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triangle has area and height dimensions, given that they are binomial factors of the area. Find the base and height when x = 10 and b = 2, if the height is greater than the length. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up If you see any quadratic expression in the form m2 – c2, you can use the difference of two squares to factor it as (m + c)(m – c), without needing to do all the math. 326326326326326 Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 TTTTTopicopicopicopicopic 6.8.26.8.2 6.8.26.8.2 6.8.2 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- dededededegggggrrrrree pol ynomials..... TTTTThese hese hese ynomials ynomials ee pol ee pol ee polynomials hese hese ynomials ee pol lude lude hniques inc tectectectectechniques inc hniques inc lude finding a hniques include lude hniques inc common factor for all terms in a polynomial, recognizing the difference of two squares, gnizing perfectectectectect gnizing perf and rrrrrecoecoecoecoecognizing perf gnizing perf gnizing perf binomials..... binomials binomials es of es of squar squar es of binomials squares of binomials es of squar squar What it means for you: You’ll learn about how to factor special quadratics called perfect square trinomials. Key words: perfect square trinomial quadratic binomial factor Don’t forget: If this seems a bit unfamiliar, take a look at Topic 6
.4.1 on special binomial products. rinomials rinomials ect Square e e e e TTTTTrinomials ect Squar ect Squar PPPPPerferferferferfect Squar rinomials rinomials ect Squar ect Square e e e e TTTTTrinomials PPPPPerferferferferfect Squar rinomials rinomials ect Squar ect Squar rinomials ect Squar rinomials Perfect square trinomials are quadratic expressions of the form (m + c)2 or (m – c)2. ect Square e e e e TTTTTrinomial rinomial rinomial ect Squar a Binomial is a Perferferferferfect Squar ect Squar a Binomial is a P he Square ofe ofe ofe ofe of a Binomial is a P a Binomial is a P he Squar TTTTThe Squar he Squar rinomial rinomial ect Squar a Binomial is a P he Squar You can use one of two equations to work out the square of a binomial: (m + c)2 = m2 + 2mc + c2 Square of binomial = First term squared + (m + c)2 m2 Twice the product of both terms 2mc + Second term squared c2 Or, if the second term in the binomial is subtracted: (m – c)2 = m2 – 2mc + c2 Square of binomial = First term squared – (m – c)2 m2 Twice the product of both terms 2mc + Second term squared c2 ect Square e e e e TTTTTrinomials rinomials rinomials ect Squar actor Perferferferferfect Squar ect Squar actor P actor P tions to F tions to F Use the Equa Use the Equa rinomials tions to Factor P Use the Equations to F rinomials ect Squar actor P tions to F Use the Equa Use the Equa Example Example Example Example Example 11111 Factor x2 + 2xy + y2. Solution Solution Solution Solution Solution Substitute x2 for m2, 2xy for 2mc, and y2 for c2 in the first equation above: (m + c)2 = m2 + 2mc + c2 �
�� x2 + 2xy + y2 = (x + y)2 Sometimes you need to factor each term in the expression to get it into the correct form. Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 327327327327327 Example Example Example Example Example 22222 Factor 4x2 – 12xy + 9y2. Solution Solution Solution Solution Solution Factor each term: 4x2 – 12xy + 9y2 4x2= (2x)2 12xy = 2(2x·3y) 9y2= (3y)2 So: 4x2 – 12xy + 9y2 = (2x)2 – 2(2x·3y) + (3y)2 Substitute (2x)2 for m2, 2(2x·3y) for 2mc, and (3y)2 for c2 in the second perfect square trinomial equation: (m – c)2 = m2 – 2mc + c2 ﬁ (2x)2 – 2(2x·3y) + (3y)2 = (2x – 3y)2 Guided Practice Factor each expression completely. 1. m2 + 2m + 1 3. 25y2 + 10y + 1 5. 9x2 – 6x + 1 7. 16x2 – 24x + 9 9. 25k2 – 20kt + 4t2 2. 4r2 – 4ry + y2 4. 9k2 + 6ky + y2 6. k2 + 4k + 4 8. 4r2x2 + 4rkx + k2 10. m2r2 – 6mr + 9 Independent Practice Factor each expression completely. 1. 16a2b2 + 24ab + 9 3. 9c2d2 + 36cd + 36 5. 162x2y2m + 144xym + 32m 6. 108k3z2 – 180k2z + 75k 2. 25x2y2 – 40xya + 16a2 4. 49m2n2 – 70mn + 25 Find the radius of each of the circles below, given that the area, A, is the product of a binomial squared and p. 7. A = (pa2 + 2pa + p) ft2 9. A = (49pm2
– 56pmn + 16pn2) ft2 10. A = (81px2y2 – 90pxyz + 25pz2) ft2 8. A = (25py2 + 30py + 9p) ft2 The volume, V, of each cylinder below is the product of the height, p, and the radius squared. Find the radius in each case: 11. V = (98x2p + 84xp + 18p) cm3, height = 2 cm 12. V = (147pa2 – 84pab + 12pb2) m3, height = 3 m 13. V = (36px3 + 48px2m + 16xm2) m3, height = 4x m 14. V = (243x4p – 270px3b + 75x2b2) ft3, height = 3x2 ft ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The phrase “perfect square trinomials” makes this Topic sound much harder than it actually is. They’re really just a special case of the normal quadratic equations that you know and love. 328328328328328 Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 TTTTTopicopicopicopicopic 6.8.36.8.3 6.8.36.8.3 6.8.3 California Standards: Students applpplpplpplpply basic y basic y basic Students a Students a 11.0: 11.0: y basic 11.0: Students a 11.0: y basic Students a 11.0: fffffactoring tec hniques to hniques to actoring tec actoring tec hniques to actoring techniques to hniques to actoring tec second- second- second- and simple thirdsecond- second- ynomials..... TTTTThese dededededegggggrrrrree pol hese hese ynomials ynomials ee pol ee pol hese ee polynomials hese ynomials ee pol lude finding lude finding hniques inc hniques inc tectectectectechniques inc lude finding hniques include finding lude
finding hniques inc or all or all actor f actor f a common f a common f actor for all a common factor f or all a common f or all actor f a common f ynomial, ynomial, ms in a pol ms in a pol terterms in a pol terter ynomial, ms in a polynomial, ter ynomial, ms in a pol recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll group like terms to factor polynomials. Key words: common factor like terms Check it out: It’s difficult to see any common factors in the expression given in Example 1, so it’s a good idea to group terms together as a first step. ouping ouping y Gr y Gr actoring b actoring b FFFFFactoring b ouping y Grouping actoring by Gr ouping y Gr actoring b FFFFFactoring b ouping ouping y Gr y Gr actoring b actoring b y Grouping actoring by Gr ouping y Gr ouping actoring b Grouping like terms means that you can more easily see whether there are common factors in a polynomial — then you can factor them out. actorsssss actor actor ms to See Common F oup Like e e e e TTTTTerererererms to See Common F ms to See Common F oup Lik GrGrGrGrGroup Lik oup Lik ms to See Common Factor actor ms to See Common F oup Lik Sometimes you need to group terms together before you can see any common factors in an expression — then you can use the distributive property, ab + ac = a(b + c), to factor them out. Example Example Example Example Example 11111 Factor by grouping 3y + 5ty – 6k –10tk. Solution Solution Solution Solution Solution Group 3y + 5ty and –6k – 10tk together in parentheses: (3y + 5ty) + (–6k – 10tk) 3y and 5ty have a common factor of y. –6k and –10tk have a common factor of –2k. Factor out the common factors: (3y + 5ty) + (–6k –10tk) = y(3 + 5t) – 2k(3 + 5t) Now you can see there’
s another common factor to factor out: (3 + 5t) Using the distributive property: y(3 + 5t) – 2k(3 + 5t) = (y – 2k)(3 + 5t) Example Example Example Example Example 22222 Factor completely 8rt – 6ckt + 3ckm – 4rm. Solution Solution Solution Solution Solution Rearrange the expression and group in parentheses: 8rt – 6ckt + 3ckm – 4rm = (8rt – 4rm) + (–6ckt + 3ckm) 8rt and –4rm have a common factor of 4r. 6ckt and 3ckm have a common factor of 3ck. Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 329329329329329 Example 2 continueduedueduedued Example 2 contin Example 2 contin Example 2 contin Example 2 contin Take out the common factors: (8rt – 4rm) + (–6ckt + 3ckm) = 4r(2t – m) + 3ck(–2t + m) = 4r(2t – m) – 3ck(2t – m) Using the distributive property: 4r(2t – m) – 3ck(2t – m) = (4r – 3ck)(2t – m) Another common factor to factor out is (2t – m). Guided Practice Factor each expression by grouping. 1. km + 2k – 2m – 4 3. 3x + 9 – 4kx – 12k 5. 1 – k + t – tk 7. 6ky + 15t – 10y – 9kt 9. 6rx2 + 15xy – 2rxy – 5y2 2. tx – ty – mx + my 4. 1 + 3y – 5k – 15ky 6. kt – 2k + 3t – 6 8. 8hx + 10h – 12tx – 15t 10. 10tx – 3k – 15t + 2kx Independent Practice Factor each expression completely. 1. 2x2 + x + 8x + 4 3. 4x2 + 14x + 14x + 49 5. 12n2 + 21n + 8n + 14 7. n2 – 16n + 20n – 320 9. 2x2 + 5xy + 4xy + 10
y2 11. 12a2 + 9ab – 28ab – 21b2 13. 4a2 – 6ab + 6ab – 9b2 2. 6x2 + 9x + 4x + 6 4. 2x2 + 5x + 6x + 15 6. 6x2 + 8x + 15x + 20 8. 3c2 – c + 6c – 2 10. 3m2 + 3mn – mn – n2 12. 2x3 + 2x2y + 3xy2 + 3y3 14. 4b2 – 20bx – 2xb + 10x2 Find a value of? so that the expression will factor into two binomials. 15. 20n2 – 25n +?n – 20 17.?c2 – 12c + 2c – 4 16. 8xy – 4xz + 4wy –?wz 18. 3a2 –?a + 6a – 2 19. The area of a rectangle is the product of two binomials (with integer coefficients). If the area of the rectangle is (3a2 + a + 3a + 1) m2, find the dimensions of the rectangle. 20. The area of a square is the square of a binomial. If the area of the square is (4x2 + 2x + 2x + 1) in.2, find the side length of the square. 21. The area of a circle is the product of p and the radius squared. If the radius is a binomial and the area of the circle is (9px2 + 15pxb + 15pxb + 25pb2) in2, find the radius. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That’s the end of a Section full of neat little ways of making math a lot less painful. You’ll often need to use the methods for difference of two squares, perfect square trinomials, and factoring by grouping — so look back over the Topics in this Section to make sure you understand them. 330330330330330 Section 6.8 Section 6.8 Section 6.8 — More on Quadratics Section 6.8 Section 6.8 Chapter 6 Investigation s trianglelelelele s triang s triang ascal’ ascal’ PPP
PPascal’ ascal’s triang s triang ascal’ PPPPPascal’ s trianglelelelele s triang s triang ascal’ ascal’ ascal’s triang s triang ascal’ Pascal’s triangle was originally developed by the ancient Chinese. However, the French mathematician Blaise Pascal was the first person to discover the importance of all the patterns it contains. 1 Part 1: Look at the numbers in the triangle. Write down a rule that could be used to predict each number in the triangle. Use your rule to predict the next row of the triangle. 1 1 1 4 Part 2: Find and simplify the following: • (x + 1)2 (x + 1)3 (x + 1)4 1 1 6 Hint: this is just (x + 1)(x + 1)2 1 1 1 2 33 6 10 20 15 4 1 5 10 5 15 1 1 6 1 How are your answers linked to Pascal’s Triangle? Use the triangle to predict the expansion of (x + 1)5. Test your prediction. Extension Investigate other patterns in Pascal’s Triangle. Here are some ideas: Look at the numbers in the diagonal lines of the triangle. Investigate “hockey stick” shapes, like the one shown on the right. Find the sum of each row of numbers. What do you notice? Find rows in which the second number is prime. What is special about these rows? 2 33 6 10 20 15 5 15 10 Open-ended Extension If there are two children in a family, there can either be two girls, two boys, or a girl and a boy. The probability of each combination and the ratios of the probabilities are shown in the table: noitanibmoC First child Girl Boy Second child Result Girl Boy Girl Boy Two girls One of each One of each Two boys ytilibaborP seitilibaborpfooitaR slrig2 yob1,lrig1 syob2 1 4 1 2 1 4 1 2 1 Investigate the link between Pascal’s Triangle and the probabilities of having different combinations of boys and girls in families with different numbers of children. Remember, it doesn’t matter which order the boys and girls are born in. ound Up ound Up RRRRRound Up RRRRRound Up ound Up ound Up ound Up ound Up ound Up o
und Up Although it just looks like a funny pile of shapes and numbers, there are a lot of real-life problems that can be solved using the patterns in Pascal’s Triangle. estigaaaaationtiontiontiontion — Pascal’s Triangle 331331331331331 estigestig estig pter 6 Invvvvvestig pter 6 In ChaChaChaChaChapter 6 In pter 6 In pter 6 In Chapter 7 Quadratic Equations and Their Applications Section 7.1 Solving Quadratic Equations....................... 333 Section 7.2 Completing the Square............................... 342 Section 7.3 The Quadratic Formula and Applications.... 355 Section 7.4 Quadratic Graphs........................................ 362 Section 7.5 The Discriminant......................................... 372 Section 7.6 Motion Tasks and Other Applications.......... 379 Investigation The Handshake Problem............................. 386 332332332332332 Topic 7.1.1 California Standards: 11.0: Students apply basic factoring techniques to second- and simple thirddegree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll solve quadratic equations by factoring. Key words: quadratic factor zero property Section 7.1 Solving Quadratic Solving Quadratic Equations by Factoring Equations by Factoring In this Topic you’ll use all the factoring methods that you learned in Chapter 6 to solve quadratic equations. Quadratic Equations Have Degree 2 Quadratic equations contain a squared variable, but no higher powers — they have degree 2. These are all quadratic equations, as the highest power of the variable is 2: (i) x2 – 3x + 2 = 0 (ii) 4x2 + 12x – 320 = 0 (iii) y2 + 4y – 7 = 2y2 – 2y The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers, and a is not 0. For example, in (i) above, a = 1, b = –3, and c = 2, while in (ii), a = 4, b
= 12, and c = –320. Example (iii) above is a quadratic in y, while the others are quadratics in x. Guided Practice The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in these equations. 1. –x2 + 5x – 6 = 0 3. 4x2 – 12x + 9 = 0 5. –x2 – 4x – 4 = 0 7. 6y2 + 28y + 20 = 5 – 6y2 9. 4(x2 – 5x) = –25 11. 7x(7x + 4) + 4x3 + 3 = 2(2x3 + 1) 2. 6x2 + 31x + 35 = 0 4. 16x2 – 8x + 1 = 0 6. 64x2 + 48x + 9 = 0 8. 4x2 + 6x + 1 = 3x2 + 8x 10. 3x(3x + 4) + 8 = 4 Solving is Finding Values That Make the Equality True An equation is a statement saying that two mathematical expressions are equal. For example, 7 + 2 = 9, 4x + 2 = 14, and x2 – 3x + 2 = 0 are all equations. If an equation contains a variable (an unknown quantity), then solving the equation means finding possible values of the variable that make the equation a true statement. Section 7.1 — Solving Quadratic Equations 333 Example 1 Find a solution of the equation x2 – 3x + 2 = 0. Solution If you evaluate the above equation using, say, x = 3, then you get: 32 – (3 × 3) + 2 = 0 This is not a true statement (since the left-hand side equals 2). So x = 3 is not a solution of the equation. But if instead you substitute x = 1, then you get: 12 – (3 × 1) + 2 = 0 This is a true statement. So x = 1 is a solution of the equation x2 – 3x + 2 = 0. Guided Practice Determine which of the two values given is a solution of the equation. 12. –x2 + 5x – 6 = 0 for x = –3 and x = 3 13. 6x2 + 31x + 35 = 0 for x = –
5 3 and x = 3 14. 4x2 – 12x + 9 = 0 for x = 1 and x = 2 2 15. 16x2 – 8x + 1 = 0 for x = – 1 4 and x = 1 16. –x2 – 4x – 4 = 0 for x = 2 and x = –2 17. 64x2 + 48x + 9 = 0 for x = – 3 5 3 4 8 and x = 3 8 Zero Property — if xy = 0, then x = 0 or y = 0 (or Both) One way to solve a quadratic equation is to factor it and then make use of the following property of zero: Check it out: This says that if the product of two expressions is zero (so you get zero when you multiply them together), then at least one of those expressions must itself be zero. Zero Property If the product mc = 0, then either: (i) m = 0, c = 0, (ii) (iii) both m = 0 and c = 0. Example 2 Solve x2 + 2x – 15 = 0 by factoring. Solution x2 + 2x – 15 = (x – 3)(x + 5) So if (x – 3)(x + 5) = 0, then by the zero property, either (x – 3) = 0 or (x + 5) = 0. So either x = 3 or x = –5. 334 Section 7.1 — Solving Quadratic Equations Don’t forget: See Chapter 6 for more information on factoring. Example 3 Solve 2x2 + 3x – 20 = 0 by factoring. Solution 2x2 + 3x – 20 = (2x – 5)(x + 4) = 0 So either x = 5 2 or x = –4. Guided Practice Solve each of these quadratic equations by using the zero property. 18. (2x + 7)(3x + 5) = 0 20. 49x2 – 1 = 0 22. 4x2 + 8x + 3 = 0 24. 4x2 – 11x – 3 = 0 26. 2x2 + 11x + 12 = 0 28. 3x2 – 17x – 28 = 0 19. (x – 5)(x – 1) = 0 21. 64a2 – 25 = 0 23. 2x2 – 17x – 9 = 0 25. 10x2 –
x – 2 = 0 27. 10x2 – 27x + 5 = 0 29. 2x2 – x – 28 = 0 Using Factoring to Solve Quadratic Equations 1) First arrange the terms in the quadratic equation so that you have zero on one side. 2) Then factor the nonzero expression (if possible). 3) Once done, you can use the zero property to find the solutions. Example 4 Solve x2 – 6x – 7 = 0. Solution The right-hand side of the equation is already zero, so you can just factor the left-hand side: x2 – 6x – 7 = (x + 1)(x – 7) So (x + 1)(x – 7) = 0. Using the zero property, either x + 1 = 0 or x – 7 = 0. So either x = –1 or x = 7. Section 7.1 — Solving Quadratic Equations 335 Example 5 Solve x2 + 2x – 11 = –3. Solution This time, you have to arrange the equation so you have zero on one side. By adding 3 to both sides, the right-hand side becomes 0. x2 + 2x – 11 = –3 ﬁ x2 + 2x – 8 = 0 Now you can factor the left-hand side: x2 + 2x – 8 = (x + 4)(x – 2) = 0 Since you have two expressions multiplied to give zero, you can use the zero property. That is, either x + 4 = 0 or x – 2 = 0. So either x = –4 or x = 2. Example 6 Solve 3x2 + 168 = 45x. Solution Once again, the first thing to do is get zero on one side: 3x2 + 168 = 45x ⇒ 3x2 – 45x + 168 = 0 The left-hand side can be factored, which means you can rewrite this as: 3(x2 – 15x + 56) = 0, or 3(x – 7)(x – 8) = 0 So using the zero property, either x – 7 = 0 or x – 8 = 0. So either x = 7 or x = 8. Guided Practice Solve each of these equations. 30. x2 – 2x – 15 = 0 32. k2 + 10k + 24 = 0 34. 8k2 – 14k = 49 36. 6y2
+ 28y + 20 = 5 – 6y2 38. 4(x2 – 5x) = –25 40. x(x + 4) + 9 = 5 42. 6x(3x – 4) – 7 = –15 44. 7x(7x + 2) + 4x3 + 3 = 2(2x3 + 1) 45. (2x + 9)2(x + 3)(x + 1)–1(x + 3)–1(x + 1) = 0 46. 2x(3x + 3) + 4(x + 1) = 1 + 2x + 2x2 31. x2 – 7x – 18 = 0 33. 4m2 + 4m – 15 = 0 35. 15k2 + 28k = –5 37. 4x2 + 6x + 1 = 3x2 + 8x 39. 3x(3x + 4) + 8 = 4 41. x(x – 5) + 3 = –3 43. 3 = 2 – 12x(3x – 1) 336 Section 7.1 — Solving Quadratic Equations Independent Practice The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in the quadratic equations below. 1. 4x2 + 20x + 9 = 0 3. 2x2 + 5x = 35 + 14x 5. y(2y + 7) = 9(2y + 7) 2. x2 – 9x + 8 = 0 4. x(2x + 3) = 5(2x + 3) 6. (x + 2)(x – 2) = 3x Use the zero product property to solve these equations. 7. (2y + 9)(2y – 3) = 0 8. (2a + 5)(a – 11) = 0 9. (y – 3)2(y – 5)(y – 3)–1 = 0 10. (y – 4)3(2y – 9)2(y – 4)–3(y – 7)(2y – 9)–1 = 0 Solve the following equations. 11. 4x2 + 20x + 9 = 0 13. 2x2 + 5x = 35 + 14x 15. y(2y + 7) = 9
(2y + 7) 17. 2x(2x – 5) = 3(2x – 5) 12. x2 – 9x + 8 = 0 14. x(2x + 3) = 5(2x + 3) 16. (x + 2)(x – 2) = 3x 18. 2x(3x – 1) + 7 = 7(2 – 3x) 19. The product of two consecutive positive numbers is 30. Find the numbers. 20. The product of two consecutive positive odd numbers is 35. Find the numbers. 21. The area of a rectangle is 35 ft2. If the width of the rectangle is x ft and the length is (3x + 16) ft, find the value of x. 22. The area of a rectangle is 75 cm2. If the length of the rectangle is (4x + 25) cm and the width is 2x cm, find the dimensions of the rectangle. 23. Eylora has x pet goldfish and Leo has (4x – 25). If the product of the numbers of Eylora’s and Leo’s goldfish is 21, how many goldfish does Leo have? 24. Scott fixed x computers and Meimei fixed (5x – 7) computers. If the product of the number each fixed is 6, who fixed more computers? Round Up Round Up Don’t forget that you need to rearrange the equation until you’ve got a zero on one side before you can factor a quadratic. Not all quadratics can be factored like this, as you’ll see in the next Topic. Section 7.1 — Solving Quadratic Equations 337 Topic 7.1.2 California Standards: 11.0: Students apply basic factoring techniques to second- and simple thirddegree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll solve quadratic equations by taking square roots. Key words: quadratic factor square root zero property Quadratic Equations Quadratic Equations — Taking Square Roots — Taking Square Roots Some quadratics can be solved by taking square roots. But to use this method properly, you have to remember something
about squares and square roots. The Square Root Method If you square the numbers m and –m, you get the same answer, since m2 = (–m)2 (= p, say). If you take the square root of p, there are two possible answers, m or –m. In other words, if m2 = p, then m = ± p. Example 1 uses the above property to find the two possible solutions of a quadratic equation. Example 1 Solve the equation x2 = 25. Solution Take the square root of both sides to get x = ±5. In Example 1, you need only put the “±” on one side of the equation. Here’s why: Take the square root of both sides of the original equation to get x 2 25=. But when you take square roots, you have to allow for both sides to be either positive or negative. So actually, there are four possibilities here: x = 5, –x = 5, x = –5, or –x = –5 However, x = 5 and –x = –5 are the same, as are –x = 5 and x = –5. So in fact it’s enough to put the “±” sign on just one side of the equation. 338 Section 7.1 — Solving Quadratic Equations Check it out: This is the same problem as in Example 1. You Can Also Solve this Equation by Factoring You could also solve the above equation using the method of factoring from the previous pages. But to use the factoring method, you have to have an equation of the form: “something = 0” — then you can use the zero property. Example 2 Solve the equation x2 = 25 by factoring. Solution x2 = 25 x2 – 25 = 0 (x + 5)(x – 5) = 0 So using the zero property, either x + 5 = 0 or x – 5 = 0. So either x = 5 or x = –5. That is, x = ±5. Guided Practice Find the square roots of the expressions below. Show your work. 1. 49 4. 128 7. x2 + 4x + 4 2. 64 5. 4t2 8. a2 + 16a + 64 3. 256 6. 16t4 9. 4y2 + 12y + 9 Solve the equations below by finding the square root. 10.
x2 = 4 13. 3a2 = 75 11. x2 = 9 14. a2 = 81 12. 2x2 = 32 15. 5x2 = 180 16. Use the zero product property and factoring to verify your answers to Exercises 10–15 above. More Square Root Examples Example 3 Find the solution set of 3x2 – 7 = 101. Solution Here, you can get x2 on its own on one side of the equation, with no x’s on the other side. This allows you to take the square root, as above. 3x2 – 7 = 101 3x2 = 108 x2 = 36 x = ± 36 That is, x = ±6 — or x Œ {6, –6} Section 7.1 — Solving Quadratic Equations 339 Check it out: x = ±6 and x Œ {6, –6} mean the same thing. Check it out: You can’t get x2 on its own because if you multiply out these parentheses, you get x2 – 14x + 49. So if you put x2 on one side, you’ll have 14x on the other side, which is no good. Example 4 Solve (x – 7)2 = 64. Solution This time, you can’t get x2 on its own (with no x’s on the other side), but you already have (x – 7)2 alone, which is just as good. (x – 7)2 = 64 x – 7 = ± 64 So x = 15 or x = –1. Check Your Answers by Using the Original Equation You can always check your answers by substituting your solutions into the original equation to see if you get a true statement. This example shows how you’d check the solution reached in Example 4. Example 5 Show that x = 15 and x = –1 are solutions of the equation (x – 7)2 = 64. Solution Do this by substituting x = 15 and x = –1 into the equation (x – 7)2 = 64, and seeing if you get true statements. Put x = 15: (15 – 7)2 = 64 82 = 64 64 = 64 Put x = –1: (–1 – 7)2 = 64 (–8)2 = 64 64 = 64 These are both true statements, so x = 15 and x = –1 are both solutions of the equation.
Guided Practice Find the square roots of the expressions below. 17. c2 + 6c + 9 20. 9x2 – 24x + 16 23. 49 + 28y + 4y2 18. x2 + 14x + 49 21. 9x2 + 30x + 25 24. 4x2 + 4bx + b2 19. x2 – 6x + 9 22. 25 – 30k + 9k2 25. k2 – 8kx + 16x2 Solve the following equations by using the square root method. 26. k2 = 1 29. m2 = 432 32. (k – 6)2 = 72 27. x2 = 49 30. (x + 3)2 = 81 33. (v + 7)2 = 147 28. p2 = 125 31. (x – 5)2 = 121 34. x2 + 4x + 4 = 36 340 Section 7.1 — Solving Quadratic Equations A Couple More Examples Example 6 Solve 4x2 – 20x + 25 = 9 by taking square roots of both sides. Solution 4x2 – 20x + 25 = 9 4 2x x− 20 + = 25 9 2 = x − ( 2 9 5 ) 2x – 5 = ±3 Since (2x – 5)2 = 4x2 – 20x + 25 So 2x = 8 or 2x = 2 — which means that x = 4 or x = 1. Example 7 Solve y2 = 75, giving your answer in its simplest form. Don’t forget: See Topics 1.3.2 and 1.3.3 for more information on expressions involving radicals. Solution y2 = 75 = ± y = ± y 75 ⋅ 25 3 ⋅ 3 = ± 25 y y = ±5 3 Independent Practice Find the square roots of the expressions below. 1. 25k2 – 60kr + 36r2 2 12 Solve the following equations by using the square root method. k 4 m 3. 25 25 20 20 + + − + p p 4 4 m 2 2 16 + 9 4. 4x2 – 4x + 1 = 16 5. 9x2 + 12x + 4 = 169 6. k2 – 14k + 49 = 9 16 7. 4x2 – 12x + 9 = 16 8. 9x2 – 6x + 1 = 4 9. The sides of a square are each (2
x – 16) cm long. Find the value of x that would give a square with an area of 108 cm². 10. The product of the number of CDs that Donna and Keisha have is 16a2 + 56a + 49. If both have the same number of CDs, find how many CDs Donna has, in terms of a. Round Up Round Up Don’t forget that square roots result in two possible solutions. Also, no matter how you’ve solved a quadratic, it’s a good idea to check your solutions by substituting them back into the original equation. Section 7.1 — Solving Quadratic Equations 341 Topic 7.2.1 Section 7.2 Completing the Square Completing the Square California Standard: 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll form perfect square trinomials by adding numbers to binomial expressions. Key words: completing the square perfect square trinomial binomial Check it out: The trinomial (with 3 terms) can be written as the square of a binomial (a binomial has 2 terms). “Completing the square” is another method for solving quadratic equations — but before you solve any equations, you need to know how completing the square actually works. Writing Perfect Square Trinomials as Perfect Squares An expression such as (x + 1)2 is called a perfect square — because it’s (something)2. In a similar way, an expression such as x2 + 2x + 1 is called a perfect square trinomial (“trinomial” because it has 3 terms). This is because it can be written as a perfect square: x2 + 2x + 1 = (x + 1)2 Any trinomial of the form x2 + 2dx + d 2 is a perfect square trinomial, since it can be written as the square of a binomial: x2 + 2dx + d 2 = (x + d)2 Converting Binomials to Perfect Squares The binomial expression x2 + 4x is not a perfect square — it can’t be written as the square of a binomial. However, it can be turned into a perfect square trinomial if you add a constant (a number) to the expression. Example 1 Convert x2 + 4x to a perfect square
trinomial. Solution To do this you have to add a number to the original expression. First look at the form of perfect square trinomials, and compare the coefficient of x with the constant term (the number not followed by x or x2): x2 + 2dx + d 2 = (x + d)2 The coefficient of x is 2d, while the constant term is d 2. So the constant term is the square of half of the coefficient of x. To convert x2 + 4x to a perfect square trinomial, add the square of half of 4 — that is, add 22 = 4, to give x2 + 4x + 4 = (x + 2)2 342 Section 7.2 — Completing the Square Don’t forget: b is the coefficient of x. Check it out: The solution is equivalent to (x + 4)2. Check it out: In Example 3, y2 – 12y + 36 could also be written as (y – 6)2. Guided Practice By adding a constant, convert each of these binomials into a perfect square trinomial. 1. x2 + 14x 3. x2 + 2x 5. y2 + 20y 2. x2 – 12x 4. x2 – 8x 6. p2 – 16p Completing the Square for x2 + bx Completing the square for x2 + bx To convert x2 + bx into a perfect square trinomial, add ⎛ ⎜⎜⎜ The resulting trinomial is x ⎝ ⎞ 2 ⎠. b+ ⎟⎟⎟2 ⎛ ⎜⎜⎜ ⎝ b 2 ⎞ 2 ⎟⎟⎟. ⎠ Example 2 Form a perfect square trinomial from x2 + 8x. Solution Here, b = 8, so to complete the square you add ⎛ ⎜⎜⎜ ⎝ 8 2 ⎞ 2 ⎟⎟⎟. ⎠ This gives you x2 + 8x + 16. Example 3 What must be added to y2 – 12y to make it a perfect square trinomial? Solution This time, b = –12. To complete the square you add −⎛ ⎜⎜
⎜ ⎝ 12 2 ⎞ 2 ⎟⎟⎟ ⎠ = (–6)2 = 36. So 36 must be added (giving y2 – 12y + 36). Section 7.2 — Completing the Square 343 Don’t forget: Remember the negative sign in front of the 10 coefficient. Example 4 Suppose x2 – 10x + c is a perfect square trinomial, and is equal to (x + k)2. What are the values of c and k? Solution Here the coefficient of x is –10. So to form a perfect square trinomial, the constant term has to be the square of half of –10, so c = (–5)2 = 25. Therefore x2 – 10x + c = x2 – 10x + 25 = (x + k)2. Now multiply out the parentheses of (x + k)2. (x + k)2 = x2 + 2kx + k2 This has to equal x2 – 10x + 25, which gives x2 – 10x + 25 = x2 + 2kx + k2. Equate the coefficients of x: the coefficient of x on the left-hand side is –10, while on the right-hand side it is 2k. So –10 = 2k, or k = –5. Comparing the constant terms in a similar way, you find that 25 = k2, which is also satisfied by k = –5. Guided Practice Find the value of k that will make each expression below a perfect square trinomial. 7. x2 – 7x + k 9. x2 + 6mx + k 8. q2 + 5q + k 10. d 2 – 2md + k Form a perfect square trinomial from the following expressions by adding a suitable term. 11. x2 + 10x 13. y2 + 2y 15. y2 – 18y 17. y2 + 12y 19. 6x + 9 21. 25 – 20y 23. 4a2 + 12ab 12. x2 – 16x 14. x2 + bx 16. a2 – 2a 18. y2 + 36y 20. 1 – 8x 22. 4y2 + 4yb 24. 9a2 + 16b2 The quadratics below are perfect square trinomials. Find the value of c and k
to make each statement true. 25. x2 – 6x + c = (x + k)2 27. 4x2 + 12x + c = (2x + k)2 29. 4a2 – 4ab + cb2 = (2a + kb)2 26. x2 + 16x + c = (x + k)2 28. 9x2 + 30x + c = (3x + k)2 30. 9a2 – 12ab + cb2 = (3a + kb)2 344 Section 7.2 — Completing the Square If the Coefficient of x2 isn’t 1, Add a Number With an expression of the form ax2 + bx, you can add a number to make an expression of the form a(x + k)2. Example 5 If 3x2 – 12x + m is equal to 3(x + d)2, what is m? Solution Multiply out the parentheses of 3(x + d)2 to get: 3(x + d)2 = 3x2 + 6dx + 3d2 So 3x2 – 12x + m = 3x2 + 6dx + 3d2 Equate the coefficients of x, and the constants, to get: –12 = 6d and m = 3d2 The first equation tells you that d = –2. And the second tells you that m = 3(–2)2, or m = 12. So 3x2 – 12x + 12 = 3(x – 2)2. Completing the square for ax2 + bx The expression ax2 + bx can be changed to a trinomial of the form a(x + k)2. ⎛ ⎜⎜⎜ ⎝ b 4 a ⎛ ⎜⎜⎜ The resulting trinomial is a x ⎝ To do this, add ⎞ 2 ⎟⎟⎟ ⎠ ⎞ ⎟⎟⎟2 Example 6 Convert 2x2 + 10x to a perfect square trinomial. Solution Here, a = 2 and b = 10, so you add This gives you 2 2x x+ 10 25 +. 2 ⎛ ⎜⎜⎜ ⎝ 10 2 1 2 ⎞ 2 ⎟�
�⎟ = ⎠ ⎛ ⎜⎜⎜ ⎝ 1 2 100 4 ⎞ ⎟⎟⎟ = ⎠ 25 2 Section 7.2 — Completing the Square 345 Check it out: + 2x x+ 10 2 can also be 25 2 ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟. ⎠ 5 2 written as 2 x + Guided Practice The quadratics below are of the form a(x + d)2. Find the value of m and d in each equation. 31. 5x2 + 10x + m = 5(x + d)2 33. 2x2 – 28x + m = 2(x + d)2 35. 4x2 + 32x + m = 4(x + d)2 37. 20x2 – 20x + m = 5(2x + d)2 39. 27x2 + 36x + m = 3(3x + d)2 32. 4x2 – 24x + m = 4(x + d)2 34. 3x2 – 30x + m = 3(x + d)2 36. 20x2 + 60x + m = 5(2x + d)2 38. 27x2 + 18x + m = 3(3x + d)2 40. 16x2 – 80x + m = 4(2x + d)2 Add a term to convert each of the following into an expression of the form a(x + k)2. 41. 2x2 – 12x 43. 6y2 – 60y 45. 5x2 + 245 47. 36x + 12 42. 3a2 + 12a 44. 4x2 – 48x 46. 8x2 + 2 48. 120x + 100 Independent Practice Find the value of c that will make each expression below a perfect square trinomial. 1. x2 + 9x + c 3. x2 + 12xy + c 2. x2 – 11x + c 4. x2 – 10xy + c Complete the square for each quadratic expression below. 5. x2 – 6x 7. b2 – 10b 9. c2 – 12bc 6. a2 – 14a 8. x2 + 8xy 10.
x2 + 4xy Find the value of m and d in each of the following. 11. 5x2 – 40x + m = 5(x + d)2 13. 3x2 – 6x + m = 3(x + d)2 15. 4x2 + 24x + m = 4(x + d)2 12. 2x2 + 20x + m = 2(x + d)2 14. 3x2 – 30x + m = 3(x + d)2 16. 7x2 – 28x + m = 7(x + d)2 Add a term to convert each of the following into an expression of the form a(x + k)2. 17. 3x2 – 30x 19. 18x2 – 48x 21. 27x2 + 12 18. 2x2 + 8x 20. 5x2 + 180 22. 36x2 + 196 23. The length of a rectangle is twice its width. If the area can be found by completing the square for (18x2 + 60x) ft2, find the width of the rectangle. Round Up Round Up OK, so now you know how to add a number to a binomial to make a perfect square trinomial. In the next Topic you’ll learn how to convert any quadratic expression into perfect square trinomial form — and then in Topic 7.2.3 you’ll use this to solve quadratic equations. 346 Section 7.2 — Completing the Square Topic 7.2.2 California Standards: 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll convert any quadratic expression into perfect square trinomial form. Key words: completing the square perfect square trinomial binomial More on Completing More on Completing the Square the Square In Topic 7.2.1, you converted binomial expressions like ax2 + bx to perfect square trinomials by adding a number. In this Topic you’ll take a more general quadratic expression like ax2 + bx + c and write this in the form a(x + k) 2 + m. Writing x2 + bx + c in the form (x + k)2 + m In earlier Topics, you converted an expression of the form x2 + bx to a ⎛ ⎜�
��⎜ ⎝ perfect square trinomial by adding ⎞ 2 ⎟⎟⎟ to it. ⎠ b 2 Example 1 Convert x2 + 4x to a perfect square trinomial. Solution Here b = 4, so to convert this to a perfect square trinomial, you add ⎛ ⎜⎜⎜ ⎝ 4 2 ⎞ 2 ⎟⎟⎟ = =. 2 4 ⎠ 2 So x2 + 4x + 4 = (x + 2)2. Here x2 + 4x + 4 is a perfect square trinomial. Another way to think about this is to say that your original expression was equal to an expression of the form (x + k)2 + m. Check it out: Example 2 is almost the same problem as in Example 1 — it’s just asked in a different way. Example 2 Express x2 + 4x in the form (x + k)2 + m. Solution x2 + 4x + 4 = (x + 2)2. Therefore x2 + 4x = (x + 2)2 – 4. Guided Practice Express each of the following in the form (x + k)2 + m. 1. x2 + 6x 4. x2 + 24x 7. x2 + 18x 2. x2 + 20x 5. x2 – 22x 8. x2 – 14x 3. x2 + 12x 6. x2 – 10x 9. x2 – 16x Section 7.2 — Completing the Square 347 More general quadratic expressions x2 + bx + c can also be written in the form (x + k)2 + m. Writing x2 + bx + c in the form (x + k)2 + m 1) First take just the first two terms (x2 + bx) and convert this to a perfect square trinomial by adding the square of half of b: 2 x + + bx ⎛ ⎜⎜⎜ ⎝ b 2 ⎞ ⎛ 2 ⎟⎟⎟ = + ⎜⎜⎜ x ⎠ ⎝ ⎞ 2 ⎟⎟⎟ ⎠ b 2 2) Rewrite this in the form
x 2 ⎛ ⎜⎜⎜ + = + ⎝ bx x 3) Add c to both sides: x 2 + + = + c bx x ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟ − ⎠ ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟ ⎠ b 2 b 2 ⎞ 2 ⎟⎟⎟ − ⎠ ⎛ ⎜⎜⎜ ⎝ ⎞ 2 ⎟⎟⎟ + ⎠ b 2 b 2 c Check it out: Compare this to (x + k)2 + m, and you see that ⎛ b= − ⎜⎜⎜ ⎝ 2 b= 2 and m + c. ⎞ 2 ⎟⎟⎟ ⎠ k Example 3 Express x2 + 4x + 1 in the form (x + k)2 + m. Solution ⎛ 1) Again, b = 4, so add 4 ⎜⎜⎜ ⎝ 2 expression to find a perfect square trinomial — that is, x2 + 4x + 4 = (x + 2)2. ⎞ 2 ⎟⎟⎟ = 22 = 4 to the first two terms of your ⎠ So the first two terms of your original expression can be expressed: x2 + 4x = (x + 2)2 – 4 2) Now add c (= 1) to both sides of this equation to get: x2 + 4x + 1 = (x + 2)2 – 3 Example 4 Write x2 – 6x + 3 in the form (x + k)2 + m. Solution 1) Here, b = –6, so add −⎛ 6 ⎜⎜⎜ ⎝ 2 ⎞ 2 ⎟⎟⎟ ⎠ = (–3)2 = 9 to x2 – 6x for a perfect square trinomial — that is, x2 – 6x + 9 = (x – 3)2. So the first two terms of the original quadratic can
be expressed: x2 – 6x = (x – 3)2 – 9 2) Add c (= 3) to both sides of this equation to get: x2 – 6x + 3 = (x – 3)2 – 6 348 Section 7.2 — Completing the Square Guided Practice Express the following in the form (x + k)2 + m [or a(x + k)2 + m]. 10. x2 + 4x + 8 12. x2 + 8x + 5 14. x2 + 3x + 5 16. x2 + 4x + 5 18. x2 – 20x + 20 20. 4x2 + 16x + 4 11. x2 + 6x + 14 13. x2 – 12x + 8 15. x2 – 5x – 7 17. x2 – 6x + 30 19. x2 + 22x + 54 21. 2x2 + 20x + 25 Writing ax2 + bx + c in the Form a(x + k)2 + m This is the most general case you can meet — writing ax2 + bx + c as an expression of the form a(x + k)2 + m. The method’s very similar to the one used in the previous examples. Example 5 Write 2x2 + 6x + 7 in the form a(x + k)2 + m. Solution 1) Factor your expression by taking the coefficient of x2 outside ( parentheses: 2x2 + 6x + ) The first two terms in parentheses (x2 + 3x) can be made into a perfect square trinomial in the usual way — by adding ⎛ ⎜⎜⎜ ⎝ ⎞ 2 9 ⎟⎟⎟ =. ⎠ 4 3 2 This means that x2 + 3x + )3 — that is, x2 + 3x = x +( )3 = x +( 2 2 So x2 + 3x + +( 2 )3 2 9 4 – 9 4. or, combining the fractions, x2 + 3x + = x +( 7 2 2 )3 2 + 5 4 3) Therefore the original expression (2x2 + 6x + 7) can be written: 2 ⎡ x +( ⎢ ⎢ ⎣ +⎡ ⎢ ⎣ 2 ⎤ �
�� + ⎦ 3 2 5 2 Section 7.2 — Completing the Square 349 Check it out: Now you can deal with the expression in parentheses exactly as before. Check it out: This is exactly the same process as in Examples 3 and 4 on the previous page. Check it out: Completing the square can be used to find the highest point of an object’s path as it flies through the air. You’ll see this in more detail in Sections 7.4 and 7.6. Example 6 Write 4x2 – 4x + 3 in the form a(x + k)2 + m. Solution 1) Factor the expression: 4x2 – 4x + 3 = 4(x2 – x + 3 4 ) 2) Convert the first two terms from inside the parentheses into a perfect square trinomial: x2 – x + 1 4 = (x – 1 2 )2 So rewriting the above, you get: x2 – x = (x – Add 3 4 to both sides: x2 – x + 3 4 = (x – 1 2 )2 + )2 – 1 4 1 2 1 2 3) Finally, use your expression from 1) above: 4x2 – 4x + 3 = 4 x −( ⎡ ⎢ ⎢ ⎣ (x – 1 2 )2 + 2 Guided Practice Express the following in the form (x + k)2 + m [or a(x + k)2 + m]. 22. 2x2 – 9x + 4 24. 3x2 – 12x + 14 26. –x2 – 3x + 4 28. –3x2 – 9x + 21 23. 2x2 + 9x + 5 25. –x2 + 2x + 3 27. –2x2 + 6x + 10 29. –4x2 + 120x – 80 Independent Practice Express the following in the form (x + k)2 + m. 1. x2 + 4x 3. x2 + 10x 5. x2 – 1 x 4 2. x2 – 6x 4. x2 – 20x 6. x2 + 2 x 3 Express the following in the form of a(x + k)2 + m 7. –5x2 – 90x – 150 9. 4x2 – 3x + 12 11. 6x2 + 48x + 16 8. 2x2 +
5x + 10 10. 5x2 + 20x + 4 12. 3x2 + 42x + 49 13. A square has an area of x2 + 14x + k. Find the value of k. 14. A circle has an area of px2 + 18px + kp. Find the value of k. Round Up Round Up There’s been a lot of build-up to actually solving quadratic equations using the completing the square method — but it’s coming up next, in Topic 7.2.3. 350 Section 7.2 — Completing the Square Topic 7.2.3 California Standards: 14.0: Students solve a quadratic equation by factoring or completing the square. What it means for you: You’ll use completing the square to solve quadratic equations. Key words: completing the square perfect square trinomial binomial Check it out: The aim is to get a perfect square involving x on one side of the equation. Solving Equations by Solving Equations by Completing the Square Completing the Square This is what the whole Section has been leading up to. The process of completing the square is a really useful method that can help solve quadratic equations. You can Solve Quadratics by Completing the Square The best way to show this is with an example: Example 1 Solve x2 – 10x + 21 = 0 by completing the square. Solution To do this, you apply some of the techniques from the earlier parts of this chapter. 1) Take the constant (= 21) onto the right-hand side of the equation. Then convert what remains on the left-hand side (x2 – 10x) to a perfect square trinomial by adding the square of half the coefficient of x (to both sides of the equation) — so add (–5)2 = 25. x2 – 10x = –21 x2 – 10x + 25 = –21 + 25 x2 – 10x + 25 = 4 2) Now you can use the square root method to solve the equation. x2 – 10x + 25 = 4 (x – 5)2 = 4 x – 5 = ±2 x – 5 = 2 or x – 5 = –2 x = 7 or x = 3 Section 7.2 — Completing the Square 351 Example 2 Solve the following quadratic equation: 4x2 – 9x + 2 = 0 Solution The best thing
to do here is to divide the equation by 4 first. Then you need to solve: x2 – 9 4 x + 1 2 = 0 1) x2 – x2 – 9 4 9 4 x = – 1 2 Move the constant to the other side x + 2⎛ ⎞ 9 ⎟⎟⎟⎟ = – ⎜⎜⎜ ⎠ ⎝ 8 1 2 + 2⎛ ⎞ 9 ⎟⎟⎟⎟ = ⎜⎜⎜ ⎠ ⎝ 8 49 64 Add a number to both sides to get a perfect square trinomial x−⎛ ⎜⎜⎜ ⎝ 2 ⎞ ⎟⎟⎟⎟ ⎠ 9 8 = 49 64 Rewrite the perfect square trinomial as a square So now you have to solve: x−⎛ ⎜⎜⎜ ⎝ 2 ⎞ ⎟⎟⎟⎟ ⎠ 9 8 = 49 64 2) Use the square root method. x−⎛ ⎜⎜⎜ ⎝ 2 ⎞ ⎟⎟⎟⎟ ⎠ 9 8 = 49 64 or x – 9 8 = – x = 16 8 = 2 or x = 2 8 = 7 8 1 4 Guided Practice Solve the following equations by completing the square. 1. x2 – 6x + 5 = 0 3. x2 – 10x – 200 = 0 5. x2 + 14x + 33 = 0 7. x2 – 2x – 8 = 0 9. x2 + 18x – 19 = 0 2. x2 + 4x – 5 = 0 4. x2 – 12x – 64 = 0 6. x2 – 8x + 12 = 0 8. x2 – 16x + 39 = 0 10. x2 – 20x – 44 = 0 11. x2 + 22x + 21 = 0 12. 2x2 – 12x + 10 = 0 13. 3a2 + 12a – 15 = 0 14. 2y2 + 20y + 18 = 0 352 Section 7.2 — Completing the Square Check it out: You could complete the square at
this stage, but it’s easier if the coefficient of x2 is 1. Example 3 Solve 5x2 – 2x – 3 = 3x2 – 6x + 13. Solution Rearrange the equation first so that all like terms are combined: 5x2 – 2x – 3 = 3x2 – 6x + 13 2x2 + 4x – 16 = 0 Now divide through by 2 so that the coefficient of x2 is 1. x2 + 2x – 8 = 0 Take the constant to the other side — then convert what’s left to a perfect square. x2 + 2x = 8 x2 + 2x + 1 = 8 + 1 (x + 1)2 = 9 Use the completed square to solve the original equation. (x + 1)2 = 9 x + 1 = ±3 So x + 1 = 3 or x + 1 = –3 x = 2 or x = –4 Example 4 Find t if 24t – 3t 2 – 48 = –9. Solution 24t – 3t2 – 48 = –9 3t2 – 24t + 39 = 0 t2 – 8t + 13 = 0 t2 – 8t = –13 ( ) = –13 + 8 2 (t – 4)2 = –13 + 16 (t – 4)2 = 3 t2 – 8t + 8 2 2 2 ( ) Add a number to both sides to get a perfect square trinomial t – 4 = ± 3 So t = 4 + 3 or t = 4 – 3 Section 7.2 — Completing the Square 353 Guided Practice Solve each of these quadratic equations using the method of completing the square: 15. x2 – 2x – 15 = 0 16. x2 + 2x – 24 = 0 17. x2 + 14x + 45 = 0 18. x2 – 15x + 56 = 0 19. 5x2 + 12x + 38 = 2x2 + 36x + 2 20. –150 + 2b2 = 20b 21. 5a(a + 2) – 120 = 0 22. 5y(y + 2) – 13 = 2(y2 – y + 1) 23. 5x2 + 44x – 49 = 2x(1 – x) 24. 4x(x – 11) – 4 = 4(x – 12) 25. 6a(a + 3) = 6(a –
1) + 216 26. 5y(y – 6) + 51 = 131 27. 7b2 + 14b – 32 = 5b2 + 26b 28. 3(x2 – 5) – 9 = 111 – 12x 29. 8(x + 6)2 – 128 = 6(x + 6)2 30. 4(x + 5)2 – 200 = 2(x + 5)2 Independent Practice Solve by completing the square. 1. x2 + 2x – 3 = 0 3. a2 + 8a – 9 = 0 2. y2 – 4y – 12 = 0 4. b2 + 10b – 24 = 0 5. x2 – 12x + 20 = 0 6. 3x2 – 42x + 39 = 0 7. 2a2 + 12a + 10 = 0 8. 5b2 + 20b – 105 = 0 9. 7d2 – 14d – 21 = 0 10. 2x2 + 48x – 50 = 0 11. 3c2 + 66c + 63 = 0 12. 3y(y – 8) + 2 = 254 13. 6c(c – 4) = 2(18c + 168) 14. 5(a2 + 14) = 5 – 70a 15. y2 – 2py – 15p2 = 0 16. x2 – 3kx + 2k2 = 0 17. An expression for the area of a rectangle is (2x2 + 4x) cm2 and is equal to 6 cm2. Find the value of x. 18. An expression for the area of a circle is (4py2 – 16py) ft2 and is equal to 20p ft2. Find the value of y. Round Up Round Up Make sure you understand all the examples — then once you understand the method, get plenty of practice. Completing the square means first forming a perfect square trinomial, and then using the new form to solve the original equation. 354 Section 7.2 — Completing the Square Topic 7.3.1 Section 7.3 The Quadratic Formula The Quadratic Formula California Standards: 19.0: Students know the quadratic formula and are familiar with its proof by completing the square. 20.0: Students use the quadratic formula to find the roots of a seconddegree polynomial and to solve quadratic equations. What it means for you: You’ll
use the quadratic formula to solve quadratic equations — and you’ll derive the quadratic formula itself. Key words: quadratic formula completing the square You can also use the quadratic formula to solve quadratic equations. It works every time. Quadratic Equations can be in Any Variable The standard form for a quadratic equation is ax2 + bx + c = 0 (a π 0) Any quadratic equation can be written in this form by, if necessary, rearranging it so that zero is on one side. A lot of the quadratic equations you will see may contain a variable other than x — but they are still quadratic equations like the one above, and can be solved in the same way. Example 1 Find the solutions of the following quadratic equations: a) x2 – 4x + 3 = 0 b) y2 – 4y + 3 = 0 Solution a) Here, a = 1, b = –4, and c = 3. This equation factors to give (x – 3)(x – 1) = 0 So using the zero property: (x – 3) = 0 or (x – 1) = 0, or x = 3 or x = 1 b) Here, the variable is y rather than x — but that does not affect the solutions. Again a = 1, b = –4, and c = 3, so it is the same equation as in a), and will have the same solutions: y = 3 or y = 1 You can see that the two quadratic equations are really the same — only the variables have changed. Section 7.3 — The Quadratic Formula and Applications 355 The Quadratic Formula Solves Any Quadratic Equation The solutions to the quadratic equation ax2 + bx + c = 0 are given by the quadratic formula: − ± b x = − 4 ac 2 b 2 a You can derive the quadratic formula by completing the square: 2 ax ax + + = bx 2 0 c + = − c bx +⎡ ⎢ x ⎣ +⎡ ⎢⎢ x ⎣ +⎡ ⎢ ac b 2 4 a − 4 ac 2 a 4 b 2 2 Dividing the equation by a Completing the square 2 Rearranging ± ac a Include positive and negative roots = − + b So x ac −2
4 b 2 a = − − b or x ac − ac −2 b 2 a Examples Using the Quadratic Formula Example 2 Solve x2 – 5x – 14 = 0 using the quadratic formula. Solution Start by writing down the values of a, b, and c: a = 1, b = –5, and c = –14 Now very carefully substitute these values into the quadratic formula ac 14 ) ± 5 = x + 56 = 25 2 ± 5 81 2 = 9 ±5 2 So x = + 5 9 2 = 7 or x = − 5 9 2 = − 2 Check it out: The quadratic formula gives you the solutions of any quadratic equation in terms of a, b, and c. It’s sometimes called the general solution of the quadratic equation ax2 + bx + c = 0. Don’t forget: See Section 7.2 for a reminder about completing the square. Don’t forget: Be very careful if there are minus signs. Check it out: You can check these answers by substituting them into the original equation: 72 – (5 × 7) – 14 = 0 and (–2)2 – (5 × (–2)) – 14 = 0. So both solutions are correct. 356 Section 7.3 — The Quadratic Formula and Applications It’s really important to practice with the quadratic formula — you have to be able to use it correctly. Example 3 Solve 2x2 – 3x – 2 = 0 using the quadratic formula. Solution a = 2, b = –3, and c = –2 Putting these values into the quadratic formula... = x = x − ± b − 4 ac 16 4 = ± 3 25 4 = ± 33 5 4 So x = + 3 5 4 = 2 or Example 4 Solve 2x2 – 11x + 13 = 0. Solution a = 2, b = –11, and c = 13 Putting these values into the quadratic formula... = x = x = x ac 2 4 − − ± b b 2 a − − ± − ) 11 ( ( ) 11 ⋅ 2 2 − 121 104 4 2 − ⋅ ⋅ ) 4 2 13 ( 11 = ±± 17 4 ± 11 So x = +11 4 17 or x = 17 −11 4 Section 7.3 — The Quadratic Formula and Applications 357 Check it out
: Leave these answers as radical expressions unless you’re told otherwise. Guided Practice Use the quadratic formula to solve each of the following equations. 1. x2 – 2x – 143 = 0 3. x2 + 2x – 1 = 0 5. 2x2 – 5x + 2 = 0 7. 2x2 – 7x – 3 = 0 9. 18x2 + 3x – 1 = 0 2. 2x2 + 3x – 1 = 0 4. x2 + 3x + 1 = 0 6. 3x2 – 2x – 3 = 0 8. 6x2 – x – 1 = 0 10. 4x2 – 5x + 1 = 0 11. The equation 2x2 – 7x – 4 = 0 factors to (2x + 1)(x – 4) = 0. Using the zero product property we can find that x = – 1 2 or x = 4. Verify this using the quadratic formula. 12. The height of a triangle is 4 ft more than 4 times its base length. If the triangle’s area is 5 2 ft2, find the length of its base. Independent Practice Use the quadratic formula to solve each of the following equations. 1. 5x2 – 11x + 2 = 0 3. 7x2 + 6x – 1 = 0 5. 10x2 + 7x + 1 = 0 7. 5x2 – 2x – 3 = 0 9. 4t2 + 7t – 2 = 0 11. 2x2 – x = 1 13. 2x2 + 7x = 4 15. 4x2 – 13x + 3 = 0 17. 25x2 – 9 = 0 19. 10x2 + 1 = 7x 2. 2x2 + 7x + 3 = 0 4. x2 – 7x + 5 = 0 6. 3y2 – 8y – 3 = 0 8. 4x2 + 3x – 5 = 0 10. 6m2 + m – 1 = 0 12. 3x2 – 5x = 2 14. 4x2 + 17x = 15 16. 4x2 – 1 = 0 18. 4x2 + 15x = 4 20. 16x2 + 3 = 26x Solve these equations by factoring and using the zero product property, then verify the solutions by solving them with the quadratic formula. 21. x2 + 4x
+ 4 = 0 23. x2 – x – 12 = 0 25. 6x2 + 29x = 5 22. 4y2 – 9 = 0 24. 2x2 – 3x – 9 = 0 26. 7x2 + 41x = 6 27. The length of a rectangle is 20 cm more than 4 times its width. If the rectangle has an area of 75 cm2, find its dimensions. 28. The equation h = –14t2 + 12t + 2 gives the height of a tennis ball t seconds after being hit. How long will the ball take before it hits the ground? Round Up Round Up The quadratic formula looks quite complicated, but don’t let that put you off. If you work through the derivation of the formula on p356 then you should see exactly why it contains all the elements it does. 358 Section 7.3 — The Quadratic Formula and Applications Topic 7.3.2 California Standards: 20.0: Students use the quadratic formula to find the roots of a seconddegree polynomial and to solve quadratic equations. What it means for you: You’ll model real-life problems using quadratic equations and then solve them using the quadratic formula. Key words: quadratic formula completing the square Check it out: The mathematical model in Example 1 is the statement that the numbers should be represented by x and (x + 9). It’s very important to be clear here to avoid confusion later. Check it out: In fact, this equation factors. But if you didn’t spot that, use the quadratic formula as in the example. Check it out: Careful — you haven’t finished yet. Now you need to go back to your initial model and interpret the solutions of your quadratic. Applications of Quadratics Applications of Quadratics Sometimes you will have to make a “mathematical model” first, and then solve the equations you get from it. Then when you get your solutions, you have to interpret them. Modeling Means Writing Your Own Equations Here’s an example of modeling a real-life situation as a quadratic equation: Example 1 The difference between a pair of numbers is 9. Find all such pairs of numbers that have a product of 220. Solution There are two different numbers in each pair — call the lower number x. Then the higher number is x + 9. Write the information
from the question in the form of an equation: x(x + 9) = 220 This is a quadratic equation, so rearrange it to the form ax2 + bx + c = 0. x(x + 9) = 220 x2 + 9x = 220 x2 + 9x – 220 = 0 Write down a, b, and c: a = 1, b = 9, and c = –220 Now you can use the quadratic formula ac 2 b 2 a 2 ) ( 9 220 − ± + 81 880 2 20 or x = =22 = −40 2 = 2 − ± 9 2 11. − So x = 961 == 31 − ±9 2 So there are two possible values for x (where x is the lower of the two numbers): x = –20 or x = 11 The higher of the two numbers is found by adding 9 to each of these values. So there are two possible pairs of numbers, and they are: –20 and –11 and 11 and 20 Section 7.3 — The Quadratic Formula and Applications 359 Example 2 Find the dimensions of the rectangle whose length is 7 inches more than twice its width, and whose area is 120 in2. Solution Let x = width in inches. Then the length is 2x + 7 inches. From the question: x(2x + 7) = 120 Rearrange this quadratic into standard form: 2x2 + 7x – 120 = 0 Now use the formula with a = 2, b = 7, and c = –120 ac − ⋅ 4 2 ⋅ 2 2 + 960 2 b a 2 2 7 49 4 ⋅ − ( 120 ) = − ± 7 1009 4 = − ±7. 31 76 4 So x = 6.19 or x = –9.69 But the width cannot be negative, so you can ignore x = –9.69. So the width must be x = 6.2 in. (to 1 decimal place), and the length is then 2x + 7 = (2 × 6.19) + 7 = 19.4 in. (to 1 decimal place). Guided Practice 1. The difference between two numbers is 7. Find all possible pairs of such numbers if the product of the two numbers is 198. 2. Find the dimensions of a rectangular garden whose length is 10 meters more than three times its width, if the area is 77 m2. 3. Twice the square of a number
is equal to eight times the number. Find the number. 4. The sum of the squares of two consecutive odd integers is 74. Find the numbers. 5. The sum of the squares of two consecutive even integers is 340. Find the possible numbers. 6. The length of a rectangular field is 10 meters less than four times its width. Find the dimensions if its area is 750 square meters. 7. When 15 and 19 are each increased by t, the product of the resulting numbers is 837. Find the value(s) of t. 8. A mother is three times as old as her daughter. Four years ago the product of their ages was 256. Find their current ages. 360 Section 7.3 — The Quadratic Formula and Applications Independent Practice 1. A man is five times older than his son. In three years’ time, the product of their ages will be 380. Find their ages now. 2. Lorraine is 10 years older than Ahanu. In three years’ time the product of their ages will be 600. Find Ahanu and Lorraine’s ages now. 3. A picture of 10 inches by 7 inches is in a frame whose area (including the space for the picture) is 154 square inches. Find the dimensions of the frame if the gap between the edge of the picture and the frame is the same all the way around. x x 10 in. 7 in. x x 4. Jennifer has a picture of her boyfriend Zach measuring 10 inches by 8 inches. She frames the picture in a frame that has an area of 224 square inches (including the space for the picture). Find the dimensions of the picture frame if the gap between the edge of the picture and the frame is the same all the way around. 5. A wire of length 50 feet is bent to form a rectangular figure that has no overlap. If the area of the figure formed is 144 square feet, find the dimensions of the figure. 6. A piece of wire 22 yards long is bent to form a rectangular figure whose area is 28 square yards. Find the dimensions of the figure, given that there is no overlap in the wire. 7. Show that the sum of the solutions of 4x2 – 4x – 3 = 0 is equal to 1 (= − b a ). 8. Show that the product of the solutions of x2 – 7x + 10 = 0 is equal to 10 (= ). c a 9. Using the quadratic formula, show that the sum of the
solutions of the general quadratic equation ax2 + bx + c = 0 is equal to − b a product of the roots is, and that the. c a Round Up Round Up Quadratic equations pop up a lot in Algebra I. If you know the quadratic formula then you’ll always be able to solve them by just substituting the values into the formula. Section 7.3 — The Quadratic Formula and Applications 361 Topic 7.4.1 California Standards: 21.0: Students graph quadratic functions and know that their roots are the x-intercepts. What it means for you: You’ll learn about the shape of various quadratic graphs. Key words: quadratic parabola concave vertex line of symmetry root Section 7.4 Graphs of Graphs of Quadratic Functions Quadratic Functions So far in this Chapter you’ve solved quadratic equations in several different ways. In this Section you’ll see how the graphs of quadratic functions can be plotted using the algebraic methods you’ve already seen. The Graphs of Quadratic Functions are Parabolas If you plot the graph of any quadratic function, you get a curve called a parabola. The graphs of y = ax2 (for various values of a) on the right show the basic shape of any quadratic graph. y = x2 a ( = 1) y = –x2 a ( = –1) -3 -4 • The parabola’s either a u-shaped or n-shaped curve depending on the sign of a. The graph of y = ax2 is concave up (u-shaped — it opens upwards) when a > 0, but concave down (n-shaped — it opens downwards) when a < 0. y y = ax2 y = x3 2 a ( = 3) 5 4 3 2 1 -2 -1 0 1 2 3 x 4 -1 -2 -3 -4 -5 1 2 y = – x2 1 a ( = – ) 2 All quadratic graphs have one vertex (maximum or minimum point). For the curves shown above, the vertex is at the origin (0, 0). All quadratic graphs have a vertical line of symmetry. For the graphs above, the line of symmetry is the y-axis. Notice that a bigger value of ΩaΩ results in a
steeper (narrower) parabola. For example, the graph of y = 3x2 is steeper than the graph of y = x2. The basic shape of all quadratic graphs (that is, for any quadratic function y = ax2 + bx + c) is very similar to the ones above. They’re all concave up or concave down depending on the sign of a (concave up if a > 0 and concave down if a < 0). However, the graph can be stretched or squashed, and in a different place relative to the x- and y-axes, depending on the exact values of a, b, and c. 362 Section 7.4 — Quadratic Graphs Guided Practice Match the equations with their graphs on the right. B 1. y = –3x2 x2 – 2 2. y = 1 4 3. y = 2x2 + 3 x2 – 1 4. y = – 1 2 5. y = 2x2 A –6 –5 –4 –3 –1 0 1 –2 –3 –4 –5 –6 –7 y = ax2 + c is Like y = ax2 but Moved Up or Down by c This diagram shows the graphs of y = x2 + c, for three values of c: y = x 2 + c y = x2 y = x2 + 1 y 5 4 3 2 1 -4 -3 -2 -1 0 1 2 3 -1 -2 -3 -4 -5 y = x2 – 4 The top and bottom parabolas in the diagram are both the same shape as the graph of y = x2. The only differences are: (i) the graph of y = x2 + 1 is 1 unit x 4 higher up the y-axis. (ii) the graph of y = x2 – 4 is 4 units lower down the y-axis. The graph of y = x2 – 4 crosses the x-axis when y = 0 — that is, when x2 – 4 = 0 (or x = ±2). In fact, the x-intercepts of any quadratic graph y = ax2 + bx + c are called the roots of the function, and they correspond to the solutions of the equation ax2 + bx + c = 0. The graph of y = x2 + 1 does not cross the x-axis at
all. This is because x2 + 1 = 0 does not have any real solutions. So the graph of a quadratic function may cross the x-axis twice (y = x2 – 4), may touch the x-axis in one place (y = x2), or may never cross it (y = x2 + 1). It all depends on how many roots the quadratic function has. However, the graph will always have a y-intercept — the graph will always cross the y-axis at some point. Section 7.4 — Quadratic Graphs 363 Guided Practice Describe the graphs of the quadratics below in relation to the graph of y = x2. 6. y = x2 + 1 8. y = 2x2 + 2 7. y = x2 – 3 9. y = 1 4 x2 – 5 10. y = –x2 + 1 11. y = –2x2 – 4 The graphs in Exercises 12 and 13 are transformations of the graph of y = x2. Find the equation of each graph. 12. 132 –1 0 1 –2 x 1 2 –2 –1 0 1 –2 –3 –4 Independent Practice Match the equations with their graphs on the right. B A 1. y = x2 – 1 2. y = –x2 – 1 3. y = 3x2 4. y = – 1 4 x2 5. y = –x2 + 3 –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 1 –2 –3 –4 –5 – Describe the graphs of the quadratics below in relation to the graph of y = x2. 6. y = 1 2 x2 + 1 8. y = –2x2 + 3 7. y = –4x2 9. y = 1 3 x2 Round Up Round Up Now you know how the a and c parts of the equation y = ax2 + c affect the graph. In the next Topic you’ll learn how to draw some quadratic graphs yourself. 364 Section 7.4 — Quadratic Graphs Topic 7.4.2 California Standards: 21.0: Students graph quadratic functions and know that their roots are the x-intercepts. 22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect
the x-axis in zero, one, or two points. What it means for you: You’ll graph quadratic functions by finding their roots. Key words: quadratic parabola intercept vertex line of symmetry root Drawing Graphs of Drawing Graphs of Quadratic Functions Quadratic Functions In this Topic you’ll use methods for finding the intercepts and the vertex of a graph to draw graphs of quadratic functions. Find the Roots of the Corresponding Equations In general, a good way to graph the function y = ax2 + bx + c is to find: (i) the x-intercepts (if there are any) — this involves solving a quadratic equation, (ii) the y-intercept — this involves setting x = 0, (iii) the vertex. Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (i) To find the x-intercepts of the graph of y = x2 – 3x + 2, you need to solve: x2 – 3x + 2 = 0 This quadratic factors to give: (x – 1)(x – 2) = 0 Using the zero property, x = 1 or x = 2. So the x-intercepts are (1, 0) and (2, 0). (ii) To find the y-intercept, put x = 0 into y = x2 – 3x + 2. This gives y = 2, so the y-intercept is at (0, 2). (iii) The x-coordinate of the vertex is always halfway between the x-intercepts. So the x-coordinate of the vertex is given by And the y-coordinate of the vertex is: ⎛ ⎜⎜⎜ ⎝ 3 2 So the vertex is at,−( 3 2 1 4 ). ⎟⎟⎟ − ×( 3 2 ⎞ ⎠ )+ = − 2 3 2 1 4 Also, the parabola’s line of symmetry passes through the vertex. So here, the line of symmetry is the line x = 3 2. Section 7.4 — Quadratic Graphs 365 Example 1 continued The next function is the same as in the previous example, only multiplied by –2. The coefficient of x2 is negative
this time, so the graph is concave down. (i) To find the x-intercepts of the graph of y = –2x2 + 6x – 4, you need to solve: –2x2 + 6x – 4 = 0 This quadratic factors to give –2(x – 1)(x – 2) = 0. So using the zero property, x = 1 or x = 2. This means the x-intercepts are at (1, 0) and (2, 0). (ii) Put x = 0 into y = –2x2 + 6x – 4 to find the y-intercept. The y-intercept is (0, –4). (iii) The vertex is at x =. 3 2 ⎛ ⎜⎜⎜ So the y-coordinate of the vertex is at: –2 × ⎝,( Therefore the coordinates of the vertex are 3 2 ×( 2 ⎞ ⎟⎟⎟⎟ + 6 3 ⎠ 2 ), and the line of symmetry is again x = Here are both graphs drawn on the same axes: Check it out: In the equation y = x2 – 3x + 2 the coefficient of x2 is positive (= 1), so the parabola will be u-shaped2 -1 0 1 2 3 x 4 -1 -2 -3 - line of symmetry 366 Section 7.4 — Quadratic Graphs Guided Practice Exercises 1–4 are about the quadratic y = x2 – 1. 1. Find the x–intercepts (if there are any). 2. Find the y–intercepts (if there are any). 3. Find the vertex. 4. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. Exercises 5–8 are about the quadratic y = (x – 1)2 – 4. 5. Find the x–intercepts (if there are any). 6. Find the y–intercepts (if there are any). 7. Find the vertex. 8. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. Independent Practice For each of the quadratics in Exercises 1–7, follow these steps: i) Find the x–intercepts (
if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. 1. y = x2 – 2x 3. y = –4x2 – 4x + 3 5. y = x2 + 4x + 4 7. y = –9x2 – 6x + 3 2. y = x2 + 2x – 3 4. y = x2 – 4 6. y = –x2 + 4x + 5 Describe the characteristics of quadratic graphs of the form y = ax2 + bx + c that have the following features, or say if they are not possible. 8. No x-intercepts 10. Two x-intercepts 12. No y-intercepts 14. Two y-intercepts 9. One x-intercept 11. Three x-intercepts 13. One y-intercept 15. Three y-intercepts 16. Which quadratic equation has the following features? Vertex (3, –4), x-intercepts (1, 0), (5, 0), and y-intercept (0, 5) 17. Which quadratic equation has the following features? Vertex (0, 16), x-intercepts (4, 0), (–4, 0), and y-intercept (0, 16) Round Up Round Up A quadratic function has the general form y = ax2 + bx + c (where a π 0). When you draw the graph of a quadratic, the value of a determines whether the parabola is concave up (u-shaped) or concave down (n-shaped), and how steep it is. Changing the value of c moves the graph in the direction of the y-axis. Note that if a = 0, the function becomes y = bx + c, which is a linear function whose graph is a straight line. Section 7.4 — Quadratic Graphs 367 Topic 7.4.3 California Standards: 21.0: Students graph quadratic functions and know that their roots are the x-intercepts. 22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero,
one, or two points. What it means for you: You’ll graph quadratic functions by first completing the square of the equation. Key words: quadratic completing the square parabola intercept vertex line of symmetry root Quadratic Graphs and Quadratic Graphs and Completing the Square Completing the Square If there are no x-intercepts, then it’s impossible to find the vertex by saying that the vertex is halfway between the x-intercepts (like you saw in Topic 7.4.2). But you can use the method of completing the square. Write the Equation in the Form y = (x + k)2 + p Example 1 Sketch the graph of y = x2 – 6x + 10. Solution You could try to find the x-intercepts by factoring the equation: x2 – 6x + 10 = 0 This time, the left-hand side doesn’t factor. So to find the solutions you could try the quadratic formula with a = 1, b = –6, and c = 10. − ± b = x 4 ac − 10 6 ± −4 2 However, since you cannot take the square root of a negative number, this tells you that the quadratic function has no real roots — the equation can’t be solved using real numbers. This means that the graph of y = x2 – 6x + 10 never crosses the x-axis. But this doesn’t mean that you can’t find the vertex — you just have to use a different method. The trick is to write the equation of the quadratic in the form y = (x + k)2 + p — you need to complete the square. So take the first two terms of the quadratic, and add a number to make a perfect square. 368 Section 7.4 — Quadratic Graphs Example 1 continued x2 – 6x + 2 )6 −( 2 x2 – 6x + 10 = (x – 3)2 – + 10 = (x – 3)2 )6 −( 2 2 x2 – 6x + 10 = (x – 3)2 + 1 Therefore the function you need to sketch is y = (x – 3)2 + 1. Now, the minimum value that (x – 3)2 takes is 0 (since a squared number cannot be negative). Therefore the minimum value of y = (x – 3)
2 + 1 is 0 + 1 = 1. This minimum value occurs at x = 3 (the value for x where (x – 3)2 = 0). So the coordinates of the vertex of the parabola are (3, 1). As before, the line of symmetry passes through the vertex — so the line of symmetry is x = 3. The graph of the quadratic function y = (x + k)2 + p has its vertex at (–k, p). The line of symmetry of the graph is x = –k. Check it out: The coefficient of x2 is positive, so this parabola is concave up. Example 1 continued The graph of y = x2 – 6x + 10 = (x – 3)2 + 1: line of symmetry = 3 x y 6 5 4 3 2 1 vertex (3, 1) -1 1 2 3 4 5 6 7 -1 x 8 Section 7.4 — Quadratic Graphs 369 Don’t forget: See Section 7.2 for more about completing the square. Check it out: The coefficient of x2 is negative, so this is a concavedown parabola. Find the Vertex by Completing the Square Example 2 Complete the square for 4x2 – 12x + 11. Then find the vertex and line of symmetry of y = 4x2 – 12x + 11. Solution This is a concave-up parabola, since the coefficient of x2 is positive. 4x2 – 12x + 11 = 4 = 4 = 4 2 2 2 11 4 ⎡ ⎢ x ⎣ ⎡ −( ⎢ x ⎢ ⎣ ⎡ −( x⎢⎢ ⎢ ⎣ ⎤ − + ⎥ 3 x ⎦ ) −( ) + ⎤ ⎥ ⎥ ⎦ 11 4 = 4 x −( 2 ) + 2 3 2 2 )3 2 is 0, the minimum value of Since the minimum value of x −( 4 x −( )3 2 2 + 2 must be 0 + 2 = 2. This minimum value occurs at x = 3 2. So the vertex of the graph of y = 4x2 – 12x + 11 is at and the line of symmetry is x = 3 2. 2,( 3 2 ), As before, put x = 0 to find the y
-intercept — this is at y = 11. Example 3 Write 4x – x2 – 7 in the form a(x + k)2 + m, and sketch the graph. Solution 4x – x2 – 7 = –x2 + 4x – 7 Factor out –1 to make completing the square easier. = –[x2 – 4x + 7] = –[(x – 2)2 + 3] = –(x – 2)2 – 3 But (x – 2)2 is never negative — the minimum value it takes is 0. So –(x – 2)2 can never be positive, and the maximum value it can take is 0. This means that the maximum value of –(x – 2)2 – 3 must be –3, which it takes when x – 2 = 0 — that is, at x = 2. So the vertex of the graph is at (2, –3). And the line of symmetry is x = 2. As always, find the y-intercept by putting x = 0. This is at y = –7. y 1 -1 0 -1 1 2 3 4 5 x 6 -2 -3 -4 -5 -6 -7 -8 -9 370 Section 7.4 — Quadratic Graphs Guided Practice Sketch the graph of each function below, stating the y-intercept and x-intercepts (where appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry. 1. y = x2 – 12x + 20 3. y = x2 – 2x – 3 5. y = –x2 – 2x + 3 7. y = –2x2 – 8x + 10 9. y = x2 – 4x + 12 2. y = x2 + 8x + 12 4. y = x2 – 4x – 5 6. y = –x2 – x + 6 8. y = 2x2 + x – 6 10. y = 3x2 + 6x + 6 Independent Practice In Exercises 1–2, use the information that a ball was thrown vertically into the air from a platform 3 2 m above sea level. The relationship between the height in meters above sea level, h, and the number of seconds since the ball was thrown, t, was found to be h = –5t2 + 6t + 3 2. 1. After how many
seconds did the ball reach its maximum height? 2. What was the ball’s maximum height above sea level? The first 8 seconds in the flight of a paper airplane can be modeled by the quadratic h = 1 8 t2 – t + 4, where h is the height in feet and t is the time in seconds. Use this information to answer Exercises 3–4. 3. In the first 8 seconds of its flight, when did the airplane reach its minimum height? 4. What was the minimum height of the plane in the first 8 seconds of its flight? A ball is thrown vertically into the air from a platform. The relationship between the ball’s height in meters, h, and the number of seconds, t, since the ball was thrown was found to be h = –5t2 + 10t + 15. Use this information to answer Exercises 5–8. 5. After how many seconds did the ball reach its maximum height? 6. What was the maximum height of the ball? 7. At what height was the ball initially thrown? 8. When did the ball hit the ground? Round Up Round Up Take a look at Section 7.2 if all this stuff about completing the square seems unfamiliar. Completing the square is a really useful way of graphing quadratics because it gives you the vertex of the graph straightaway. Section 7.4 — Quadratic Graphs 371 Topic 7.5.1 California Standards: 22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. What it means for you: You’ll learn what the discriminant is, and how you can use it to tell how many roots a quadratic equation has. Key words: discriminant quadratic intercept root Section 7.5 Quadratic Equations Quadratic Equations and the Discriminant and the Discriminant In this Section you’ll meet a particular part of the quadratic formula that tells you how many solutions a quadratic equation has. This Section carries straight on from the previous Section on graphing quadratic functions. Use the Quadratic Formula to Find x-Intercepts A general quadratic equation has the form: ax2 + bx + c = 0 The solutions to this equation are given by the quadratic formula: − ± b
= x 4 ac −2 b 2 a The quadratic formula can be used to help draw the graph of a quadratic function y = ax2 + bx + c. By finding where y = 0 (that is, by solving ax2 + bx + c = 0), you can find the x-intercepts of the parabola. But it’s sometimes impossible to get an answer from the quadratic formula. When b2 – 4ac is negative, the square root in the formula cannot be taken. This means the graph never crosses the x-axis. Don’t forget: See Section 7.3 for more on the quadratic formula. = x b − ± b 2 4– 2 a ac When b2 – 4ac < 0, there are no x-intercepts — the parabola never crosses the x-axis. In fact, the value of b2 – 4ac tells you a lot about the solutions of a quadratic equation, and about the x-intercepts of the corresponding quadratic graph. b2 – 4ac Tells You How Many Roots a Quadratic Has For the quadratic equation ax2 + bx + c = 0, the expression b2 – 4ac is called the discriminant. The discriminant’s used to determine the number of roots of the function y = ax2 + bx + c (and so the number of x-intercepts of the graph of y = ax2 + bx + c). 372 Section 7.5 — The Discriminant b2 – 4ac > 0 Means the Quadratic Has 2 Distinct Roots If b2 – 4ac > 0 (if b2 – 4ac is positive), then the equation ax2 + bx + c = 0 has two distinct real solutions, and the function y = ax2 + bx + c has two distinct real roots. In other words, the roots are real and unequal. Example 1 Describe the nature of the solutions of the equation 2x2 + 3x – 2 = 0. Then find the values of the roots, and sketch the graph of y = 2x2 + 3x – 2. Solution First write down the values of a, b, and c: 2x2 + 3x – 2 = 0 so a = 2, b = 3, and c = –2 So the discriminant is: b2 – 4ac = 32 –
[4 × 2 × (–2)] = 9 – (–16) = 9 + 16 = 25 Since b2 – 4ac is positive, the equation 2x2 + 3x – 2 = 0 has two distinct (unequal) real solutions. This in turn means that the function y = 2x2 + 3x – 2 has two real roots — its graph crosses the x-axis in two places. To work out the actual values of the roots, use the quadratic formula: 4 ac −2 b 2 a 25 − ± 2 or 1 2 So the graph of y = 2x2 + 3x – 2 meets the x-axis in two places: (–2, 0) and ( 1 2, 0). To sketch the graph, you also need the y-intercept: y = 2x2 + 3x – 2 = 2(0)2 + 3(0) – 2 = –2 y 5 4 3 2 1 x -4 -3 -2 -1 -1 1 2 3 -2 -3 -4 Section 7.5 — The Discriminant 373 Check it out: There are two distinct (unequal), real roots — so there are two distinct x-intercepts on the graph. Example 2 Find the nature of the solutions of the quadratic equation x2 + 5x + 2 = 0. Solution Again, the best thing to do first is write down the values of a, b, and c: x2 + 5x + 2 = 0 so a = 1, b = 5, and c = 2 Then the discriminant is: b2 – 4ac = 52 – (4 × 1 × 2) = 25 – 8 = 17 The discriminant is positive, so there are two unequal, real solutions of x2 + 5x + 2 = 0. You could use the quadratic formula to find the actual solutions: x = − ± 5 17 2 = –5 + 17 2 or –5 – 17 2 So the x-intercepts of the graph of y = x2 + 5x + 2 are at: ⎛ ⎜⎜⎜ ⎝ –5+ 17 2 ⎞ ⎟⎟⎟ and ⎠ ⎛ ⎜⎜⎜ ⎝ 0, –5– 17 2 ⎞ ⎟⎟⎟ ⎠ 0, Guided
Practice Describe the nature of the solutions of each quadratic equation, and find the values of the solutions. 1. x2 + x – 12 = 0 3. x2 + 5x + 4 = 0 5. 2x2 + 5x + 2 = 0 7. 3x2 + 7x – 6 = 0 9. 2(5x2 + 1) = 9x 11. x(10x + 7) = –1 2. x2 + 2x – 3 = 0 4. 3x2 – 7x + 4 = 0 6. x2 + 3x – 1 = 0 8. 2x2 + 9x = 5 10. 6(2x2 + 1) + 17x = 0 12. 2x(4x + 7) = –3 Use the x- and y–intercepts of the quadratics below to decide in which quadrant the vertex of each equation would be. 13. y = x2 + x – 12 15. y = –x2 + 2x + 3 14. y = x2 + 2x – 3 16. y = 3x2 – 12x – 15 374 Section 7.5 — The Discriminant b2 – 4ac = 0 Means the Quadratic Has 1 Double Root If b2 – 4ac = 0, then the solutions of ax2 + bx + c = 0 (and the roots of y = ax2 + bx + c) are real and equal. Graphically, equal roots (also called a double root) mean the graph of the quadratic function just touches the x-axis, but does not cross it. Example 3 Determine the nature, and find the value(s), of the solution(s) of the equation x2 – 6x + 9 = 0. Sketch the graph of y = x2 – 6x + 9. Solution First write down the values of a, b, and c: x2 – 6x + 9 = 0 so a = 1, b = –6, and c = 9 So the discriminant is: b2 – 4ac = (–6)2 – 4 × 1 × 9 = 36 – 36 = 0 This time, the discriminant is zero, so y = x2 – 6x + 9 has a double root. In other words, its graph just touches the x-axis without actually crossing it. As always, to work out where this double root is, use the quad
ratic formula: − 4 ac 1 1 2 3 4 5 -1 x So the graph of y = x2 – 6x + 9 touches the x-axis at (3, 0). Guided Practice Describe the nature of the solutions of the quadratic equations, and find the value(s) of the solution(s). 17. x2 + 4x + 4 = 0 19. x2 + 6x = –9 21. 5x(5x – 6) = –9 23. 4x2 = –5(4x + 5) 25. 8x(x + 3) + 18 = 0 18. 9x2 – 6x + 1 = 0 20. 3x(x – 10) + 75 = 0 22. 8x(2x + 3) = –9 24. 4x(x + 1) + 1 = 0 26. 24x(2x + 9) + 243 = 0 Section 7.5 — The Discriminant 375 Check it out: b2 – 4ac = 0, so the quadratic formula simplifies to −b a2. b2 – 4ac < 0 Means the Quadratic Has No Real Roots If b2 – 4ac < 0 (if b2 – 4ac is negative), then ax2 + bx + c = 0 has no real solutions, and y = ax2 + bx + c has no real roots. If there are no real roots, then there are no x-intercepts — the graph of the quadratic does not intersect (cross or touch) the x-axis. Example 4 Describe the graphs of: (a) y = x2 + 2x + 3, and (b) y = –2x2 + 4x – 5 Solution (a) Here a = 1, b = 2, and c = 3, so: b2 – 4ac = 22 – 4 × 1 × 3 = 4 – 12 = –8 So the graph of y = x2 + 2x + 3 never intersects the x-axis. Since a > 0, the graph is concave up (u-shaped) and stays above the x-axis. -5 -4 -3 -2 (b) Here a = –2, b = 4, and c = –5, so: b2 – 4ac = 42 – 4 × (–2) × (–5) = 16 – 40 = –24 y
(ab) = –2 ² + 4 – 5 y x -1 6 5 4 3 2 1 -1 -2 -3 -4 -5 -6 -7 So the graph of y = –2x2 + 4x – 5 never intersects the x-axis either. But this time, since a < 0, the graph is concave down (n-shaped) and stays below the x-axis. Guided Practice Use the discriminant to verify that there are no real number solutions for the quadratic equations below. 27. x2 + 3x + 4 = 0 29. 3x2 + 8 = 9x 31. 2x(2x + 3) + 3 = 0 33. 3x2 + 5x + 3 = 0 28. 5x2 – 4x + 3 = 0 30. 2x2 = 7(x – 1) 32. 3x2 + 1 = 3x 34. 2x(x – 3) = –5 376 Section 7.5 — The Discriminant Check it out: If k = 0, the x2 term would disappear, leaving a linear equation. The graph crosses the x-axis in two places, so the equation can’t be linear. Using the Discriminant Example 5 Suppose the graph of y = kx2 + x – 6 intersects the x-axis at two distinct points (where k is some constant). What are the possible values of k? Solution Since the graph of y = kx2 + x – 6 has two distinct x-intercepts, the discriminant (b2 – 4ac) must be positive. So write down your a, b, and c: a = k, b = 1, c = –6 Now use the fact that the discriminant is positive: b2 – 4ac > 0 12 – 4k(–6) > 0 1 + 24k > 0 24k > –1 k > – 1 24 So k can be any real number greater than – 1 24, but not zero: – 1 24 < k < 0 or k > 0. Example 6 At how many points does the graph of y = x2 – 2x + 3 intersect the x-axis? Solution Here, a = 1, b = –2, c = 3 So b2 – 4ac = (–2)2 – 4 × 1 × 3 = 4 – 12 = –8 The discriminant is negative, so y = x2
– 2x + 3 = 0 has no real roots, and so the graph of y = x2 – 2x + 3 does not intersect the x-axis. Example 7 Find the values of k for which y = 5x2 – 3x + k has a double root. Solution As always, it’s a good idea to begin by writing down your a, b, and c: a = 5, b = –3, c = k So b2 – 4ac = (–3)2 – 4 × 5 × k = 9 – 20k The quadratic has a double root where the discriminant equals zero. This is the case when 9 – 20k = 0, or when k = 9 20. So y = 5x2 – 3x + k has a double root only when k = 9 20. Section 7.5 — The Discriminant 377 Example 8 Describe the nature of the roots of y = x2 + 2 x + 1 2. Solution This time, a = 1, b = 2, and c = 1 2, so b2 – 4ac = ( 2 ) The discriminant is zero, so the function has a double root. This root is at x − ± b = − 4 ac Independent Practice Determine the number of roots of the following functions, and find the values of any real roots. 1. y = x2 – 2x – 3 3. y = x2 – 3x – 1 5. y = 4x2 – 4x + 1 7. y = 2x2 – 4x + 3 9. y = 2x2 + x – 6 11. y = x2 – x + 5 2. y = x2 + 4x + 3 4. y = x2 + x – 1 6. y = x2 – 8x + 16 8. y = x2 + x + 3 10. y = 4x2 – 12x + 9 12. y = x2 – 2x – 2 13. Find the possible values of k if y = 3x2 – kx + 3 has a double root. 14. Find the possible values of p if 2x2 – 5x + p = 0 has two real solutions. 15. If x = –1 is a root of y = 3x2 + x – k, find k and the other root. 16. If x = 1 2 is a solution to kx2 + 9x – 5 = 0, find
both solutions. 17. Find the possible values of p if y = px2 – 7x – 7 has no real roots. State the number of times that the graphs of the following quadratic functions intercept the x-axis: 18. y = x2 – 3x – 28 20. y = 4x2 + 2x + 1 22. y = 5x2 + 3x + 1 19. y = 4x2 + 4x + 1 21. y = 2x2 – x – 1 23. Find all possible values of k when y = 3x² – 2kx + k has a double root. 24. When y = 3x² – 2kx + k has a root of x = 2, find k and the other root. Round Up Round Up Remember — the discriminant is just one part of the quadratic formula. If all you need to know is how many roots a function has, you don’t need to use the full formula — just the discriminant. 378 Section 7.5 — The Discriminant Topic 7.6.1 Section 7.6 Motion Tasks Motion Tasks California Standards: 23.0: Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. What it means for you: You’ll model objects under the force of gravity using quadratic equations, and then solve the equations. Key words: quadratic gravity vertex parabola intercept completing the square Quadratic equations have applications in real life. In particular, you can use them to model objects that are dropped or thrown up into the air. Quadratic Functions Describe Motion Under Gravity Example 1 The height of a stone thrown up in the air is modeled by the equation h = 80t – 16t 2, where t represents the time in seconds since the stone was thrown and h is the height of the stone in feet. After how many seconds is the stone at a height of 96 feet? Explain your answer. Solution The stone reaching a height of 96 feet is represented by h = 96, so you need to solve 80t – 16t 2 = 96. Rewriting this in the form ax2 + bx + c = 0 (using t instead of x) gives: 16t2 – 80t + 96 = 0 t2 – 5t + 6 = 0 Divide through by 16 (t – 2)(t – 3) = 0 Factor the quadratic
equation t – 2 = 0 or t – 3 = 0 Solve using the zero property t = 2 or t = 3 So the stone is at a height of 96 feet after 2 seconds (on the way up), and again after 3 seconds (on the way down). Example 2 Use the same information from Example 1. After how many seconds does the stone hit the ground? Explain your answer. Solution When the stone hits the ground, h = 0. So solve 80t – 16t 2 = 0. 5t – t 2 = 0 t(5 – t) = 0 t = 0 or t = 5 Divide through by 16 Solve using the zero property But t = 0 represents when the stone was thrown, so the stone must land at t = 5 — after 5 seconds. Section 7.6 — Motion Tasks and Other Applications 379 Example 3 Use the same information from Example 1. Calculate h at t = 7. Explain your answer. Solution At t = 7, h = (80 × 7) – (16 × 72) = 560 – 784 = –224 Negative values of h suggest that the stone is beneath ground level. This can’t be true — the height can’t be less than zero feet. But the stone landed after 5 seconds — so after t = 5, the function h = 80t – 16t 2 doesn’t describe the motion of the stone. Example 4 Use the same information from Example 1. What is the maximum height of the stone? Justify your answer. Solution The graph of h = 80t – 16t 2 is a parabola, and the maximum height reached is represented by the vertex of the parabola, which you can find by completing the square. Check it out: The formula h = 80t – 16t2 only describes the stone’s motion for a certain range of t. It can’t describe what happens before the stone is thrown or after it lands. Don’t forget: See Section 7.4 for more about plotting quadratic graphs. h = 80t – 16t2 = –16(t2 – 5t) ⎡ = –16 t −( ⎢ ⎢ ⎣ 5 2 2 ) − ⎤ ⎥ ⎥ ⎦ 25 4 = –16 t −( 2 )5 2 + 100 y 100 90 80 70 60 50 40 30 20 10 The stone reaches its maximum height after half its flight
time — at t = 2.5 s. x 1 2 3 4 5 So the maximum height is 100 feet (which is reached at t = 5 2 = 2.5 s). Guided Practice 1. In a Physics experiment, a ball is thrown into the air from an initial height of 24 meters. Its height h (in meters) at any time t (in seconds) is given by h = –5t2 + 10t + 24. Find the maximum height of the ball and the time t at which it will hit the ground. 2. A firework is propelled into the air from the ground. Its height after t seconds is modeled by h = 96t – 16t2. The firework needs to explode at a height of 128 feet from the ground. After how long will it first reach this height? If the firework fails to explode, when will it hit the ground? 380 Section 7.6 — Motion Tasks and Other Applications Plotting a Graph Makes Problems Easier to Understand Example 5 The height above the ground in feet (h) of a ball after t seconds is given by the quadratic function h = –16t2 + 32t + 48. Explain what the h-intercept and t-intercepts mean in this situation, and find the maximum height reached by the ball. Check it out: From the graph, you can see that the ball reaches its maximum height before half its flight time has passed. Solution To get a clearer picture of what everything means, it helps to draw a graph. The intercept on the vertical axis (the h-axis) is found by putting t = 0: h = 48. h 80 70 60 50 40 30 20 10 The intercepts on the horizontal axis (the t-axis) are found by solving h = 0 — that is, –16t2 + 32t + 48 = 0. –1 0 –2 1 2 3 t t2 – 2t – 3 = 0 (t – 3)(t + 1) = 0 t = 3 or t = –1 In this situation, the intercept on the vertical axis (the h-intercept) represents the initial height of the ball when it was thrown (at t = 0). So here, the ball was thrown from 48 feet above the ground. The intercepts on the horizontal axis (the t-intercepts) represent the times at which the ball was at ground level. However, the function only describes the motion of the ball between t =
0 (when it was thrown) and t = 3 (when it lands). So the t-intercept at t = 3 represents the point when the ball lands. The t-intercept at t = –1 doesn’t have any real-life significance here. To find the maximum height, you need to find the vertex of the parabola — so complete the square: –16t2 + 32t + 48 = –16[t2 – 2t – 3] = –16[(t – 1)2 – 4] = –16(t – 1)2 + 64 The vertex of this parabola occurs where t = 1, and so the vertex is at (1, 64). This means the maximum height of the ball is 64 feet. Section 7.6 — Motion Tasks and Other Applications 381 Check it out: Put t = 1 into the quadratic to find the maximum value. Independent Practice A baseball is hit from homebase. Its height in meters is modeled by the equation h = 25t – 5t2, where t is the time in seconds. 1. After how many seconds will the ball be at a height of 20 meters? 2. What height will the baseball reach before it starts descending? A rocket is fired into the air. Its height in feet at any time is given by the equation h = 1600t – 16t2, where t is the time in seconds. 3. Find the height of the rocket after 2 seconds. 4. After how many seconds will the rocket be 30,000 feet above the ground? 5. After how many seconds will the rocket hit the ground? 6. As a skydiver steps out of a plane, she drops her watch. The distance in feet, h, that the watch has fallen after t seconds is given by the equation h = 16t2 + 4t. After how many seconds will the watch have fallen 600 feet? The height in feet of an object projected upwards is modeled by the equation h = 100t – 16t2. 7. How long after being projected is the object 100 feet above the ground? 8. What is the greatest height reached by the object? The area of a rectangle is given by the formula A = x(20 – x) cm, where x is the width. 9. Find x when A = 84 cm2. 10. What value of x maximizes the area A? James and Mei are each standing on diving boards, and each throw
a ball directly upwards. The height of each ball above the pool in feet, h, is plotted against the time in seconds, t, since it was thrown. 11. The height of James’s ball can be calculated using the equation h = –16t² + 30t + 10. From what height above the pool does James throw his ball? 12. The height of Mei’s ball can be calculated using the equation h = –16t² + 32t + 20. After how many seconds does her ball reach its maximum height? 13. Calculate the difference in maximum heights of the balls, to 1 decimal place. Round Up Round Up Plotting a graph is often a good idea when you’re working on motion problems — because then you can see at a glance when an object reaches the ground, or when it reaches its maximum height. 382 Section 7.6 — Motion Tasks and Other Applications Topic 7.6.2 California Standards: 23.0: Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. What it means for you: You’ll model money problems using quadratic equations, and then solve the equations. Key words: economic profit quadratic vertex parabola completing the square Economic Tasks Economic Tasks As well as the motion tasks you saw in Topic 7.6.1, you can use quadratic equations to model real-life problems involving money. Applications of Quadratics to Economics The best way to introduce quadratic equations modeling money problems is to show you an example: Example 1 The owner of a restaurant wishes to graph the annual profit of his restaurant against the number of people he employs. He calculates that the annual profit in thousands of dollars (P) can be modeled by the formula P = –0.3x2 + 4.5x, where x is the number of people employed. According to the owner’s formula, how many full-time members of staff does the restaurant have to employ to make a profit of $15,000? Solution You have a formula for the profit P, and you have to find when this equals 15 (since the formula gives you the profit in thousands of dollars). So you need to solve the quadratic equation –0.3x2 + 4.5x = 15. Rewriting this in the form ax2 + bx + c = 0 gives: 0.3x2 – 4.
5x + 15 = 0 x2 – 15x + 50 = 0 (x – 10)(x – 5) = 0 x = 10 or x = 5 Divide through by 0.3 Solve using the zero property This means that the restaurant can employ either 5 people or 10 people and make a profit of $15,000. Section 7.6 — Motion Tasks and Other Applications 383 Example 2 Use the same information from Example 1. According to the owner’s formula, how many full-time members of staff should the restaurant employ to make maximum profit? Solution To find the maximum profit, you need to find the maximum value of the quadratic P = –0.3x2 + 4.5x. To do this, you can complete the square: P = –0.3x2 + 4.5x = –0.3(x2 – 15x ⎡ ⎛ ⎢ ⎜⎜⎜ ⎢ ⎝ ⎣ ⎛ ⎜⎜ x⎜⎜ ⎝ x − 15 2 2 ⎞ ⎟⎟⎟ ⎠ − 15 2 2 ⎞ ⎟⎟⎟ ⎠ 2 ⎞ ⎟⎟⎟ ⎠ ⎤ ⎥ ⎥ ⎦ − 15 2 2 ⎞ ⎛ ⎟⎟⎟ ⎜⎜⎜ ⎠ ⎝ + (. 0 3 + ( 3 10 225 4 ) ) 225 4 = −. 0 3 ⎛ ⎜⎜⎜ ⎝ x − 15 2 = −. 0 3 ⎛ ⎜⎜⎜ ⎝ − x 2 ⎞ ⎟⎟⎟ ⎠ 15 2 + 135 8 So the vertex of the parabola is at ( 15 2, 135 8 ), which (in theory) means that the restaurant should employ 7.5 people to make the maximum possible profit of $16,875. Clearly, the restaurant can’t employ 7.5 people — a good idea now is to draw the graph so that you can answer this question more realistically. Find the x-intercepts by solving P = 0: P = –0.3x2
+ 4.5x = –0.3(x2 – 15x) = –0.3x(x – 15) = 0 at x = 0 and x = 15 So the graph looks like this 20 P (7.5, 16.875) 15 10 10 11 12 13 14 15 16 Number of employees x 384 Section 7.6 — Motion Tasks and Other Applications Check it out: Work out the profit like this: (–0.3 × 72) + (4.5 × 7) = (–0.3 × 82) + (4.5 × 8) = 16.8 Example 2 continued You can see from the symmetry of the graph (the line of symmetry is x = 7.5) that the maximum possible profit while employing a whole number of people is at x = 7 and x = 8, at which points the profit is $16,800. So, if the restaurant employs more than 8 people, profits decrease, possibly because there is not enough work for more than 8 people to do efficiently. Guided Practice 1. The profit p in cents per 10-minute period earned from driving a taxicab is given by p = 80x – 3x2, where x is the speed in mph. What speed would yield a profit of 512 cents per 10-minute period? 2. An investor kept track of her portfolio profit, P, at time, t, measured in years after she began investing. If P = 4000t2 – 28000t + 3000 represents her profit, after how many years will she have made $150,000 profit? 3. The amount of money a customer is willing to spend at a store is related to t, the number of minutes they have to wait before being served. If M = –t2 + 8t + 17 represents the money a customer spends, how long will it take before a customer decides to leave the store without spending any money? Independent Practice Leo produces x pounds of salsa. The ingredients cost 0.1x2 – 30 dollars and he makes 2x dollars revenue from the sale of his salsa. 1. What is Leo’s maximum possible profit? 2. How many pounds of salsa would Leo need to sell to break even? The value in dollars, V, of a certain stock can be modeled by the equation V = –16t2 + 88t + 101, where t represents the time in months. 3. What was the original value of the stock? 4. What was the
maximum value of the stock? 5. When did the stock reach the maximum value? 6. When did the stock become worthless? The value, V, of Juan’s investment portfolio can be modeled by the equation V = 16t2 – 256t + 16,000, where t is the time in months. 7. What was the original value of Juan’s portfolio? 8. What was the minimum value of Juan’s portfolio? 9. When will Juan’s investment portfolio be worth $16,576.00? Round Up Round Up Usually when you graph quadratic problems involving money, the vertex of the graph shows you the point where there’s maximum profit. Section 7.6 — Motion Tasks and Other Applications 385 Chapter 7 Investigation he Handshake Pre Pre Pre Pre Proboboboboblemlemlemlemlem he Handshak he Handshak TTTTThe Handshak he Handshak TTTTThe Handshak he Handshake Pre Pre Pre Pre Proboboboboblemlemlemlemlem he Handshak he Handshak he Handshak This Investigation shows that Math techniques can be useful everywhere — even at parties. There are 45 people at a party. If each person shakes hands with every other person, how many handshakes will there be altogether? Suppose there are n people at a party. How many handshakes will there be in terms of n? Things to think about: • How many handshakes does each person make? How many handshakes would there be if there were smaller groups of people? Are there any patterns? Extension 1) The triangular numbers form the following sequence: 1, 3, 6, 10, 15, 21,... If the first triangular number is 1, what will the nth triangular number be? 2) There are n people at a round table. Each person shakes hands with everyone else at the table, except for the person on their left and the person on their right. How many handshakes will there be in total? 3) Three diagonals of a hexagon are shown on the right. (The sides of the shape are not diagonals.) How many diagonals will a regular polygon with n sides have? Explain your answer. Open-ended Extension 1) At a high school reunion everyone shakes hands with everyone else once. Towards the end of the evening, a math teacher arrives and shakes hands with just the students that he actually taught. There
were 3107 handshakes in total. How many people were at the reunion before the math teacher arrived and how many of the people present had the math teacher taught? Is your answer the only one possible? Explain your reasoning. 2) Look back at the original handshake problem. Change the problem in some way and investigate the effect on the number of handshakes. For example, you could investigate the number of handshakes if each person shakes both the left and right hands of each other person with their own left and right hand. How many handshakes would there be for groups of extraterrestrials with h hands? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up In this Investigation you’ve recorded results, identified patterns and applied what you’ve learned to new situations. There’s a lot you can say about even quite simple-looking math problems. estigaaaaationtiontiontiontion — The Handshake Problem estigestig estig pter 7 Invvvvvestig pter 7 In 386386386386386 ChaChaChaChaChapter 7 In pter 7 In pter 7 In Chapter 8 Rational Expressions and Functions Section 8.1 Rational Expressions.......................................... 388 Section 8.2 Multiplying and Dividing Rational Expressions.... 394 Section 8.3 Adding and Subtracting Rational Expressions.... 403 Section 8.4 Solving Equations with Fractions........................ 412 Section 8.5 Relations and Functions...................................... 420 Investigation Transforming Functions....................................... 435 387387387387387 Topic 8.1.1 California Standards: 12.0: Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms. What it means for you: You’ll find out about the conditions for rational numbers to be defined. Key words: rational numerator denominator undefined Don’t forget: Take another look at Chapter 1 to remind yourself about rational numbers. Check it out: p is called the numerator and q is called the denominator. Section 8.1 Fractions and Fractions and Rational Expressions Rational Expressions In this Topic you’ll find out about the necessary conditions for rational numbers to be defined. Rational Expressions Can Be Written as Fractions A rational expression is any expression that can be written in
the form of a fraction — that means it has a numerator and a denominator. Examples of rational expressions are. Rational expressions are written in the form p q, where q π 0. An Expression is Undefined if the Denominator is Zero If the denominator is equal to zero, then the expression is said to be undefined (see Topic 1.3.4). So, for example, x 2 x − 1 + 1 Example 1 is defined whenever x is not equal to –0.5. Determine the value of x for which the expression 7 x + is undefined. 2 Solution It is undefined when the denominator x + 2 equals zero. This means that the expression is undefined when x = –2. Example 2 Determine the value(s) of x for which the expression 2 2x − is undefined. 4 Solution It’s undefined when the denominator x2 – 4 equals zero. So, solve x2 – 4 = 0 to find the values of x: x2 – 4 = (x – 2)(x + 2) = 0 x – 2 = 0 or x + 2 = 0 ﬁ x = 2 or x = –2 Therefore, 388 Section 8.1 — Rational Expressions 2 2x − is undefined when x = ±2. 4 Example 3 Determine the value(s) of x for which the expression x 7 8 x− + is undefined. 15 2 x Solution Factor the denominator to give: x 7 )( 3 x − ) 5 − ( x If the denominator equals zero, the expression is undefined. This happens when either (x – 3) or (x – 5) equals zero. So, the expression is undefined when x = 3 or x = 5. Guided Practice Determine the value(s) of the variables that make the following rational expressions undefined. − 3 2 + 4 8 3 1y − + − 2. x x 3. 1. 4 x x 5 4 12 5. − 2 11 − 3 y 4 2y + 6. k 4 + 26 11 k 5 − 2 k 1 30 + 7 28 8 45 9. 3 a 3 Independent Practice Determine the value(s) of the variables that make the following rational expressions undefined. x 1. 23. 2 m 4 − 18 k 5. 3 k 3 6 13 + 2 x 5 30 x 32 7. Jane states that the rational expression 3 x + 3 x − − is defined 2 1 x 1 2 when x
is any real number. Show that Jane is incorrect. Round Up Round Up This Topic about the limitations on rational numbers will help you when you’re dealing with fractions in later Topics. In Topic 8.1.2 you’ll simplify rational expressions to their lowest terms. Section 8.1 — Rational Expressions 389 Topic 8.1.2 California Standards: 12.0: Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms. What it means for you: You’ll learn about equivalent fractions and how to simplify fractions to their lowest terms. Key words: equivalent rational simplify common factor Check it out: The two shapes on the far right represent x being divided up into the fractions 5 and 10 6 12. Equivalent Fractions Equivalent Fractions Saying that two rational expressions are equivalent is just a way of saying that two fractions represent the same thing. Equivalent Fractions Have the Same Value A ratio is a comparison of two numbers, often expressed by a fraction — for example, a b. A proportion is an equality between two ratios. Four quantities a, b, c, and d are in proportion if a b = c d. Fractions like these that represent the same rational number or expression are often called equivalent fractions 10 12 You can determine whether two fractions are equivalent by using this rule: The rational expressions a b and c d are equivalent if ad = bc. Example 1 Prove that x 5 6 and x 10 12 are equivalent. Solution 5x·12 = 60x and 6·10x = 60x This is ad in the rule above This is bc in the rule above So, the two rational expressions are equivalent. Guided Practice Prove that the following pairs of rational expressions are equivalent. 1. m 54 6 and m 18 2 2. 1 3 and 2 6 x x 3. 12 − x 3 9 and 4 − x 3 390 Section 8.1 — Rational Expressions Check it out: All of the common factors have been canceled, leaving the simplest equivalent fraction. Don’t forget: The GCF of two numbers is the largest number that is a factor of both of them. is the simplest form of 56 64x Check it out: 7 8x because 1 is the only common factor of the numerator and denominator. Simplify Fractions by Canceling Common Factors A rational expression can be written in its lowest terms by reducing it to the simplest equivalent fraction. This is done by
factoring both the numerator and denominator and then canceling the common factors — that means making sure its numerator and denominator have no common factors other than 1. For example: 66 78 1 ⋅ 6 11 = ⋅ 1 6 13 = 11 13 Example 2 Reduce the expression 56 64x to its lowest terms. Solution The greatest common factor (GCF) of 56 and 64 is 8. This means that: 56 x 64 = So, 56 64x and 7 8x are equivalent fractions. Numbers are not the only things that can be canceled — variables can be canceled too. For example: mc cv = ⋅ Guided Practice Reduce each of the following rational expressions to their lowest terms. 4. 21 28 7. d 10 30 5. 12 18 8. − 4 10 2 m c 3 2 m c 6. bx b x2 9 )( 5 5 )( − + ) m m ) 10. 11. + + 5 15 x x 3 12. 2 ( m m + )( Section 8.1 — Rational Expressions 391 Check it out: The greatest common factor of (x + 3) and 6 is 1. So, x + 3 is the simplest form. 6 Check it out: Note that m – 3 = –1(3 – m) so you can substitute this into the expression. Some Harder Examples to Think About Factoring the numerator and denominator is the key to doing this type of question. Breaking down a complicated expression into its factors means you can spot the terms that will cancel. Example 3 Simplify the expression 2 x 6 x − 9 −. 18 Solution Factor the numerator and denominator, then cancel common factors: 2 x 6 x 1 − 9 − = 18 ( x − + 3 )( Example 4 Simplify the expression Solution Factor the numerator and denominator completely )( m m 3 )( 1 − m 3 3 )( ( − ⋅ − m m 1 3 )( ( )) = − ) Cancel the common factor (3 – m) Example 5 Reduce this expression to its lowest terms: Solution Factor both the numerator and denominator 10 x 2 − + x 15 21 10 x 2 − + x 15 21 x = = ) 15 ) 21 10 ⋅ − + 1 )( ( x x x 5 + ⋅ + 1 1 )( ( = Cancel the common factors 392 Section 8.1 — Rational Expressions Guided Practice 13. Show how you
can simplify the rational expression Simplify the following rational expressions. 14. 4 2 k − − k 16 15. 2 − 2 m c − 2 mc c 16 20 − 2 3 + k k 26 − − 3 13 k 10 )( m m m )( + 5 ) + 5 ) 17. 18. 2 m 2 k + − 6 k mk + − 2 mk m 2 2 19. Independent Practice 1. Simplify 2 2 k 2 c 2 − + ck 3 ck. Simplify ( ). Reduce each of the following rational expressions to their lowest terms. 2 16 x − xv 2 2 − v − 24 4. 3. 2 v 2 y 3 y 18 − − 2 y 21 y 12 + − 30 48 6. 2 x 4 2 x 10 − − 4 x 24 + + 50 60 x 7. 3 k + 2 2 + 10 k + − 2 k 21 k 35. 2 2 x x + + 5 4 xy xy − − 2 2 14 21 y y 9. 10. a 2 − a 3 12 − − 2 3 a 10 11. 12. Matthew simplified 2 2x 2 − − x + 6 x 21 in this way 13 ab ab )( +. Explain the error that Matthew has made and then simplify the expression correctly. Round Up Round Up If you were to condense everything from this Section into a couple of points, they would be: • Rational expressions are the same as fractions and are undefined if the denominator equals zero. • A rational expression can be reduced to its lowest equivalent fraction by dividing out common factors of the numerator and denominator. Section 8.1 — Rational Expressions 393 TTTTTopicopicopicopicopic 8.2.18.2.1 8.2.18.2.1 8.2.1 California Standards: Students Students 13.0: 13.0: Students add, subtract, 13.0: Students 13.0: Students 13.0: ultiplyyyyy, and divide rrrrraaaaational mmmmmultipl tional tional ultipl ultipl tional tional ultipl eeeeexprxprxprxprxpressions and functions essions and functions..... essions and functions essions and functions essions and functions e both e both Students solv Students solv Students solve both e both Students solv e both Students solv y and y and tionall tionall computa computa y and tionally and computation
all y and tionall computa computa hallenging hallenging ptually cy cy cy cy challenging ptuall ptuall conce conce hallenging conceptuall hallenging conce ptuall conce prprprprproboboboboblems b y using these y using these lems b lems b y using these lems by using these y using these lems b hniques..... hniques hniques tectectectectechniques hniques What it means for you: You’ll multiply rational expressions by factoring and canceling. Key words: rational nonzero common factor Section 8.2 Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying Multiplying essions essions tional Expr tional Expr RRRRRaaaaational Expr essions tional Expressions essions tional Expr RRRRRaaaaational Expr essions essions tional Expr tional Expr tional Expressions essions tional Expr essions In Section 8.1 you learned about simplifying rational expressions. In this Topic you’ll learn to multiply rational expressions, but then you’ll use the same simplification methods to express your solutions in their simplest forms. essions essions tional Expr ying by Ry Ry Ry Ry Raaaaational Expr tional Expr ying b ying b Multipl Multipl essions tional Expressions Multiplying b essions tional Expr ying b Multipl Multipl Given any nonzero expressions m, c, b, and v mb cv m c = That is, the product of two rational expressions is the product of the numerators divided by the product of the denominators. These expressions can often be quite complicated, so simplify them as much as you can before multiplying. First, factor the numerators and denominators (if possible). Then find any factors that are common to both the top line (the numerator) and the bottom line (the denominator) and cancel them before multiplying. Example Example Example Example Example 11111 Simplify ab + 4 b 2 a 5 a ⋅ 10 + 2 a b b. Solution Solution Solution Solution Solution Step 1: Factor the numerators and denominators ) ) ⋅ Step 2: Cancel all the common factors and multiply 2a 394394394394394 Section 8.2 Section 8
.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 Example Example Example Example Example 22222 Multiply and simplify Solution Solution Solution Solution Solution Step 1: Factor the numerators and denominators if possible ( ) 33 2 2 a a )( 3 2 a )( − 2 a a )( ( − ⋅ − 2 1 )( ( − + + − ( ( = = a a ) ⋅ ⋅ Step 2: Cancel all the common factors and multiply − )( ( 2 a a − ⋅ − 1 2 1 )( ( = − 1 2 Guided Practice Multiply and simplify the rational expressions. 1. 14 15 mc ck ⋅ 30 28 kp pt +. 5 a ⋅ ab + 4 b b 10 + 2 a b 5. 2 a − v t p ⋅ tv −5 t 5 v 2. − ( p 2 ( + 2 )( p + 3 ) p 1 ) ⋅ 4. ( p + p + 1 )( 3 + p 4 ) 6 10 + + 11 3 4 You can extend this concept to the multiplication of any number of rational expressions, with any number of variables. Example Example Example Example Example 33333 Simplify 14 16 x − − x + 2 12 Solution Solution Solution Solution Solution Factor the numerators and denominators, and cancel all the common factors and multiply )( − + 1 x ( 2 ( )( −7 2x 2 or )( − x2 7 ) 1 3 )) 1 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 395395395395395 Check it out: Either of these answers is acceptable. The second is the simpler of the two forms though. Example Example Example Example Example 44444 Simplify 12 15 ⋅ ax 2 + + a 8 15 a − + − a 3 4 x ⋅ 12 2 x x 2 + + x 9 18 + +. 6 x 9 Solution Solution Solution Solution Solution Factor the numerators and denominators, then cancel all the common factors and multiply. Check it out: You can leave your answer in either “factored” or “multiplied out” form — they’re both equally valid. = = 1 ( x 2 a ( + )( 3 x − 3 a )( + )
( 3 )( + + )( 6 3 (( ) + ) )( − 3 )( a + 6 ) − 3 ) ax or 18 Guided Practice Multiply and simplify the rational expressions. 7 20 8 12 3 2 x x + + xy 3 + − 2 xy. 2 − x y − − xy ) Independent Practice Multiply and simplify the rational expressions ab + − x + − 2 x 2 + ab 5 + 2 ab 7 + a 6 + a 3 2 3. 5. 2 a 2 b 2 + − ab 2 b + − 3 ab + − ab 6 a − + 3 ab + ak + ak 27 + + 6 x 60 ) 8 ⋅ 6 ) 12 ( ( x x + + )( 1 )(( 15 36 18 20 42 ⋅ 2 x 7. 3 − 2 x + 10 x 9 + 24 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Usually the most difficult thing when solving problems like these is factoring the numerators and denominators. Look for “difference of two squares” expressions, perfect squares, and minus signs that you can factor outside the parentheses. 396396396396396 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 TTTTTopicopicopicopicopic 8.2.28.2.2 8.2.28.2.2 8.2.2 California Standards: Students Students 13.0: 13.0: Students add, subtract, 13.0: Students 13.0: Students 13.0: multiply, and dididididivide r vide raaaaational tional tional vide r vide r tional tional vide r eeeeexprxprxprxprxpressions and functions essions and functions..... essions and functions essions and functions essions and functions e both e both Students solv Students solv Students solve both e both Students solv e both Students solv y and y and tionall tionall computa computa y and tionally and computationall y and tionall computa computa ptually cy cy cy cy challenging hallenging hallenging ptuall ptuall conce conce conceptuall hallenging conce hallenging ptuall conce prprprprpro
boboboboblems b y using these y using these lems b lems b y using these lems by using these y using these lems b hniques..... hniques hniques tectectectectechniques hniques What it means for you: You’ll divide rational expressions by factoring and canceling. Key words: rational reciprocal common factor viding viding DiDiDiDiDividing viding viding viding viding DiDiDiDiDividing viding viding essions essions tional Expr tional Expr RRRRRaaaaational Expr essions tional Expressions essions tional Expr RRRRRaaaaational Expr essions essions tional Expr tional Expr tional Expressions essions tional Expr essions Dividing by rational expressions is a lot like multiplying — you just have to do an extra step first. ocal ocal ecipr ecipr y the R y the R ying b ying b viding is the Same as Multipl DiDiDiDiDividing is the Same as Multipl viding is the Same as Multipl ocal eciprocal y the Recipr ying by the R viding is the Same as Multiplying b ocal ecipr y the R ying b viding is the Same as Multipl Given any nonzero expressions m, c, b, and v: m ÷ = ⋅ = c mv cb m c b v v b That is, to divide m c by b v, multiply by the reciprocal of b v. You can extend this concept to the division of any rational expression. Don’t forget: See Topic 1.2.2 for more on reciprocals. Suppose you pick a number such as 10 and divide by, say, 1 2 The question you’re trying to answer is “How many times does 1 2. go into 10?” or “How many halves are in 10?” Division Equivalent to 10 ÷ 1 2 = 20 Check it out: The middle numbers in these two columns are reciprocals of each other. 10 ÷ 10 ÷ 10 ÷ 1 3 1 4 1 n = 30 = 40 10 × 2 = 20 10 × 3 = 30 10 × 4 = 40 = 10n 10 × n = 10n So, 10 divided by a fraction is equivalent to 10 multiplied
by the reciprocal of that fraction. Dividing anything by a rational expression is the same as multiplying by the reciprocal of that expression. So you can always rewrite an expression a ÷ b in the form a (where b is any nonzero expression). ⋅ =1 b a b Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 397397397397397 actorsssss actor actor ou Should Still Cancel Common F YYYYYou Should Still Cancel Common F ou Should Still Cancel Common F ou Should Still Cancel Common Factor actor ou Should Still Cancel Common F Example Example Example Example Example 11111 2 − k 25 2 k ÷ (k + 5). Simplify Solution Solution Solution Solution Solution 2 − k 25 2 k ÷ (k + 5) can be written as 2 − k 25 2 k ÷ k + 5 1 Rewrite the division as a multiplication by the reciprocal of the divisor. = 2 − k 25 2 k 1 k + · 5 Factor as much as you can. − ( k = )( Cancel any common factors between the numerators and denominators. − ( k = 5 )( Check your answer. Multiply your answer by (k + 5): = )( 25 2 k Example Example Example Example Example 22222 Simplify Solution Solution Solution Solution Solution Rewrite the division as a multiplication by the reciprocal of the divisor Factor all numerators and denominators. − − + − = − ⋅ )( 2 2 )( ) Cancel any common factors between the numerators and denominators )( 2 )( 398398398398398 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 ✓ Guided Practice Divide and simplify each expression. 1. 2 2 2 b c d 3 bcd ÷ 2 bcd abc 3 10 ÷ 2 x − − x 4 − 2 25 )( b − 1 ) + ( b 2. 4 Check it out: It makes a difference which order you do the divisions because division is not associative, which means: (a ÷ b) ÷ c π a ÷ (b ÷ c) Check it out: In practice, this means that you can rewrite each division as a multiplication by the reciprocal of the divisor. At Once At Once
essions essions Expr Expr vide Long Strings of vide Long Strings of ou Can Di YYYYYou Can Di ou Can Di At Once essions At Once Expressions vide Long Strings of Expr ou Can Divide Long Strings of At Once essions Expr vide Long Strings of ou Can Di Just like multiplication, you can divide any number of rational expressions at once, but it makes a big difference which order you do things in. If there are no parentheses, you always work through the calculation from left to right, so that Example Example Example Example Example 33333 Simplify. Solution Solution Solution Solution Solution Rewrite each division as a multiplication by the reciprocal of the divisor + − 22 Factor all numerators and denominators )( + )( ( )( 1 x x + 1 ) Cancel any common factors between the numerators and denominators )( + )( + )( Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 399399399399399 Parentheses override this order of operations, so you need to simplify any expressions in parentheses first: ⎛ c ⎜⎜⎜ ÷ ÷ ⎝ d a b ⎞ ⎟⎟⎟⎟ ⎠ Guided Practice − + − 3 k − 2 10 k 2 k Divide and simplify each expression 14 ⎜⎜⎜ ⎝ ) ( 12 14 + − )( + ) 3 ÷ 8 18 x ( x ( 6. 2 7. 8. ÷ 6 − 2 ⎞ ⎟⎟⎟⎟ ⎠ ) 4 ) 1 x + + 2 2 − 36 the Same TimeTimeTimeTimeTime t the Same t the Same vide a vide a y and Di y and Di ou Can Multipl YYYYYou Can Multipl ou Can Multipl vide at the Same y and Divide a ou Can Multiply and Di t the Same vide a y and Di ou Can Multipl Say you have an expression like this to simplify: a b c ÷ × d e f Again, you work from left to right, and anywhere you get a division, multiply by the reciprocal, so Example Example Example Example Example 44444 Simplify 2 p 2 p 2 + − 2 pq q − − 2
pq 3 q 2 × 2 p pq − + − pq 3 pq + 2 q. Solution Solution Solution Solution Solution Rewrite any divisions as multiplications by reciprocals. = 2 p 2 p 2 + − pq 2 q − − 2 pq 3 q 2 × 2 p pq − + pq pq + 2 q Factor all numerators and denominators. − + 2 q p )( 3 q p )( q p )( + )( q ) Cancel any common factors between the numerators and denominators. − p q p ( q p )( + )( q p 3 q p )( − + 400400400400400 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8. )( 1 q ) 1 our Stepspspspsps our Ste ustify All All All All All YYYYYour Ste our Ste ustify ustify ou Can J e Sure e e e e YYYYYou Can J ou Can J e Sur MakMakMakMakMake Sur e Sur ou Can Justify our Ste ustify ou Can J e Sur Example Example Example Example Example 55555 21 ÷ 2 a t + 2 2 a t + − at − at 12 t 14 t 2 = − a 2 +. Justify your work. a 2 1 Show that a 2 2 a Solution Solution Solution Solution Solution ÷ 2 a t + 2 2 a t + − at − at 12 t 14 t 2 t with left-hand side t with left-hand side StarStarStarStarStart with left-hand side t with left-hand side t with left-hand side t vision vision di di Definition of Definition of vision division Definition of di vision di Definition of Definition of opertytytytyty oper oper e pre proper e pre pr Distributiutiutiutiutivvvvve pr Distrib Distrib oper Distrib Distrib opertytytytyty oper oper e pre proper e pre pr Distributiutiutiutiutivvvvve pr Distrib Distrib oper Distrib Distrib CommCommCommCommCommutautautautautatititititivvvvve and e and e and e and e and associa associa associatititititivvvvve pr e pre proper e pre pr oper oper opertiestiestiestiesties ass
ocia associa oper ofofofofof m m m m multiplica ultiplica ultiplica ultiplicationtiontiontiontion ultiplica InInInInInvvvvverererererse and se and se and se and se and identity pr identity pr oper oper opertiestiestiestiesties identity proper identity pr identity pr oper Distrib Distrib Distributiutiutiutiutivvvvve pr e pre proper e pre pr oper oper opertytytytyty Distrib Distrib oper 21 − + + − 221 a 4 − − )( ( a a 1 2 − + )( ( )( 1 − a )( − at 12 14 + − t at + − a 6 + − 2 a ) 77 1 ) − a 1 )( 7 ) +11 a ) )( 1 − − ) 3 ) 3 )( )( ( + − − − )( 1 )( = Guided Practice Simplify these rational expressions. 2 t 10 12 11 14 21 ÷ 12 10 × 15 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 401401401401401 Independent Practice Divide and simplify each expression. 2 2 − k + − m km − + km 3 2 2 m 2. 2 2 k 3 16 − 2 x 4. 6. 2 m ÷ − mv 10 10 11 12 3 ÷ 2 − x 16 + + 12. 2 y 13 Simplify these rational expressions. 14 15. − + 2 v 2 − vw 4 + vw vw + − vw 2 vw 55 4 vw vw − + w 3 w 8 2 2 4 16. 17. m 2 + 2 m n 2 + mn 2 − 3 mn mn 5 mn + − + mn 2 − 3 mn n 2 ÷ ⎛ ⎜⎜⎜⎜ ⎝ mn 5 mn + − mn 3 mn 2 + n − 22 n ÷ 2 − + 2 mn mn mn 3 mn 2 + n − 22 n ⎞ ⎟⎟⎟⎟ ⎠ 2 ÷ − + 2 mn mn 2 n 2 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up
ound Up ound Up It’s really important that you can justify your work step by step, because division of rational expressions can involve lots of calculations that look quite similar. 402402402402402 Section 8.2 Section 8.2 Section 8.2 — Multiplying and Dividing Rational Expressions Section 8.2 Section 8.2 Topic 8.3.1 California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll add and subtract rational numbers with the same denominators. Key words: common denominator rational common factor Section 8.3 Fractions with Fractions with Identical Denominators Identical Denominators In Section 8.2 you dealt with multiplying and dividing rational expressions, so you can probably guess that adding and subtracting is next. First up are fractions with the same denominators, which are the easiest kind to add and subtract. Common Denominators Make Things Easier To add or subtract two rational expressions with the same denominator, just add or subtract the numerators, then put the result over the common denominator. Here’s a simple numerical example Reduce your answer to its simplest form. Add the numerators and divide the answer by the common denominator. Subtract the numerators and divide the answer by the common denominator. In general, given any expressions m, c, and v, where v is nonzero and Example 1 Simplify. Solution Add the numerators and divide by the common denominator Section 8.3 — Adding and Subtracting Rational Expressions 403 Example 2 Simplify, justifying each step. Solution Subtract the numeratorsk t = Guided Practice Simplify each expression. Divide the difference by the common denominator, t Use distributive property to simplify –(2k – 5) Commutative and associative properties of addition Combine like terms + 3 x + 4 x 2. 5 2 − 3 −. 3. 5 Cancel Any Common Factors Example 3 Add and simplify m 5 4 − m+ 7 4 4. Solution Add the numerators and divide by the common denominator = 12 m 4 1 m( 4 3 1 4 = 3m – 1 = − ) 1 Factor the numerator and cancel any common factors 404 Section 8.3 — Adding and Subtracting Rational Expressions Example 4 Subtract and simplify Solution
Subtract the numerators and divide by the common denominator )( Guided Practice Simplify each expression. 7. 3 + + 12 10. 2 4 b 9 Independent Practice Simplify each expression. 1 + + 11 x + + 2 4 x 28 3 2 35 9 x 2 3 + + 18 13 + + 4 17 12 − 12 6. ⎛ ⎜⎜⎜ ⎝ − 1 2 x − 2 x 4 1 ⎛ ⎞ ⎜⎜⎜⎜ ⎟⎟⎟ ⎠ ⎝ ⎞ ⎟⎟⎟⎟⎟ ⎠ ⎛ ⎜⎜⎜⎜ ⎝ 7 ⎞ ⎛ ⎟⎟⎟⎟ ⎜⎜⎜⎜ ⎠ ⎝ ( 23 10 20 + + xx + 23 ⎟⎟⎟⎟20 ⎠ Round Up Round Up When subtracting algebraic fractions, be sure to distribute the minus sign over the whole parenthesis that follows it, changing any plus signs to minus signs and vice versa. And make sure you reduce your answer to its simplest form — look for “difference of two squares” patterns. Section 8.3 — Adding and Subtracting Rational Expressions 405 Topic 8.3.2 California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll add and subtract rational numbers with different denominators. Key words: common denominator rational common factor least common multiple Don’t forget: See Topic 1.3.4 for more on equivalent fractions. Fractions with Fractions with Different Denominators Different Denominators For you to be able to add or subtract two rational expressions, they must have a common denominator. If the two fractions you start with have different denominators, you need to find equivalent fractions with a common denominator before you can add or subtract them. You Need to Find a Common Denominator In general, you can find a common denominator by simply multiplying the denominators together. You then convert each fraction to an equivalent fraction with this common denominator. Example
1 Add 2 3 1 +. 5 Solution Multiply the denominators to get a common denominator (3◊5). Convert each fraction into an equivalent fraction with the common denominator and respective numerators 2◊5 and 1◊3 10 = + 15 3 15 Once you have two fractions with the same denominator, add the numerators and divide by the common denominator. = 13 15 So — given any expressions m, c, b, and v, where c and v are nonzero and b c − = v c mv + bc cv mv − bc cv 406 Section 8.3 — Adding and Subtracting Rational Expressions Example 2 Subtract. Solution Find a common denominator: 4 × 5 = 20 Convert each fraction to an equivalent fraction with 20 as the denominator ) = 20 5 − m 20 and ) 12 = 20 + m 20 Don’t forget: See Topic 1.3.6 for more on the LCM. In Examples 1 and 2, the LCM of the denominators is simply their product. Subtract the equivalent fractions. 20 − m 20 5 − 12 + m 20 20 = ( 20 m + m 20 ) − − ) 5 ( 12 20 20 m − − 5 12 20 − m 20 Distribute the minus sign and simplify 25 8 m− 20 = = This method works well with fairly simple examples like these, but when you have more complicated problems to deal with, it can become a lengthy process. A better alternative is to find the least (or lowest) common multiple — the LCM. Guided Practice Use the method of multiplying denominators to simplify the following. 1. 4 − 16 12 Simplify each expression by first finding the LCM of the denominators. 5. 2 c − − c 5 8 24 + 2 8 c 3 + − c 2 12 2 3 y + 6. 17 10 + y 20 10 2 2 x 9. + + x 15 + 14 10 + 11 5 Section 8.3 — Adding and Subtracting Rational Expressions 407 Finding the Least Common Multiple (LCM) The least common multiple of two or more denominators is the smallest possible number (or simplest expression) that’s divisible by all denominators. To find the LCM: Factor the denominators completely. Multiply together the highest power of each factor that appears in any of the denominators. Any factor that appears in more than one denominator should only be
included once in the LCM. Using the LCM as your common denominator makes the problem as simple as possible. Adding or Subtracting Fractions with Different Denominators When adding or subtracting fractions: 1) Find the least common multiple of the denominators. 2) Convert each fraction into an equivalent fraction with the LCM as its denominator. 3) Add or subtract the numerators and divide by the common denominator (LCM). 4) Factor both the numerator and denominator if possible, and cancel any common factors. Check it out: The highest power of x is x2. The highest power of y is y. The highest power of (x + 1) is (x + 1)2. So the LCM is x2y(x + 1)2. Example 3 Simplify − Solution Step 1: Find the LCM of x2y(x + 1) and x(x + 1)2: LCM = x2y(x + 1)2 Step 2: Convert each fraction to an equivalent fraction with x2y(x + 1)2 as the denominator )( )( ( xy ⋅ ( xy ( xy 408 Section 8.3 — Adding and Subtracting Rational Expressions Example 3 continued Step 3: Add the equivalent fractions. − y x 2 )( + + + xy y x )( 1 ) )+ 21 2 ( xx y x xy = xy − − + + y 2 ( x y x 2 + xy 2 ) 1 + ( 5 x ) Guided Practice Simplify each expression. 10. 4 − 6 2 k + 12 3 − 4 k 12 ) ( 11. 13 Add Lots of Fractions Using a Common Denominator If you want to add or subtract more than two fractions at once, you have to put all the numerators over a common denominator. Example 4 Simplify Solution Step 1: Find the LCM of 2x + 6, 4x – 4, and x2 + 2x – 3. Factor each denominator: 2x + 6 = 2 · (x + 3) 4x – 4 = 4 · (x – 1) = 22 · (x – 1) x2 + 2x – 3 = (x – 1)(x + 3) LCM = 4(x – 1)(x + 3) Section 8.3 — Adding and Subtracting Rational Expressions 409 Check it out: The
highest power of 2 is 22 = 4. The highest power of (x – 1) is (x – 1). Highest power of (x + 3) is (x + 3). So the LCM is 4(x – 1)(x + 3). Example 4 continued Step 2: Convert each fraction to an equivalent fraction with 4(x – 1)(x + 3) as the denominator − ( 10 x − )( − )( )( )( + x 1 ) + = + 20 x ( x− x )( 1 1 ) + ( 44 ) 3 ) 3 Step 3: Add or subtract the equivalent fractions. − ) 3 ( 4 + x ( 20 − x 1 )( ) 1 + ) 3 x == = = = − 20 + + x ( 4 − 4 x 1 )( ( + − ( 10 − ) 1 x ( 10 + − 1 x x )( − + ) 1 ( 4 x ) 3 10 ( 4 ( 4 x x 12 x 20 ) 3 4 x ( − x )( 1 20 ( ) 3 − + + − x 4 10 + x −− ( 4 )( 1 − − x 18 6 + − x x 1 )( − + x ( 6 − )( − x )( 1 ( 22 ) Guided Practice Simplify each expression. 14 16 15 17 18 Check it out: Simplify your answer as much as possible — factor the numerator and denominator and cancel any common factors. 19 410 Section 8.3 — Adding and Subtracting Rational Expressions Independent Practice Simplify each expression. 1 16 + − x 4 5 2 a 3 10 − 11 − 10 ab 5 2y 6. − + xy x − y 1 − + 1 xy 2 y 7 −( 1 ) t 2 9. 2 t 4 x + x 5 15 − − + 1 3 x x 8. 10. 2 y 3 + + 7 y 12 − 2 + − 3 y 4 2 y 11 12 13 14 15. y 2 + 2 x xy − 2 3 y xy + 16. 2 k 17 18. 2 x 19 20 21 22 Round Up Round Up Remember — you can only add or subtract fractions when they’ve got the same denominators. That means that the first step when you’re dealing with fractions with different denominators is to find a common denominator. Then you can use the same methods you saw in Topic 8.3.1. Section 8.3 —
Adding and Subtracting Rational Expressions 411 Topic 8.4.1 Section 8.4 Solving Fractional Equations Solving Fractional Equations California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll solve equations involving rational numbers. Key words: common denominator rational common factor limitation undefined least common multiple This Topic uses all the skills that you’ve learned in the earlier Topics in this Section — but now you’ll be dealing with equations rather than just expressions. There are Limitations on Fraction Equation Solutions A fractional equation is undefined when the denominator of any of its expressions equals zero. This means that there are limitations on the possible solutions of a fractional equation. For example, for the fractional equation 1 − x 3 1 = 2 + the limitations 1 x on x are that 3x – 1 π 0 and x + 1 π 0, that is x πππππ 1 3 and x π π π π π –1. Guided Practice State the limitations on the possible solutions of these equations. 1. 5 − − x 6 2x = 1 x 2 13. 2 x Solve Using the LCM of the Denominators When solving fractional equations where the rational expressions have different denominators, multiply the entire equation by the least common multiple (LCM) of the denominators. This allows all the denominators to be canceled out, which makes the equation easier to solve. Example 1 Solve the fractional equation + Solution To solve for x, find the least common multiple of the denominators. In this example, the LCM of the denominators is simply the denominators multiplied together, i.e. 8(3x – 1)(x + 1). 412 Section 8.4 — Solving Equations with Fractions Example 1 continued Multiply the entire equation by the LCM of the denominators, then divide out the common factors, which gives − )( )( − )( 1 1 + 1 ) 1 x Multiplying by the LCM of the denominators means that all the denominators in the equation cancel, leaving: 3(3x – 1)(x + 1) + 8(x + 1) = 2·8(3x – 1) (9x2 + 6x –
3) + (8x + 8) = 48x – 16 9x2 + 14x + 5 = 48x – 16 9x2 – 34x + 21 = 0 (9x – 7)(x – 3) = 0 x = 7 9 or x = 3 Guided Practice Check it out: This is a quadratic equation, so you could use the quadratic formula (see Topic 7.3.1) to work out the solutions instead of factoring. Check it out: Make sure that these solutions are allowed — that x π 1 3 and x π –1. Solve each of the following fractional equations. − 3 2 7. 3 6. = − − + 1 1 x x 10 − = x x 5 6 2 3 8. a − = 1 5 + a 3 10. 5 x + = 1 11 − x 1 11 Always Check that the Solutions Work It’s a good idea to check the solutions you have worked out to make sure that they are correct. The best way to check a solution is to put it back into the original equation and verify that both sides of the equation are equal. In the problem on the last page, the solutions for the fractional equation + were found to be x = 7 9 or x = 3. Check these solutions by putting them back into the equation, one at a time. Section 8.4 — Solving Equations with Fractions 413 Check it out: Remember that dividing a number by a fraction is the same as multiplying that number by the reciprocal of the fraction, for example: ÷ = × = a a c b ac b b c Example 1 continued First, check 16 16 7 9 is valid. Both sides are equal, which means that the solution x = Now, check x = 3 = Both sides are equal, which means that the solution x = 3 is correct. If you put the solution into the equation and find that the two sides are not equal, then your solution is incorrect. That means you’ll have to go back and check each stage of your work. Example 2 Solve = −, first stating any values of x for which the 1 equation is undefined. Solution First take a look at the equation and figure out the limitations. The equation is undefined when either 2x + 1 = 0 or x – 1 = 0. This happens when x = – 1 2 or x = 1. So the limitations on the solution are that x cannot equal – 1 2 or 1
. Step 1: Find the least common multiple of the denominators: 2x + 1 and x – 1. The LCM for these denominators is (2x + 1)(x – 1). Step 2: Now, multiply both sides of the equation by (2x + 1)(x – 1) to eliminate the denominators from the rational expressions − )( )( 1 1 = − ⋅ ( 1 2 x + )( 1 x − ) 1 414 Section 8.4 — Solving Equations with Fractions Check it out: The whole equation can be divided by the common factor 2. Check it out: Make sure both solutions are allowed, that is x π – 1 2, x π 1. Check it out: Both of these solutions are correct because the equation balances when the solutions are substituted for x. Example 2 continued Step 3: Reduce the equation to its lowest terms: 3·(x – 1) – 4·(2x +1) = –1·(2x + 1)(x – 1) 3x – 3 – 8x – 4 = –1·(2x2 – x – 1) –5x – 7 = –2x2 + x + 1 2x2 – 6x – 8 = 0 x2 – 3x – 4 = 0 Step 4: Then factor to find the solutions: (x – 4)(x + 1) = 0 So, x – 4 = 0 or x + 1 = 0 So the solutions are: x = 4 or x = –1 Now, check your solutions to make sure that they work. Put x = 4 into the equation: Put x = –1 into the equation Guided Practice Solve each of the following fractional equations. 12 36 + − 2 x 8 2 x 13 +( 2 c 1 ) 2 c 14. + 4 15 15 12 16 Section 8.4 — Solving Equations with Fractions 415 Independent Practice Solve each of the following fractional equations. 3. y 3 y 5. 6 1 + = x 12 2 x 4. Solve for y the following equation: − 1 1 7. Use your solution to Exercise 6 to find the value of y2 – 1. = − 2 3 y y 8 1 + 10 y − y 2 + + 8. Find the value of 5k – 7 if. One number is 8 less than another. The quotient formed by dividing the higher number by the lower number is Find the values of the
two numbers. 15 11. 10. One integer is 5 less than another. Find the numbers if the sum of their reciprocals is 17 66. 11. The denominator of a fraction is 3 more than its numerator. If the sum of the fraction and its reciprocal is 29 10, find the fraction. 12. The denominator of a fraction is 2 more than the numerator. A second fraction is created by subtracting 2 from the numerator and subtracting 3 from the denominator of the first fraction. Find the two fractions if the first fraction subtracted from the second fraction is 5 88. 13. Solve the following system of equations by first rearranging them so that they are not in fractional form 15 x + 18 = 11 x + + 4 y 3 2 y Round Up Round Up The main thing to learn here is that fractional equations can be solved by first multiplying the entire equation by the least common multiple of the denominators, then canceling all common factors. The equation you’re left with can then be solved using methods from earlier Chapters. Remember to check that all your solutions actually work by putting them into the original equation. 416 Section 8.4 — Solving Equations with Fractions Topic 8.4.2 California Standards: 13.0: Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. What it means for you: You’ll model real-life situations and solve equations involving rational numbers. Key words: common denominator rational common factor limitation undefined least common multiple Don’t forget: Remember: Time = Distance Speed Check it out: You need to convert the time from minutes to hours because in the problem, speed is measured in miles per hour. Applications of Applications of Fractional Equations Fractional Equations Now it’s time to solve some more fractional equations. This time, the problems are about real-life situations, so you’ll have to model them as equations before you can start with the math working. You Might Be Given a Word Problem to Solve Here’s one long example showing a model of a real-life problem. Example 1 In a bike race, Natasha cycles along flat ground for 4 miles against a 2 mile per hour wind. After four miles, she turns around and follows the same route back to the start, this time with the wind behind her. If the journey takes 50 minutes, find how fast
Natasha would travel if there were no wind. (Assume that she would travel at a constant speed without a wind.) Solution Step 1 — Write the equation. The time Natasha takes to complete the race, can be written as: Timethere + Timeback = 50 minutes Using this, an equation for Natasha’s race time can be written in terms of distance and speed. Distance Speed there there + Distance Speed back back = Time pp total tri 4 − s 2 + 4 + s 2 50 = = 60 5 6 Where: s = Natasha’s speed in no wind s – 2 = Natasha’s speed against the wind s + 2 = Natasha’s speed with the wind behind her s π ±2 (for the rational expressions to be defined) Section 8.4 — Solving Equations with Fractions 417 Example 1 continued Step 2 — Multiply by the LCM. The LCM of the denominators is 6(s – 2)(s + 2). Multiply the equation by the LCM of the denominators to give: 24(s + 2) + 24(s – 2) = 5(s – 2)(s + 2) 24s + 48 + 24s – 48 = 5s2 – 20 5s2 – 48s – 20 = 0 Step 3 — Factor and Solve for x. Now, factor the equation and solve for x: Don’t forget: Remember to check that the solutions work when put back into the original equation. (5s + 2)(s – 10) = 0 So, s = – 2 5 or s = 10 So Natasha’s speed if there were no wind would be 10 miles per hour. Although – 2 5 is a valid solution of the algebraic problem, it isn’t a correct answer for this example because speed can only have a positive value. Guided Practice 1. On Monday, a distribution company shipped a load of oranges in crates, with a total weight of 124 lb. On Tuesday it shipped another load of oranges, also with a total weight of 124 lb. However, on Tuesday there was one crate fewer than on Monday, so each crate was 1 8 2. Rose spent $2.40 on pens. If each pen had cost 4 cents more, she would have been able to buy 10 fewer pens for the same money. How many pens did Rose buy? lb heavier. How many crates were shipped on Monday? 3. Dajanique bought x boxes
of candy for a total of $1.26. She kept four boxes and sold the rest for a total of $1.40. If she sold each box for 3 cents more than it cost her, how many boxes did she buy? 4. A teacher spent $8.40 on sets of chapter tests. If each set of tests had been 2 cents less, the teacher would have gotten two extra sets for the same price. How many sets did the teacher get? 418 Section 8.4 — Solving Equations with Fractions Independent Practice 1. Vanessa bought a set of military medals through an antique dealer for $120. She gave two of the medals to her father and resold the rest, charging $4 more per medal than she paid. If Vanessa sold the medals for a total of $156, how much did she charge for each medal? 2. Hearst Castle is 180 miles from Ventura by road. Two coaches leave Ventura at the same time, both heading for Hearst Castle, but one averages 5 mph faster than the other. If the faster coach reaches Hearst Castle half an hour earlier than the slower coach, what is the average speed of the faster coach? 3. A cyclist completes a 150-mile race in a certain amount of time. She completes another 150-mile race a month later, but this time it takes an hour longer to cover the same distance and her average speed is 5 mph less than in the first race. Find the average speed for the cyclist during her first race. 4. Juan’s band produced a number of CDs to sell at a gig. The batch of CDs cost $170.00 to make and each CD was sold for $2.50 more than it cost to produce. Juan gave 2 CDs to his friends, but sold the rest for a total of $198.00. How many CDs did Juan produce? 5. Chen bought a bag of groceries weighing 15 pounds. His friend, Jo, bought a bag of groceries that also weighed 15 pounds, but contained one less item. The average weight per item for Jo’s groceries was ½ pound more than for Chen’s. How many items were in Jo’s grocery bag? 6. José ordered a box of fruits for his market stand costing $6. When his order arrived, José discovered that 20 fruits were rotten and threw them in the trash. He sold all of the remaining fruits for a total of $8, charging 3 cents more for each fruit than he paid for it.
How many fruits did José order to begin with? 7. A wholesaler bought a batch of T-shirts for $77.00. She gave two of the T-shirts to her daughters and then sold the rest for a total of $90. If the wholesaler sold each T-shirt for $2 more than it cost her, how much did she pay for each T-shirt? Round Up Round Up The only difference between solving word problems and answering straightforward algebra questions is that you have to write the equations yourself for a word problem. After that, they are solved in exactly the same way. Section 8.4 — Solving Equations with Fractions 419 Topic 8.5.1 Section 8.5 Relations Relations California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 17.0: Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression. What it means for you: You’ll find out what relations are, and some different ways of showing relations. Key words: relation ordered pair domain range Check it out: In an ordered pair (x, y), the first variable, x, is usually called the independent variable, and the second, y, is the dependent variable — it depends on the first. For example, in the relation y = 4x2 – 3x, the values of y in the range depend on the values of x in the domain. Check it out: The order you write the elements of the domain or range isn’t important, so you could write the range from Example 1 as {2, 4, 5, 7, 8}. Also, if an element is connected to more than one element in the other set, you only need to write it down once. Relations in Math are nothing to do with family members. They’re useful for describing how the x and y values of coordinate pairs are linked. A Relation is a Set of Ordered Pairs Before you can define a relation, you need to understand what “ordered pairs” are: An ordered pair is just two numbers or letters, (or anything else) written in the form (x, y). If x and y are both real numbers, ordered pairs can be plotted as points on a coordinate plane, where the first number in the ordered pair represents the x-coord
inate and the second number represents the y-coordinate. y This point represents the ordered pair (–2, 1). x 1 2 3 3 2 1 –1 –1 –2 –3 –3 –2 A relation is any set of ordered pairs. Relations are represented using set notation, and can be named using a letter: for example: m = {(1, 4), (2, 8), (3, 12), (4, 16)}. Every relation has a domain and a range. Domain: the set of all the first elements (x-values) of each ordered pair, for example: domain of m = {1, 2, 3, 4} Range: the set of all the second elements (y-values) of each ordered pair, for example: range of m = {4, 8, 12, 16} An important point to note is that there may or may not be a reason for the pairing of the x and y values. Looking at the relation m, above, you can see that the x and y values are related by the equation y = 4x — but not all relations can be described by an equation. Example 1 State the domain and range of the relation r = {(1, 4), (3, 7), (3, 5), (5, 8), (9, 2)}. Solution Domain = {1, 3, 5, 9} Range = {4, 7, 5, 8, 2} 420 Section 8.5 — Relations and Functions Example 2 State the domain and range of the relation f = {(a, 2), (b, 3), (c, 4), (d, 5)}. Solution Domain = {a, b, c, d} Range = {2, 3, 4, 5} Ordered pairs are not the only way to represent relations. Guided Practice State the domain and range of each relation. 1. f(x) = {(1, 1), (–2, 1), (3, 5), (–3, 10), (–7, 12)} 2. f(x) = {(–1, –1), (2, 2), (3, –3), (–4, 4)} 3. f(x) = {(1, 2), (3, 4), (5, 6), (7, 8)} 4. f(x) = {(a, b), (c, d), (e, f), (
g, h)} 5. f(x) = {(–1, 0), (–b, d), (e, 3), (7, –f)} 6. f(x) = {(a, –a), (b, –b), (–c, c), ( 1 2, –j)} Mapping Diagrams Can Be Used to Represent Relations One way to visualize a relation is to use a mapping diagram. In the diagram, the area on the left represents the domain, while the area on the right represents the range. The arrows show which member of the domain is paired with which member of the range. This mapping diagram represents the relation t = {(2, v), (3, c), (6, m)}. 2 3 6 t c v m Domain Range Guided Practice State the domain and range of each relation. 7. 8. 2 4 6 8 10 12 4 36 100 16 64 144 1 2 –3 0 –1 9. a e i o u –8 –6 5 12 10 Section 8.5 — Relations and Functions 421 You Can Use Input-Output Tables In input-output tables, the input is the domain and the output is the range. This table represents the relation {(1, 1), (2, 3), (3, 6), (4, 10), (5, 15)}. Input Output 1 2 3 4 5 1 3 6 10 15 Guided Practice State the domain and range of each relation. 11. Input Output Input Output 12. 1 5 12 32 6 0.3 –2 –1 2 3 8 5 8 13 13. Input Output –3 –1 0 1 –26 0 1 2 Relations Can Be Plotted as Graphs Relations can be plotted on a coordinate plane, where the domain is represented on the x-axis and the range on the y-axis. Graphs are most useful when you have continuous sets of values for the domain and range, so that you can connect points with a smooth curve or straight line. This graph represents the relation {(x, y = x)} with domain {–2 £ x £ 8}. y 8 6 4 2 –2 2 4 6 8 x –2 Guided Practice State the domain and range of each relation. 14. y (–2, 3) 15. y (3, 9) x (2, –3) (–9, –3) x Check it out: The filled circles mean that the domain includes the values –2
and 8. A hollow circle would mean that the value was not included, and an arrowhead would mean that the domain continued to infinity in that direction. So, for example: y domain = {–2 < x < 8} 8 6 4 2 –2 2 4 6 8 –2 y ∞ domain = {– 8 6 ∞x < < } 4 2 –2 2 4 6 8 –2 x x 422 Section 8.5 — Relations and Functions Independent Practice Define each of the following terms. 1. Relation 2. Range 3. Domain In Exercises 4–15, state the domain and range of each relation. 4. {(x, x2) : x Œ {–1, 2, 4}} 5. y (0, 2) 6. y (0, 1) (–2, 0) x (2, 0) (–1, 0) x (1, 0) (0, –2) (0, –1) ( ) 7. f x = {(, {,., }} 8. f(x) = {(x, x2 – 1) : x Œ {0, 1, 2}} 9. f(x) = {(x, –x2 + 3) : x Œ {–2, –1, 0, 1, 2}} 10. f(x) = {(x, 11. f(x) = {(x, x x − {–1, 0, 2}} ) : x Œ {–1, 0, 2, 3}} 12. y (–2, 4) 13. (–4, 6) y (3, 2) x (–3, –2) (4, 2) x (2, –2) 14 15. 1 –3 a e –1 4 d a b Round Up Round Up The important thing to remember is that a relation is just a set of ordered pairs showing how a domain set and a range set are linked. Section 8.5 — Relations and Functions 423 Topic 8.5.2 California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 18.0: Students determine whether a relation defined by a graph, a set of ordered pairs, or a symbolic expression is a function and justify the conclusion. What it means for you: You’
ll find out what functions are, and you’ll say whether particular relations are functions. Key words: function relation ordered pair domain range Functions Functions A function is a special type of relation. Functions Map from the Domain to the Range A relation is any set of ordered pairs — without restriction. A function is a type of relation that has the following restriction on it: A function is a set of ordered pairs (x, y) such that no two ordered pairs in the set have the same x-value but different y-values. That is, each member of the domain maps to a unique member of the range. Example 1 Determine whether each of the following relations is a function or not. Justify your answers. a) k = {(0, 0), (1, 1), (2, 4), (3, 9)} b) m = {(1, 2), (2, 5), (1, 4), (3, 6)} c) p = {(–1, –3), (0, –1), (1, 1), (2, 3)} d) v = {(–2, 5), (–1, 5), (0, 5), (1, 5), (5, 5)} Solution a) k is a function. No two different ordered pairs have the same x-value. b) m is not a function. The ordered pairs (1, 2) and (1, 4) have the same first entry, but different second entries. c) p is a function. No two different ordered pairs have the same x-value. d) v is a function. No two different ordered pairs have the same x-value. Guided Practice State whether each relation in Exercises 1–4 is a function or not. Explain your reasoning. 1. m = {(1, 1), (2, 8), (3, 27)} 2. b = {(a, 1), (b, 2), (c, 3), (a, 4)} 3. v = {(7, 1), (7, 2), (7, 7)} 4. t = {(1, 7), (2, 7), (7, 7)} 424 Section 8.5 — Relations and Functions Functions Can Be Represented in Different Ways Relations do not have to be written as lists of ordered pairs for you to be able to identify functions. Example 2 Determine whether each of the mappings below represents a function.
a) b) c) 2 5 7 11 a b c k n 2 5 6 7 11 a b c k n 2 5 6 7 11 a b c n Domain Range Domain Range Domain Range Solution a) and b) are NOT functions, since 7 is mapped to two different values — (7, a) and (7, c) have the same x-value. c) IS a function. Each member of the domain only maps onto one member of the range. Guided Practice The mapping on the right shows the relation g(x). 5. State the domain and range of the relation. 6. Is the relation a function? Explain your answer. 7. Find g(0). The mapping on the right shows the relation h(x). 8. State the domain and range of the relation. 9. Is this relation a function? Explain your answer. 10. Find h(–3). –4 –3 2 5 p q r s t Don’t forget: Remember that this isn’t always the case — some functions cannot be expressed as an equation. Functions are Often Written as Equations Some functions can be expressed as an equation. For this to be possible, there must be a reason for the pairing between each member of the domain and each member of the range. The equation represents the way the members of the domain and range are paired. Section 8.5 — Relations and Functions 425 Example 3 Express the following function as an equation: f = {(1, 1), (2, 4), (3, 9), (4, 16)} Check it out: The symbol Œ is shorthand for “is a member of the set.” It just means that x takes only those values. Solution The relationship between the x-values and y-values is y = x2. The domain of the function is x Œ {1, 2, 3, 4}. The function can be written: f = {(x, y) such that y = x2 and x ŒŒŒŒŒ {1, 2, 3, 4}} You will often see functions written in the form y = x2, without any domain specified. By convention, you then take the domain to be all values of x for which the function is defined. Guided Practice Express the following functions in terms of an equation. 11. f = {(–4, 0), (–3, 1), (0, 4), (1, 5),
(2, 6)} 12. g = {(–2, 5), (0, 1), (1, 2), (2, 5)} 13. h = {(–5, –4.5), (–3, –2.5), (1, 1.5), (3, 3.5), (5, 5.5)} 14. f = {(–3, –27), (–2, –8), (–1, –1), (0, 0), (1, 1), (2, 8)} 15. g = {(–2, 8), (0, 0), (1, 2), (2, 8), (3, 18)} 16. h = {(–3, 17), (–1, 1), (0, –1), (1, 1), (2, 7)} The Vertical Line Test Shows if a Graph is a Function By definition, a function cannot have any two ordered pairs that have the same first coordinate but different second coordinates, i.e. for each value of x there is only one possible value of y. Graphically, this means that no vertical line can intersect the graph of a function at more than one point. Vertical Line Test to determine whether a graph represents a function: Simply hold a straightedge parallel to the y-axis at the far left-hand side of the graph, then move it horizontally along the graph from left to right. If, at any position along the x-axis, it is possible for you to draw a vertical line that intersects with the graph more than once, then the graph does not represent a function. 426 Section 8.5 — Relations and Functions Example 4 Use the vertical line test to determine whether the following graphs represent functions. y y y x x x Solution The first two graphs pass the vertical line test — you cannot draw a vertical line that intersects with either graph at more than one point, so they represent functions. The third graph does not represent a function — the vertical line test fails. Guided Practice In Exercises 17–20, use the vertical line test to determine whether each graph represents a function or not. 17. 19. y y 18. 20. x x y y x x Section 8.5 — Relations and Functions 427 Independent Practice 1. Define a function and give an example. In Exercises 2–7, use the given relation with the domain x = {–2, –1, 0, 1
, 2} to generate sets of ordered pairs. Use them to determine whether the relation is a function or not. 2. m = {(x, x2 – 4)} 3. t = {(x, x + 2)} 4. k = {(x, y = (x – 2)(x + 2)} 5. p = {(x, y = ± 4 2− x } 6. j = {(x, y = 2x – 1)} 7. b = {(x, y = x ± (3x – 4)} 8. In the equation x2 + y2 = 9, is y a function of x? Explain your reasoning. Using the graph on the right, answer Exercises 9–12 about relation f. 9. State the domain and range of the relation. 10. Is the relation a function? 11. Find the value(s) of f(0). 12. Find the value(s) of f(3). 3 2 1 –1 –( ) 1 2 3 x Use the vertical line test to determine if the graphs below are functions. 13. y 14. y x x For Exercises 15–16, find the range (y) for each relation when the domain is {–4, –2, 0, 2, 4}, and determine whether the relation is a function. 15. y = 1 2 16. y = x – 6 x + 1 For Exercises 17–19, find the domain (x) for each relation when the range is {–6, –3, 0, 3, 6}, and determine whether the relation is a function. 17. y = 1 3 x – 2 18. y = x + 5 19. y = ± x 2 20. Are all quadratics of the form y = ax2 + bx + c and y = –ax2 + bx + c functions? Explain your answer. 21. Are circles functions? Explain your answer. Round Up Round Up Functions are special types of relations. That means that all functions are relations — but not all relations are functions. A relation is only a function if it maps each member of the domain to only one member of the range. 428 Section 8.5 — Relations and Functions Topic 8.5.3 California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 17.0
: Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression. What it means for you: You’ll see some different ways of representing functions. Key words: function domain range Check it out: The notation m(x) is read as “m of x” or “the value of function m at x.” Function Notation Function Notation Functions are often written using function notation. Think of a Function as a “Rule” or “Machine” The function m is a rule that assigns to each value x of its domain a distinct value m(x) of its range. Think of m as a machine that processes x according to some rule and outputs the value m(x). Suppose m(x) = x2. You could draw the following diagram to represent how the function processes any given value of x. Function machine: m(x) = x2 Input x = 3 Domain m x( ) = x2 2= 3 Output m x( ) = 9 Range m(x) = x2 takes a value of x from the domain, in this case x = 3, squares it, 32 = 9, and outputs a value of the range, 9. When you are referring to a particular value of a function, you replace the x with the relevant number — so in the example above, the value of m when x = 3 would be denoted m(3). Similarly, you can substitute expressions into a function, so for example you could find m(x + 3) = (x + 3)2 by replacing x with (x + 3) in the function. Section 8.5 — Relations and Functions 429 Example 1 Given the function f (x) = 2x2 – 1, find a) f (–1), b) f (3), and c) f (x + a). Solution a) f (–1) = 2 · (–1) (–1) = 1 b) f (3) = 2 · 32 – 1 = 2 · 9 – 1 = 18 – 1 = 17 \ f (3) = 17 c) f (x + a) = 2 · (x + a)2 – 1 = 2(x2 + 2ax + a2) – 1 = 2x2 + 4ax + 2a2 – 1 Example 2 If P(x) = x3 + 1, find the
range of P(x) when the domain of P(x) is the set {–2, –1, 0, 1, 2}. Solution The range is the set of all values of P(x) for which x Œ {–2, –1, 0, 1, 2}. So, range = {P(–2), P(–1), P(0), P(1), P(2)} = {(–2)3 + 1, (–1)3 + 1, (0)3 + 1, (1)3 + 1, (2)3 + 1} = {–7, 0, 1, 2, 9} Guided Practice For Exercises 1–6, let f (x) = 3x – 1 and g(x) = x2 – 2. Find each of the values indicated. 1. f (–2) 4. f (a – 1) 2. f ⎛ ⎜⎜⎜ ⎝ ⎞ ⎟⎟⎟ ⎠ 1 3 3. g(a) 5. f (3) · g (–2) 6. g(k) – f (k) Independent Practice 1. The function f(t) = 5 9 (t – 32) is used to convert temperatures from degrees Fahrenheit to degrees Celsius. Find f(212). 2. Supposing f (x) = –2x2 + x – 5, evaluate f (x) for x Œ{–1, 2, b}. In Exercises 3–5, find f (x + h) for each of the given functions. 3. f (x) = 2x – 1 5. f (x) = –3x + h 4. f (x) = x2 – x – 2 Use the functions f(x) = 2x + 1 and g(x) = x2 + 2x + 1 to find the value of each expression in Exercises 6–9. 6. f(–1) + g(2) 8. f(x + h) + g(x + h) 7. g(–1) – f(3) 9. f(x + h) – g(x + h) Round Up Round Up There’s nothing too complicated here — the main thing is to recognize that if you see the notation f(something), then
it’s likely to be a function. 430 Section 8.5 — Relations and Functions Topic 8.5.4 California Standards: 16.0: Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. 17.0: Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression. What it means for you: You’ll show whether two functions are equal. Key words: function ordered pair domain range More on Functions More on Functions This Topic’s all about the special rules for telling whether two functions are equal. Equality of Functions A function f is equal to another function g if, and only if, the set of ordered pairs of f is identical to the set of ordered pairs of g. It isn’t enough for the two functions to be represented by equivalent equations [for example, y = 2x + 4 and y = 2(x + 2)]. For the two sets of ordered pairs to be identical, the two functions must also have the same domain (and therefore the same range). Example 1 a) Determine whether the following two functions are equal: m = {(x, y), such that y = x2 – 4 and x Œ {–1, 0, 2, 3}} b = {(x, y), such that y = (x – 2)(x + 2) and x Œ {–1, 0, 2, 3}} b) State the range of m. Solution a) Find each set of ordered pairs by substituting each value of x into the equation for y. m = {(–1, [–1]2 – 4), (0, 02 – 4), (2, 22 – 4), (3, 32 – 4)} = {(–1, –3), (0, –4), (2, 0), (3, 5)} b = {(–1, [–1 – 2][–1 + 2]), (0, [0 – 2][0 + 2]), (2, [2 – 2][2 + 2]), (3, [3 – 2][3 + 2])} = {(–1, –3), (0, –4), (2, 0), (3, 5)} Since each ordered pair for m is in b and vice versa, then the functions m and b
are equal. b) The range of m = range of b = {–3, –4, 0, 5} Section 8.5 — Relations and Functions 431 Example 2 The domain of both of the following functions is the set of all real numbers. p = {(x, y = x3 + 3x2 + 3x + 1)} q = {(x, y = (x + 1)3} Determine whether the two functions are equal. Solution Both functions have the same domain, so if the equations that generate the y-values are equal, then the functions must be equal. Expand (x + 1)3: (x + 1)3 = (x + 1)(x + 1)(x + 1) = (x2 + 2x + 1)(x + 1) = x3 + 3x2 + 3x + 1 So, p and q are equal. Guided Practice Determine whether the pairs of relations/functions below are equal. 1. A = {(x, y), such that y = 2x + 8 and x Œ {–1, 0, 1, 2}} B = {(x, y), such that 2y = 4x + 16 and x Œ {–1, 0, 1, 2}} 2. A = {(x, y), such that y = x2 and x Œ {1, 4, 9, 16}} B = {(x, y), such that y2 = x and x Œ {1, 4, 9, 16}} Determine whether the pairs of relations/functions below are equal. The domain of each relation/function is the set of real numbers. 3. M = {(x, y = (x + 4)2)} N = {(x, y = x2 + 4x + 4)} 4. M = {(x, y = x2 + 1)} N = {(x, y = (x + 1)2)} 5. M = {(x, y = x4(x + 1)(2x – 3))} N = {(x, y = 2x6 – x5 – 3x4)} 6. M = {(x, y = x(3x – 2)(3x + 1))} N = {(x, y = 9x3 + 3x2 – 2x)} More Function Examples Example 3 Suppose v(x) State any restrictions
on x. b) Find v(–3). Solution a) The function is undefined when its denominator (x – 1) is 0, so x πππππ 1. b) Substitute x = –3: 432 Section 8.5 — Relations and Functions 1 = + 9 1 − 4 10 = − = − 4 5 2 2 v(–3) = )− + 3 ( − − 3 1 So, v(–3) = − 5 2 Example 4 Determine the range of the function represented by the graph on the right. Explain your answer. f x( ) x 1 (–, –3) 2 –3 Solution The lowest value of f(x) is –3. The values of f(x) get infinitely large as x gets larger (in both the positive and negative x directions) — as indicated by the arrowheads on the graph. Range = {f(x) : f(x) ≥≥≥≥≥ –3} Example 5 Check it out: A colon means “such that,” so you would read this solution as “the set of all f(x) such that f(x) is greater than or equal to –3.” Determine the domain of f (x) = 2 for which f (x) is defined. Explain your thinking. 6 x −, given that it contains all real x Solution Any expression n is defined for all real values of n ≥ 0. So, for the function to be defined, 2x – 6 ≥ 0. ﬁ 2x ≥ 6 ﬁ x ≥ 3 So, the domain of the function = {x : x ŒŒŒŒŒ R and x ≥≥≥≥≥ 3} Don’t forget: R is the set of all real numbers. Example 6 Given that g(x) = x2 – 1, find ( in terms of x and h. Solution = = ⎡ ⎣⎢ ( x + − 2 h ) ⎦⎥ − −⎡ ⎤ x 1 ⎣⎢ 2 ⎤ 1 ⎦⎥ 2 x + 2 xh Substitute in the functions g(x) and g(x + h) 1 Expand and simplify + 2 = xh h 2xx h = 2x + Section 8.5 — Relations and Functions 433
Guided Practice In Exercises 7–10, find the values in terms of h (and x, where appropriate). 7. Supposing f (x) = x2 – 2, find f (h + 2) – f (2). 8. If f (x) = 2x2 + 4x – 6, find f (h – 3) + f (2h). ( ) f x ( f x 9. Supposing f (x) = x2, find + −. 10. If m(x) = 2x – 3, find the value of Determine the domain for the functions in Exercises 11 and 12, given that they contain all real numbers for which the functions are defined. 11 16 4 12 Independent Practice For the functions in Exercises 1–4, state any restrictions and find f(–1). 1. f(x) = 1 x + 3x + 2 3. f(x) = 2 x + 6 2. f(x) = 2 2 x x 3 4. f(x) = x 2 − − 9 x 6 − + 11 x 4 1− Determine the range of the functions in Exercises 5–6. 6. 1 2 3 x 4 –4 –3 –2 5. –4 –3 –2 y 4 3 2 1 –1 –1 –2 –3 –4 y 4 3 2 1 –1 –1 –2 –3 –4 1 2 3 x 4 7. If f(x) = x2 + 2x – 1, find f(x + h) – f(x). 8. If g(x) = 2x2 + x – 4, find g(x + h) – g(x). 9. If f(x) = (x + 1)2 + x, find f(2h) – f(2). ( ) f x ( f x 10. If f(x) = x2 + 2x, find. + − ). 11. If f(x) = x2 – 2x + 1, find 12. Give an example of two equal functions. Round Up Round Up There are two methods to determine whether two functions are equal. If you’re given a small number of values for the domain, you can substitute each value of the domain into both functions and see if you get identical ordered pairs. The alternative is to show that the two functions can be represented by equivalent equations —
and if the domains are also equal, the functions will be equal. 434 Section 8.5 — Relations and Functions Chapter 8 Investigation ming Functions ming Functions ansforororororming Functions ansf ansf TTTTTrrrrransf ming Functions ming Functions ansf ansforororororming Functions TTTTTrrrrransf ming Functions ming Functions ansf ansf ming Functions ansf ming Functions In this Investigation you’ll practice interpreting information about functions using graphs. On a set of axes, draw the graph of the following function: Investigate the functions: = = f f {(, x y ) such that y = x and x ∈ R } {(, x y ) such that y = + x, a x ∈ R, and a ∈ R }, and f = {(, x y ) such that y = + x, a x ∈ R, and a ∈ R }. Write down a general rule for what happens. Extension 1) The graph of the function f = {(, x y ) such that y = x } original function y-axis can be stretched or compressed. The diagram on the right shows a horizontal compression and stretch. x-axis Investigate how the function must be changed to stretch or compress the graph. Write down a general rule. 2) Investigate how the function f = {(, x y ) such that y = x } must be changed so that the graph is reflected in the x-axis. Open-ended extension 1) The graph on the right has been produced by performing a series of different transformations on the graph of f = {(, x y ) such that y = x }. Identify the new function that is represented by the graph. 2) Perform a series of transformations on the graph of =, recording each one. such that y, x y {( = } x ) f Identify the function that your graph represents. -8 -7 -6 -5 -4 -3 -1 0 -1 -2 -3 -4 -5 -6 -7 - Check that you are correct by creating a table of x- and y-values for your function and plotting the points. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The rules about transforming the absolute value function also apply to other graphs of functions — so this Investigation is really

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