at every point (x, 1) where x < 2 Relative maximum AND minimum at every point on the graph y 1 −3 −2 −1 −1 1 2 3 4 5 x Relative minimum at every point (x, −3) where x ≤ 0 Relative maximum at every point (x, −3) where x < 0 y 2 1 −2 −1 −1 −2 −3 −4 1 2 x 2.2 Absolute Value Functions 187 31. Re-write f (x) = \|x + 2\| − x as f (x) = −2x − 2 2 if if x < −2 x ≥ −2 No zeros No x-intercepts y-intercept (0, 2) Domain (−∞, ∞) Range [2, ∞) Decreasing on (−∞, −2] Constant on [−2, ∞) Absolute minimum at every point (x, 2) where x ≥ −2 32. Re-write f (x) = \|x + 2\| − \|x\| as x < −2 if if −2 ≤ x < 0 x ≥ 0 if −2 2x + 2 2 f (x) =    f (−1) = 0 x-intercept (−1, 0) y-intercept (0, 2) Domain (−∞, ∞) Range [−2, 2] Increasing on [−2, 0] Constant on (−∞, −2] Constant on [0, ∞) Absolute minimum at every point (x, −2) where x ≤ −2 33. Re-write f (x
) = \|x + 4\| + \|x − 2\| as −2x − 2 6 2x + 2 x < −4 if if −4 ≤ x < 2 x ≥ 2 if f (x) =    No zeros No x-intercept y-intercept (0, 6) Domain (−∞, ∞) Range [6, ∞) Decreasing on (−∞, −4] Constant on [−4, 2] Increasing on [2, ∞) Absolute minimum at every point (x, 6) where −4 ≤ x ≤ 2 No absolute maximum Relative minimum at every point (x, 6) where 35. f (x) = \|\|x\| − 4\| No absolute maximum Relative minimum at every point (x, 2) where x ≥ −2 Relative maximum at every point (x, 2) where x > −2 y 3 2 1 −3 −2 −1 1 2 x Absolute maximum at every point (x, 2) where x ≥ 0 Relative minimum at every point (x, −2) where x ≤ −2 and at every point (x, 2) where x > 0 Relative maximum at every point (x, −2) where x < −2 and at every point (x, 2) where 3 −2 −1 −1 −2 −4 ≤ x ≤ 2 Relative maximum at every point (x, 6) where −5 −4 −3 −2 −1 1 2 3 x 188 Linear and Quadratic Functions 2.3 Quadratic Functions You may recall studying quadratic equations in Intermediate Algebra. In this section, we review those equations in the context of our next family of functions: the quadratic functions. Deﬁnition 2.5. A quadratic function is a function of the form f (x) = ax2 + bx + c, where a, b and c are real numbers with a = 0. The domain of a quadratic function is (−∞, ∞). The most basic quadratic function is f (x) = x2, whose graph appears below. Its shape should look familiar from Intermediate Algebra – it is called a parabola. The point (0, 0) is called the vertex of the parabola. In this case, the vertex is a relative minimum and is also the where the absolute minimum value of f can be
found. y 4 3 2 1 (2, 4) (1, 1) (−2, 4) (−1, 1) −2 −1 1 (0, 0) f (x) = x2 2 x Much like many of the absolute value functions in Section 2.2, knowing the graph of f (x) = x2 enables us to graph an entire family of quadratic functions using transformations. Example 2.3.1. Graph the following functions starting with the graph of f (x) = x2 and using transformations. Find the vertex, state the range and ﬁnd the x- and y-intercepts, if any exist. 1. g(x) = (x + 2)2 − 3 2. h(x) = −2(x − 3)2 + 1 Solution. 1. Since g(x) = (x + 2)2 − 3 = f (x + 2) − 3, Theorem 1.7 instructs us to ﬁrst subtract 2 from each of the x-values of the points on y = f (x). This shifts the graph of y = f (x) to the left 2 units and moves (−2, 4) to (−4, 4), (−1, 1) to (−3, 1), (0, 0) to (−2, 0), (1, 1) to (−1, 1) and (2, 4) to (0, 4). Next, we subtract 3 from each of the y-values of these new points. This moves the graph down 3 units and moves (−4, 4) to (−4, 1), (−3, 1) to (−3, −2), (−2, 0) to (−2, 3), (−1, 1) to (−1, −2) and (0, 4) to (0, 1). We connect the dots in parabolic fashion to get 2.3 Quadratic Functions 189 y 4 3 2 1 (−2, 4) (−1, 1) y (2, 4) (−4, 1) 1 (0, 1) (1, 1) (−3, −2) (−1, −2) −4 −3 −2 −1 x −1 −2 −1 (0, 0) 1 2 x −3 (−2, −3) f (x) = x2 −−−−−−−−−−−−→ g
(x) = f (x + 2) − 3 = (x + 2)2 − 3 From the graph, we see that the vertex has moved from (0, 0) on the graph of y = f (x) to (−2, −3) on the graph of y = g(x). This sets [−3, ∞) as the range of g. We see that the graph of y = g(x) crosses the x-axis twice, so we expect two x-intercepts. To ﬁnd these, we set y = g(x) = 0 and solve. Doing so yields the equation (x + 2)2 − 3 = 0, or (x + 2)2 = 3. Extracting square roots gives x + 2 = ± 3. Our x-intercepts are (−2 − 3, 0) ≈ (−0.27, 0). The y-intercept of the graph, 3, 0) ≈ (−3.73, 0) and (−2 + (0, 1) was one of the points we originally plotted, so we are done. 3, or x = −2 ± √ √ √ √ 2. Following Theorem 1.7 once more, to graph h(x) = −2(x − 3)2 + 1 = −2f (x − 3) + 1, we ﬁrst start by adding 3 to each of the x-values of the points on the graph of y = f (x). This eﬀects a horizontal shift right 3 units and moves (−2, 4) to (1, 4), (−1, 1) to (2, 1), (0, 0) to (3, 0), (1, 1) to (4, 1) and (2, 4) to (5, 4). Next, we multiply each of our y-values ﬁrst by −2 and then add 1 to that result. Geometrically, this is a vertical stretch by a factor of 2, followed by a reﬂection about the x-axis, followed by a vertical shift up 1 unit. This moves (1, 4) to (1, −7), (2, 1) to (2, −1), (3, 0) to (3, 1), (4, 1) to (4, −1) and (5, 4
) to (5, −7). y (3, 1) 2 1 (2, −1) 3 4 5 (4, −1) x 1 −1 −2 −3 −4 −5 −6 (2, 4) (1, 1) (1, −7) (5, −7) y 4 3 2 1 (−2, 4) (−1, 1) −2 −1 (0, 0) 1 2 x f (x) = x2 −−−−−−−−−−−−→ h(x) = −2f (x − 3) + 1 = −2(x − 3)2 + 1 The vertex is (3, 1) which makes the range of h (−∞, 1]. From our graph, we know that there are two x-intercepts, so we set y = h(x) = 0 and solve. We get −2(x − 3)2 + 1 = 0 190 Linear and Quadratic Functions which gives (x − 3)2 = 1 add 3 to each side,2 we get x = 6± 2 6+ 2, 0 √ 2 √ 2 2. Extracting square roots1 gives x − 3 = ± 2, so that when we √ 6− 2 ≈ (2.29, 0) and, 0 2. Hence, our x-intercepts are √ 2 ≈ (3.71, 0). Although our graph doesn’t show it, there is a y-intercept which can be found by setting x = 0. With h(0) = −2(0 − 3)2 + 1 = −17, we have that our y-intercept is (0, −17). A few remarks about Example 2.3.1 are in order. First note that neither the formula given for g(x) nor the one given for h(x) match the form given in Deﬁnition 2.5. We could, of course, convert both g(x) and h(x) into that form by expanding and collecting like terms. Doing so, we ﬁnd g(x) = (x + 2)2 − 3 = x2 + 4x + 1 and h(x) = −2(x − 3)2 + 1 = −2x2 + 12x − 17. While these ‘simpliﬁed
’ formulas for g(x) and h(x) satisfy Deﬁnition 2.5, they do not lend themselves to graphing easily. For that reason, the form of g and h presented in Example 2.3.2 is given a special name, which we list below, along with the form presented in Deﬁnition 2.5. Deﬁnition 2.6. Standard and General Form of Quadratic Functions: Suppose f is a quadratic function. The general form of the quadratic function f is f (x) = ax2 + bx + c, where a, b and c are real numbers with a = 0. The standard form of the quadratic function f is f (x) = a(x − h)2 + k, where a, h and k are real numbers with a = 0. It is important to note at this stage that we have no guarantees that every quadratic function can be written in standard form. This is actually true, and we prove this later in the exposition, but for now we celebrate the advantages of the standard form, starting with the following theorem. Theorem 2.2. Vertex Formula for Quadratics in Standard Form: For the quadratic function f (x) = a(x − h)2 + k, where a, h and k are real numbers with a = 0, the vertex of the graph of y = f (x) is (h, k). We can readily verify the formula given Theorem 2.2 with the two functions given in Example 2.3.1. After a (slight) rewrite, g(x) = (x + 2)2 − 3 = (x − (−2))2 + (−3), and we identify h = −2 and k = −3. Sure enough, we found the vertex of the graph of y = g(x) to be (−2, −3). For h(x) = −2(x − 3)2 + 1, no rewrite is needed. We can directly identify h = 3 and k = 1 and, sure enough, we found the vertex of the graph of y = h(x) to be (3, 1). To see why the formula in Theorem 2.2 produces the vertex, consider the graph of the equation y = a(x − h)2 + k. When we substitute x = h, we get
y = k, so (h, k) is on the graph. If x = h, then x − h = 0 so (x − h)2 is a positive number. If a > 0, then a(x − h)2 is positive, thus y = a(x − h)2 + k is always a number larger than k. This means that when a > 0, (h, k) is the lowest point on the graph and thus the parabola must open upwards, making (h, k) the vertex. A similar argument 1and rationalizing denominators! 2and get common denominators! 2.3 Quadratic Functions 191 shows that if a < 0, (h, k) is the highest point on the graph, so the parabola opens downwards, and (h, k) is also the vertex in this case. Alternatively, we can apply the machinery in Section 1.7. Since the vertex of y = x2 is (0, 0), we can determine the vertex of y = a(x−h)2 +k by determining the ﬁnal destination of (0, 0) as it is moved through each transformation. To obtain the formula f (x) = a(x − h)2 + k, we start with g(x) = x2 and ﬁrst deﬁne g1(x) = ag(x) = ax2. This is results in a vertical scaling and/or reﬂection.3 Since we multiply the output by a, we multiply the y-coordinates on the graph of g by a, so the point (0, 0) remains (0, 0) and remains the vertex. Next, we deﬁne g2(x) = g1(x − h) = a(x − h)2. This induces a horizontal shift right or left h units4 moves the vertex, in either case, to (h, 0). Finally, f (x) = g2(x) + k = a(x − h)2 + k which eﬀects a vertical shift up or down k units5 resulting in the vertex moving from (h, 0) to (h, k). In addition to verifying Theorem 2.2, the arguments in the two preceding paragraphs have also shown us the role of the number a in the graphs of quadratic functions. The graph of
y = a(x−h)2+k is a parabola ‘opening upwards’ if a > 0, and ‘opening downwards’ if a < 0. Moreover, the symmetry enjoyed by the graph of y = x2 about the y-axis is translated to a symmetry about the vertical line x = h which is the vertical line through the vertex.6 This line is called the axis of symmetry of the parabola and is dashed in the ﬁgures below. vertex a < 0 vertex a > 0 Graphs of y = a(x − h)2 + k. Without a doubt, the standard form of a quadratic function, coupled with the machinery in Section 1.7, allows us to list the attributes of the graphs of such functions quickly and elegantly. What remains to be shown, however, is the fact that every quadratic function can be written in standard form. To convert a quadratic function given in general form into standard form, we employ the ancient rite of ‘Completing the Square’. We remind the reader how this is done in our next example. Example 2.3.2. Convert the functions below from general form to standard form. Find the vertex, axis of symmetry and any x- or y-intercepts. Graph each function and determine its range. 1. f (x) = x2 − 4x + 3. 2. g(x) = 6 − x − x2 3Just a scaling if a > 0. If a < 0, there is a reﬂection involved. 4Right if h > 0, left if h < 0. 5Up if k > 0, down if k < 0 6You should use transformations to verify this! 192 Solution. Linear and Quadratic Functions 1. To convert from general form to standard form, we complete the square.7 First, we verify that the coeﬃcient of x2 is 1. Next, we ﬁnd the coeﬃcient of x, in this case −4, and take half of it to get 1 2 (−4) = −2. This tells us that our target perfect square quantity is (x − 2)2. To get an expression equivalent to (x − 2)2, we need to add (−2)2 = 4 to the x2 − 4x to create a perfect square trinomial, but to keep the balance, we
must also subtract it. We collect the terms which create the perfect square and gather the remaining constant terms. Putting it all together, we get f (x) = x2 − 4x + 3 2 (−4) = −2.) = x2 − 4x + 4 − 4 + 3 (Add and subtract (−2)2 = 4 to (x2 + 4x).) = x2 − 4x + 4 − 4 + 3 (Group the perfect square trinomial.) = (x − 2)2 − 1 (Factor the perfect square trinomial.) (Compute 1 Of course, we can always check our answer by multiplying out f (x) = (x − 2)2 − 1 to see that it simpliﬁes to f (x) = x2 − 4x − 1. In the form f (x) = (x − 2)2 − 1, we readily ﬁnd the vertex to be (2, −1) which makes the axis of symmetry x = 2. To ﬁnd the x-intercepts, we set y = f (x) = 0. We are spoiled for choice, since we have two formulas for f (x). Since we recognize f (x) = x2 − 4x + 3 to be easily factorable,8 we proceed to solve x2 − 4x + 3 = 0. Factoring gives (x − 3)(x − 1) = 0 so that x = 3 or x = 1. The x-intercepts are then (1, 0) and (3, 0). To ﬁnd the y-intercept, we set x = 0. Once again, the general form f (x) = x2 − 4x + 3 is easiest to work with here, and we ﬁnd y = f (0) = 3. Hence, the y-intercept is (0, 3). With the vertex, axis of symmetry and the intercepts, we get a pretty good graph without the need to plot additional points. We see that the range of f is [−1, ∞) and we are done. 2. To get started, we rewrite g(x) = 6 − x − x2 = −x2 − x + 6 and note that the coeﬃcient of x2 is −1, not 1. This means our ﬁrst step is to factor out the
(−1) from both the x2 and x terms. We then follow the completing the square recipe as above. g(x) = −x2 − x + 6 = (−1) x2 + x + 6 x2 + x + 1 = (−1) = (−1) x2 + x + 1 4 = − x + 1 + 25 4 2 2 4 4 − 1 + 6 + (−1) − 1 4 (Factor the coeﬃcient of x2 from x2 and x.) + 6 (Group the perfect square trinomial.) 7If you forget why we do what we do to complete the square, start with a(x − h)2 + k, multiply it out, step by step, and then reverse the process. 8Experience pays oﬀ, here! 2.3 Quadratic Functions 193 2 + 25 4, we get the vertex to be − 1 From g(x) = − x + 1 and the axis of symmetry to 2 be x = − 1 2. To get the x-intercepts, we opt to set the given formula g(x) = 6 − x − x2 = 0. Solving, we get x = −3 and x = 2, so the x-intercepts are (−3, 0) and (2, 0). Setting x = 0, we ﬁnd g(0) = 6, so the y-intercept is (0, 6). Plotting these points gives us the graph below. We see that the range of g is −∞, 25 4 2, 25. 0, 3) x = 2 (1, 0) (3, 0) −1 1 2 3 4 5 x −1 (2, −1) f (x) = x2 − 4x + 3 y (0, 6) − 1 2, 25 4 6 5 4 3 2 (−3, 0) x = 1 2 (2, 0) −3 −2 −1 1 2 x g(x) = 6 − x − x2 With Example 2.3.2 fresh in our minds, we are now in a position to show that every quadratic function can be written in standard form. We begin with f (x) = ax2 + bx + c, assume a = 0, and complete the square in complete generality. f (x) = ax2 + bx + c x + c (Factor out
coeﬃcient of x2 from x2 and x.) x2 + = a x2 + = a x2 + = a b a b a b a b2 4a2 + c x + x + b2 4a2 − b2 4a2 − a = a x + 2 b 2a + 4ac − b2 4a b2 4a2 + c (Group the perfect square trinomial.) (Factor and get a common denominator.) so that Comparing this last expression with the standard form, we identify (x − h) with x + b 2a h = − b. As such, we have derived a vertex formula for the general form. We summarize both vertex formulas in the box at the top of the next page. 2a. Instead of memorizing the value k = 4ac−b2, we see that f − b 2a = 4ac−b2 4a 4a 194 Linear and Quadratic Functions Equation 2.4. Vertex Formulas for Quadratic Functions: Suppose a, b, c, h and k are real numbers with a = 0. If f (x) = a(x − h)2 + k, the vertex of the graph of y = f (x) is the point (h, k). If f (x) = ax2 + bx + c, the vertex of the graph of y = f (x) is the point − b 2a, f −. b 2a There are two more results which can be gleaned from the completed-square form of the general form of a quadratic function, f (x) = ax2 + bx + c = a x + 2 b 2a + 4ac − b2 4a We have seen that the number a in the standard form of a quadratic function determines whether the parabola opens upwards (if a > 0) or downwards (if a < 0). We see here that this number a is none other than the coeﬃcient of x2 in the general form of the quadratic function. In other words, it is the coeﬃcient of x2 alone which determines this behavior – a result that is generalized in Section 3.1. The second treasure is a re-discovery of the quadratic formula. Equation 2.5. The Quadratic Formula: If a, b and c are real numbers with a
= 0, then the solutions to ax2 + bx + c = 0 are −b ± x = √ b2 − 4ac 2a. Assuming the conditions of Equation 2.5, the solutions to ax2 + bx + c = 0 are precisely the zeros of f (x) = ax2 + bx + c. Since f (x) = ax2 + bx + c = a x + 2 b 2a + 4ac − b2 4a the equation ax2 + bx + c = 0 is equivalent to a x + 2 b 2a + 4ac − b2 4a = 0. Solving gives 2.3 Quadratic Functions 195 a x + 2 b 2a + 4ac − b2 4a = 0 2 b 2a 2a = − 4ac − b2 4a b2 − 4ac 4a 1 a b2 − 4ac 4a2 b2 − 4ac 4a2 = = = ± = ± √ b2 − 4ac 2a √ b2 − 4ac 2a b 2a ± x = − x + 2 b 2a x + x + b 2a b 2a extract square roots −b ± x = √ b2 − 4ac 2a In our discussions of domain, we were warned against having negative numbers underneath the b2 − 4ac is part of the Quadratic Formula, we will need to pay special square root. Given that attention to the radicand b2 − 4ac. It turns out that the quantity b2 − 4ac plays a critical role in determining the nature of the solutions to a quadratic equation. It is given a special name. √ Deﬁnition 2.7. If a, b and c are real numbers with a = 0, then the discriminant of the quadratic equation ax2 + bx + c = 0 is the quantity b2 − 4ac. The discriminant ‘discriminates’ between the kinds of solutions we get from a quadratic equation. These cases, and their relation to the discriminant, are summarized below. Theorem 2.3. Discriminant Trichotomy: Let a, b and c be real numbers with a = 0. If b2 − 4ac < 0, the equation ax2 + bx + c = 0 has no real solutions. If b2 − 4ac = 0, the equation ax
2 + bx + c = 0 has exactly one real solution. If b2 − 4ac > 0, the equation ax2 + bx + c = 0 has exactly two real solutions. The proof of Theorem 2.3 stems from the position of the discriminant in the quadratic equation, and is left as a good mental exercise for the reader. The next example exploits the fruits of all of our labor in this section thus far. 196 Linear and Quadratic Functions Example 2.3.3. Recall that the proﬁt (deﬁned on page 82) for a product is deﬁned by the equation Proﬁt = Revenue − Cost, or P (x) = R(x) − C(x). In Example 2.1.7 the weekly revenue, in dollars, made by selling x PortaBoy Game Systems was found to be R(x) = −1.5x2 + 250x with the restriction (carried over from the price-demand function) that 0 ≤ x ≤ 166. The cost, in dollars, to produce x PortaBoy Game Systems is given in Example 2.1.5 as C(x) = 80x + 150 for x ≥ 0. 1. Determine the weekly proﬁt function P (x). 2. Graph y = P (x). Include the x- and y-intercepts as well as the vertex and axis of symmetry. 3. Interpret the zeros of P. 4. Interpret the vertex of the graph of y = P (x). 5. Recall that the weekly price-demand equation for PortaBoys is p(x) = −1.5x + 250, where p(x) is the price per PortaBoy, in dollars, and x is the weekly sales. What should the price per system be in order to maximize proﬁt? Solution. 1. To ﬁnd the proﬁt function P (x), we subtract P (x) = R(x) − C(x) = −1.5x2 + 250x − (80x + 150) = −1.5x2 + 170x − 150. Since the revenue function is valid when 0 ≤ x ≤ 166, P is also restricted to these values. 2. To ﬁnd the x-intercepts, we set P (x) = 0 and solve −
1.5x2 + 170x − 150 = 0. The mere thought of trying to factor the left hand side of this equation could do serious psychological damage, so we resort to the quadratic formula, Equation 2.5. Identifying a = −1.5, b = 170, and c = −150, we obtain x = = = = √ −b ± b2 − 4ac 2a −170 ± 1702 − 4(−1.5)(−150) √ −170 ± 2(−1.5) 28000 −3 √ 170 ± 20 70 3 170−20 3 √ 70 170+20 3 √ 70. To ﬁnd the y-intercept, we set We get two x-intercepts: x = 0 and ﬁnd y = P (0) = −150 for a y-intercept of (0, −150). To ﬁnd the vertex, we use the fact that P (x) = −1.5x2 + 170x − 150 is in the general form of a quadratic function and 2(−1.5) = 170 appeal to Equation 2.4. Substituting a = −1.5 and b = 170, we get x = − 170 3. and, 0, 0 2.3 Quadratic Functions 197 3 To ﬁnd the y-coordinate of the vertex, we compute P 170 and ﬁnd that our vertex 3 is 170. The axis of symmetry is the vertical line passing through the vertex so it is 3, 14000 the line x = 170 3. To sketch a reasonable graph, we approximate the x-intercepts, (0.89, 0) and (112.44, 0), and the vertex, (56.67, 4666.67). (Note that in order to get the x-intercepts and the vertex to show up in the same picture, we had to scale the x-axis diﬀerently than the y-axis. This results in the left-hand x-intercept and the y-intercept being uncomfortably close to each other and to the origin in the picture.) = 14000 3 y 4000 3000 2000 1000 10 20 30 40 50 60 70 80 90 100 110 120 x 3. The zeros of P are the solutions to P (x) = 0, which we have found to be approximately
0.89 and 112.44. As we saw in Example 1.5.3, these are the ‘break-even’ points of the proﬁt function, where enough product is sold to recover the cost spent to make the product. More importantly, we see from the graph that as long as x is between 0.89 and 112.44, the graph y = P (x) is above the x-axis, meaning y = P (x) > 0 there. This means that for these values of x, a proﬁt is being made. Since x represents the weekly sales of PortaBoy Game Systems, we round the zeros to positive integers and have that as long as 1, but no more than 112 game systems are sold weekly, the retailer will make a proﬁt. 4. From the graph, we see that the maximum value of P occurs at the vertex, which is approximately (56.67, 4666.67). As above, x represents the weekly sales of PortaBoy systems, so we can’t sell 56.67 game systems. Comparing P (56) = 4666 and P (57) = 4666.5, we conclude that we will make a maximum proﬁt of $4666.50 if we sell 57 game systems. 5. In the previous part, we found that we need to sell 57 PortaBoys per week to maximize proﬁt. To ﬁnd the price per PortaBoy, we substitute x = 57 into the price-demand function to get p(57) = −1.5(57) + 250 = 164.5. The price should be set at $164.50. Our next example is another classic application of quadratic functions. Example 2.3.4. Much to Donnie’s surprise and delight, he inherits a large parcel of land in Ashtabula County from one of his (e)strange(d) relatives. The time is ﬁnally right for him to pursue his dream of farming alpaca. He wishes to build a rectangular pasture, and estimates that he has enough money for 200 linear feet of fencing material. If he makes the pasture adjacent to a stream (so no fencing is required on that side), what are the dimensions of the pasture which maximize the area? What is the maximum area? If an average alpaca needs 25 square
feet of grazing area, how many alpaca can Donnie keep in his pasture? 198 Linear and Quadratic Functions Solution. It is always helpful to sketch the problem situation, so we do so below. river w pasture w l We are tasked to ﬁnd the dimensions of the pasture which would give a maximum area. We let w denote the width of the pasture and we let l denote the length of the pasture. Since the units given to us in the statement of the problem are feet, we assume w and l are measured in feet. The area of the pasture, which we’ll call A, is related to w and l by the equation A = wl. Since w and l are both measured in feet, A has units of feet2, or square feet. We are given the total amount of fencing available is 200 feet, which means w + l + w = 200, or, l + 2w = 200. We now have two equations, A = wl and l + 2w = 200. In order to use the tools given to us in this section to maximize A, we need to use the information given to write A as a function of just one variable, either w or l. This is where we use the equation l + 2w = 200. Solving for l, we ﬁnd l = 200 − 2w, and we substitute this into our equation for A. We get A = wl = w(200 − 2w) = 200w − 2w2. We now have A as a function of w, A(w) = 200w − 2w2 = −2w2 + 200w. Before we go any further, we need to ﬁnd the applied domain of A so that we know what values of w make sense in this problem situation.9 Since w represents the width of the pasture, w > 0. Likewise, l represents the length of the pasture, so l = 200 − 2w > 0. Solving this latter inequality, we ﬁnd w < 100. Hence, the function we wish to maximize is A(w) = −2w2 +200w for 0 < w < 100. Since A is a quadratic function (of w), we know that the graph of y = A(w) is a parabola. Since the coeﬃcient of w2 is −2, we know that this parabola opens downwards. This
means that there is a maximum value to be found, and we know it occurs at the vertex. Using the vertex formula, we ﬁnd w = − 200 2(−2) = 50, and A(50) = −2(50)2 + 200(50) = 5000. Since w = 50 lies in the applied domain, 0 < w < 100, we have that the area of the pasture is maximized when the width is 50 feet. To ﬁnd the length, we use l = 200 − 2w and ﬁnd l = 200 − 2(50) = 100, so the length of the pasture is 100 feet. The maximum area is A(50) = 5000, or 5000 square feet. If an average alpaca requires 25 square feet of pasture, Donnie can raise 5000 25 = 200 average alpaca. We conclude this section with the graph of a more complicated absolute value function. Example 2.3.5. Graph f (x) = \|x2 − x − 6\|. Solution. Using the deﬁnition of absolute value, Deﬁnition 2.4, we have f (x) = − x2 − x − 6, x2 − x − 6, if x2 − x − 6 < 0 if x2 − x − 6 ≥ 0 The trouble is that we have yet to develop any analytic techniques to solve nonlinear inequalities such as x2 − x − 6 < 0. You won’t have to wait long; this is one of the main topics of Section 2.4. 9Donnie would be very upset if, for example, we told him the width of the pasture needs to be −50 feet. 2.3 Quadratic Functions 199 2 − 1 2 − 6 = − 25 Nevertheless, we can attack this problem graphically. To that end, we graph y = g(x) = x2 − x − 6 using the intercepts and the vertex. To ﬁnd the x-intercepts, we solve x2 − x − 6 = 0. Factoring gives (x − 3)(x + 2) = 0 so x = −2 or x = 3. Hence, (−2, 0) and (3, 0) are x-intercepts. The y-intercept (0, −6) is found by setting x = 0. To plot the vertex, we ﬁnd x =
− b 2, and y = 1 4 = −6.25. Plotting, we get the parabola seen below on the left. To obtain 2 points on the graph of y = f (x) = \|x2 − x − 6\|, we can take points on the graph of g(x) = x2 − x − 6 and apply the absolute value to each of the y values on the parabola. We see from the graph of g that for x ≤ −2 or x ≥ 3, the y values on the parabola are greater than or equal to zero (since the graph is on or above the x-axis), so the absolute value leaves these portions of the graph alone. For x between −2 and 3, however, the y values on the parabola are negative. For example, the point (0, −6) on y = x2 − x − 6 would result in the point (0, \| − 6\|) = (0, −(−6)) = (0, 6) on the graph of f (x) = \|x2 − x − 6\|. Proceeding in this manner for all points with x-coordinates between −2 and 3 results in the graph seen below on the right. 2a = − −1 2(13−2−1 −1 1 2 3 x −3−2−1 −1 1 2 3 x −2 −3 −4 −5 −6 −2 −3 −4 −5 −6 y = g(x) = x2 − x − 6 y = f (x) = \|x2 − x − 6\| If we take a step back and look at the graphs of g and f in the last example, we notice that to obtain the graph of f from the graph of g, we reﬂect a portion of the graph of g about the x-axis. We can see this analytically by substituting g(x) = x2 − x − 6 into the formula for f (x) and calling to mind Theorem 1.4 from Section 1.7. f (x) = −g(x), g(x), if g(x) < 0 if g(x) ≥ 0 The function f is deﬁned so that when g(x) is negative (i.e., when its graph is below the x-axis), the graph of f is its refection across the
x-axis. This is a general template to graph functions of the form f (x) = \|g(x)\|. From this perspective, the graph of f (x) = \|x\| can be obtained by reﬂecting the portion of the line g(x) = x which is below the x-axis back above the x-axis creating the characteristic ‘∨’ shape. 200 Linear and Quadratic Functions 2.3.1 Exercises In Exercises 1 - 9, graph the quadratic function. Find the x- and y-intercepts of each graph, if any exist. If it is given in general form, convert it into standard form; if it is given in standard form, convert it into general form. Find the domain and range of the function and list the intervals on which the function is increasing or decreasing. Identify the vertex and the axis of symmetry and determine whether the vertex yields a relative and absolute maximum or minimum. 1. f (x) = x2 + 2 2. f (x) = −(x + 2)2 3. f (x) = x2 − 2x − 8 4. f (x) = −2(x + 1)2 + 4 5. f (x) = 2x2 − 4x − 1 6. f (x) = −3x2 + 4x − 7 7. f (x) = x2 + x + 1 8. f (x) = −3x2 + 5x + 4 9.10 f (x) = x2 − 1 100 x − 1 In Exercises 10 - 14, the cost and price-demand functions are given for diﬀerent scenarios. For each scenario, Find the proﬁt function P (x). Find the number of items which need to be sold in order to maximize proﬁt. Find the maximum proﬁt. Find the price to charge per item in order to maximize proﬁt. Find and interpret break-even points. 10. The cost, in dollars, to produce x “I’d rather be a Sasquatch” T-Shirts is C(x) = 2x + 26, x ≥ 0 and the price-demand function, in dollars per shirt, is p(x) = 30 − 2x, 0 ≤ x ≤ 15. 11. The cost, in

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.