Solution Manual for Calculus I with Precalculus, 3rd Edition

NOT FOR SALE C H A P T E R 2 Polynomial and Rational Functions Section 2.1 Quadratic Functions and Models ……………………………………………….174 Section 2.2 Polynomial Functions of Higher Degree…………………………………….188 Section 2.3 Polynomial and Synthetic Division ……………………………………………201 Section 2.4 Complex Numbers……………………………………………………………………215 Section 2.5 The Fundamental Theorem of Algebra ………………………………………220 Section 2.6 Rational Functions……………………………………………………………………237 Review Exercises …………………………………………………………………………………………..252 Chapter Test ……………………………………………………………………………………………….270 Problem Solving ……………………………………………………………………………………………272 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE C H A P T E R 2 Polynomial and Rational Functions Section 2.1 Quadratic Functions and Models 1. polynomial x 4 8. f x 2 opens upward and has vertex 4, 0 . Matches graph (c). 2. nonnegative integer; real x 2 2 opens upward and has vertex 0, 2 . 9. f x 3. quadratic; parabola Matches graph (b). 4. axis x 1 10. f x 5. positive; minimum 4 x 2 11. f x x 2 7. f x 2 opens upward and has vertex 1, 2 . Matches graph (a). 6. negative; maximum 2 2 opens upward and has vertex 2, 0 . 2 x 2 2 4 opens downward and has vertex 2, 4 . Matches graph (f). Matches graph (e). x 4 12. f x 2 opens downward and has vertex 4, 0 . Matches graph (d) 13. (a) y 1 x2 2 (b) 18 x 2 y (c) (d) 3 x 2 y y y y y 5 6 5 6 4 4 4 4 3 3 2 −6 −4 1 1 2 x 4 −2 2 6 3 Vertical shrink Vertical shrink and reflection in the x-axis x2 1 (b) y 1 −1 2 (c) (d) x2 3 y y y 5 4 10 8 4 3 8 6 3 2 6 4 2 1 x x − 3 −2 −1 −1 1 2 –2 2 3 3 Vertical shift one unit upward 6 Vertical stretch and reflection in the x-axis x2 3 y y –3 4 3 Vertical stretch x2 1 y x 2 x −3 −2 −1 −6 –1 x − 6 −4 − 2 1 −4 x –3 –2 –1 14. (a) y 3 2 x 2 y –2 Vertical shift one unit downward − 6 −4 −2 −2 x 2 4 6 Vertical shift three units upward − 6 –4 –2 4 6 −4 Vertical shift three units downward INSTRUCTOR USE ONLY 174 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 x 1 15. (a) y 2 (b) y 5 4 3 2 3x y 1 (c) Quadratic Function Functions and Models 2 1x 3 y y 3 (d) y x 3 y 175 2 y 5 8 10 4 6 8 3 4 x x –2 –1 1 2 3 −3 −2 −1 −1 4 –1 Horizontal shift one unit to the right 12 x 2 16. (a) y 2 x 1 2 −6 −2 −2 3 6 2 x −8 −6 −4 −2 −2 −4 Horizontal shrink and a vertical shift 1 Horizontal stretch and a vertical shift three one unit upward 12 x 2 (c) y 2 4 Horizontal shift three units to the left units downward 2 1 4 6 y y 8 6 6 4 4 2 6 x −8 −6 −4 x −6 −4 −2 2 8 10 −4 −6 −8 Horizontal shift two units to the right, vertical shrink Horizontal shift two units to the left, vertical shrink each y -value is multiplied by 12 , reflection in the each y -value is multiplied by 12 , reflection in x-axis, and vertical shift one unit upward x-axis, and vertical shift one unit downward 2 ª1 x 1 º 3 ¬2 ¼ (b) y 2 ¬ª2 x 1 º¼ 4 (d) y y y 7 10 8 6 4 4 3 2 x −8 −6 −4 2 6 8 1 x −4 −4 − 3 − 2 −1 −1 −6 1 2 3 4 Horizontal shift one unit to the right, horizontal stretch each x-value is multiplied by 2 , and vertical Horizontal shift one unit to the left, horizontal shrink shift three units downward four units upward 1 x2 17. f x Vertex: 0, 8 y 4 0 2 x −4 −3 −2 0 1 x r1 x 2 x-intercepts: 1, 0 , 1, 0 −1 1 2 3 4 x2 8 0 −3 2 8 x r2 x 6 0 Find x-intercepts: −2 −4 y Axis of symmetry: x or the y-axis 3 Find x-intercepts: 1 x2 x2 8 18. f x Vertex: 0, 1 Axis of symmetry: x or the y-axis each x-value is multiplied by 12 , and vertical shift 4 2 x −8 −6 −4 2 4 6 8 −10 2 x-intercepts: 2 2, 0 , 2 2, 0 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 176 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x2 7 19. f x x 4 23. f x y Vertex: 0, 7 Axis of symmetry: x or the y-axis 12 x2 7 0 x 4 3 x 4 2 r 3 x 4 x −8 −6 −4 −2 2 4 6 8 4 r 2 3 −7 −6 3 −4 −3 −1 −1 x 1 −2 −3 −4 x x-intercepts: 4 y Vertex: 0, 12 1 2 2 12 x 3 Find x-intercepts: 6 x-intercepts: none 20. f x 4 4 Axis of symmetry: x 4 0 y 2 Find x-intercepts: x2 7 3 Vertex: 4, 3 14 0 2 3, 0 , 4 3, 0 14 Axis of symmetry: x or the y-axis 0 6 Find x-intercepts: 12 x 2 2 8 y Vertex: 6, 8 4 14 2 0 x 12 x2 x r2 −8 −6 −2 2 6 8 Axis of symmetry: x 6 12 10 Find x-intercepts: 3 0 1 2 x 4 2 x 6 2 x 6 2 8 6 8 4 2 x-intercepts: 2 3, 0 , 2 3, 0 21. f x x 6 24. f x 8 8 x −2 −2 2 x 2 8 x 16 25. h x y x 4 1 x2 4 2 x 20 x –1 1 2 3 4 –2 0 –3 2 8 –5 x r 8 Axis of symmetry: x 4 16 Find x-intercepts: 12 2 8 x 4 0 x 4 0 x 4 r2 2 x-intercepts: 4 x –4 4 8 12 16 1 2 x-intercept: 4, 0 2 2, 0 , 2 2, 0 x2 2 x 1 26. g x 22. f x y Vertex: 4, 0 2 1 Find x-intercepts: 8 10 12 14 2 3 0 –4 –3 6 x-intercepts: none 2 1 x 0 4 2 Vertex: 0, 4 Axis of symmetry: x or the y-axis 4 16 14 x 2 14 x 2 16 x 1 2 y Vertex: 1, 0 6 y Vertex: 0, 16 Axis of symmetry: x 1 5 18 Axis of symmetry: x or the y-axis Find x-intercepts: 16 14 x 2 0 x2 64 x r8 4 Find x-intercepts: 0 12 x 1 9 6 2 −9 −6 −3 −3 x 3 6 9 2 1 x 1 0 x 1 3 3 0 x –4 –3 –2 –1 x-intercept: 1, 0 x-intercepts: 8, 0 , 8, 0 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 27. f x 5 4 1· 1 5 § 2 ¨x x ¸ 4¹ 4 4 © x 2 2 x 1 1 5 2 x 1 2 1· § ¨x ¸ 1 2¹ © Axis of symmetry: x 1 y Find x-intercepts: 5 1 2 Axis of symmetry: x Find x-intercepts: 4 x2 2 x 5 0 3 x 2x 5 0 1 x 2 x 0 –2 1r x 6 Vertex: 1, 6 §1 · Vertex: ¨ , 1¸ ©2 ¹ 5 4 177 x2 2 x 5 29. f x x2 x x2 x Quadratic Function Functions and Models –1 1 2 3 2r 4 20 2 1r 6 x-intercepts: 1 15 2 6, 0 , 1 6, 0 y Not a real number 6 No x-intercepts 28. f x 1 4 9· 9 1 § 2 ¨ x 3x ¸ 4 4 4 © ¹ x 2 3x x –4 2 6 –2 –4 2 3· § ¨x ¸ 2 2¹ © 30. f x x2 4x 1 x 2 4 x 4 4 1 § 3 · Vertex: ¨ , 2 ¸ © 2 ¹ x 2 Axis of symmetry: x 1 4 Find x-intercepts: x 4 x 1 0 x 4x 1 0 2 2 3 r 9 1 x 2 3 r 2 § 3 x-intercepts: ¨ © 2 5 2 Axis of symmetry: x 0 x 2 Vertex: 2, 5 3 2 Find x-intercepts: x 2 3x x2 4 x 1 2 · § 3 2, 0 ¸, ¨ ¹ © 2 · 2, 0 ¸ ¹ x-intercepts: 2 5, 0 , 2 4 r 16 4 2 2 r 5 5, 0 y 5 y 4 4 2 3 1 2 x 1 –6 –5 x –5 –4 –3 –2 –1 1 2 –3 –2 –1 1 2 –2 –3 –2 –3 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 178 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 4 x 2 4 x 21 31. h x 34. f x 1· § §1· 4¨ x 2 x ¸ 4¨ ¸ 21 4¹ © © 4¹ 13 x 2 9 x 6 13 x 2 9 x 81 13 81 6 4 4 2 1· § 4¨ x ¸ 20 2¹ © y 13 x 92 §1 · Vertex: ¨ , 20 ¸ ©2 ¹ Vertex: Axis of symmetry: x 1 2 10 –4 4 8 0 4r x 16 336 24 2 x2 x 1 8 10 −4 x 2 9 x 18 0 x 3 x 6 0 x2 2 x 3 x 1 2 4 1 −8 7 x-intercepts: 3, 0 , 1, 0 −5 36. f x 2 1· 7 § 2¨ x ¸ 4¹ 8 © y x 2 x 30 x 2 x 30 6 §1 7· Vertex: ¨ , ¸ ©4 8¹ x 2 x 14 14 30 5 4 Axis of symmetry: x x 12 3 1 4 1 Find x-intercepts: x –3 –2 –1 1 2 0 1r 2 35 121 4 −10 10 Vertex: 12 , 121 4 3 12 Axis of symmetry: x 18 22 −80 x-intercepts: 6, 0 , 5, 0 Not a real number 37. g x No x-intercepts x 2 8 x 11 x 4 2 5 14 Vertex: 4, 5 1 2 x 2 x 12 4 4 Axis of symmetry: x 1 x 2 8 x 16 4 14 16 12 x-intercepts: 4 r −18 −6 y Vertex: 4, 16 38. f x 4 Axis of symmetry: x 4 x –8 4 Find x-intercepts: 1 x 2 2 x 12 4 8 x 2 10 x 14 x 2 10 x 25 25 14 16 x 5 0 x 8 x 48 0 x 4 x 12 0 2 12 5, 0 2 1 x 4 16 4 4 or x 6 −6 Axis of symmetry: x 2 x 4 5 1· § §1· 2¨ x ¸ 2¨ ¸ 1 4¹ © © 16 ¹ 33. f x −2 −2 9 2 Vertex: 1, 4 1 · § 2¨ x 2 x ¸ 1 2 ¹ © x x 0 35. f x 2x x 1 2 9 3 , 2 4 13 x 2 3 x 6 No x-intercepts 2 y x-intercepts: 3, 0 , 6, 0 Not a real number 32. f x 34 Find x-intercepts: x –8 2 Axis of symmetry: x 20 Find x-intercepts: 4 x 2 4 x 21 13 x 2 3 x 6 –12 11 5 −20 Vertex: 5, 11 10 –16 –20 12 2 Axis of symmetry: x x-intercepts: 5 r 5 11, 0 −15 INSTRUCTOR USE ONLY x-intercepts: x-intercepts: 4, 0 , 12, 0 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 39. f x 2 x 2 16 x 31 2 x 4 2 1 −6 4 x-intercepts: 4 r 12 2, 0 12 4 x 2 6 x 41 1 Because the graph passes through 0, 3 , −12 4 x 2 24 x 41 2 a x 2 f x Axis of symmetry: x 179 44. 2, 1 is the vertex. 48 Vertex: 4, 1 40. f x Quadratic Function Functions and Models 3 a0 2 3 4a 1 4 4a 1 a. 2 1 2 x 2 So, y 1. 4 x 6 x 9 36 41 2 4 x 3 2 5 45. 2, 2 is the vertex. 0 0 6 Axis of symmetry: x 0 a 1 2 2 a. −20 1 2 x 4x 2 2 1 x 2 2 2 3 4 So, y Vertex: 2, 3 2 Axis of symmetry: x x-intercepts: 2 r 2 2 Because the graph passes through 1, 0 , 3 No x-intercepts 41. g x a x 2 y Vertex: 3, 5 −8 4 6, 0 2 2 2 x 2 2 2. 46. 2, 0 is the vertex. a x 2 f x 2 0 a x 2 2 −4 Because the graph passes through 3, 2 , 3 2 x 6x 5 5 42. f x 3 2 x 6x 9 5 27 3 5 2 3 x 3 42 5 5 −14 Axis of symmetry: x x-intercepts: 3 r 10 3 a x 1 2 −10 14, 0 Because the graph passes through 1, 0 , 0 a11 4 4a 1 a. So, y 4 1 x 1 a. 2 2x 2 . a x 2 2 5 Because the graph passes through 0, 9 , a0 2 4 4a 1 a. 2 5 1x 2 So, f x 2 5 x 2 2 5. 48. 4, 1 is the vertex. a x 4 f x 2 2 47. 2, 5 is the vertex. 9 4 2 2 f x 43. 1, 4 is the vertex. y a3 2 So, y 6 Vertex: 3, 42 5 2 4 x 1 2 4. 2 1 Because the graph passes through 2, 3 , 3 a2 4 3 4a 1 4 4a 1 a. So, f x 2 1 x 4 2 1. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 180 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 49. 1, 2 is the vertex. a x 1 f x 2 54. 2 Because the graph passes through 1, 14 , 14 a 1 1 14 4a 2 16 4a 4 a. 2 2 2 4 x 1 So, f x a x 2 2 2. f x a x 52 3 2 34 Because the graph passes through 2, 4 , 4 a 2 52 4 81 a 34 4 81 a 4 2 34 a. 2 19 x 52 81 So, f x 34 . 55. 52 , 0 is the vertex. Because the graph passes through 0, 2 , 2 is the vertex. 19 4 19 81 50. 2, 3 is the vertex. f x 5 , 34 2 3 a x 52 f x 2 2 a0 2 2 4a 3 Because the graph passes through 72 , 16 , 3 1 4a 16 3 a 72 52 14 a. 16 3 a. 14 x 2 So, f x 2 3. a x 5 f x 2 56. 6, 6 is the vertex. 12 15 a7 5 3 4a Ÿ a 12 3 . 4 52. 2, 2 is the vertex. a x 2 f x 2 Because the graph passes through 1, 0 , 0 a 1 2 0 a 2 2 a. So, f x 2 2 2 2x 2 a x 14 2 6 61 a 10 6 3 2 1 a 6 100 92 1 a 100 450 a. 2 6 450 x 6 2 6. x2 4 x 57. f x 4 x-intercepts: 0, 0 , 4, 0 2. 53. 14 , 32 is the vertex. f x 3 2 So, f x 2 2 61 3 ,2 , Because the graph passes through 10 2 3 x 5 12. 4 So, f x a x 6 f x Because the graph passes through 7, 15 , 2 2 16 x 52 . 3 So, f x 51. 5, 12 is the vertex. 2 0 x2 4 x 0 x x 4 x 0 or −4 8 −4 x 4 The x-intercepts and the solutions of f x 0 are the same. 32 Because the graph passes through 2, 0 , 0 a 2 14 32 49 a Ÿ a 16 2 32 24 . 49 2 INSTRUCTOR USE ONLY So, f x 24 x 1 49 4 23 . © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 14 2 x 2 10 x 58. f x 63. f x 2 x 2 10 x 0 2 x x 5 2 x 0 Ÿ x x 5 −1 g x ª¬ x 1 º¼ x 3 x2 2 x 3 5 x2 2x 3 0 are the same. a x 1 x 3 Note: f x x 2 9 x 18 x 2 9 x 18 0 x 3 x 6 x 3 or x 64. f x −8 ª¬ x 5 º¼ x 5 x 5 x 5 16 x 2 25, opens upward −4 6 The x-intercepts and the solutions of f x 0 are the same. g x f x , opens downward g x x 2 25 a x 2 25 Note: f x x 8 x 20 2 60. f x x-intercepts: 2, 0 , 10, 0 0 x 2 8 x 20 0 x 2 x 10 has x-intercepts 1, 0 and 3, 0 for all real numbers a z 0. 12 x-intercepts: 3, 0 , 6, 0 0 opens downward x 1 x 3 0 The x-intercepts and the solutions of f x 59. f x opens upward x2 2x 3 6 −6 0 Ÿ x ª¬ x 1 º¼ x 3 181 x 1 x 3 x-intercepts: 0, 0 , 5, 0 0 Quadratic Function Functions and Models has x-intercepts 5, 0 and 5, 0 for all real numbers a z 0. 10 −4 12 65. f x x 0 x 10 opens upward x 10 x 2 −40 x 2 0 Ÿ x 2 x 10 0 Ÿ x 10 g x x 0 x 10 opens downward x 10 x 2 The x-intercepts and the solutions of f x Note: f x 0 are the a x 0 x 10 ax x 10 has x-intercepts 0, 0 and 10, 0 for all real same. numbers a z 0. 2 x 7 x 30 2 61. f x x-intercepts: 52 , 0 , 6, 0 0 2 x 2 7 x 30 0 2x 5 x 6 x 52 or x 10 66. f x −5 x 2 12 x 32, opens upward −40 6 g x f x , opens downward g x x 2 12 x 32 Note: f x The x-intercepts and the solutions of f x 67. f x 7 x 2 12 x 45 10 x-intercepts: 15, 0 , 3, 0 0 7 x 2 12 x 45 10 0 x 15 x 3 10 −18 0 Ÿ x 15 x 3 0 Ÿ x 3 ª¬ x 3 º¼ ª x 12 º 2 ¬ ¼ opens upward x 3 x 12 2 4 x 3 2x 1 2 x2 7 x 3 −60 x 15 a x 4 x 8 has x-intercepts 4, 0 and 8,0 for all real numbers a z 0. 0 are the same. 62. f x x 4 x 8 10 g x 2 x2 7 x 3 opens downward = 2×2 7 x 3 Note: f x a x 3 2 x 1 has x-intercepts INSTRUCTOR USE E ONLY The x-intercepts and the solutions of f x same. same 0 are the 3,, 0 and 12 , 0 for all real numbers a z 0. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 182 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 2 ª x 52 º x 2 ¬ ¼ 68. f x x 2 3x 3 72. y 2 x 52 x 2 (a) From the graph it appears that there are no xintercepts. 2 x 2 12 x 5 (b) There are no x-intercepts and there are no real solutions to the equation x 2 3 x 3 0. 2 x 2 x 10, opens upward g x f x , opens downward g x 2 x x 10 (c) x 2 3 x 3 0 x 3x 3 0 2 2 a x 52 Note: f x x 2 has x-intercepts 73. f x x2 4 x 5 32 4 1 3 3 r 2 3 2 Not a real number Ÿ No x-intercepts 52 , 0 and 2, 0 for all real numbers a z 0. 69. y 3 r x ax 2 bx c (a) x-intercepts: 5, 0 , 1, 0 b · § a¨ x 2 x ¸ c a ¹ © (b) The x-intercepts and the solutions of the equation are the same. § b b2 b2 · a¨ x 2 x ¸ c 2 a 4a 4a 2 ¹ © (c) 0 x2 4x 5 0 x 5 x 1 x 5 or x § § b2 · b b2 · a¨ 2 ¸ c a¨ x 2 x 2¸ a 4a ¹ © © 4a ¹ 1 2 The x-intercepts are 5, 0 and 1, 0 . b · b2 4ac § a¨ x ¸ 2a ¹ 4a 4a © 2×2 5x 3 b · 4ac b 2 § a¨ x ¸ 2a ¹ 4a © 2 70. y (a) From the graph it appears that the x-intercepts are 1 , 0 and 3, 0 . 2 § b 4ac b 2 · The vertex is ¨ , ¸. 4a © 2a ¹ (b) The x-intercepts and solutions of 2 x 2 5 x 3 0 are the same. § b · f ¨ ¸ © 2a ¹ 2×2 5x 3 0 2x 1 x 3 0 (c) 1 or 2 x § b2 · b2 a¨ 2 ¸ c © 4a ¹ 2a 3 Ÿ The x-intercepts are x 1 ,0 2 and 3, 0 . 71. y (a) From the graph it appears that the x-intercept is 1, 0 . (b) The x-intercept and the solution to x 2 2 x 1 0 are the same. (c) x 2 2 x 1 0 x 2x 1 0 x 1 2 x 1 x b2 2b 2 4ac 4a 4a 4a b 2 4ac 4a x2 2 x 1 2 2 § b · § b · a ¨ ¸ b¨ ¸ c © 2a ¹ © 2a ¹ 4ac b 2 4a § b § b ·· Thus, the vertex occurs at ¨ , f ¨ ¸ ¸. 2 a © 2a ¹ ¹ © 74. (a) Yes, it is possible for a quadratic equation to have only one x-intercept. That happens when the vertex is the x-intercept. 0 0 1 Ÿ The x-intercept is at 1, 0 . (b) Yes. If the vertex is above the x-axis and the parabola opens upward, or if the vertex is below the x-axis and the parabola opens downward, then the graph of the quadratic equation will have no x-intercepts. Examples: f x x 2 4; g x x2 1 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 75. Let x the first number and y the second number. 78. Let x 110 Ÿ y 110 x. The product is P x xy x 110 x 110 x x 2 . x 2 110 x P x x 2 110 x 3025 3025 ª x 55 ¬ x 55 2 2 3025º ¼ 3025 x y S Ÿ y The product is P x 42 21 3 and y the second number. S x. the first number and y The maximum value of the product occurs at the vertex of P x and is 147. This happens when x 21 The maximum value of the product occurs at the vertex y 55. of P x and is 3025. This happens when x 76. Let x the first number and y Then the sum is 79. xS x xy Sx x 2 . 7. So, the numbers are 21 and 7. x y y Sx x 2 P x x 2 Sx x 2x 2 y § S S · ¨ x 2 Sx ¸ 4 4 ¹ © 2 2 2 (a) A x 50 x x 50 x xy Domain: 0 x 50 The maximum value of the product occurs at the vertex of P x and is S 2 4. This happens when y 100 y S· S2 § ¨ x ¸ 2¹ 4 © x 183 the second number. 42 x Then the sum is x 3 y 42 Ÿ y . 3 § 42 x · xy x¨ The product is P x ¸. © 3 ¹ 1 P x x 2 42 x 3 1 x 2 42 x 441 441 3 1ª 1 2 2 x 21 147 x 21 441º ¼ 3¬ 3 Then the sum is x y Quadratic Function Functions and Models (b) A 700 S 2. 560 420 77. Let x the first number and y the second number. 280 140 Then the sum is x 2y 24 Ÿ y The product is P x P x x 24 x . 2 xy 10 § 24 x · x¨ ¸. © 2 ¹ 1 x 2 24 x 2 1 x 2 24 x 144 144 2 1ª 1 2 2 x 12 72 x 12 144º ¼ 2¬ 2 The maximum value of the product occurs at the vertex of P x and is 72. This happens when x 12 and y 24 12 2 20 30 40 50 (c) The area is maximum (625 square feet) when x y 25. The rectangle has dimensions 25 ft u 25 ft. 80. Let x length of rectangle and y 2x 2 y y (a) A x 36 18 x x 18 x xy Domain: 0 x 18 (b) (c) The area is maximum (81 square meters) when x y 9 meters. The rectangle has dimensions 9 meters u 9 meters. A 100 6. So, the numbers are 12 and 6. width of rectangle. 80 60 40 20 INSTRUCTOR USE ONLY x 4 8 12 16 20 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 184 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions 4 24 x2 x 12 9 9 81. y 82. y 24 9 2 4 9 b 2a The vertex occurs at 3. The maximum height is y 3 4 2 24 3 3 12 9 9 16 feet. 16 2 9 x x 1.5 2025 5 (a) The ball height when it is punted is the y-intercept. y 16 9 2 0 0 1.5 2025 5 (b) The vertex occurs at x 1.5 feet b 2a § 3645 · The maximum height is f ¨ ¸ © 32 ¹ 95 2 16 2025 3645 . 32 2 16 § 3645 · 9 § 3645 · ¨ ¸ ¨ ¸ 1.5 2025 © 32 ¹ 5 © 32 ¹ 6561 6561 1.5 64 32 6561 13,122 96 64 64 64 6657 feet | 104.02 feet. 64 (c) The length of the punt is the positive x-intercept. 0 16 2 9 x x 1.5 2025 5 95 r x 95 2 4 1.5 16 2025 32 2025 | 1.8 r 1.81312 0.01580247 x | 0.83031 or x | 228.64 The punt is about 228.64 feet. 800 10 x 0.25 x 2 83. C 0.25 x 2 10 x 800 The vertex occurs at x b 2a The cost is minimum when x 10 2 0.25 20. 12 p 2 150 p 86. R p (a) R $4 R $6 20 fixtures. R $8 230 20 x 0.5 x 2 84. P 20 2 0.5 20. Because x is in hundreds of dollars, 20 u 100 2000 dollars is the amount spent on advertising that gives maximum profit. 150 $4 $408 12 $6 2 150 $6 $468 12 $8 2 150 $8 $432 p b 2a 25 p 1200 p (a) R 20 $14,000 thousand $14,000,000 R 25 $14,375 thousand $14,375,000 R 30 $13,500 thousand $13,500,000 150 2 12 $6.25. Revenue is maximum when price $6.25 per pet. The maximum revenue is R $6.25 2 85. R p 2 (b) The vertex occurs at b 2a The vertex occurs at x 12 $4 12 $6.25 2 150 $6.25 $468.75. (b) The revenue is a maximum at the vertex. b 2a 1200 2 25 R 24 14,400 24 The unit price that will yield a maximum revenue of $14,400 thousand is $24. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE 87. (a) (b) (c) x Quadratic Function Functions and Models 1 200 4 x 3 4 50 x 3 8 x 50 x 4x 3y 200 Ÿ y A ª4 º 2 x « 50 x » ¬3 ¼ y x Section 2.1 2 xy 8 x 50 x 3 185 3 x This area is maximum when x 25 feet and 100 1 y 33 feet. 3 3 A 5 600 10 1066 15 1400 20 1600 25 1666 30 1600 2 3 8 x 50 x 3 8 x 2 50 x 3 8 2 x 50 x 625 625 3 8ª 2 x 25 625º ¼ 3¬ 8 5000 2 x 25 3 3 (d) A 2 3 The maximum area occurs at the vertex and is 5000 3 square feet. This happens when x 25 feet 200 4 25 and y 2000 3 100 3 feet. 1 50 feet by 33 feet. 3 The dimensions are 2 x (e) They are all identical. 0 60 0 This area is maximum when x 100 1 y 33 feet. 3 3 x 88. (a) (d) Area of rectangular region: A y 1 y 2 Distance around two semicircular parts of track: §1 · d 2S r 2S ¨ y ¸ S y ©2 ¹ (b) Radius of semicircular ends of track: r (c) Distance traveled around track in one lap: S y 2 x 200 d Sy y 89. (a) Revenue Let y m 1 33 feet 3 25 feet and y x 25 feet and 200 2 x xy § 200 2 x · x¨ ¸ S © ¹ 1 200 x 2 x 2 S 2 S 2 S 2 S x 2 100 x x 2 100 x 2500 2500 x 50 2 5000 The area is maximum when x 200 2 50 100 y . 200 2 x S S S 50 and S number of tickets sold price per ticket attendance, or the number of tickets sold. 100, 20, 1500 y 1500 100 x 20 y 1500 y 100 x 2000 100 x 3500 R x y x R x 100 x 3500 x R x 100 x 2 3500 x INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 186 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions (b) The revenue is at a maximum at the vertex. 3500 2 100 b 2a 17.5 100 17.5 R 17.5 2 3500 17.5 $30,625 A ticket price of $17.50 will yield a maximum revenue of $30,625. Area of rectangle Area of semicircle 90. (a) Area of window xy 1 2 S radius 2 xy 1 § x· S¨ ¸ 2 © 2¹ xy 2 Sx2 8 To eliminate the y in the equation for area, introduce a secondary equation. perimeter of rectangle perimeter of semicircle Perimeter 1 circumference 2 1 2y x 2S ˜ radius 2 § x· 2y x S ¨ ¸ © 2¹ 2y x 16 16 16 y 8 Sx 1 x 2 4 Substitute the secondary equation into the area equation. xy Area S x2 8 S x · S x2 1 § x¨ 8 x ¸ 2 4 ¹ 8 © 8x S x2 1 2 S x2 x 2 4 8 8x 1 2 S x2 x 2 8 1 64 x 4 x 2 S x 2 8 (b) The area is maximum at the vertex. 8x Area 1 2 S x2 x 2 8 § 1 S· 2 ¨ ¸ x 8x 8¹ © 2 b 2a x y 8 8 | 4.48 § 1 S· 2¨ ¸ 8¹ © 2 S 4.48 1 4.48 | 2.24 2 4 The area will be at a maximum when the width is about 4.48 feet and the length is about 2.24 feet. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 91. (a) Quadratic Function Functions and Models 187 4200 0 55 0 (b) The maximum annual consumption occurs at the point 16.9, 4074.813 . 4075 cigarettes 1966 o t 16 The maximum consumption occurred in 1966. After that year, the consumption decreases. It is likely that the warning was responsible for the decrease in consumption. (c) Annual Consumption per smoker Annual consumption in 2005 ˜ total population total number of smokers in 2005 1487.9 296,329,000 59,858,458 7365.8 About 7366 cigarettes per smoker annually Number of cigarettes per year Number of days per year Daily Consumption per smoker 7366 365 | 20.2 About 20 cigarettes per day 92. (a), (b) 95. True. The negative leading coefficient causes the parabola to open downward, making the vertex the maximum point on the graph. 7 0 7 0 y 0.0408 x 2 0.715 x 2.82 (c) The model is a good fit to the actual data. (d) The greatest sales occurred in the year 2007. (e) Sales will be at a maximum at the vertex. x b 2a 0.715 | 8.76 2 0.0408 96. True. The positive leading coefficient causes the parabola to open upward, making the vertex the minimum point on the graph. x 2 bx 75, maximum value: 25 97. f x The maximum value, 25, is the y-coordinate of the vertex. Find the x-coordinate of the vertex: x Sometime during 2008. (f ) 2011 o Use x y 11 11. 0.0408 11 2 0.715 11 2.82 | 5.75 Sales in the year 2011 will be about $5.75 billion. 93. True. The equation 12 x 1 0 has no real solution, so the graph has no x-intercepts. 2 g x is 54 , 71 . 4 b 2 1 b 2 f x x 2 bx 75 §b· f¨ ¸ © 2¹ §b· §b· ¨ ¸ b¨ ¸ 75 2 © ¹ © 2¹ 25 2 b2 b2 75 4 2 400 b2 4 b2 r20 b 100 94. True. The vertex of f x is 54 , 53 and the vertex of 4 b 2a INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 188 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x 2 bx 16, maximum value: 48 98. f x x 2 bx 25 , minimum value: –50 100. f x The maximum value, 48, is the y-coordinate of the vertex. The minimum value, –50, is the y-coordinate of the vertex. Find the x-coordinate of the vertex: Find the x-coordinate: b 2a x b 2 1 b 2 x b 2a f x x 2 bx 16 §b· f¨ ¸ © 2¹ §b· §b· ¨ ¸ b¨ ¸ 16 2 © ¹ © 2¹ § b· f ¨ ¸ © 2¹ b2 b2 16 4 2 50 b 21 b 2 x 2 bx 25 f x 2 2 § b· § b· ¨ ¸ b¨ ¸ 25 2 © ¹ © 2¹ 25 256 b2 4 b2 100 b2 b2 25 4 2 b 2 4 b2 r16 b r10 b 48 64 x 2 bx 26 , minimum value: 10 99. f x The minimum value, 10, is the y-coordinate of the vertex. Find the x-coordinate of the vertex: x b 2a b 21 b 2 f x x 2 bx 26 § b· f ¨ ¸ © 2¹ 2 64 r8 b 16 ax 2 bx c has two real zeros, then by the 103. If f x Quadratic Formula they are x 2 b b 26 4 2 b2 4 b2 10 y value is adjusted by a factor of a, and the parabola becomes narrower or wider. Every point on the parabola is shifted up k units. 102. Conditions (a) and (d) are preferable because profits would be increasing. § b· § b· ¨ ¸ b¨ ¸ 26 © 2¹ © 2¹ 2 101. The graph of f x is moved h units to the right. Every b r b 2 4ac . 2a The average of the zeros of f is b b b 2 4ac 2a 2 b 2 4ac 2a 2b 2a 2 b . 2a This is the x-coordinate of the vertex of the graph. Section 2.2 Polynomial Functions of Higher Degree 1. continuous 9. f x 2 x 3 is a line with y-intercept 0, 3 . Matches graph (c). 2. Leading Coefficient Test 10. f x 3. x n x 2 4 x is a parabola with intercepts 0, 0 and 4, 0 and opens upward. Matches graph (g). 4. n; n 1 5. (a) solution; (b) x a ; (c) x-intercept 11. f x 2 x 2 5 x is a parabola with x-intercepts 0, 0 and 52 , 0 and opens downward. Matches graph (h). 6. repeated; multiplicity 7. touches; crosses 8. standard 12. f x 2 x3 3 x 1 has intercepts 0, 1 , 1, 0 , 12 12 3, 0 and 12 12 3, 0 . INSTRUCTOR USE ONLY Matches graph (f). © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.2 14 x 4 3 x 2 has intercepts 0, 0 and 13. f x Polynomial Functions of Hi Higher Degree H 18. y r2 3, 0 . Matches graph (a). 189 x5 x 1 (a) f x 5 y 13 x3 x 2 43 has y-intercept 0, 43 . 14. f x 4 3 Matches graph (e). 2 1 x 4 2 x3 has intercepts 0, 0 and 2, 0 . 15. f x x – 4 –3 1 2 3 4 Matches graph (d). –3 –4 1 x 5 2 x 3 9 x has intercepts 5 5 16. f x 0, 0 , 1, 0 , 1, 0 , 3, 0 , 3, 0 . Matches graph (b). Horizontal shift one unit to the left x5 1 (b) f x 17. y x3 y x 4 (a) f x 3 y 4 4 3 3 2 2 1 x −2 1 2 4 5 x 6 –4 –3 –2 1 2 3 4 −2 −3 –3 −4 –4 Vertical shift one unit upward Horizontal shift four units to the right (b) f x x3 4 1 12 x5 (c) f x y y 4 2 3 1 2 x −3 −2 1 2 3 4 x −2 –4 –3 –2 −3 2 3 4 –3 −6 –4 Reflection in the x-axis, vertical shrink each y -value is multiplied by 12 , and vertical shift one unit upward Vertical shift four units downward (c) f x 1 x3 4 y (d) f x 12 x 1 4 5 y 3 2 4 1 3 x −4 −3 −2 2 3 2 4 1 −2 x −3 1 –5 –4 –3 –2 2 3 −4 –3 Reflection in the x-axis and a vertical shrink each y -value is multiplied by 14 (d) f x x 4 3 –4 Refection in the x -axis, vertical shrink each y -value is multiplied by 12 , and y 4 horizontal shift one unit to the left 2 1 x −2 1 2 3 4 5 6 −2 −3 −4 −5 −6 INSTRUCTOR USE ONLY Horizontal shift four units to the right and vertical shift four units downward © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 190 NOT FOR SALE Chapter 2 19. y Polynomial ynomial and Rational Function Functions x4 x 3 (a) f x 4 (b) x4 3 f x y f x 4 x4 y 6 4 6 5 3 5 4 2 3 f (c) y 1 3 x 2 –4 –3 –2 1 2 3 2 4 1 x –5 –4 –3 –2 –1 1 2 x 3 –4 –3 –2 –2 –4 Horizontal shift three units to the left (e) f x 2 3 4 –2 Vertical shift three units downward 1 x 14 2 (d) f x 1 –1 4 2x y 1 Reflection in the x-axis and then a vertical shift four units upward (f ) 4 1x 2 f x y 2 y 6 6 5 4 3 2 1 x –4 –3 –2 –1 1 2 3 −4 −3 −2 −1 −1 4 Horizontal shift one unit to the right and a vertical shrink each y -value is multiplied by 12 x6 (a) f x 1 2 3 −4 −3 4 1 3 4 18 x 6 Vertical shift two units downward and a horizontal stretch each y -value Vertical shift one unit upward and a horizontal shrink each y -value is multiplied by 16 (b) x 2 f x y 6 4 1 is multipied by 16 (c) f x x6 5 y y 4 3 3 2 2 1 x 1 x –4 –3 –2 x −1 −1 −2 –2 20. y x 2 –1 3 −4 −3 −2 x 4 –5 –4 –2 1 2 1 −1 2 3 4 3 –2 –3 –4 –4 Horizontal shift two units to the left and a vertical shift four units downward Vertical shrink each y -value is multiplied by 18 and reflection in the x-axis (d) f x 14 x 6 1 (e) 1x 4 f x y 6 2 Vertical shift five units downward (f ) f x 2x y 6 1 1 2 y 4 3 2 x –4 –3 –2 –1 2 3 4 –2 –3 –4 Reflection in the x-axis, vertical shrink each y -value is multiplied by 14 , and x −8 − 6 −2 2 6 8 − 4 − 3 −2 −1 4 −2 −4 Horizontal stretch each x-value is multiplied by 4 , and vertical shift two units downward x 3 Horizontal shrink each x -value is multiplied by 12 , and vertical shift one unit downward INSTRUCTOR USE ONLY vertical shift one unit upward © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.2 21. f x 1 x3 4 x 5 The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right. 2 x 3x 1 2 Degree: 2 Leading coefficient: 2 The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right. 3 2 t 3t 6 4 Leading coefficient: 3 4 The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 78 s 3 5s 2 7 s 1 30. f s Degree: 3 Leading coefficient: 78 The degree is odd and the leading coefficient is negative. 5 72 x 3x 2 The graph rises to the left and falls to the right. Degree: 2 Leading coefficient: 3 3x 3 9 x 1; g x 31. f x The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 24. h x 191 Degree: 2 Leading coefficient: 15 23. g x 29. f x Degree: 3 22. f x Polynomial Functions of Hi Higher Degree H 8 g f −4 1 x6 Degree: 6 Leading coefficient: 1 The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 3×3 4 −8 13 x3 3 x 2 , g x 32. f x 13 x3 6 25. f x 2.1x 4 x 2 5 g 3 f −9 Degree: 5 Leading coefficient: 2.1 −6 The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. 26. f x 33. f x 4 x5 7 x 6.5 x 4 4 x3 16 x ; g x x4 12 Degree: 5 Leading coefficient: 4 The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right. 27. f x 9 −8 f − 20 34. f x 6 2 x 4 x 2 5 x3 8 g 3x 4 6 x 2 , g x 3×4 5 Degree: 3 Leading coefficient: 5 f The degree is odd and the leading coefficient is negative. g −6 6 The graph rises to the left and falls to the right. 3x 4 2 x 5 4 Degree: 4 3 Leading coefficient: 4 The degree is even and the leading coefficient is positive. −3 28. f x The graph rises to the left and rises to the right. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 192 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 35. f x x 2 36 38. f x x 2 10 x 25 (a) 0 x 2 36 (a) 0 x 2 10 x 25 x 6 x 6 0 0 x 6 0 x 6 x 6 (b) x 2 5 Zero: x x 6 x 5 5 has a multiplicity of 2 (even multiplicity). Turning points: 1 (the vertex of the parabola) Zeros: r6 (c) 25 (b) Each zero has a multiplicity of one (odd multiplicity). Turning points: 1 (the vertex of the parabola) (c) − 25 15 −5 6 − 12 12 1 x2 1 x 2 3 3 3 39. f x (a) 0 − 42 36. f x 81 x 2 (a) 0 81 x 2 1 2 x x 2 3 1 x 2 3 x 1 2, x Zeros: x 9 x 9 x 0 1 2 x 13 x 23 3 1 (b) Each zero has a multiplicity of 1 (odd multiplicity). 9 x 0 9 x 0 9 x x 9 Turning points: 1 (the vertex of the parabola) (c) 4 Zeros: r 9 −6 (b) Each zero has a multiplicity of one (odd multiplicity). 6 −4 Turning points: 1 (the vertex of the parabola) (c) 90 1 2 5 3 x x 2 2 2 40. f x − 15 (a) For 15 −9 t 6t 9 2 37. h t (a) 0 t 6t 9 2 Zero: t (b) t 3 5 r 2 37 4 Zeros: x 5 ,c 2 3 . 2 2 §5· § 1 ·§ 3 · ¨ ¸ 4¨ ¸¨ ¸ © 2¹ © 2 ¹© 2 ¹ 1 3 has a multiplicity of 2 (even multiplicity). 10 1 ,b 2 5 r 2 x t 3 0, a 2 Turning points: 1 (the vertex of the parabola) (c) 1 2 5 3 x x 2 2 2 5 r 37 2 (b) Each zero has a multiplicity of 1 (odd multiplicity). Turning points: 1 (the vertex of the parabola) (c) −6 3 12 −2 −8 4 −5 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.2 41. f x 3 x 3 12 x 2 3 x (a) 0 3 x3 12 x 2 3 x Zeros: x 3x x 2 4 x 1 2r 0, x Polynomial Functions of Hi Higher Degree H 44. f x x 4 x3 30 x 2 (a) 0 x 4 x3 30 x 2 0 x 2 x 2 x 30 0 x2 x 6 x 5 x2 0 x 6 0 x 5 0 x 0 x 6 x 5 3 (by the Quadratic Formula) (b) Each zero has a multiplicity of 1 (odd multiplicity). Turning points: 2 (c) 8 Zeros: x −6 5 6, x 6 (b) The multiplicity of x 0 is 2 (even multiplicity). The multiplicity of x 6 is 1 (odd multiplicity). The multiplicity of x 5 is 1 (odd multiplicity). − 24 Turning points: 3 42. g x 5x x2 2x 1 (a) 0 5x x2 2x 1 0 x x2 2x 1 For x 2 2 x 1 x 0, x 193 2 r (c) 0, a 2 60 −9 2 2, c 1, b 9 1. 4 1 1 −300 t 5 6t 3 9t 45. g t 21 1r tt 3 2 t t 4 6t 2 9 t 3 r 3 t t2 3 2 2 2 Zeros: x 1r 0, x Zeros: t 2 (b) Each zero has a multiplicity of 1 (odd multiplicity). (b) t 0, t 0 has a multiplicity of 1 (odd multiplicity). r t Turning points: 2 (c) t 5 6t 3 9t (a) 0 2r 8 2 3 each have a multiplicity of 2 (even multiplicity). 16 Turning points: 4 −6 (c) 6 6 −9 − 16 9 t 3 8t 2 16t 43. f t −6 (a) 0 t 8t 16t 0 t t 2 8t 16 0 tt 4 t 4 t 0 t 4 0 t 4 t 0 t 4 t Zeros: t 0, t 4 3 2 46. (a) f x x5 x3 6 x 0 x x4 x2 6 0 0 x x2 3 x2 2 4 Zeros: x 0, r 2 (b) Each zero has a multiplicity of 1 (odd multiplicity). (b) The multiplicity of t 0 is 1 (odd multiplicity). The multiplicity of t 4 is 2 (even multiplicity). Turning points: 2 (c) 6 Turning points: 2 10 (c) −9 9 INSTRUCTOR USE ONLY −6 −9 9 −2 − 2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 194 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 47. f x 3x 4 9 x 2 6 50. f x x 3 4 x 2 25 x 100 (a) 0 3x 4 9 x 2 6 (a) 0 x 2 x 4 25 x 4 0 3 x 3x 2 0 x 2 25 x 4 0 3 x2 1 x2 2 0 x 5 x 5 x 4 4 2 No real zeros (c) r 5, 4 Zeros: x (b) Turning points: 1 (b) Each zero has a multiplicity of 1 (odd multiplicity). 21 Turning points: 2 (c) −6 140 6 −3 −9 48. f x 2 x 4 2 x 2 40 (a) 0 2 x 4 2 x 2 40 51. y 0 2 x 4 x 2 20 (a) 0 2 x2 4 x2 5 Zeros: x r 4 x3 20 x 2 25 x 12 −2 5 (b) x-intercepts: 0, 0 , 52 , 0 Turning points: 3 20 −6 (c) 0 4 x3 20 x 2 25 x 0 x 4 x 2 20 x 25 0 x 2x 5 x 0, 52 6 − 60 x3 3 x 2 4 x 12 49. g x (a) 0 x3 3 x 2 4 x 12 x2 4 x 3 Zeros: x r 2, x 6 −4 (b) Each zero has a multiplicity of 1 (odd multiplicity). (c) 9 − 20 x2 x 3 4 x 3 x 2 x 2 x 3 (d) The solutions are the same as the x-coordinates of the x-intercepts. 52. y 4 x3 4 x 2 8 x 8 (a) 3 2 2 −3 3 (b) Each zero has a multiplicity of 1 (odd multiplicity). Turning points: 2 (c) 4 −8 − 11 7 (b) −16 1, 0 , 2, 0 , 2, 0 (c) 0 4 x3 4 x 2 8 x 8 0 4 x2 x 1 8 x 1 0 4×2 8 x 1 0 4 x2 2 x 1 x r 2, 1 (d) The solutions are the same as the x-coordinates of the x-intercepts. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.2 53. y x5 5 x3 4 x (a) Polynomial Functions of Higher Hi H Degree 59. f x 195 x 0 x 4 x 5 x x 2 9 x 20 4 x3 9 x 2 20 x −6 6 Note: f x ax x 4 x 5 has zeros 0, 4, and 5 for all real numbers a z 0. −4 (b) x-intercepts: 0, 0 , r1, 0 , r2, 0 (c) 0 60. f x x x 2 11x 10 x5 5 x3 4 x x3 11x 2 10 x 0 x x2 1 x2 4 0 x x 1 x 1 x 2 x 2 x 0, r1, r 2 (d) The solutions are the same as the x-coordinates of the x-intercepts. 54. y 1 x3 x 2 9 4 (a) 12 −18 Note: f x 61. f x x 4 4 x3 9 x 2 36 x Note: f x 18 62. f x x x4 5×2 4 x 5 5 x3 4 x 0, r 3 Note: f x x 0 x 8 63. f x ax x 8 has zeros 0 and 8 for all real 3º ¼ ªx 1 ¬ 3 ºª ¼¬ x 1 3 º¼ 2 3 2 x2 2x 2 x2 7 x ax x 7 has zeros 0 and 7 for all real numbers a z 0. Note: f x 1 64. f x x 2 x 6 x 2 4 x 12 a x 2 x 6 has zeros 2 and 6 for a x 2 2 x 2 has zeros 3 and 1 3 for all real numbers a z 0. x 2 ªx 4 ¬ x 2 ª¬ x 4 x 2ªx 4 ¬ all real numbers a z 0. 58. f x 3 ºª x 1 ¼¬ x2 2x 1 3 x 0 x 7 Note: f x ªx 1 ¬ x 1 numbers a z 0. 57. f x ax x 2 x 1 x 1 x 2 has zeros 2, 1, 0, 1, and 2 for all real numbers a z 0. x 8x Note: f x x 1 x 0 x 1 x 2 x x2 4 x2 1 2 56. f x x 2 x x 2 x 1 x 1 x 2 1 x3 x 2 9 4 Note: f x a x 4 4 x3 9 x 2 36 x has zeros 4, 3, 3, and 0 for all real numbers a z 0. (d) The solutions are the same as the x-coordinates of the x-intercepts. 55. f x x 4 x 3 x 3 x 0 x 4 x2 9 x (b) x-intercepts: 0, 0 , 3, 0 , 3, 0 x ax x 1 x 10 has zeros 0, 1, and 10 for all real numbers a z 0. −12 (c) 0 x 0 x 1 x 10 2 x x 4 2 5 ºª x 4 ¼¬ 5 ºª ¼¬ x 4 5º ¼ 5 º¼ 5º ¼ 5x 2 x 4 2 10 x 4 x 5 x 8 x 16 x 5 x 2 x 16 x 32 10 x x 20 x3 10 x 2 27 x 22 3 2 Note: f x a x 4 x 5 has zeros 4 and 5 for Note: f x 2 2 a x3 10 x 2 27 x 22 has zeros INSTRUCTOR USE ONLY all real numbers a z 0. 2, 4 5, and 4 5 forr all real numbers a z 00. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 196 Chapter 2 65. f x NOT FOR SALE Polynomial ynomial and Rational Function Functions x 3 x 3 68. f x x2 6 x 9 zero x 66. f x x 2 x 2 11x 28 a x 2 6 x 9 , a z 0, has degree 2 and Note: f x 3. x 12 x 6 a x 2 18 x 72 , a z 0, has degree 2 and zeros x 12 and 6. 2, 4, and 7. x 0 x 3 x 3 x x zeros x x x2 4 x 5 70. f x x 4x 5x 2 Note: f x ax x 2 4 x 5 , a z 0, has degree 3 and zeros x 0, 5, and 1. 3 x 3 x3 3x a x3 3x , a z 0, has degree 3 and Note: f x x 0 x 5 x 1 3 degree 3 and zeros x 69. f x Note: f x x3 9 x 2 6 x 56 a x3 9 x 2 6 x 56 , a z 0, has Note: f x x 2 18 x 72 67. f x x2 x4 x7 0, 3, and 3. x 0 x 2 2 ª x 2 2 º ¬ ¼ x x 2 2 x 2 2 x x2 8 x3 8 x Note: f x a x3 8 x has these zeros for all real numbers a z 0. 71. f x x 1 ª¬ x 2 º¼ ª x 1 ¬ x 1 x 2 ª¬ x 1 x2 x 2 ª x 1 ¬ 2 3 ºª x 1 ¼¬ 3 ºª ¼¬ x 1 3º ¼ 3 º¼ 3º ¼ x2 x 2 x2 2x 2 x 4 x3 6 x 2 2 x 4 Note: f x 72. f x a x 4 x3 6 x 2 2 x 4 has these zeros for all real numbers a z 0. x 3 ª¬ x 2 º¼ ª x 2 ¬ x 3 x 2 ª¬ x 2 x2 x 6 ª x 2 ¬ 2 5 ºª x 2 ¼¬ 5 ºª ¼¬ x 2 5º ¼ 5 º¼ 5º ¼ x2 x 6 x2 4 x 1 x 4 5 x3 3 x 2 25 x 6 Note: f x 73. f x a x 4 5 x3 3 x 2 25 x 6 has these zeros for all real numbers a z 0. x4 x 4 x5 4 x 4 or f x x3 x 4 2 x5 8 x 4 16 x3 or f x x2 x 4 3 x5 12 x 4 48 x3 64 x 2 or f x x x 4 4 x5 16 x 4 96 x3 256 x 2 256 x Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x 0 and 4. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.2 x 1 74. f x 2 x 4 x 7 x 8 2 Polynomial Functions of Hi Higher Degree H x5 17 x 4 79 x3 11x 2 332 x 224 or f x x 1 x 4 or f x x 1 x 4 x 7 or f x x 1 x 4 x 7 x 8 x 7 x 8 x5 22 x 4 169 x3 496 x 2 208 x 896 2 x 8 x5 25 x 4 223 x3 787 x 2 532 x 1568 2 x5 26 x 4 241×3 884 x 2 640 x 1792 1, 4, 7, and 8. Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x x3 25 x 75. f x x x 5 x 5 x 2 10 x 16 78. g x x 2 x 8 (a) Falls to the left; rises to the right (a) Falls to the left; falls to the right (b) Zeros: 0, 5, 5 (b) Zeros: 2, 8 (c) (c) x 2 1 0 1 2 f x 42 24 0 24 42 (d) 197 (d) y x 1 3 5 7 9 g x 7 5 9 5 7 y 10 48 8 (−5, 0) −2 2 6 (5, 0) (0, 0) − 8 −6 4 6 x 4 8 − 24 2 − 36 (2, 0) − 48 4 x4 9 x2 76. f x x2 x 3 x 3 (b) Zeros: 0, 2 x 2 1 0 1 2 f x 24 8 0 8 24 (d) x2 x 2 (a) Falls to the left; rises to the right (b) Zeros: 3, 0, 3 (c) x 10 x3 2 x 2 79. f x (a) Rises to the left; rises to the right (8, 0) 6 (c) y x 1 0 1 2 1 2 3 f x 3 0 83 1 0 9 (d) y 15 4 10 (− 3, 0) 5 (0, 0) −4 −2 − 1 1 (3, 0) 2 3 2 x 4 1 (0, 0) (2, 0) −4 −3 −2 −1 3 x 4 −20 −25 1 t 12 7 4 2 1 t 2 2t 15 4 77. f t t f t 2 x 4 2 x x2 (a) Rises to the left; falls to the right (b) No real zeros (no x-intercepts) (c) 8 x3 80. f x (a) Rises to the left; rises to the right (b) Zero: 2 1 0 1 2 3 4.5 3.75 3.5 3.75 4.5 (d) The graph is a parabola with vertex 1, 72 . y (c) x 2 1 0 1 2 f x 16 9 8 7 0 (d) y 14 8 12 10 6 6 4 2 INSTRUCTOR NSTR STR TR USE SE ONLY 2 t –44 – –2 2 −44 −33 −2 2 −1 1 −1 (2, 0) 1 3 x 4 4 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 198 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 3×3 15 x 2 18 x 81. f x 3x x 2 x 3 4 x3 4 x 2 15 x 82. f x (a) Falls to the left; rises to the right x 4 x 2 4 x 15 (b) Zeros: 0, 2, 3 x 2x 5 2x 3 (c) x 0 1 2 2.5 3 3.5 (a) Rises to the left; falls to the right f x 0 6 0 1.875 0 7.875 (b) Zeros: 32 , 0, 52 (d) (c) y 7 6 5 4 3 2 3 2 1 0 1 2 3 f x 99 18 7 0 15 14 27 (0, 0) (4, 0) 2 6 (d) y (2, 0) (0, 0) −3 −2 −1 x 20 (3, 0) 1 16 x 4 5 6 12 −2 8 (− 23, 0) 4 (0, 0) ( 25, 0) 1 3 −4 −3 −2 5 x 2 x3 83. f x x 4 x2 5 x (a) Rises to the left; falls to the right (d) y (b) Zeros: 0, 5 (c) 2 5 (−5, 0) −15 x 5 4 3 2 1 0 1 f x 0 16 18 12 4 0 6 (0, 0) x −10 5 10 −20 48 x 2 3 x 4 84. f x 3x 2 x 2 16 (a) Rises to the left; rises to the right (d) (b) Zeros; 0, r 4 (c) y (− 4, 0) 100 x 4 3 2 1 0 1 2 3 4 5 f x 675 0 189 144 45 0 45 –144 –189 0 675 −6 −2 x −200 −300 x2 x 4 85. f x (a) Falls to the left; rises to the right (d) y (b) Zeros: 0, 4 (c) 2 –4 x 1 0 1 2 3 4 5 f x 5 0 3 8 9 0 25 –2 (0, 0) (4, 0) 2 6 x 8 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.2 1 x3 x 4 2 3 86. h x Polynomial Functions of Hi Higher Degree H 89. f x (a) Falls to the left; rises to the right x3 16 x 199 x x 4 x 4 32 (b) Zeros: 0, 4 −6 (c) x 1 0 1 2 3 4 5 h x 25 3 0 3 32 3 9 0 125 3 (d) 6 −32 Zeros: 0 of multiplicity 1; 4 of multiplicity 1; and 4 of y multiplicity 1. 14 12 10 90. f x 8 1 x4 2 x2 4 6 6 4 (0, 0) (4, 0) –4 –2 2 4 6 2 14 t 2 87. g t x 8 10 12 t 2 −9 2 −6 Zeros: 2.828 and 2.828 of multiplicity 1; 0 of multiplicity 2 (a) Falls to the left; falls to the right (b) Zeros: 2, 2 (c) 9 91. g x t 3 2 1 0 1 2 3 g t 25 4 0 94 4 94 0 25 4 1 x 12 x 3 5 2x 9 14 y (d) (− 2, 0) (2, 0) −12 t –3 –1 –1 1 2 18 3 −6 –2 Zeros: 1 of multiplicity 2; 3 of multiplicity 1; 92 of multiplicity 1 –5 –6 92. h x 2 3 1 x 1 x 3 10 88. g x 2 2 1 x 2 3x 5 5 21 (a) Falls to the left; rises to the right (b) Zeros: 1, 3 (c) −12 x 2 1 0 1 2 4 g x 12.5 0 2.7 3.2 0.9 2.5 (d) 12 −3 Zeros: 2, 53 , both with multiplicity 2 y 6 4 2 (−1, 0) –6 –4 –2 (3, 0) 4 6 x 8 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 200 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions l ˜w˜h 95. (a) Volume x 4 ; f x is even. 93. f x height length y 5 4 x width 2 36 2 x 36 2 x x So, V x 3 36 2 x x 36 2 x . (b) Domain: 0 x 18 2 1 The length and width must be positive. x –3 –2 –1 1 2 3 (c) –1 (a) g x f x 2 Vertical shift two units upward g x f x 2 f x 2 g x Even (b) g x f x 2 Horizontal shift two units to the left Box Height Box Width Box Volume, V 1 36 2 1 1ª¬36 2 1 º¼ 2 36 2 2 2 ª¬36 2 2 º¼ 3 36 2 3 3ª¬36 2 3 º¼ 4 36 2 4 4 ª¬36 2 4 º¼ 5 36 2 5 5ª¬36 2 5 º¼ 2 6 36 2 6 6 ª¬36 2 6 º¼ 2 7 36 2 7 7 ª¬36 2 7 º¼ 2 Neither odd nor even (c) g x f x x 4 x4 Reflection in the y-axis. The graph looks the same. Even (d) g x f x x Reflection in the x-axis (e) g x (d) f 1156 2 2 2 2048 2700 3136 3380 3456 3388 The volume is a maximum of 3456 cubic inches when the height is 6 inches and the length and width are each 24 inches. So the dimensions are 6 u 24 u 24 inches. 4 Even 1 x 2 2 3600 1 4 x 16 Horizontal stretch Even (f ) g x 0 1 f 2 x 1 4 x 2 The maximum point on the graph occurs at x Vertical shrink f x3 4 x3 4 4 4 x3 4 x3 , x t 0 l ˜w˜h 96. (a) Volume 8 x 12 x 6 x f D f x (b) x ! 0, f f x f x4 x 4 24 2 x 24 4 x x 2 12 x ˜ 4 6 x x Neither (h) g x 6. This agrees with the maximum found in part (c). Even (g) g x 18 0 12 x ! 0, 6 x ! 0 x 12 x 6 Domain: 0 x 6 4 (c) V 720 16 x 600 Even 480 360 94. R 1 x3 600 x 2 100,000 240 120 x The point of diminishing returns (where the graph changes from curving upward to curving downward) occurs when x 200. The point is 200, 160 which 1 2 3 4 5 6 x | 2.5 corresponds to a maximum of 665 cubic inches. corresponds to spending $2,000,000 on advertising to obtain a revenue of $160 million. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 97. False. A fifth-degree polynomial can have at most four turning points. x 1 98. True. f x 6 Polynomial and Synthetic Synth Division 201 100. (a) Degree: 3 Leading coefficient: Positive (b) Degree: 2 has one repeated solution. Leading coefficient: Positive 99. True. A polynomial of degree 7 with a negative leading coefficient rises to the left and falls to the right. (c) Degree: 4 Leading coefficient: Positive (d) Degree: 5 Leading coefficient: Positive Section 2.3 Polynomial and Synthetic Division 1. f x is the dividend; d x is the divisor: q x is the 9. y1 quotient: r x is the remainder x2 2 x 1 , y2 x 3 (a) and (b) 2. improper; proper x 1 2 x 3 3 −9 9 3. improper 4. synthetic division −9 5. Factor x 1 (c) x 3 x 2 2 x 1 x 2 3x x 1 x 3 2 6. Remainder 7. y1 x2 and y2 x 2 x 2 4 x 2 x 2 x 2 x2 0x 0 x2 2x 2 x 0 2 x 4 4 So, 8. y1 x2 x 2 10. y1 4 and y1 x 2 x 2 x 4 3x 2 1 and y2 x2 5 x 3x 1 So, x2 5 2 x 1 x4 x2 1 , y2 x2 1 x2 x2 8 (a) and (b) y2 . 39 x2 5 y2 . 6 6 −2 x2 (c) x 0 x 1 x 0 x x 2 0 x 1 x 4 0 x3 x 2 1 39 and y1 x 8 2 x 5 2 2 and y1 x 3 1 x2 1 −6 x2 8 2 4 x 5 x 3x 2 1 x4 5×2 8 x 2 1 8 x 2 40 39 4 x2 2x 1 x 3 So, y 2. 2 4 3 So, x4 x2 1 x2 1 x2 1 and y1 x2 1 y2 . 2x 4 11. x 3 2 x 2 10 x 12 2 x2 6x 4 x 12 4 x 12 0 2 x 2 10 x 12 x 3 2 x 4, x z 3 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. 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All Rights Reserved. 202 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions 5x 3 x 2 3x 9 12. x 4 5 x 17 x 12 5 x 2 20 x 3 x 12 3 x 12 0 17. x 3 x 0 x 2 0 x 27 x3 3x 2 3x 2 0 x 3x 2 9 x 9 x 27 9 x 27 0 2 5 x 2 17 x 12 x 4 3 5 x 3, x z 4 x x3 27 x 3 3x 1 2 13. 4 x 5 4 x3 7 x 2 11x 5 4 x3 5 x 2 12 x 2 11x 12 x 2 15 x 4x 5 4x 5 0 4 x3 7 x 2 11x 5 4x 5 x 2 5 x 25 x 2 3x 1, x z 18. x 5 x3 0 x 2 0 x 125 x3 5 x 2 5 x 2 0 x 5 x 2 25 x 25 x 125 25 x 125 0 5 4 x3 125 x 5 2x 4x 3 2 14. 3x 2 6 x3 16 x 2 17 x 6 12 x 2 17 x 12 x 2 8 x 7x 3 x 2 9x 6 9x 6 0 2 x 2 4 x 3, x z 2 3 x3 3x 2 1 15. x 2 x 4 5 x3 6 x 2 x 2 x 4 2 x3 3×3 6 x 2 3×3 6 x 2 16. x 3 x 4 x 2 x3 3x 2 1, x z 2 3 x 12 x 3x 7 x 2 3x 7 x 2 21x 18 x 12 18 x 54 42 3 2 x3 4 x 2 3 x 12 x 3 11 x 2 4 20. 2 x 1 8 x 5 8x 4 9 8x 5 2x 1 x3 9 x2 1 x 2 7 x 18 3 7 4 9 2x 1 x 21. x 2 0 x 1 x3 0 x 2 0 x 9 x3 0 x 2 x x 9 x 2 x 2 0 x 4 5 x3 6 x 2 x 2 x 2 x 2 5 x 25, x z 5 7 19. x 2 7 x 3 7 x 14 11 6 x3 4 x 2 6 x3 16 x 2 17 x 6 3x 2 x 2 3 x 9, x z 3 x 9 x2 1 x2 22. x3 0 x 2 0 x 1 x5 0 x 4 0 x3 0 x 2 0 x 7 x5 0 x 4 0 x3 x 2 7 x2 x5 7 x3 1 x 2 7 x 18 x x2 x2 7 x3 1 42 x 3 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 2x 8 23. x 2 0 x 1 2 x3 8 x 2 3x 9 2 x3 0 x 2 2 x 8 x 2 x 9 8 x 2 0 x 8 x 1 2 x3 8 x 2 3x 9 x2 1 2x 8 29. Polynomial and Synthetic Synth Division 3 6 x 1 x2 1 24. x 2 x 3 x 4 5 x3 0 x 2 20 x 16 x 4 x3 3 x 2 6 x3 3 x 2 20 x 6 x3 6 x 2 18 x 9 x 2 2 x 16 9 x 2 9 x 27 7 x 11 30. x 1 x 3 3 6 31. x 1 27. 5 17 3 3 15 3 2 10 25 2 5 0 3x 17 x 15 x 25 x 5 3 28. 3 2 5 5 18 7 32. 2 6 3 2 0 5 x3 18 x 2 7 x 6 x 3 7 12 192 2 32 199 8 9 18 8 0 18 0 9 0 18 32 18 0 32 0 16 0 9 x3 18 x 2 16 x 32 x 2 1 1 34. 6 3 75 250 10 100 250 10 25 0 0 72 18 12 72 2 12 0 3x 16 x 72 x 6 35. 4 2 5 5 36. 2 0 8 20 56 224 14 56 232 2 5 5 3x 2 2 x 12, x z 6 6 5x 6x 8 x 4 3 x 2 10 x 25, x z 10 16 3 5 x 2 14 x 56 0 6 8 10 20 52 10 26 44 5 x3 6 x 8 x 2 199 x 6 9 x 2 16, x z 2 0 x 3 75 x 250 x 10 248 x 3 4 x 2 9, x z 2 16 9 3 5 x 2 3 x 2, x z 3 2 x 2 2 x 32 2 9 6 9 20 4 x 8 x 9 x 18 x 2 2 3x 2 2 x 5, x z 5 15 14 4 33. 10 17 x 5 x 2x 1 6 x 2 25 x 74 12 4 25 15 248 2 3 26. x 2 2 x 1 2 x3 4 x 2 15 x 5 2 x3 4 x 2 2 x 17 x 5 2x 222 74 2 2x 2 75 25 2 x 14 x 20 x 7 x 6 6×2 8x 3 2 x3 4 x 2 15 x 5 18 2 7 x 11 x 6x 9 2 x x 3 x 1 26 3 2 x 3 25. x3 3x 2 3 x 1 x 4 0 x3 0 x 2 0 x 0 x 4 3×3 3x 2 x 3×3 3x 2 x 0 3×3 9 x 2 9 x 3 6×2 8x 3 x4 1 6 x 3 7 x 2 x 26 x 3 x2 6 x 9 x 4 5 x3 20 x 16 x2 x 3 7 6 203 232 x 4 44 x 2 INSTRUCTOR USE ONLY 5 x 2 10 x 26 © 2012 Cengage Learning. 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All Rights Reserved. 204 Chapter 2 37. 6 NOT FOR SALE Polynomial ynomial and Rational Function Functions 10 50 60 60 360 2160 10 10 60 360 1360 0 10 x 4 50 x 3 800 x 6 38. 3 1 1 10 x 3 10 x 2 60 x 360 13 0 0 120 80 3 48 144 16 48 144 432 312 936 856 x 13 x 120 x 80 x 3 5 39. 4 8 1 1 9 1 0 0 512 8 64 512 8 64 0 0 0 729 9 81 729 9 81 0 x 729 x 9 3 2 x 4 16 x 3 48 x 2 144 x 312 856 x 3 45. x 2 8 x 64, x z 8 1 41. 1360 x 6 1 2 4 46. 3 0 6 12 24 48 3 6 12 24 48 3 x 4 x 2 0 3 2 3 3 0 3 x 3 6 x 2 12 x 24 2 3 3 3 x x 2 4 43. 6 48 x 2 0 0 6 12 24 48 6 12 24 48 4 1 1 1 1 2 0 0 180 0 6 36 216 6 36 36 216 216 4 7 15 14 30 0 4 x 2 14 x 30, x z 4 0 5 9 2 1 2 3 4 3 4 9 8 49 8 3x 2 1 3 49 x 2 4 8 x 12 x3 x 2 14 x 11, k 1 48 3 x 6 x 12 x 24 x 2 180 x x x 6 1 0 3 1 44. 0 2 3×3 4 x 2 5 3 x 2 47. f x 42. 15 4 x3 16 x 2 23 x 15 1 x 2 x 2 9 x 81, x z 9 0 23 16 4 x3 512 x 8 40. 800 0 1 14 11 4 12 8 3 2 3 4 f x x 4 x 2 3x 2 3 f 4 43 42 14 4 11 3 48. f x x3 5 x 2 11x 8, k 2 –2 216 x3 6 x 2 36 x 36 x 6 1 1 5 11 8 2 14 6 7 3 2 5 f x x 2 x2 7 x 3 2 3 6 f 2 2 –6 11 2 3 1 3 5 3 x 2 x 2 x3 x 1 3 5 2 2 1 2 11 2 8 2 11 x 1 INSTRUCTOR USE ONLY x 2 3x 6 © 2012 Cengage Learning. 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NOT FOR SALE Section 2.3 49. f x 23 15 x 4 10 x3 6 x 2 14, k 10 6 0 14 10 0 4 83 0 6 4 34 3 15 15 23 f 23 4 15 23 10 23 3 x3 3 x 2 2 x 14, k 51. f x 2 6 23 f x x 2 2 f 14 50. f x 1 5 10 x 22 x 3x 4, k 22 10 10 1 5 2 3 4 2 4 75 20 7 13 5 x 15 10 x 2 20 x 7 13 5 f 15 10 15 53. f x 1 3 22 15 54. f x 2 3 4 f 1 4 4 3 10 2 3 4 2 4 3 2 2 3 0 4 1 3 61 3 x3 8 x 2 10 x 8, k 3 3 2 12 1 2 8 10 8 2 4 2 8 23 2 8 4 2 0 2 3 2 2 3 2 8 2 ªx 2 3 ¬ 3 3 2 2 2 x 3 2º 8 ¼ 2 2 14 8 5 5 1 5 2 5 4 5 2 5 5 10 1 2 5 2 5 6 x 5 ªx 2 2 ¬ 5 3 2 5 x 2 5º 6 ¼ 5 2 5 5 4 6 3 4 0 2 6 3 2 x 2 6 3 ª4 x 2 2 4 3 x 2 2 3 º ¬ ¼ 3 f x 2 3 2 3 4 3 2 13 5 12 x 1 2 6 f x 3 f 2 3 15 4 4 x3 6 x 2 12 x 4, k 4 f 1 2 14 x3 2 x 2 5 x 4, k f x f x 2 34 3 52. f x 3 3 1 205 2 3 1 2 x 23 15 x3 6 x 4 34 3 f x Polynomi Polynomial al and Synthetic Synth Division 2 ª3x 2 2 3 2 x 8 4 2 º ¬ ¼ 2 3 82 2 2 10 2 2 8 0 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. 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All Rights Reserved. 206 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Functions 2 x3 7 x 3 55. f x (a) Using the Remainder Theorem: f 1 21 3 71 3 (b) Using the Remainder Theorem: 2 f 2 Using synthetic division: 1 2 7 3 2 2 5 2 5 2 0 2 3 2 2 7 2 3 Using synthetic division: –2 7 3 –4 8 –2 –4 1 1 2 0 2 Verify using long division: Verify using long division: 2x 2x 5 x 1 2x 0 x2 7 x 3 2 x3 2 x 2 2×2 7 x 2×2 2x 5x 3 5x 5 2 2×2 4x 1 x 2 2 x 0 x2 7 x 3 2 x3 4 x 2 4×2 7 x 4×2 8x x 3 x 2 1 (d) Using the Remainder Theorem: 2 3 3 (c) Using the Remainder Theorem: 3 §1· f¨ ¸ © 2¹ §1· §1· 2¨ ¸ 7¨ ¸ 3 2 © ¹ © 2¹ 1 4 Using synthetic division: 1 2 2 0 1 2 1 f 2 22 13 4 1 4 72 3 7 2 0 4 8 2 2 4 1 5 3 3 5 Using synthetic division: 2 7 1 2 13 2 1 3 Verify using long division: 2×2 4 x 1 x 2 2x 0 x2 7 x 3 2 x3 4 x 2 4×2 7 x 4×2 8x x 3 x 2 5 3 Verify using long division: 13 2 1 x 2 x3 0 x 2 7 x 3 2 2 x3 x 2 2×2 x x2 7 x 1 x2 x 2 13 x 3 2 13 13 x 2 4 1 4 INSTRUCTOR USE ONLY © 2012 Cengage Learning. 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Section 2.3 Polynomi Polynomial al and Synth Synthetic Division 207 2 x6 3x 4 x2 3 56. g x (a) Using the Remainder Theorem: g 2 22 6 32 4 2 2 (b) Using the Remainder Theorem: 3 175 g1 Using synthetic division: 2 2 2 0 1 0 4 8 22 44 86 172 4 11 22 43 86 175 0 3 1 2 x 4 x 11x 22 x 43x 86 x 2 2 x 6 0 x5 3 x 4 0 x3 x 2 0 x 3 2 x6 4 x5 4 x5 3 x 4 4 x5 8 x 4 11x 4 0 x3 11x 4 22 x3 22 x3 x 2 22 x3 44 x 2 43 x 2 0 x 43 x 2 86 x 86 x 3 86 x 172 175 3 2 (c) Using the Remainder Theorem: g3 23 33 3 2 2 2 3 0 1 0 3 6 18 63 6 21 63 189 188 564 564 1692 1695 Verify using long division: 2x 3 7 0 3 0 1 0 3 2 2 5 2 5 5 5 4 4 4 4 7 2 1 6 3 1 4 1 2 3 7 4 –1 2 2 0 3 0 1 0 3 2 2 5 2 5 5 5 4 –4 –4 4 7 Verify using long division: 6 x 21x 63 x 188 x 564 4 3 2 x 3 2 x 0 x 3x 0 x x 0x 3 2 x 6 6 x5 6 x5 3 x 4 6 x 4 18 x 4 21x 4 0 x3 21x 4 63 x3 x2 63×3 3 63x 189 x 2 188 x 2 0 x 188 x 2 564 x 564 x 3 564 x 1692 1695 5 2 Using synthetic division: 3 6 1 2 x 5 2 x 4 5 x3 5 x 2 4 x 4 x 1 2 x 6 0 x5 3x 4 0 x3 x 2 0 x 3 2 x 6 2 x5 2 x5 3x 4 2 x5 2 x 4 5 x 4 0 x3 5 x 4 5 x3 5 x3 x 2 5 x3 5 x 2 4×2 0x 4×2 4x 4x 3 4x 4 7 g 1 1695 0 5 4 (d) Using the Remainder Theorem: Using synthetic division: 3 31 Verify using long division: 4 4 2 2 Verify using long division: 6 6 Using synthetic division: 3 5 21 3 2 2 x5 2 x 4 5 x3 5 x 2 4 x 4 x 1 2 x 0 x 5 3 x 4 0 x3 x 2 0 x 3 2 x6 2 x5 2 x5 3x 4 2 x5 2 x 4 5 x 4 0 x3 5 x 4 5 x3 5 x3 x 2 5 x3 5 x 2 4×2 0x 4×2 4x 4 x 3 4 x 4 7 6 INSTRUCTOR USE ONLY © 2012 Cengage Learning. 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All Rights Reserved. 208 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Functions x3 5 x 2 7 x 4 57. h x (a) Using the Remainder Theorem: h3 3 3 53 2 (b) Using the Remainder Theorem: 73 4 35 h2 Using synthetic division: 3 1 1 2 3 52 2 72 4 Using synthetic division: 5 7 4 3 6 39 2 13 35 2 1 1 5 7 4 2 6 26 3 13 22 Verify using long division: Verify using long division: x 2 x 13 x 3 x 5×2 7 x 4 x3 3x 2 2 x 2 7 x 2 x 2 6 x 13 x 4 13 x 39 35 x 2 3 x 13 x 2 x 5×2 7 x 4 x3 2 x 2 3x 2 7 x 3x 2 6 x 13 x 4 13 x 26 22 (c) Using the Remainder Theorem: (d) Using the Remainder Theorem: 2 3 h 2 2 3 5 2 2 3 7 2 4 10 Using synthetic division: –2 22 h 5 5 3 5 5 2 7 5 4 211 Using synthetic division: 1 5 7 2 14 14 1 7 7 10 –5 4 Verify using long division: 1 5 7 4 5 50 215 1 10 43 211 Verify using long division: x2 7 x 7 x 2 10 x 43 x 2 x3 5 x 2 7 x 4 x3 2 x 2 7 x 2 7 x 7 x 2 14 x 7x 4 7 x 14 10 x 5 x3 5 x 2 7 x 4 x3 5 x 2 10 x 2 7 x 10 x 2 50 x 43 x 4 43 x 215 211 4 x 4 16 x3 7 x 2 20 58. f x (a) Using the Remainder Theorem: f 1 41 4 16 1 3 (b) Using the Remainder Theorem: 7 1 20 Using synthetic division: 1 4 4 15 f 2 4 2 4 16 2 3 7 2 2 20 240 Using synthetic division: 16 7 0 4 12 5 20 5 12 5 5 15 Verify using long division: 4 x 3 12 x 2 5 x 5 x 1 4 x 4 16 x 3 7 x 2 0 x 20 4 x 4 4 x3 12 x 3 7 x 2 12 x 3 12 x 2 5 x 2 0 x 5 x 2 5 x 5 x 20 5 x 5 15 2 4 4 16 7 0 20 8 48 110 24 55 110 220 240 Verify using long division: 4 x3 24 x 2 55 x 110 x 2 4 x 4 16 x3 7 x 2 0 x 20 4 x3 8 x3 24 x3 7 x 2 24 x3 48 x 2 55 x 2 0 x 55 x 2 110 x 110 x 20 110 x 220 240 24 INSTRUCTOR USE ONLY © 2012 Cengage Learning. 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NOT FOR SALE Section 2.3 (c) Using the Remainder Theorem: f 5 45 4 16 5 3 75 2 4 20 695 7 0 20 20 20 135 675 4 27 135 695 Verify using long division: 4x 3 2 1 1 0 7 6 2 4 6 2 3 0 x3 7 x 6 4 4 10 10 4 16 10 3 7 10 4 0 20 40 560 5670 56 567 5670 56,700 56,720 1 1 4 x 27 x 135 4 x3 56 x 2 567 x 62. 2 3 48 80 41 32 32 6 48 48 9 0 6 48 x3 80 x 2 41x 6 x 23 48 x 2 48 x 9 x 23 4 x 3 12 x 3 3x 2 4 x 3 4 x 1 0 28 48 4 16 48 4 12 0 Zeros: 23 , 34 , 14 63. 3 x 4 x 2 4 x 12 3 Zeros: 4, 2, 6 1 2 2 2 15 27 1 10 14 20 0 2x 1 x 2 x 5 64. 6 3 2 3 0 2 3 3 2 3 2 0 x 2 Zeros: 3, x 12 2 x 2 14 x 20 6 3 x 2 x 3x 6 7 3 3 2 3 2 1 1 2 x 3 15 x 2 27 x 10 2 10 3 2 3 1 1 x 4 x 6 x 2 61. 5670 7 x2 0x 20 x 10 4 x 4 16 x3 4 x 4 40 x3 56 x3 7 x 2 56 x3 560 x 2 567 x 2 0x 567 x 2 5670 x 20 5670 x 5670 x 56,700 56,720 x 2 x2 2x 3 x3 28 x 48 56,720 Verify using long division: Zeros: 2, 3, 1 4 20 7 x 2 x 3 x 1 60. 2 16 2 x 5 4 x 4 16 x3 7 x 2 0 x 20 4 x 4 20 x3 4 x3 7 x 2 4 x 3 20 x 2 27 x 2 0 x 27 x 2 135 x 135 x 20 135 x 675 695 59. f 10 Using synthetic division: 16 4 209 (d) Using the Remainder Theorem: Using synthetic division: 5 Polynomial and Synthetic Synth Division Zeros: 12 , 2, 5 2 1 2 1 2 2 4 2 2 2 2 4 2 2 2 0 2 2 2 2 2 2 2 2 0 1 x 2x 2x 4 3 x 2 Zeros: 2, 3 x 2 3, 2 1 2 3 x 2 x 2 x 2 INSTRUCTOR USE ONLY 2, 2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 210 Chapter 2 65. 1 NOT FOR SALE Polynomial ynomial and Rational Functions 3 1 3 1 1 3 1 3x 3 2 x 2 19 x 6; 68. f x 1 3 1 3 2 Factors: x 3 , x 2 3 1 3 0 (a) 2 3 1 3 1 3 1 3 –1 ªx 1 ¬ 3, 1 3 3 ºª x 1 ¼¬ 2 3 º x 1 ¼ 3 x 1 3 3 3 2 19 6 9 21 6 7 2 0 3 3 0 x 1 x 1 Zeros: 1, 1 2 2 1 x3 3x 2 2 0 7 2 6 2 1 0 Both are factors of f x because the remainders 3 are zero. 66. 2 2 1 13 3 (b) The remaining factor is 3x 1 . 2 5 7 3 5 3 1 1 5 6 3 5 (c) f x 0 1 1 5 6 3 5 2 5 63 5 3 0 1 5 5 1 x 3 x 2 13 x 3 x2 Zeros: 2 5, 3 5 x2 3×3 2 x 2 19 x 6 3x 1 x 3 x 2 (d) Zeros: 13 , 3, 2 (e) 35 5 x3 −4 2 1 2 2 x 4 4 x3 15 x 2 58 x 40; 69. f x 1 5 2 Factors: x 5 , x 4 4 6 2 (a) 3 1 0 2 2 3 1 2 1 1 0 5 4 are zero. (e) 2x 1 x 2 x 1 15 58 40 5 5 50 1 10 8 40 0 1 10 8 4 12 8 3 2 0 1 1 Both are factors of f x because the remainders (b) The remaining factor is 2 x 1 . (d) Zeros: 4 1 1 Both are factors of f x because the remainders (c) f x 3 −10 2 x 3 x 2 5 x 2; Factors: x 2 , x 1 67. f x (a) 5, 2 are zero. (b) x 2 3 x 2 1 , 2, 1 2 x 1 x 2 The remaining factors are x 1 and x 2 (c) f x 7 x 1 x 2 x 5 x 4 (d) Zeros: 1, 2, 5, 4 (e) −6 6 20 −6 6 −1 −180 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 8 x 4 14 x 3 71x 2 10 x 24; 70. f x 2 4 Factors: 2 x 1 , 3x 2 14 71 10 16 60 22 24 24 8 30 11 12 0 8 30 11 12 32 8 12 2 3 0 8 8 211 6 x 3 41x 2 9 x 14; 71. f x Factors: x 2 , x 4 (a) Polynomial and Synthetic Synth Division (a) 12 6 6 2 3 41 9 14 3 19 14 38 28 0 6 38 28 4 28 6 42 0 Both are factors of f x because the remainders Both are factors of f x because the remainders are zero. are zero. (b) 8 x 2 2 x 3 4x 3 2x 1 (b) 6 x 42 The remaining factors are 4 x 3 and 2 x 1 . (c) 6 x 7 This shows that 4x 3 2x 1 x 2 x 4 f x (d) Zeros: 34 , 12 , 2, 4 (e) so f x 1 ·§ 2· § ¨ x ¸¨ x ¸ 2 ¹© 3¹ © f x 2 x 1 3x 2 6x 7, x 7. 40 −3 The remaining factor is x 7 . 5 (c) f x x 7 2 x 1 3x 2 1 2 (d) Zeros: 7, , 2 3 −380 (e) 320 −9 3 − 40 10 x 3 11x 2 72 x 45; 72. f x Factors: 2 x 5 , 5 x 3 (a) 52 3 5 11 72 45 25 90 45 10 36 18 0 10 36 18 6 18 30 0 10 10 (b) 10 x 30 10 x 3 This shows that so f x 5 3· § ·§ ¨ x ¸¨ x ¸ 2 5 © ¹© ¹ f x 2 x 5 5x 3 10 x 3 , x 3. The remaining factor is x 3 . Both are factors of f x because the remainders are zero. (c) f x (e) x 3 2x 5 5x 3 5 3 (d) Zeros: 3, , 2 5 100 −4 4 INSTRUCTOR USE ONLY −80 − 80 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 212 Chapter 2 Polynomial ynomial and Rational Function Functions (d) Zeros: r 4 3, 3 2 x3 x 2 10 x 5; 73. f x Factors: 2 x 1 , x (a) 1 2 2 1 10 5 1 0 5 0 10 0 2 5 (e) 5 2 −8 10 2 5 10 2 5 0 (a) The zeros of f are x 2 and x | r2.236. (b) An exact zero is x 2. (c) are zero. 2x x 3 2 x 2 5 x 10 75. f x Both are factors of f x because the remainders (b) 2 x 2 5 8 −240 0 2 60 2 1 5 1 f x 1· § ¨x ¸ x 2¹ © This shows that so f x 2x x (d) Zeros: 5, (e) 5, 10 0 5 0 5 x 5 x3 4 x 2 2 x 8 76. g x 5 . 5 2x 1 1 2 4, x | 1.414, x | 1.414. (a) The zeros of g are x (b) x 4 is an exact zero. (c) 1 4 4 14 2 8 4 0 8 0 2 0 x 4 x2 2 f x x 4 x 6 −6 2 t 3 2t 2 7t 2 77. h t x 3x 2 48 x 144; 2 x 3 74. f x (a) The zeros of h are t Factors: x 4 3 , x 3 (a) 0 x 2 x 5 1 −6 2 5. 5 5 x 10 x 2 x The remaining factor is x (c) f x 5 2 f x 5 x 2x 1 x 5, 2 3 2. (b) An exact zero is t 3 48 144 3 0 144 0 48 0 0 48 ht 4 3 48 4 3 0 By the Quadratic Formula, the zeros of t 2 4t 1 are 2 r 3. Thus, 1 1 4 3 2, t | 3.732, t | 0.268. 1 1 Both are factors of f x because the remainders are zero. (c) 2 1 1 ht 2 7 2 2 8 2 4 1 0 t 2 t 2 4t 1 t 2 ªt 2 ¬ 3 ºªt 2 ¼¬ 3 º. ¼ (b) The remaining factor is x 4 3 . (c) f x x 4 3 x 4 3 x 3 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 s 3 12 s 2 40 s 24 78. f s 6, s | 0.764, s | 5.236 (a) The zeros of f are s (b) s 6 is an exact zero. (c) 1 6 1 81. 12 40 24 6 36 24 6 4 0 5. Thus, f s s 6 ªs 3 ¬ So, 5 ºªs 3 ¼¬ 5 º. ¼ 82. x 7 x 10 x 14 x 24 x 5 79. h x 4 3 2 (a) The zeros of h are x 0, x 3, x (c) 4 1 8 4, 1 1 1 10 14 24 4 12 8 24 3 2 6 0 x x 4 x 3 x 83. (a) The zeros of a are x 2 x 3, x 3, x 2 (c) –3 6 1.5, 3. 11 51 99 27 18 87 108 27 0 29 6 9 36 3 2 2 0 4×2 2 x 2 2 2×2 x 1 2 x 2 x 1, x z 1 64 64 8 56 64 7 8 0 2 1 1 3 . 2 x 2 7 x 8, x z 8 2 1 1 x 4 6 x3 11x 2 6 x x 1 x 2 6 11 6 0 1 5 6 0 5 6 0 0 1 x | 0.333. (b) An exact zero is x 3 x 4 6 x 3 11x 2 6 x x 2 3x 2 6 x 4 11x 3 51x 2 99 x 27 80. g x 6 x x 64 x 64 x 8 3 x 4 x 4 3 x3 2 x 2 6 x h x 3 x 3 x 2 64 x 64 x 8 4. 7 1 4 x3 8 x 2 x 3 2x 3 x | 1.414, x | 1.414. (b) An exact zero is x 8 4 x3 8 x 2 x 3 3 x 2 By the Quadratic Formula, the zeros of s 2 6 s 4 are 3 r 4 4 s 6 s 6s 4 213 4 x3 8 x 2 x 3 2x 3 3 2 2 f s Polynomi Polynomial al and Synth Synthetic Division 5 6 0 2 6 0 3 0 0 x 4 6 x3 11x 2 6 x x 1 x 2 x 2 3 x, x z 2, 1 x 3 6 x 29 x 36 x 9 3 a x 2 x 3 x 3 2 x 3 3x 1 84. x 4 9 x3 5 x 2 36 x 4 x2 4 2 1 1 2 1 1 x 4 9 x3 5 x 2 36 x 4 x 2 x 2 9 5 36 4 2 22 34 4 11 17 2 0 11 17 2 2 18 2 9 1 0 x 4 9 x3 5 x 2 36 x 4 x2 4 INSTRUCTOR USE ONLY x 2 9 x 1,, x z r2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 214 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x2n 6 xn 9 (c) Year Actual Value Estimated Value 0 23.2 23.4 1 24.2 23.7 2 23.9 23.8 3 23.9 24.1 4 24.4 24.6 5 25.6 25.7 x 2n x n 3 6 28.0 27.4 86. x n 2 x3n 3 x 2 n 5 x n 6 x 3n 2 x 2 n x2n 5xn x2n 2 xn 3x n 6 3x n 6 0 7 29.8 30.1 85. x 3 x 9 x 27 x 27 x3n 3 x 2 n 6 x 2 n 27 x n 6 x 2 n 18 x n 9 x n 27 9 x n 27 0 n 3n 2n n x3n 9 x 2 n 27 x n 27 xn 3 x 2 n 6 x n 9, x n z 3 x 3n 3 x 2 n 5 x n 6 xn 2 (d) 2010 o t 10 1 1 4 3 c 5 45 210 9 42 c 210 1 1 1.81 22.3 0.181 2.23 45.7 92. (a) and (b) 65 0 7 25 A | 0.0576t 3 0.913t 2 0.28t 30.7 Year Actual Value Estimated Value 0 0 2 1 c 0 30.5 30.7 2 4 8 20 42 1 32.2 31.8 2 4 10 21 c 42 2 34.2 34.5 3 38.0 38.2 4 42.7 42.7 5 47.9 47.7 6 52.7 52.8 7 57.6 57.6 (d) 2010 o t 10 To divide evenly, c 42 must equal zero. So, c must equal 42. 91. (a) and (b) 35 0 0.349 No, because the model will approach infinity quickly. (c) 2 23.4 In 2010, the amount of money supporting higher education is about $45.7 billion. To divide evenly, c 210 must equal zero. So, c must equal 210. 90. 0.42 A 10 | $45.7 x 2 n x n 3, x n z 2 88. You can check polynomial division by multiplying the quotient by the divisor. This should yield the original dividend if the multiplication was performed correctly. 5 0.168 0.0349 0.0349 87. A divisor divides evenly into a dividend if the remainder is zero. 89. 10 7 10 0 0.0576 A | 0.0349t 3 0.168t 2 0.42t 23.4 0.0576 0.913 0.28 30.7 0.576 3.37 36.5 0.337 3.65 67.2 In 2010, the amount of money spent on health care is about $67.2 billion. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.4 93. False. If 7 x 4 is a factor of f , then 74 is a zero 100. If x 3 is a factor of f x of f . then f 3 f 3 94. True. 1 2 6 6 1 92 45 184 4 48 3 2 45 0 92 48 4 90 0 184 96 0 2 x 1 x 1 x 2 x 3 3x 2 x 4 f x 95. True. The degree of the numerator is greater than the degree of the denominator. k is a zero of f x , then x k is a factor 96. True. If x of f x , and f k 215 x3 kx 2 2kx 12 0. 3 0 3 k 3 2 2k 3 12 27 9k 6k 12 15 3k 5 k x 1 101. (a) x 1 x 2 0 x 1 x2 x x 1 x 1 0 x2 1 x 1 0. x 1, x z 1 x2 x 1 97. False. (b) x 1 x3 0 x 2 0 x 1 x3 x 2 x2 0x x2 x x 1 x 1 0 To divide x 4 3 x 2 4 x 1 by x 2 using synthetic division, the set up would be: 2 Comp Complex Numbers 1 3 0 1 4 A zero must be included for the missing x3 term. x3 1 x 1 x k q x r 98. f x (a) k 2, r 5, q x x3 any quadratic ax 2 bx c x 2 x2 5 3, r (b) k 1, q x x3 2 x 2 5 any quadratic ax 2 bx c where a 0. One example: x 3 x2 1 f x 99. If x 4 is a factor of f x then f 4 f 4 0 x3 3x 2 1 x3 kx 2 2kx 8, x4 1 x3 x 2 x 1, x z 1 x 1 xn 1 x n 1 x n 2 ” x 1, x z 1 x 1 0. 4 3 k 4 2 2k 4 8 64 16k 8k 8 56 8k 7 k x2 x 1 (c) x 1 x 4 0 x3 0 x 2 0 x 1 x 4 x3 x3 0 x 2 x3 x 2 x2 0 x x2 x x 1 x 1 0 where a ! 0. One example: f x x 2 x 1, x z 1 102. (a) f 3 0 because x 3 is a factor of f . (b) Because f x is in factored form, it is easier to evaluate directly. Section 2.4 Complex Numbers 1. (a) iii 2. 1; 1 (b) i (c) ii 3. principal square 4. complex conjugates INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 216 NOT FOR SALE Chapter 2 5. a bi Polynomial ynomial and Rational Function Functions 12 7i a 12 b 7 6. a bi 20. 3 2i 6 13i 3 11i 21. 2 13 4i a 13 b 4 22. 8 5 Ÿ a 6 b 3 8 Ÿ b 5 6 Ÿ a 18 4 3 2i 8 3 2i 4 3 2i 4 23. 13i 14 7i 13i 14 7i 14 20i 24. 32 52 i 0 5 11 i 3 3 32 52 i 53 11 i 3 96 15 i 10 22 i 6 6 6 52 9. 8 25 8 5i 10. 2 27 2 1 7i 6 6 25. 27i 5 ˜ 10 2 3 3i 75 2 11. 80 4 5i 26. 12. 4 2i 27. 1 i 3 2i 13. 0.09 75i 5i 10i 50i 2 5 2 1 2 75i 2 5 2 75 3 2i 3i 2i 2 3i 2 5 i 0.09i 28. 7 2i 3 5i 0.3i 14. 14 2 2 2i 5 5 2i 6 5i 5 Ÿ b 2b 50 5 8i a 1 8. a 6 2bi 8 5 3 3 2i 7. a 1 b 3 i a 6 3 2i 6 13i 14 0i 15. 10i i 2 16. 4i 2 2i 21 35i 6i 10i 2 21 41i 10 14 11 41i 10i 1 1 10i 29. 12i 1 9i 12i 108i 2 4 1 2i 12i 108 4 2i 108 12i 17. 7 i 3 4i 30. 8i 9 4i 10 3i 72i 32i 2 32 72i 18. 13 2i 5 6i 8 4i 31. 19. 9 i 8 i 1 32. 3 10 5 7 14 10i 14 10i 14 10i 2 14 10 3 5i 7 10i 21 3 10i 7 5i 21 24 50i 2 50 7 5 3 10 i 21 5 2 7 5 3 10 i 33. 6 7i 2 36 84i 49i 2 34. 5 4i 2 25 40i 16i 2 36 84i 49 25 40i 16 13 84i 9 40i INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.4 35. The complex conjugate of 9 2i is 9 2i. 9 2i 9 2i 46. 81 4i 2 2i 5 2 i 2i 85 1 5i 1 5i is 1 15 37. The complex conjugate of 2 5i 2 5i 20i 2 38. The complex conjugate of 6 39. 3 i ˜ i i 40. 41. 6 44. 3i i 2 14 2i ˜ 2i 2i 28i 4i 2 8 16i 2i ˜ 2i 2i 3i 2 6 20 2 5i is 2 5i . 47. i 3 8i 2i 3 2i i 2i 3 2i 3 8i 3 2i 3 8i 3i 8i 2 6i 4i 2 9 24i 6i 16i 2 20 6 is 4i 2 9i 9 18i 16 4 9i 25 18i ˜ 25 18i 25 18i 6. 28i 4 13 13i 1 i2 100 72i 225i 162i 2 625 324 62 297i 62 297 i 949 949 949 7i 13 13i 2 16i 32i 2 4i 2 48. 1i 3 4 i i 3i 9 40i ˜ 9 40i 9 40i 21 i 31 i 1i 1i 1 i 4 i 3i i4i 4 i 4i i 2 3i 4i i 2 5 1 4i ˜ 1 4i 1 4i 5 20i 1 16i 2 5 20 i 17 17 8 4i 3i 16 40i 25i 2 2 3 1i 1i 13 13 i 2 2 6 12i 7i 14i 2 1 4i 2 20 5i 4i 5 27i 120i 2 120 27i 81 1600 1681 120 27 i 1681 1681 45. 2i 2i 3i 1i 13 ˜ 1i 1i 4 5i 2i 2 i 5 2 i 6 6 7i 1 2i ˜ 42. 1 2i 1 2i 43. 5i. 1 5i 2 5i 217 4i 2i 2 10 5i 4 i2 12 9i 5 12 9 i 5 5 81 4 36. The complex conjugate of 1 Comp Complex Numbers 49. x 2 2 x 2 0; a 2 r 2 x 1, b 2 2, c 2 41 2 21 2 r 4 2 2 r 2i 2 1ri 2 2i 3 3i 11 1 5i 2 1 5 i 2 2 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 218 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 50. 4 x 2 16 x 17 16 r x 0; a 16 2 4, b 16, c 53. 1.4 x 2 2 x 10 17 7 x 10 x 50 2 4 4 17 x 16 r 16 8 16 r 4i 8 1 2 r i 2 6 r 6, c 0; a 9, b 2 4 9 37 54. 37 1296 18 6 r 36i 1 r 2i 18 3 4 2 12 r 4, c 16, b 3 55. 6i 3 i 2 4 16 3 57. i 1 2i 3 1 i 1 8i 3 72 2r 2i 6i 2i i 2 6 1 i 1 11 i 8 1 i 2i 4 3 18 6i 1 1 6i 1 r 8 1 i3 2 6 56. 4i 2 2i 3 2i 12 12 r 6 2i 6 176 32 6 0 23 4 r 4 11i 32 2 58. 60. 0; a 0 Multiply both sides by 2. 12 r 2 16 4 r 59. 3 2 x 6x 9 2 3x 2 12 x 18 x 6r 4 r 4 7 50 5 5 15 r 7 7 29 t 2 10 r 10 15 14 1500 14 6 52. 16t 2 4t 3 10 27 10 r 51. 9 x 2 6 x 37 x 0 10 r 24 0 Multiply both sides by 5. 6 8i 6 1 i ˜ i i 1 8i 2i 1 8i 8 1 1 1 8i 2i 2i 2 i i 2 1 i 3 4 1 2 1 i 1 i 2 i 1 1 i 4 2i i 8 61. a bi a bi i a 2 abi abi b 2i 2 a 2 b 2 1 1 8i ˜ 8i 8i 8i 64i 2 a 2 b2 1 i 8 which is a real number since a and b are real numbers. Thus, the product of a complex number and its conjugate is a real number. 62. 63. a1 b1i a2 b2i 3 4i 2 2i 2i 6 6 6i 6i 6i 2 6 a1 a2 b1 b2 i The complex conjugate of this sum is a1 a2 b1 b2 i. The sum of the complex conjugates is a1 b1i a2 b2i a1 a2 b1 b2 i. So, the complex conjugate of the sum of two complex numbers is the sum of their complex conjugates. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section 2.4 64. (a) 24 16 (b) 2 (c) 2i 4 24 i 4 4 16 1 1 16i 2i 2 4 4 2 i 2 x 3 66. f x i2 (b) i 25 i2 (c) i 50 i2 (d) i 67 i2 16 4 2i (d) 65. (a) i 40 2 16 1 1 2 2 16i i 4, g x 16 2 x 3 2 16 20 1 12 33 20 1 i 1 i 25 1 33 i 1 i i 219 1 12 ˜i 25 Complex Comp Numbers 4 (a) The graph of f is a parabola with vertex at the point 3, 4 . (c) If all the zeros contain i, then the graph has no x-intercepts. The a value is positive, so the graph opens upward. The graph of g is also a parabola with vertex at the point 3, 4 . The a value is negative, so the graph opens downward. f has an x-intercept and g does not because when g x 0, x is a complex number. f x 2x 3 2 4 0 2x 3 2 4 4 2x 3 2 2 x 3 r 2 x 3 (b) 3r 2 (d) If a and k have the same sign (both positive or both negative), then the graph of f has no x-intercepts and the zeros are complex. Otherwise, the graph of f has x-intercepts and the zeros are real. 2 x 2 x 3 2 4 2 x 3 2 4 4 2 x 3 2 2 x 3 r 2 x 3 g x 0 3r 2i 67. False, if b 2 x 0 then a bi a bi 68. True. a. x 4 x 2 14 That is, if the complex number is real, the number equals its conjugate. i 6 4 i 6 2 14 36 6 14 56 56 ? 56 ? 56 56 69. False. i 44 i150 i 74 i109 i 61 22 i2 1 22 i2 75 1 i2 75 1 111i i 70. (a) z1 (b) 1 z 9 16i, z2 1 1 z1 z2 37 i2 37 54 i i2 54 30 i 30 1 i 1 i 1 20 10i 1 1 9 16i 20 10i § 340 230i ·§ 29 6i · ¨ ¸¨ ¸ © 29 6i ¹© 29 6i ¹ 20 10i 9 16i 9 16i 20 10i 11,240 4630i 877 29 6i 340 230i 11,240 4630 i 877 877 INSTRUCTOR USE ONLY z © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 220 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions Section 2.5 The Fundamental Theorem of Algebra 1. Fundamental Theorem of Algebra x3 6 x 2 11x 6 17. f x 2. Linear Factorization Theorem Possible rational zeros: r1, r 2, r 3, r 6 3. Rational Zero 1 1 4. conjugate 1 5. linear; quadratic; quadratic 6 11 6 1 5 6 5 6 0 x3 6 x 2 11x 6 x 1 x2 5x 6 6. irreducible; reals 7. f x x x 6 x 1 x 2 x 3 2 The zeros are: x 8. f x 0, x 11. f x 3 6, x 3 9 6 3 2 0 So, the rational zeros are 2, 1, and 3. 8 x3 4 x 2 x 4 19. g x i , x x2 x 4 1 x 4 i x 4 x2 1 2, x 3i, x 3i x 4 x 1 x 1 So, the rational zeros are 4, 1, and 1. x 2x x 2 3 6 x 3 x 2 x 1 t 3 t 2 t 3i t 3i 3, x 7 0 x 3 x 2 3x 2 2 2 x3 9 x 2 20 x 12 20. h x Possible rational zeros: r1, r 2 Possible rational zeros: r1, r 2, r 3, r 4, r 6, r12 Zeros shown on graph: 2, 1, 1 1 14. f x 1 f x x 6 x i x i The zeros are: x 13. f x 4 5, x The zeros are: x 12. h t 1 1, x Possible rational zeros: r1, r 2, r 3, r 6 1 2, x The zeros are: x x2 x 3 x 1 x 1 3 x 2 x 4 x 5 x 8 x3 7 x 6 18. f x 3, x 0, x The zeros are: x 10. f x 6 x2 x 3 x2 1 The zeros are: x 9. g x So, the rational zeros are 1, 2, and 3. 1 x3 4 x 2 4 x 16 Possible rational zeros: r1, r 2, r 4, r 8, r16 Zeros shown on graph: 2, 2, 4 1 h x 9 20 12 1 8 12 8 12 0 x 1 x 2 8 x 12 2 x 4 17 x 3 35 x 2 9 x 45 x 1 x 2 x 6 Possible rational zeros: r1, r 3, r 5, r 9, r15, r 45, So, the rational zeros are 1, 2, and 6. 15. f x r 12 , r 32 , r 52 , r 92 , r 15 , r 45 2 2 Zeros shown on graph: 1, 32 , 3, 5 16. f x 4 x5 8 x 4 5 x3 10 x 2 x 2 Possible rational zeros: r1, r 2, r 12 , r 14 INSTRUCTOR USE ONLY Zeros shown on graph: 1, 12 , 12 , 1, 2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 t 3 8t 2 13t 6 21. h t 1 Possible rational zeros: r1, r 2, r 3, r 4, r 6, r 8, r12, r 24, 8 13 6 6 12 6 r 13 , r 23 , r 43 , r 83 , r 19 , r 92 , r 94 , r 98 2 1 0 í2 1 t 3 8t 2 13t 6 9 t 6 t 1 t 1 1 1 9 27 27 3 18 27 6 9 0 2 1 2 1 1 1 1 2 x3 3 x 2 1 1 0 x 1 2×2 x 1 2 í1 3 3 f x í19 33 í9 9 í30 9 í10 3 0 0 23 15 í25 í10 2 í5 í2 5 25 0 2 í5 í2 5 2 í3 í5 2 í3 í5 0 í3 í5 í2 5 í5 0 2 x 3 3 x 2 10 x 3 x 3 3x 1 x 3 So, the rational zeros are 3 and 13 . f x x 5 x 1 x 1 2x 5 So, the rational zeros are 5, 1, 1 and 52 . 3×3 19 x 2 33 x 9 Possible rational zeros: r1, r 3, r 9, r 13 3 í12 í4 í25 2 2x 1 So, the rational zeros are 1 and 12 . 24. f x 0 0 10 x 1 x 1 2x 1 x 1 27 í15 2 Possible rational zeros: r1, r 12 0 12 Possible rational zeros: r1, r 5, r 25, r 12 , r 52 , r 25 2 5 3 í4 2 x 4 15 x3 23x 2 15 x 25 2 x3 3x 2 1 2 8 12 x 2 x 3 9×2 4 f x 26. f x So, the rational zero is 3. 1 í24 0 54 í4 So, the rational zeros are 2, 3, 23 , and 23 . x 3 x 3 x 3 23. C x 24 í27 í27 9 4 x 2 x 3 3x 2 3x 2 x 3 x2 6x 9 f x í58 í18 9 Possible rational zeros: r1, r 3, r 9, r 27 3 3 x3 9 x 2 27 x 27 22. p x í9 9 t 6 t 2 2t 1 So, the rational zeros are 1 and 6. 221 9 x 4 9 x3 58 x 2 4 x 24 25. f x Possible rational zeros: r1, r 2, r 3, r 6 6 The Fundamental Theorem Theore of Algebra 27. z 4 z 3 z 2 3 z 6 0 Possible rational zeros: r1, r 2, r 3, r 6 1 1 1 1 1 3 í6 1 2 3 2 3 6 6 0 z 1 z 3 2 z 2 3z 6 0 z 1 z 3 z 2 0 2 So, the real zeros are í2 and 1. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 222 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 28. x 4 13 x 2 12 x 0 x x3 13 x 12 0 x3 x 2 4 x 4 31. f x (a) Possible rational zeros: r1, r 2, r 4 Possible rational zeros of x3 13 x 12: y (b) 4 r1, r 2, r 3, r 4, r 6, r12 í1 2 0 í13 í12 í1 1 í1 í12 12 0 x x 1 x 2 x 12 0 x x 1 x 4 x 3 0 1 1 x –6 –4 4 6 –4 –6 –8 (c) Real zeros: 2, 1, 2 The real zeros are 0, 1, 4, and 3. 29. 2 y 4 3 y 3 16 y 2 15 y 4 0 Possible rational zeros: r 12 , r1, r 2, r 4 1 2 2 3 í16 15 í4 2 5 í11 4 5 í11 4 0 3 x3 20 x 2 36 x 16 32. f x (a) Possible rational zeros: r1, r 2, r 4, r 8, r16, r 13 , r 23 , r 43 , r 83 , r 16 3 (b) y 10 8 6 4 2 x –4 –2 1 2 2 5 í11 4 2 7 –4 7 í4 0 –6 (c) Real zeros: 23 , 2, 4 0 y 1 y 1 2y 1 y 4 0 So, the real zeros are 4, 12 and 1. 30. x5 x 4 3×3 5 x 2 2 x 0 x x x 3x 5 x 2 0 3 8 10 12 –4 y 1 y 1 2 y2 7 y 4 4 6 2 33. f x 4 x3 15 x 2 8 x 3 (a) Possible rational zeros: r1, r 3, r 12 , r 32 , r 14 , r 34 y (b) 4 2 x Possible rational zeros of x 4 x3 3x 2 5 x 2: –6 –4 –2 6 8 10 –6 1 1 í2 4 –4 r1, r 2 1 2 1 1 í1 í3 5 í2 1 0 í3 2 0 í3 2 0 0 í3 2 í2 4 í2 í2 1 0 (c) Real zeros: 14 , 1, 3 34. f x 4 x3 12 x 2 x 15 (a) Possible rational zeros: r1, r 3, r 5, r15, r 12 , r 23 , r 52 , r 15 , r 14 , r 43 , r 54 , r 15 2 4 y (b) 15 12 x x 1 x 2 x2 2x 1 0 x x 1 x 2 x 1 x 1 0 The real zeros are 2, 0, and 1. x –9 –6 –3 6 9 12 INSTRUCTOR USE ONLY ((c)) Real zeros: 1,, 32 , 52 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 2 x 4 13 x3 21x 2 2 x 8 35. f x The Fundamental Theorem Theore of Algebra x 4 3x 2 2 39. f x r1, about r1.414 (a) Possible rational zeros: r1, r 2, r 4, r 8, r 12 (a) x (b) (b) An exact zero is x 1. 0 í3 0 2 1 1 í2 í2 1 í2 í2 0 16 1 −4 223 1 8 1 −8 (c) (c) Real zeros: 12 , 1, 2, 4 í1 4 x 4 17 x 2 4 36. f x í2 í2 í1 0 2 0 í2 0 1 (a) Possible rational zeros: r1, r 2, r 4, r 12 , r 14 (b) 1 1 x 1 x 1 x2 2 f x x 1 x 1 x 9 −8 8 2 x 2 t 4 7t 2 12 40. P t r 2, about r1.732 (a) t −15 (b) An exact zero is t 2. í7 0 12 2 4 í6 í12 2 í3 í6 0 An exact zero is t 2. (c) Real zeros: 2, 12 , 12 , 2 2 32 x3 52 x 2 17 x 3 37. f x 0 1 (a) Possible rational zeros: r1, r 3, r 12 , r 23 , r 14 , r 43 , 1,r 3,r 1,r 3 r 18 , r 83 , r 16 16 32 32 (b) 1 í2 2 í3 í6 í2 0 6 0 í3 0 1 6 1 −1 3 t 2 t 2 t2 3 (c) P t −2 t 2 t 2 t (c) Real zeros: 18 , 34 , 1 4 x 7 x 11x 18 38. f x 2 1 3 9 1 3 9 r ,r ,r ,r ,r ,r 2 2 2 4 4 4 (b) x 0, 3, 4, about r1.414 (b) An exact zero is x 3 8 −8 x x 4 7 x3 10 x 2 14 x 24 (a) h x (a) Possible rational zeros: r1, r 2, r 3, r 6, r 9, r18, (c) (c) Real zeros: 2, 1 r 8 4 145 8 10 14 í24 3 í12 í6 24 1 í4 í2 8 0 1 í4 í2 8 4 0 í8 0 í2 0 1 h x 3. í7 1 8 −24 3 x5 7 x 4 10 x3 14 x 2 24 x 41. h x 3 3 t x x 3 x 4 x2 2 INSTRUCTOR USE ONLY x x 3 x 4 x 2 x 2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 224 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions 44. If 3i is a zero, so is its conjugate, 3i. 6 x 4 11×3 51x 2 99 x 27 42. g x (a) x r 3, 1.5, about 0.333 (b) An exact zero is x 3 6 x 4 x 3i x 3i x 4 x2 9 3. x3 4 x 2 9 x 36 í11 í51 99 í27 18 21 í90 27 Note: f x 7 í30 9 0 real number, has the zeros 4, 3i, and 3i. 6 3. An exact zero is x í3 f x 6 45. If 5 i is a zero, so is its conjugate, 5 i. 7 í30 9 í18 33 í9 í11 3 0 6 a x3 4 x 2 9 x 36 , where a is any f x x 2 x 5i x 5i x 2 x 10 x 26 2 x3 12 x 2 46 x 52 x 3 x 3 6 x 11x 3 Note: f x x 3 x 3 3x 1 2 x 3 any nonzero real number, has the zeros 2 and 5 r i. 2 (c) g x 43. If 5i is a zero, so is its conjugate, 5i. f x 46. If 3 2i is a zero, so is its conjugate, 3 2i. x 1 x 5i x 5i f x x 3 2i x 5 x 2 6 x 13 x3 x 2 25 x 25 x3 11x 2 43 x 65 a x3 x 2 25 x 25 , where a is any Note: f x 2i is a zero, so is its conjugate, 3 a x3 11x 2 43 x 65 , where a is any nonzero real number, has the zeros 5 and 3 r 2i. nonzero real number, has the zeros 1 and r 5i. f x x 5 x 3 2i x 1 x 2 25 Note: f x 47. If 3 a x3 12 x 2 46 x 52 , where a is 2i. 3x 2 x 1 ª x 3 ¬ 2i ºª x 3 ¼¬ 2i º ¼ 3x 2 x 1 ª¬ x 3 2iºª ¼¬ x 3 2iº¼ 2 3 x 2 x 2 ª« x 3 ¬ 2 2i º» ¼ 3x 2 x 2 x 2 6 x 9 2 3 x 2 x 2 x 2 6 x 11 3x 4 17 x3 25 x 2 23x 22 Note: f x 48. If 1 f x a 3 x 4 17 x3 25 x 2 23 x 22 , where a is any nonzero real number, has the zeros 23 , 1, and 3 r 3i is a zero, so is its conjugate, 1 x 5 2 x 1 3i x 1 3i. 49. f x 3i x 2 10 x 25 x 2 2 x 4 Note: f x 3 x 4 6 x 2 27 (a) f x x2 9 x2 3 (b) f x x2 9 x (c) f x x 3i x 3i x x 8 x 9 x 10 x 100 4 2i. 3 x 3 2 a x 4 8 x3 9 x 2 10 x 100 , where a is any real number, has the zeros 5, 5, and 1 r 3 x 3 3i. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 50. f x x 6 x 2 x 3x 2 12 x 18 6×2 x4 3 2 x 3x 2 12 x 2 x3 12 x 2 18 3x 18 3x 2 0 51. f x 225 x 4 2 x3 3 x 2 12 x 18 x2 2x 3 2 The Fundamental Theorem Theore of Algebra 4 3 (a) f x x2 6 x2 2 x 3 (b) f x x 6 x 6 x2 2x 3 (c) f x x 6 x 6 x 1 2i x 1 2i Note: Use the Quadratic Formula for (c). x 4 4 x3 5 x 2 2 x 6 x2 2x 3 x 2x 2 x 4 x 5×2 2x 6 x 4 2 x3 2 x 2 2 x3 7 x 2 2 x 2 x3 4 x 2 4 x 3x 2 6 x 6 3x 2 6 x 6 0 2 4 3 (a) f x x2 2x 2 x2 2 x 3 (b) f x x 1 3 x 1 3 x2 2 x 3 (c) f x x 1 3 x 1 3 x 1 2i x 1 2i Note: Use the Quadratic Formula for (b) and (c). 52. f x x 4 3 x3 x 2 12 x 20 x 2 3x 5 x 2 4 x 4 3 x3 x 2 12 x 20 4×2 x4 3 3 x 5 x 2 12 x 3 x3 12 x 5 x 2 20 5 x 2 20 0 (a) f x x 2 4 x 2 3x 5 (b) f x § 3 29 ·§ 3 29 · x 2 4 ¨¨ x x ¸¨ ¸¸ ¸¨ 2 2 © ¹© ¹ (c) f x § 3 29 ·§ 3 29 · x 2i x 2i ¨¨ x x ¸¨ ¸¸ ¸¨ 2 2 © ¹© ¹ Note: Use the Quadratic Formula for (b). INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 226 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions x3 x 2 4 x 4 53. f x Alternate Solution: Because x Because 2i is a zero, so is í2i. 2i 1 í1 4 í4 x 2i x 2i 2i 4 2i 4 By long division, you have: 1 2i 1 2i 0 x 1 x 2 0 x 4 x3 x 2 4 x 4 x3 0 x 2 4 x x2 0x 4 x2 0x 4 0 2i 1 1 2i 1 í2i 2i 2i í1 0 f x x 2i x 2i x 1 f x 2 x3 3 x 2 18 x 27 54. f x x2 4 x 1 Because x r 3i are zeros of f x , 2 3 18 27 x 3i x 3i 6i 9i 18 í27 By long division, you have: 2 3 6i 9i 0 í3i 2 2 3 6i 9i 6i í9i 3 0 The zeros of f x are x 2x 3 f x r 3i, 32 . 2 Because x r 5i are zeros of f x , í1 49 í25 í25 x 5i x 5i 10i 5i 50 5i 25 25 By long division, you have: 1 10i 1 5i í5i 0 2 f x 1 10i 1 5i í5i 10i 5i 5i í1 í1 0 x 5i x 5i 2 x 2 x 1 x 5i x 5i 2 x 1 x 1 The zeros of f x are x r5i, 12 , 1. x 2 25 is a factor of f x . 2×2 x 1 x 0 x 25 2 x x 49 x 2 25 x 25 2 x 4 0 x3 50 x 2 x3 x 2 25 x x3 0 x 2 25 x x 2 0 x 25 x 2 0 x 25 0 2 2 r 3i, 32 . Alternate Solution: Because 5i is a zero, so is í5i. 2 x2 9 2 x 3 The zeros of f x are x 2 x 4 x3 49 x 2 25 x 25 55. f x x 2 9 is a factor of f x . x 2 0 x 9 2 x3 3 x 2 18 x 27 2 x3 0 x 2 18 x 3 x 2 0 x 27 3 x 2 0 x 27 0 x 3i x 3i 2 x 3 f x í5i 1, r 2i. Alternate Solution: Because 3i is a zero, so is 3i. 5i x 2 4 is a factor of f x . The zeros of f x are x 1, r 2i. The zeros of f x are x 3i r 2i are zeros of f x , f x 4 3 x 2 25 2 x 2 x 1 The zeros of f x are x r 5i, 12 , 1. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 The Fundamental Theorem Theore of Algebra 227 x3 7 x 2 x 87 56. g x Because 5 2i is a zero, so is 5 2i. 5 2i í7 í1 87 5 2i 14 6i í87 1 2 2i 15 6i 0 1 2 2i 15 6i 5 2i 15 6i 3 0 1 5 2i 1 The zero of x 3 is x 3. The zeros of f x are x 4 x3 23 x 2 34 x 10 57. g x Alternate Solution Because 3 r i are zeros of g x , Because 3 i is a zero, so is 3 i. 3 i 3 i 23 34 í10 12 4i 37 i 10 4 11 4i 3 i 0 4 11 4i 3 i 12 4i 3i 4 í1 4 The zero of 4 x 1 is x g x are x 3, 5 r 2i. ¬ª x 3 i ºª ¼¬ x 3 i º¼ x 3 4x 1 x 6 x 10 4 x 23 x 34 x 10 4 x3 24 x 2 40 x x 2 6 x 10 x 2 6 x 10 0 2 0 1 . The zeros of 4 3 r i, 14 . 3 2 x 2 6 x 10 4 x 1 3 r i, 14 . 3×3 4 x 2 8 x 8 Because 1 1 i2 is a factor of g x . By long division, you have: The zeros of g x are x 1 3 x 2 6 x 10 g x 58. h x ª¬ x 3 iºª ¼¬ x 3 iº¼ 3i 3i 3i is a zero, so is 1 3i. í4 8 8 3 3 3i 10 2 3i í8 3 1 3 3i 2 2 3i 0 3 1 3 3i 2 2 3i 3 3 3i 2 2 3i 2 0 3 3 The zero of 3x 2 is x 23 . The zeros of f x are x 23 , 1 r 3i. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 228 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x 4 3 x3 5 x 2 21x 22 59. f x 2i is a zero, so is 3 Because 3 ª x 3 ¬ 2i ºª x 3 ¼¬ 2i º ¼ 2i, and ªx 3 ¬ x 3 2 2iºª ¼¬ x 3 2i 2iº¼ 2 x 2 6 x 11 is a factor of f x . By long division, you have: x 2 3x 2 x 6 x 11 x 3 x 5 x 2 21x 22 2 4 3 x 4 6 x3 11x 2 3 x3 16 x 2 21x 3 x3 18 x 2 33x 2 x 2 12 x 22 2 x 2 12 x 22 0 x 2 6 x 11 x 2 3 x 2 f x x 2 6 x 11 x 1 x 2 3 r The zeros of f x are x 2i, 1, 2. x3 4 x 2 14 x 20 60. f x 63. h x Because 1 3i is zero, so is 1 3i. 1 3i 1 3i 1 By the Quadratic Formula, the zeros of f x are 4 14 20 1 3i 12 6i í20 1 3 3i 2 6i 0 1 3 3i 2 6i 1 3i 2 6i 2 0 1 x f x 2. The zeros of f x are x 2, 1 r 3i. x 2 36 64. g x x f x 65. f x x f x 1 224 2 1 r 4i. x 1 4i x 2 10 x 17 10 r 100 68 2 x 5 2 2 10 r 2 32 5 r 2 2. x 5 2 2 2 x 5 2 2 x 4 16 x2 4 x2 4 By the Quadratic Formula, the zeros of f x are 1r 64 2 x 1 4i x 5 2 r 6i. x 2 x 56 62. f x 2r 4 68 2 x 1 4i x 1 4i x 6i x 6i The zeros of f x are x 2r By the Quadratic Formula, the zeros of f x are The zero of x 2 is x 61. f x x 2 2 x 17 1r 223i 2 . x 2 x 2 x 2i x 2i Zeros: r 2, r 2i § 1 223i ·§ 1 223i · x ¨¨ x ¸¨ ¸¸ ¸¨ 2 2 © ¹© ¹ INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 66. y 4 256 f y The Fundamental Theorem Theore of Algebra x3 x 2 x 39 70. f x y 2 16 y 2 16 Possible rational zeros: r1, r 3, r13, r 39 y 4 y 4 y 4i y 4i 3 1 Zeros: r 4, r 4i 1 z2 2z 2 67. f z By the Quadratic Formula, the zeros of f z are z 2r 48 2 1 39 3 12 39 4 13 0 4r 16 52 2 2 r 3i Zeros: 3, 2 r 3i ª¬ z 1 i ºª ¼¬ z 1 i º¼ f z 1 By the Quadratic Formula, the zeros of x 2 4 x 13 are: x 1 r i. 229 x 3 x 2 3i x 2 3i f x z 1i z 1 i x3 x 6 71. h x x3 3x 2 4 x 2 68. h x Possible rational zeros: r1, r 2, r 3, r 6 Possible rational zeros: r1, r 2 1 1 1 í3 4 í2 1 í2 2 í2 2 0 2 1 are x 48 2 1 r i. x 2r x 1 x 1i x 1 i h x 1 1 3 1 5 4 6 2 3 0 4 12 2 1r 2i. 2i. x 2 ªx 1 ¬ 2i ºª x 1 ¼¬ x 2 x 1 2i x 1 2i º ¼ 2i 1 4 5 4 5 0 Possible rational zeros: r1, r 5, r 7, r 35 5 4r 16 20 2 2ri x 1 x 2i x 2i 9 27 35 5 20 35 4 7 0 By the Quadratic Formula, the zeros of x 2 4 x 7 are x Zeros: 1, 2 r i 1 1 By the Quadratic Formula, the zeros of x 2 4 x 5 g x 2 x3 9 x 2 27 x 35 72. h x Possible rational zeros: r1, r 5 are: x 6 x3 3x 2 x 5 69. g x 1 1 Zeros: 2, 1 r Zeros: 1, 1 r i h x 0 By the Quadratic Formula, the zeros of x 2 2 x 3 are By the Quadratic Formula, the zeros of x 2 2 x 2 2r 1 4 r Zeros: 5, 2 r h x 16 28 2 2 r 3i. 3i x 5 x 2 3i x 2 3i INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 230 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 5 x3 9 x 2 28 x 6 73. f x Possible rational zeros: r1, r 3, r 9 Possible rational zeros: 1 2 3 6 r1, r 2, r 3, r 6, r , r , r , r 5 5 5 5 15 9 28 6 1 2 6 10 30 0 5 5 3 2r 4 24 2 1 Zeros: , 1 r 5 3 1r 1 1 6 10 6 9 3 9 3 3 1 3 9 0 3 1 3 3 0 3 0 1 0 r i. The zeros of x 2 1 are x 5i. Zeros: 3, r i 5i. h x ª § 1 ·º ª « x ¨ 5 ¸» 5 ¬ x 1 © ¹¼ ¬ f x 1 1 By the Quadratic Formula, the zeros of 5 x 2 10 x 30 5 x 2 2 x 6 are x x 4 6 x3 10 x 2 6 x 9 76. h x 5x 1 x 1 5i ºª x 1 ¼¬ 5i x 1 5i º ¼ 77. f x 2 x 3 x i x i x 4 10 x 2 9 x2 1 x2 9 5i x i x i x 3i x 3i 2 x3 x 2 8 x 21 74. g x Zeros: r i, r 3i Possible rational roots: 1 3 7 21 r , r1, r , r 3, r , r 7, r , r 21 2 2 2 2 3 2 2 2 1 8 21 3 6 21 4 14 0 78. f x x 2 25 x 2 4 x 2i x 2i x 5i x 5i Zeros: r 2i, r 5i By the Quadratic Formula, the zeros of 2 x 2 4 x 14 4r are x 16 112 4 4r 96 4 6i x 1 x 4 4 x3 8 x 2 16 x 16 Possible rational zeros: r1, r 2, r 4, r 8, r16 2 2 1 4 8 2 1 2 1 2 4 8 2 0 8 0 4 0 1 g x x3 24 x 2 214 x 740 Possible rational zeros: r1, r 2, r 4, r 5, r10, r 20, r 37, r 74, r148, r185, r 370, r 740 6i. 6i 3· § ¨x ¸ x 1 2¹ © 75. g x 1r 79. f x 2000 3 Zeros: , 1 r 2 f x x 4 29 x 2 100 16 16 4 8 4 8 16 0 6i −20 10 −1000 Based on the graph, try x 10 1 1 10. 24 214 740 10 140 740 14 74 0 By the Quadratic Formula, the zeros of x 2 14 x 74 are x 14 r 196 296 2 The zeros of f x are x 7 r 5i. 10 and x 7 r 5i. x 2 x 2 x2 4 x 2 2 x 2i x 2i INSTRUCTOR USE ONLY 2i Zeros: 2, r 2i Zeros © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 2 s 3 5s 2 12 s 5 80. f s The Fundamental Theorem Theore of Algebra 9 x3 15 x 2 11x 5 82. f x 1 5 Possible rational zeros: r1, r 5, r , r 2 2 1 5 1 5 Possible rational zeros: r1, r 5, r , r , r , r 3 3 9 9 10 5 −10 −5 10 5 −5 −10 Based on the graph, try s 1 2 2 5 12 1 2 5 2 4 10 0 Based on the graph, try x 1 . 2 1 2r 4 20 2 9 1 and s 2 5 6 5 0 The zeros of f x are x 1 and x 1 2 r i. 3 3 2 x 4 5 x3 4 x 2 5 x 2 1 2 20 −4 4 Based on the graph, try x 2 and x 2 2 −5 Based on the graph, try x 3 . 4 20 4 15 12 24 15 32 20 0 2 64 80 8 1 2 2 2 5 4 5 2 4 2 4 1 2 1 2 0 1 2 1 1 0 1 0 2 0 The zeros of 2 x 2 2 By the Quadratic Formula, the zeros of 16 x 2 32 x 20 4 4 x 2 8 x 5 are x 6 Possible rational zeros: r1, r 2, r 3 8r 9 −5 −3 16 5 1 2 r i. 3 3 20 16 11 36 180 18 83. f x 16 x3 20 x 2 4 x 15 3 4 15 6r are x 1 r 2i. Possible rational zeros: 1 3 5 15 1 3 r1, r 3, r 5, r15, r , r , r , r , r , r , 2 2 2 2 4 4 5 15 1 3 5 15 1 3 5 15 r ,r ,r ,r ,r ,r ,r ,r ,r ,r 4 4 8 8 8 8 16 16 16 16 1. By the Quadratic Formula, the zeros of 9 x 2 6 x 5 1 r 2i. The zeros of f s are s 81. f x 9 5 By the Quadratic Formula, the zeros of 2 s 2 2 s 5 are s 231 The zeros of f x are x 1 1 r i. 2 The zeros of f x are x 3 and x 4 1r 2 x 2 1 are x 2, x 1 . 2 r i. 1 , and x 2 r i. 1 i. 2 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 232 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x5 8 x 4 28 x3 56 x 2 64 x 32 84. g x Possible rational zeros: r1, r 2, r 4, r 8, r16, r 32 10 4 y3 3 y 2 8 y 6 87. f y Possible rational zeros: r1, r 2, r 3, r 6, r 12 , r 32 , r 14 , r 34 34 −10 4 10 3 8 6 3 0 6 0 8 0 4 −10 Based on the graph, try x 2 2 2 4 y3 3 y2 8 y 6 2. 1 8 28 56 64 32 1 2 6 12 16 32 24 48 16 y 34 4 y 2 2 32 0 4 y 3 y2 2 1 6 16 24 16 1 2 4 8 8 16 8 16 0 1 4 8 8 2 4 8 2 4 0 1 So, the only real zero is 34 . 2r are x 4 16 2 Possible rational zeros: r1, r 2, r 5, r10, r 13 , r 23 , r 53 , r 10 3 2 3 1r The zeros of g x are x 3i. 2 and x 3 3 1r 3i. 4 0 3x 2 x 2 5 1 1 4 x2 9 4 x2 4 4 4 1 1 0 1 2 x 3 4 2x 3 x 2 x 2 4 x 1 4 x2 4 x 1 2 The rational zeros are r 32 and r 2. 1 2 x 3 3 x 2 23 x 12 2 90. f x Possible rational zeros: r1, r 2, r 3, r 4, r 6, r12, r 12 , 23 4 Possible rational zeros: r1, r 3, r 9, r 12 , r 32 , r 92 , r 13 , 1 r 14 , r 34 , r 94 , r 16 , r 12 f z 15 1 4 x4 25 x 2 36 4 2 12 z 3 4 z 2 27 z 9 12 0 3 1 So, the real zeros are 1 and . 2 12 10 x 4 25 x2 9 4 89. P x x 1 2x 1 3 2 10 0 0 4 x3 3x 1 86. f z 15 2 So, the only real zero is 23 . Possible rational zeros: r1, r 12 , r 14 4 2 x 23 3 x 2 15 g x 4 x3 3x 1 85. f x 3 x3 2 x 2 15 x 10 88. g x By the Quadratic Formula, the zeros of x 2 2 x 4 1 y 34 4 y 2 8 4 27 9 18 21 9 14 6 0 2 f x 3 23 12 8 20 12 5 3 0 1 x 4 2 2×2 5x 3 1 x 4 2 2x 1 x 3 The rational zeros are 3, 12 , and 4. 2 z 32 6 z 2 7 z 3 2 z 3 3z 1 2 z 3 INSTRUCTOR USE ONLY So, the real zeros are 32 , 13 , and 32 . © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 1 4 x3 x 2 4 x 1 4 1 ªx 2 4 x 1 4¬ 233 97. Zeros: 2, 12 , 3 x3 14 x 2 x 14 91. f x The Fundamental Theore Theorem of Algebra x 2 2x 1 x 3 f x 2 x3 3x 2 11x 6 1 4 x 1 º¼ 1 4x 1 4 x2 1 Any nonzero scalar multiple of f would have the same three zeros. 1 4x 1 4 x 1 x 1 Let g x and r1. The rational zeros are 14 af x , a ! 0. There are infinitely many possible functions for f. y 1 6 z 3 11z 2 3 z 2 6 92. f z 8 Possible rational zeros: r1, r 2, r 12 , r 13 , r 23 , r 16 2 11 3 2 12 2 2 1 1 0 6 6 f x 6 x2 x 1 1 x 2 6 3x 1 2 x 1 (3, 0) x −8 1 x 2 6 ( 21, 0) (−2, 0) 98. −4 4 8 12 y 50 (−1, 0) The rational zeros are 2, 13 , and 12 . 10 (1, 0) (4, 0) x 93. f x x 1 x 1 x x 1 3 2 Rational zeros: 1 x 99. Interval: f, 2 , 2, 1 , 1, 4 , 4, f Matches (d). Value of f(x): Positive Negative Negative Positive 1, x 4. x 3 2 x2 3 2x 3 4 (b) The graph touches the x-axis at x 1. 3 Irrational zeros: 1 x (c) The least possible degree of the function is 4 because there are at least four real zeros (1 is repeated) and a function can have at most the number of real zeros equal to the degree of the function. The degree cannot be odd by the behavior at r f. 2 Matches (a). x3 x x x 1 x 1 Rational zeros: 3 x 0, r1 (d) The leading coefficient of f is positive. From the information in the table, you can conclude that the graph will eventually rise to the left and to the right. x 2 x 1 (e) f x Irrational zeros: 0 4 3 x 4 2 (Any nonzero multiple of f(x) is also a solution.) x3 2 x (f ) x x2 2 x x 2 x 4 x 3x 14 x 8 Matches (b). 96. f x 2, x x3 2 (a) Zeros of f x : x Rational zeros: 0 95. f x 5 1 Irrational zeros: 0 94. f x (3, 0) 4 y (− 2, 0) 2 2 x Rational zeros: 1 x Irrational zeros: 2 x 2 0 r −3 (1, 0) −1 −4 −6 −8 −10 2 (4, 0) 3 x 5 2 Matches (c). INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 234 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 100. (a) 2, 0, 2 2 x 0 ª¬ x 2 º¼ x 2 (e) f x (b) The graph touches the x-axis at x x 2. 0 and at x 5 2 x 4 4 x3 8 x 2 (c) The least possible degree of f is 5 because there are at least 5 real zeros (0 and 2 are repeated) and a function can have at most the number of real zeros equal to the degree of the function. The degree cannot be even by the definition of multiplicity. (f ) 120 Ÿ y Volume l ˜w˜h y 14 12 10 8 6 (d) The leading coefficient of f is positive. From the information in the table, you can conclude that the graph will eventually fall to the left and rise to the right. 4x y 2 x2 x 2 x 2 101. (a) Combined length and width: (−2, 0) −4 − 3 (2, 0) (0, 0) (c) 120 4 x 1 2 3 x 4 13,500 4 x 2 30 x 4 x3 120 x 2 13,500 0 x3 30 x 2 3375 0 2 x y x 2 120 4 x 15 1 30 1 4 x 30 x 2 (b) 18,000 0 3375 15 225 3375 15 225 0 x 15 x 2 15 x 225 0 Using the Quadratic Formula, x 0 2 15, 30 15 r 15 5 . 2 0 Dimensions with maximum volume: 20 in. u 20 in. u 40 in. The value of 15 15 5 is not possible because it is 2 negative. 102. (a) (b) V 15 9− x 2x 15 x −2 Because length, width, and height must be positive, you have 0 x 92 for the domain. x (c) V 125 Volume of box 15 2 x 9 2 x x x 9 2 x 15 2 x x 9 l ˜w˜h 100 75 50 (d) 56 x 9 2 x 15 2 x 56 135 x 48 x 2 4 x3 0 4 x3 48 x 2 135 x 56 The zeros of this polynomial are 12 , 72 , and 8. 25 x 1 2 3 4 x cannot equal 8 because it is not in the domain of V. 5 Length of sides of squares removed [The length cannot equal 1 and the width cannot equal 7. The product of 8 1 7 56 so it The volume is maximum when x | 1.82. The dimensions are: length | 15 2 1.82 width | 9 2 1.82 height x | 1.82 11.36 showed up as an extraneous solution.] 5.36 So, the volume is 56 cubic centimeters when x centimeter or x 7 2 1 2 centimeters. 1.82 cm u 5.36 cm u 11.36 cm INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section 2.5 103. The Fundamental Theorem Theore of Algebra 235 76 x3 4830 x 2 320,000, 0 d x d 60 P 76 x3 4830 x 2 320,000 2,500,000 76 x 3 4830 x 2 2,820,000 0 The zeros of this equation are x | 46.1, x | 38.4, and x | 21.0. Because 0 d x d 60, we disregard x | 21.0. The smaller remaining solution is x | 38.4. The advertising expense is $384,000. P 45 x3 2500 x 2 275,000 800,000 45 x 2500 x 275,000 105. (a) Current bin: V New bin: V 0 45 x 3 2500 x 2 1,075,000 V x 0 9 x3 500 x 2 215,000 104. 3 2 2u3u4 24 cubic feet 5 24 120 cubic feet 2 x 3 x 4 x (b) x3 9 x 2 26 x 24 120 120 The zeros of this equation are x | 18.0, x | 31.5, x 9 x 26 x 96 and x | 42.0. Because 0 d x d 50, disregard x | 18.02. The smaller remaining solution is x | 31.5, or an advertising expense of $315,000. The only real zero of this polynomial is x 2. All the dimensions should be increased by 2 feet, so the new bin will have dimensions of 4 feet by 5 feet by 6 feet. 250 x 160 x 106. (a) A x 1.5 160 250 (b) 60,000 x 2 410 x 40,000 0 x 2 410 x 20,000 x 410 r 4102 4 1 20,000 3 2 60,000 410 r 21 248,100 2 410 248,100 | 44.05. 2 The new length is 250 44.05 294.05 ft and the new width is 160 44.05 so the new dimensions are 294.05 ft u 204.05 ft . x must be positive, so x (c) A x 250 2 x 160 x 2 x 570 x 20,000 570 r 0 5702 4 2 20,000 570 r 22 x must be positive, so x 204.05 ft, 60,000 2 x 0 484,900 4 570 The new length is 250 2 31.6 484,900 | 31.6. 4 313.2 ft and the new width is 160 31.6 191.6 ft, so the new dimensions are 313.2 ft u 191.6 ft. 107. C x · § 200 100¨ 2 ¸, x t 1 x 30 ¹ © x C is minimum when 3×3 40 x 2 2400 x 36000 108. (a) 0. The only real zero is x | 40 or 4000 units. 12 0 7 8 (b) A | 0.01676t 4 0.2152t 3 0.794t 2 0.44t 8.7 (c) The model is a good fit to the actual data. (d) A t 10 when t 3, 4, and 7, which corresponds to the years 2003, 2004, and 2007. (e) Yes. The degree of A is even and the leading coefficient is positive, so as t increases, A will increase. This implies that attendance will continue to ggrow. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 236 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions 119. Because 1 i is a zero of f, so is 1 i. From the graph, 1 is also a zero. 16t 2 48t 6 109. h t Let h 64 and solve for t. 64 16t 2 48t 58 16t 2 48t 6 48 r i 1408 . 32 Because the equation yields only imaginary zeros, it is not possible for the ball to have reached a height of 64 feet. R C P x f x 0.0001x 60 x 150,000 0.0001x 2 60 x 9,150,000. x 1 Because the graph rises to the left and falls to the right, a 1, and f x x3 x 2 2. 121. Because f i 0, then i and 2i are zeros of f. f 2i Because i and 2i are zeros of f, so are i and –2i. 300,000 r10,000 15i 111. False. The most complex zeros it can have is two, and the Linear Factorization Theorem guarantees that there are three linear factors, so one zero must be real. f x x4 5×2 4 122. Because f 2 3 f x . This function has the same zeros as f because it is a vertical stretch of f. The zeros of g are r1 , r2 , and r3. 0, 2 is a zero of f. Because f i f x 1 x 2 x i x i 1 x 2 x 2 1 x3 2 x 2 x 2 123. Answers will vary. Some of the factoring techniques are: 1. Factor out the greatest common factor. 2 Use special product formulas. a 2 b2 a b a b f x 5 . The graph of g x is a horizontal a 2 2ab b 2 a b 2 shift of the graph of f x five units of the right, so the a 2 2ab b 2 a b 2 zeros of g x are 5 r1 , 5 r2 , and 5 r3 . a 3 b3 a b a 2 ab b 2 a 3 b3 a b a 2 ab b 2 115. g x 116. g x f 2 x . Note that x is a zero of g if and only if 2x is a zero of f. The zeros of g are 117. g x r1 r2 r , , and 3 . 2 2 2 3 f x . Because g x is a vertical shift of the graph of f x , the zeros of g x cannot be determined. 118. g x f x . Note that x is a zero of g if and only if x is a zero of f. The zeros of g are r1 , r2 , and r3 . 0, i is a zero of f. Because i is a zero of f, so is i . f x . This function would have the same zeros as f x , so r1 , r2 , and r3 are also zeros of g x . x i x i x 2i x 2i x2 1 x2 4 112. False. f does not have real coefficients. 114. g x x 1i x3 x 2 2 Because the solutions are both complex, it is not possible to determine a price p that would yield a profit of 9 million dollars. 113. g x x 1i x2 2 x 2 x 1 2 60 r 60 0.0002 x 1 120. Because 1 i is a zero of f, so is 1 i. From the graph, 1 is also a zero. xp C 0.0001x 2 60 x 150,000 Thus, 0 x 1i x3 3x 2 4 x 2 x 140 0.0001x 80 x 150,000 9,000,000 x 1i x2 2 x 2 x 1 0 By the Quadratic Formula, t 110. f x 3. Factor by grouping, if possible. 4. Factor general trinomials with binomial factors by “guess-and-test” or by the grouping method. 5. Use the Rational Zero Test together with synthetic division to factor a polynomial. 6. Use Descartes’s Rule of Signs to determine the number of real zeros. Then find any zeros and use them to factor the polynomial. 7. Find any upper and lower bounds for the real zeros to eliminate some of the possible rational zeros. Then test the remaining candidates by synthetic division and use any zeros to factor the polynomial. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.6 237 x4 4×2 k 124. f x x Ration Rational Functions 4 r 2 4 2 41 k 4r 2 4 k 2 21 r x 2r 2r 4 k 4 k (a) For there to be four distinct real roots, both 4 k (d) For there to be four complex zeros, 2 r 4 k must be nonreal. This occurs when k ! 4. Some 5, k 6, k 7.4, etc. possible k-values are k and 2 r 4 k must be positive. This occurs when 0 k 4. So, some possible k-values are 1 k 1, k 2, k 3, k ,k 2, etc. 2 (e) g x No. This function is a horizontal shift of f x . Note that x is a zero of g if and only if x 2 is a zero of f; the number of real and complex zeros is not affected by a horizontal shift. (b) For there to be two real roots, each of multiplicity 2, 4 k must equal zero. So, k 4. (c) For there to be two real zeros and two complex zeros, 2 4 k must be positive and (f ) g x 2 4 k must be negative. This occurs when k 0. So, some possible k-values are 1 k 1, k 2, k , etc. 2 x 125. (a) f x bi x 126. (a) f x cannot have this graph because it also has a zero at x ª¬ x a bi ºª ¼¬ x a bi º¼ (b) f x x a function. Its graph is a parabola. 2 bi x 2ax a b 2 2 0. (b) g x cannot have this graph because it is a quadratic ª¬ x a biºª ¼¬ x a biº¼ 2 f 2x No. Because x is a zero of g if and only if 2x is a zero of f, the number of real and complex zeros of g is the same as the number of real and complex zeros of f. x2 b bi f x 2 (c) h x is the correct function. It has two real zeros, 2 x 2 and x 3.5, and it has a degree of four, needed to yield three turning points. (d) k x cannot have this graph because it also has a zero at x 1. In addition, because it is only of degree three, it would have at most two turning points. Section 2.6 Rational Functions 1. rational functions 3. horizontal asymptote 2. vertical asymptote 4. slant asymptote 1 x 1 5. f x (a) x f x x f x x f x 0.5 –2 1.5 2 5 0.25 0.9 –10 1.1 10 10 0. 1 0.99 –100 1.01 100 100 0.01 0.999 –1000 1.001 1000 1000 0.001 (b) The zero of the denominator is x 1, so x 1 is a vertical asymptote. The degree of the numerator is less than the degree of the denominator, so the x-axis, or y 0, is a horizontal asymptote. (c) The domain is all real numbers x except x 1. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 238 NOT FOR SALE Chapter 2 5x x 1 6. f x (a) Polynomial ynomial and Rational Function Functions x f x x f x x f x 0.5 –5 1.5 15 5 6.25 0.9 –45 1.1 55 10 5.5 0.99 –495 1.01 505 100 5.05 (b) The zero of the denominator is x 1, so x 1 is a vertical asymptote. The degree of the numerator is equal to the degree of 5 5 the denominator, so the line y 1 is a horizontal asymptote. (c) The domain is all real numbers x except x 0.999 9. f x 5005 1000 5.005 x f x x f x x f x 0.5 –1 1.5 5.4 5 3.125 0.9 –12.79 1.1 17.29 10 3.03 0.99 –147.8 1.01 152.3 100 3.0003 0.999 –1498 1.001 1502 1000 3 r1, (b) The zeros of the denominator are x so both x 1 and x 1 are vertical asymptotes. The degree of the numerator equals the degree of the denominator, so 3 y 3 is a horizontal asymptote. 1 (c) The domain is all real numbers x except x r1. 4x x2 1 x f x x f x x f x 0.5 2.6 1.5 4.8 5 0.83 0.9 –18.95 1.1 20.95 10 0.40 0.99 –199 1.01 201 100 0.04 0.999 –1999 1.001 2001 1000 0.004 4 x2 (b) The zeros of the denominator are x r1, so both x 1 and x 1 are vertical asymptotes. The degree of the numerator is less than the degree of the denominator, so the x-axis, or y 0, is a horizontal asymptote. (c) The domain is all real numbers x except x r1. 11. f x Domain: all real numbers x except x Vertical asymptote: x 0 0 Domain: all real numbers x except x ª¬Degree of N x 12. f x 4 3 Domain: all real numbers x except x Vertical asymptote: x Horizontal asymptote: y x 5 x 5 2 3 7x 3 2x 1 degree of D x º¼ 7 x 3 2x 3 Domain: all real numbers x except x 2 Vertical asymptote: x 0 ª¬Degree of N x degree of D x º¼ Horizontal asymptote: y ª¬Degree of N x 5 5 Horizontal asymptote: y ª¬Degree of N x degree of D x º¼ x 2 5 x 5 x Vertical asymptote: x 0 Horizontal asymptote: y 10. f x 1. 2 8. f x (a) 1.001 3x 2 x 1 7. f x (a) –4995 3 2 3 2 7 2 degree of D x º¼ INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.6 13. f x x3 x 1 Domain: all real numbers x except x Vertical asymptotes: x r1 Vertical asymptote: x r1 x 2 x 4 20. f x 4 x2 x 2 Vertical asymptote: x Domain: all real numbers x except x 4 1 Horizontal asymptote: y 2 Matches graph (b). 2 Horizontal asymptote: None ª¬Degree of N x 1 Matches graph (c). ª¬Degree of N x ! degree of D x º¼ Vertical asymptote: x 4 Horizontal asymptote: y Horizontal asymptote: None 14. f x degree of D x º¼ x2 9 x 3 21. g x x 3 x 3 x 3 The only zero of g x is x 15. f x 3x 2 1 2 x x 9 3 makes g x 10 x2 5 10 4 2 x 5 10 x2 5 10 4 h x 22. Vertical asymptote: None 0 Horizontal asymptote: y 3 4 degree of D x º¼ 4 x 2 5 3x x 5 x2 1 2 16. f x 3. x undefined. Domain: All real numbers x. The denominator has no real zeros. [Try the Quadratic Formula on the denominator.] ª¬Degree of N x 239 x 1 x 4 19. f x 2 Ration Rational Functions 4 x 2 30 15 2 x2 Domain: All real numbers x. The denominator has no real zeros. [Try the Quadratic Formula on the denominator.] No real solution, h x has no real zeros. Vertical asymptote: None Horizontal asymptote: y ¬ªDegree of N x 17. f x degree of D x º¼ 0 4 x 5 Vertical asymptote: x 2 x 7 x 7 x 5 Horizontal asymptote: y 0 x Matches graph (d). 18. f x 5 x 2 Vertical asymptote: x Horizontal asymptote: y Matches graph (a). 24. 2 0 1 2 9 9 is a zero of f x . g x x3 8 x2 1 x3 8 x3 8 x2 1 0 0 3 8 x 2 x x 2 x 7 2 1 x 7 1 f x 23. 3 2 is a real zero of g x . INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 240 Chapter 2 25. f x NOT FOR SALE Polynomial ynomial and Rational Function Functions x 4 x 2 16 x 4 x 4 x 4 r4 x 2 3x 4 2×2 x 1 x 1 x 4 Because x 4 is a 2x 1 x 1 Domain: all real numbers x except x Vertical asymptote: x 4 common factor of N x and D x , x 1 ,x z 4 x 4 4 is not a x 1 x2 1 x 1 x 1 x 1 1 , x z 1 x 1 asymptote of f x . ) vertical asymptote of f x . ) ¬ªDegree of N x x 2 25 x 4x 5 0 Vertical asymptote: x x5 ,x z 5 x 1 x 5 x 1 5 and x 1 1 Because x 5 is a common factor of N x and D x , x 5 is not a vertical asymptote of f x . Horizontal asymptote: y ª¬Degree of N x 28. f x 1 degree of D x º¼ x2 4 2 x 3x 2 x 2 x 2 x 2 x 1 Domain: all real numbers x except 3 1 x or x 2 3 1 (Because 2 x 3 is a 3 3 common factor of N x and D x , x is not a 2 vertical asymptote of f x . ) x 2 ,x z 2 x 1 Vertical asymptote: x Horizontal asymptote: y ª¬Degree of N x Domain: all real numbers x except x 1 and x 2 Vertical asymptote: x 1 (Because x 2 is a common 2 is not a vertical factor of N x and D x , x 1 degree of D x º¼ 1 x 2 31. f x (a) Domain: all real numbers x except x Horizontal asymptote: y 2 Horizontal asymptote: y 1 degree of D x º¼ 2 § 1· (b) y-intercept: ¨ 0, ¸ © 2¹ (c) Vertical asymptote: x asymptote of f x . ) ª¬Degree of N x 3x 1 3 ,x z 3x 1 2 2 x 3 3x 1 x5 x5 Domain: all real numbers x except x degree of D x º¼ 6 x 2 11x 3 6×2 7 x 3 2 x 3 3x 1 30. f x 2 1 2 Horizontal asymptote: y ª¬Degree of N x degree of D x º¼ 27. f x 1 1 (Because x 1 is a 2 1 is not a common factor of N x and D x , x Domain: all real numbers x except x r1 Vertical asymptote: x 1 (Because x 1 is a common factor of N x and D x , x 1 is not a vertical Horizontal asymptote: y 1 and x 2 Vertical asymptote: x 0 ª¬Degree of N x degree of D x º¼ 26. f x x 4 , x z 1 2x 1 Domain: all real numbers x except x vertical asymptote of f x . Horizontal asymptote: y 29. f x (d) x –4 y 1 2 3 –1 0 1 0 1 1 1 2 1 3 y 2 (0, 12 ) –3 1 x –1 –1 INSTRUCTOR USE S ONLY –2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.6 (d) 1 x 3 32. f x (a) Domain: all real numbers x except x 3 x 4 5 7 8 y 1 2 1 –1 1· § (b) y-intercept: ¨ 0, ¸ 3¹ © 0 1 y 1 3 1 2 1 2 6 3 4 Horizontal asymptote: y x 241 y (c) Vertical asymptote: x (d) Ration Rational Functions 2 0 2 –1 4 5 6 1 1 2 1 3 y x −2 −2 2 4 10 −4 −6 7 2x 2 x 35. C x 3 2 (0, 16 ) 2x 7 x 2 2 (a) Domain: all real numbers x except x 1 § 7 · (b) x-intercept: ¨ , 0 ¸ © 2 ¹ x 2 4 5 6 –1 –2 (0, − 13 ) § 7· y-intercept: ¨ 0, ¸ © 2¹ –3 1 x 4 33. h x 2 (c) Vertical asymptote: x Horizontal asymptote: y (a) Domain: all real numbers x except x 4 (d) 1· § (b) y-intercept: ¨ 0, ¸ 4¹ © x –3 –1 1 3 y –1 5 3 13 5 4 (c) Vertical asymptote: x 2 y Horizontal asymptote: y (d) 0 6 5 x –5 –3 –1 1 )0, ) 7 2 3 y 1 –1 1 3 1 5 1 x −6 −5 −4 7 − 2, 0 ) y 1 −1 ) 2 −2 4 3 2 −7 −6 −5 x −1 −2 )0, − ) 1 4 1 §1 · (b) x-intercept: ¨ , 0 ¸ ©3 ¹ −4 1 6 x 3x 1 x 1 (a) Domain: all real numbers x except x −3 34. g x 1 3x 1 x 36. P x 1 1 x 6 y-intercept: 0, 1 (a) Domain: all real numbers x except x Horizontal asymptote: y (c) Vertical asymptote: x Horizontal asymptote: y § 1· (b) y-intercept: ¨ 0, ¸ © 6¹ (c) Vertical asymptote: x 6 (d) 6 1 3 y x –1 0 2 3 y 2 1 5 4 6 5 4 0 (0, 1) ( 13 , 0) ––1 2 INSTRUCTOR USE ONLY N x –22 3 4 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 242 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x2 x 9 37. f x 1 40. f x 2 2 x 2 (a) Domain: all real numbers x (a) Domain: all real numbers x except x (b) Intercept: 0, 0 1· § (b) y-intercept: ¨ 0, ¸ 4¹ © (c) Horizontal asymptote: y (d) 1 y x ±1 ±2 ±3 y 1 10 4 13 1 2 (c) Vertical asymptote: x 3 2 x Horizontal asymptote: y 0 1 2 –1 1 x 0 y 2 –1 1 2t t 38. f t 2t 1 t y –1 5 2 –3 1 –1 –4 –4 –1 4 4 9 1 4 3 –3 2 1 2 1 0 –1 –4 2 x2 5x 4 x2 4 41. h x 3 2 x 1 x 4 x 2 x 2 (a) Domain: all real numbers x except x r2 (b) x-intercepts: 1, 0 , 4, 0 y-intercept: 0, 1 t 1 7 2 x 0 0 ( 12 , 0) –1 3 –2 y –2 4 9 5 2 –1 Horizontal asymptote: y –2 3 2 (0, − 14 ) §1 · (b) t-intercept: ¨ , 0 ¸ ©2 ¹ (c) Vertical asymptote: t t 1 4 1 y (a) Domain: all real numbers t except t (d) 2 (d) (0, 0) –2 2 2 –1 (c) Vertical asymptotes: x Horizontal asymptote: y 2, x 2 1 –3 (d) 4s s2 4 39. g s x –4 3 1 y 10 3 28 5 10 3 –1 1 3 4 0 2 5 0 y (a) Domain: all real numbers s 6 (b) Intercept: 0, 0 4 2 (c) Vertical asymptote: none Horizontal asymptote: y (d) 0 −6 −4 0 (1, 0) x (4, 0) 6 (0, −1) s –2 –1 0 1 2 y –1 4 5 0 4 5 1 y 4 3 2 1 −2 −1 (0, 0) 2 3 4 s INSTRUCTOR T USE ONLY −2 −3 3 −4 © 2012 Cengage Learning. 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NOT FOR SALE Section 2.6 x 4 x 2 x2 2 x 8 x2 9 42. g x r3 (a) Domain: all real numbers x except x and x (c) Vertical asymptotes: x –5 4 y 27 16 16 7 y −6 −4 Horizontal asymptote: y 2 0 2 0 8 9 8 5 4 5 0 7 16 (d) –4 y 4 2 4 0 5 9 0 –1 0 0 5 12 3 1 3 3 2 1 (4, 0) 4 6 x 2 ( 1 0, − 3 −6 ) 4 5 (2, 0) −2 −3 −4 −5 2x 1 x 3 2×2 5x 3 x 2×2 x 2 x 2 x 1 x 1 2 (a) Domain: all real numbers x except x 2, x 1 and x 1, x y (−1, 0) 2 3 9 35 4 −4 x x3 x 2 3x x x6 45. f x x3 x2 2 1, (a) Domain: all real numbers x except x x 2 § 1 · (b) x-intercept: ¨ , 0 ¸, 3, 0 © 2 ¹ (c) Vertical asymptote: x x –3 2 0 y 3 4 5 4 1, x 2, x Horizontal asymptote: y 3 2 3 and 2 Horizontal asymptote: y (c) Vertical asymptotes: x x , x z 3 x2 (b) Intercept: 0, 0 3· § y-intercept: ¨ 0, ¸ 2¹ © (d) x 6 −2 2, x (c) Vertical asymptotes: x 1 x (− 2, 0) 43. f x 3 (0, 89) 2 2, 1· § y-intercept: ¨ 0, ¸ 3¹ © 3 3, x Horizontal asymptote: y x 1, x (b) x-intercepts: 1, 0 , 2, 0 § 8· y-intercept: ¨ 0, ¸ © 9¹ (d) x 1 x 2 x 3 3 (a) Domain: all real numbers x except x (b) x-intercepts: 4, 0 , 2, 0 243 x 1 x 2 x2 x 2 x 2 x2 5x 6 44. f x x 3 x 3 Ration Rational Functions 1 (d) x –1 0 1 3 4 y 1 3 0 –1 3 2 0 1.5 3 4 48 5 0 3 10 1 y 6 4 y 2 ( ( −4 −3 (0, 0) 6 3 x −6 −4 −2 9 − 1, 0 2 (3, 0) 3 4 4 6 −4 x −6 (0, − 32( INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 244 NOT FOR SALE Chapter 2 46. f x Polynomial ynomial and Rational Function Functions 5×4 5×4 x x 12 x4 x3 2 (a) Domain: all real numbers x except x x 3 5 , x z 4 x3 4 and 5· § (b) y-intercept: ¨ 0, ¸ 3¹ © (c) Vertical asymptote: x (d) x y –2 0 –1 5 3 3x 2 ,x z 2 2x 1 x 2 2x 1 (a) Domain: all real numbers x except x 1 x 2 3 Horizontal asymptote: y 3x 2 8 x 4 2 x 2 3x 2 x 2 3x 2 48. f x 2 and §2 · (b) x-intercept: ¨ , 0 ¸ ©3 ¹ 0 2 5 7 –5 5 2 5 4 y-intercept: 0, – 2 (c) Vertical asymptote: x 1 2 y 3 2 Horizontal asymptote: y 6 4 (d) 2 x ( 2 ) 4 6 x –3 –1 0 2 3 3 y 11 5 5 –2 0 1 8 5 0, − 3 −4 −6 y 2 x2 5x 2 2×2 x 6 2x 1 x 2 47. f x 2x 1 ,x z 2 2x 3 2x 3 x 2 (a) Domain: all real numbers x except x 3 x 2 x −4 −3 − 2 − 1 ( 23 , 0) 3 4 (0, − 2) 2 and t 1 t 1 t2 1 t 1 49. f t §1 · (b) x-intercept: ¨ , 0 ¸ ©2 ¹ t 1 t 1; t z 1 (a) Domain: all real numbers t except t 1· § y-intercept: ¨ 0, ¸ 3¹ © 1 (b) t-intercept: 1, 0 y-intercept: 0, 1 3 2 (c) Vertical asymptote: x Horizontal asymptote: y (d) 1 (c) Vertical asymptote: none Horizontal asymptote: none 1 (d) x –3 –2 –1 0 y 7 3 5 –3 1 1 3 1 5 t –1 0 2 3 y 0 1 3 4 y 4 y 3 4 2 3 −4 −3 1 −5 −4 −3 − 2 1 0, − 3 ) ) 1 (0, 1) −1 1 (− 1, 0) 2 )12 , 0) 3 x t 2 3 4 −2 −3 −4 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.6 x 6 x 6 x 2 36 x 6 50. f x x 6 (a) Domain: all real numbers x except x x 6; x z 6 6 y-intercept: 0, 6 (b) f x (c) Vertical asymptote: none Horizontal asymptote: none x 1 2 3 4 y –5 –4 –3 –2 (c) x 2x x x 2 x 2 0 1 1.5 2 2.5 3 f x –1 Undef. 1 1.5 Undef. 2.5 3 g x –1 0 1 1.5 2 2.5 3 8 10 −4 −6 x2 x 2 2 –1 x 6 x2 x 2 0 and x (6, 0) 2 x and the denominator of f, neither x 0 nor x is a vertial asymptote of f. So, f has no vertical asymptotes. y −6 −4 −2 ,g x 245 Because x x 2 is a factor of both the numerator 4 2 x2 2x (a) Domain of f: All real numbers x except x x 2 Domain of g: All real numbers x (b) x-intercept: 6, 0 (d) x2 x 2 52. f x Ration Rational Functions (0, − 6) (d) −10 2 −12 −2 x2 1 ,g x x 1 51. f x x 1 −2 (a) Domain of f: all real numbers x except x 1 (e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where it does not exist. Domain of g: all real numbers x (b) f x x2 1 x 1 x 1 x 1 Because x 1 is a factor of both the numerator and 1 is not a vertical the denominator of f , x (c) x –3 –2 –1.5 –1 –0.5 0 1 f x –4 –3 –2.5 Undef. –1.5 –1 0 (d) –2.5 –2 –1.5 –1 x 2 x x 2 0 1 x asymptote. The only vertical asymptote of f is x 0 1 −4 x 2 x2 2 x 0 and Because x 2 is a factor of both the numerator and 2 is not a vertical the denominator of f , x (c) –3 1 x (a) Domain of f: All real numbers x except x x 2 Domain of g: All real numbers x except x (b) f x asymptote. So, f has no vertical asymptotes. –4 x 2 ,g x x2 2x 53. f x x 1 x 1 g x 4 2 x –0.5 0 f x –2 Undef. 2 1 g x –2 Undef. 2 1 (d) 0.5 1 1.5 2 2 3 2 3 Undef. 1 2 0. 3 1 3 1 3 2 −3 (e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where it does not exist. −3 3 −2 (e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where here it does not exist. e INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 246 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 2x 6 ,g x x 7 x 12 54. f x 2 x 4 2 (a) Domain of f: All real numbers x except x x 4 Domain of g: All real numbers x except x 3 and (c) Vertical asymptote: x x x –2 –1 1 2 y 5 2 8 –8 2 x 4 0 1 f x 1 2 1 2 2 3 2 3 g x (d) 2 y=x 4 4 as its only vertical 4 (3, 0) 4 6 x 8 −4 5 6 −6 −8 –1 Undef. Undef. 2 1 x2 5 5 x x x (a) Domain: all real numbers x except x 56. g x –1 –2 Undef. 2 1 0 (b) No intercepts 3 −1 2 (− 3, 0) −8 − 6 3 5 2 y Because x 3 is a factor of both the numerator and 3 is not a vertical the denominator of f , x x 0 Slant asymptote: y x 3 x 4 (c) 0 (b) x-intercepts: 3, 0 , 3, 0 4 (d) asymptote of f. So, f has x asymptote. 9 x x (a) Domain: all real numbers x except x 2x 6 x 2 7 x 12 2 x 3 (b) f x x2 9 x 55. h x (c) Vertical asymptote: x 0 Slant asymptote: y x 8 (d) x –3 y −3 (e) Because there are only a finite number of pixels, the utility may not attempt to evaluate the function where it does not exist. 14 3 –2 –1 1 2 3 9 2 –6 6 9 2 14 3 y 6 4 y=x 2 −6 −4 −2 −2 x 2 4 6 −4 2 x2 1 1 2x x x (a) Domain: all real numbers x except x 57. f x 0 (d) x –4 y (b) No intercepts (c) Vertical asymptote: x Slant asymptote: y 0 33 4 –2 2 4 6 9 2 9 2 33 4 73 6 2x y 6 4 2 y = 2x x –6 –4 –2 2 4 6 INSTRUCTOR USE ONLY –6 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.6 1 x2 1 x x x (a) Domain: all real numbers x except x x2 1 1 x x x (a) Domain: all real numbers x except x 58. f x 247 59. g x 0 (b) x-intercepts: 1, 0 , 1, 0 0 (b) No intercepts (c) Vertical asymptote: x 0 Slant asymptote: y x (c) Vertical asymptote: x 0 Slant asymptote: y x (d) Ration Rational Functions (d) x –6 4 2 2 y 35 6 15 4 3 2 4 3 2 6 15 4 35 6 x –4 y 2 2 4 6 5 2 5 2 17 4 37 6 17 4 y y y = −x 6 8 4 6 4 y=x 2 2 (−1, 0) (1, 0) –8 –6 –4 –2 4 x x 6 –6 –4 –2 2 4 6 8 –4 –6 –6 –8 x2 x 1 60. h x 1 x 1 x 1 (a) Domain: all real numbers x except x 1 (d) y x –4 y (b) Intercept: 0, 0 (c) Vertical asymptote: x 1 Slant asymptote: y x 1 16 5 2 2 4 6 4 3 4 16 3 36 5 8 6 4 y=x+1 2 (0, 0) x –4 2 4 6 8 –2 –4 61. f t t2 1 t 5 t 5 26 t 5 (a) Domain: all real numbers t except t 1· § (b) y-intercept: ¨ 0, ¸ 5¹ © 5 1 1 1 x 3 9 9 3x 1 (a) Domain: all real numbers x except x 1 3 (b) Intercept: 0, 0 (c) Vertical asymptote: t 5 Slant asymptote: y t 5 t –7 –6 –4 –3 0 y 25 37 –17 –5 (d) x2 3x 1 62. f x y 1 5 1 3 (c) Vertical asymptote: x Slant asymptote: y 1 1 x 3 9 x –3 –2 1 y 9 8 4 5 (d) 25 1 2 1 2 1 2 0 2 0 4 7 20 y=5−t y 15 (0, − 15( 5 −20 −15 − 10 − 5 1 t 1 1 y = 3x − 9 2 3 (0, 0) x −1 1 3 2 3 1 4 3 INSTRUCTOR USE ONLY © 2012 Cengage Learning. 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All Rights Reserved. 248 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x3 4x x 2 x 4 x 4 (a) Domain: all real numbers x except x 63. f x r2 x 1 x 1 (a) Domain: all real numbers x except x 1 (b) y-intercept: 0, 1 (b) Intercept: 0, 0 2 2 and x (c) Vertical asymptotes: x Slant asymptote: y x (d) x2 x 1 x 1 65. f x 2 (c) Vertical asymptote: x 1 Slant asymptote: y x (d) x –3 –1 1 3 y 27 5 1 3 1 3 27 5 x –4 –2 0 2 4 y 21 5 7 3 –1 3 13 3 y y 8 8 6 4 6 y=x 4 (0, 0) −8 − 6 − 4 4 x 6 y=x 2 (0, −1) 8 –4 –2 x 2 4 6 8 –4 x3 2x 8 64. g x 1 4x x 2 2×2 8 2 2 x2 5x 5 3 2x 1 x 2 x 2 2 (a) Domain: all real numbers x except x 66. f x (a) Domain: all real numbers x except x r2 5· § (b) y-intercept: ¨ 0, ¸ 2¹ © (c) Vertical asymptote: x 2 Slant asymptote: y 2x 1 (b) Intercept: 0, 0 r2 (c) Vertical asymptote: x Slant asymptote: y 1 x 2 x –6 –4 –1 1 4 6 y 27 8 8 3 1 6 1 6 8 3 27 8 (d) (d) x –6 y –3 107 8 38 5 1 3 6 7 –2 8 47 4 68 5 y y 15 8 12 6 9 4 6 (0, 0) y = 2x − 1 3 x –8 –6 –4 4 6 x 8 –9 –6 –3 y = 12 x ( 0, − 52 3 6 9 12 15 ) –9 67. f x 2 x3 x 2 2 x 1 x 2 3x 2 2x 1 x 1 x 1 x 1 x 2 2x 1 x 1 , x 2 2 x 2 3x 1 x 2 15 2x 7 , x 2 x z 1 x z 1 1 and x 2 INSTRUCTOR USE ONLY (a) Domain: all real numbers x except x © 2012 Cengage Learning. 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NOT FOR SALE Section 2.6 § 1· (b) y-intercept: ¨ 0, ¸ © 2¹ (d) §1 · x-intercepts: ¨ , 0 ¸, 1, 0 ©2 ¹ (c) Vertical asymptote: x 2 x –4 y 3 2 0 1 –28 20 1 2 0 18 12 71. f x (1, 0) −5 −4 −3 ,x z 2 x2 5x 8 x 3 x 2 Vertical asymptote: x 3 Slant asymptote: y x 2 Line: y 2×2 5x 2 x 1 9 2x 7 ,x z 2 x 1 3 ) 12 , 0) y = 2x − 7 2 x 3 3 8 −14 x 2 10 −8 72. f x (a) Domain: all real numbers x except x 2 x 1 and § 1 · x-intercepts: 2, 0 , ¨ , 0 ¸ © 2 ¹ (c) Vertical asymptote: x 1 Slant asymptote: y 2x 7 2 x2 x x 1 2x 1 1 x 1 Domain: all real numbers x except x (b) y-intercept: 0, 2 (d) x −1 −12 −18 −24 Domain: all real numbers x except x x 2 x 1 x 1 )0, 12 ) −30 −36 2 x3 x 2 8 x 4 x 2 3x 2 x 2 x 2 2x 1 x 2 2x 1 249 y –3 2x 7 Slant asymptote: y 68. f x 45 2 Ration Rational Functions Vertical asymptote: x 1 Slant asymptote: y 2x 1 Line: y 2x 1 1 6 −12 12 −10 x –3 –2 –1 0 1 2 y 5 4 0 1 2 –2 –10 3 2 3 4 28 35 2 18 y 30 73. g x 1 3x 2 x3 x2 1 3 x x2 Domain: all real numbers x except x Vertical asymptote: x 0 Slant asymptote: y x 3 Line: y x 3 1 x2 0 12 x 3 24 −12 12 18 y = 2x + 7 12 (−2, 0) −6 −2 −4 ( 1 − ,0 2 2 4 ) x 6 74. h x (0, −2) 1 x 2 2 Domain: All real numbers except x One possibility: f x 1 2 x 1 2 4 x Domain: all real numbers x except x 69. Domain: All real numbers One possibility: f x 12 2 x x 2 24 x Vertical asymptote: x 4 Slant asymptote: y 1 x 1 2 2. 1 x 2 Line: y 1 x 1 2 4 10 −16 8 −6 (Answers are not unique). 70. An asymptote is a line to which a graph gets arbitrarily close to, but does not reach, as x or y increases INSTRUCTOR USE ONLY without bound. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 250 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions x 1 x 3 75. y (a) x-intercept: 1, 0 0 x 1 x 3 x 1 1 x (b) 0 (a) 300,000 0 0 2x x 3 2x 0 x C 90 x x2 x x 3 78. y (a) 25,000 90 $225,000 100 90 1 0.04t ,t t 0 N 10 500 deer N 25 800 deer (b) The herd is limited by the horizontal asymptote: 60 N 1500 deer 0.04 82. (a) 0.25 50 0.75 x 2 x 0 2 x x 2 3x 2 0 x 1 x 2 x 1, x x 3 C 50 x 12.50 0.75 x 4 ˜ 50 x 4 50 3 x 3x 50 4 50 x 4 x 50 C C (b) Domain: x t 0 and x d 1000 50 So, 0 d x d 950. Using interval notation, the domain is >0, 950@. 2 255 p , 0 d p 100 100 p 79. C $25,000 100 50 (a) N 5 | 333 deer (a) x-intercepts: 1, 0 , 2, 0 (b) 0 25,000 50 20 5 3t 81. N 1 x x 1 x 1 r1 0 | $4411.76 (c) C o f as x o 100. No. The function is undefined for p 100. (a) x-intercepts: 1, 0 , 1, 0 (b) 100 15 C 50 1 x x 77. y 25,000 15 (b) C 15 (a) x-intercept: 0, 0 (b) 0 100 0 2x x 3 76. y 25,000 p , 0 d p 100 100 p 80. C (c) C 1.0 0.8 2,000 0.6 0.4 0.2 0 x 100 200 400 600 800 1000 0 (b) C 10 C 40 C 75 255 10 100 10 255 40 100 40 255 75 100 75 | 28.33 million dollars 170 million dollars (d) As the tank is filled, the concentration increases more slowly. It approaches the horizontal asymptote 3 of C 0.75 75%. 4 765 million dollars (c) C o f as x o 100. No. The function is undefined at p 100. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.6 84. (a) A 83. (a) Let t1 time from Akron to Columbus and t2 time from Columbus back to Akron. xt1 100 Ÿ t1 yt2 100 Ÿ t2 50 t1 t2 xy and x 4 y 2 100 x 100 y 30 30 x 4 y 2 2 y 200 So, A 2 x x 11 § 2 x 22 · x¨ ¸ © x 4 ¹ xy 2 x 22 x 4 30 x 4 . 4 100 100 x y 4 100 y 100 x 4 xy (b) Domain: Because the margins on the left and right are each 2 inches, x ! 4. In interval notation, the domain is 4, f . xy (c) So, y 25 x xy 25 y 25 x y x 25 25 x . x 25 x 4 200 4 40 4 0 (b) Vertical asymptote: x 25 Horizontal asymptote: y x 5 6 7 8 9 10 y1 Area 160 102 84 76 72 70 x 11 12 13 14 15 y1 Area 69.143 69 69.333 70 70.999 25 200 25 65 0 (d) 251 t1 t2 25 y 25 x (c) Ration Rational Functions x 30 35 40 45 50 55 60 y 150 87.5 66.7 56.3 50 45.8 42.9 (e) Sample answer: No. You might expect the average speed for the round trip to be the average of the average speeds for the two parts of the trip. (f ) No. At 20 miles per hour you would use more time in one direction than is required for the round trip at an average speed of 50 miles per hour. The area is minimum when x | 11.75 inches and y | 5.87 inches. 85. False. Polynomial functions do not have vertical asymptotes. x crosses y x2 1 which is a horizontal asymptote. 86. False. The graph of f x 0, 87. False. A graph can have a vertical asymptote and a horizontal asymptote or a vertical asymptote and a slant asymptote, but a graph cannot have both a horizontal asymptote and a slant asymptote. A horizontal asymptote occurs when the degree of N x is equal to the degree of D x or when the degree of N x is less than the degree of D x . A slant asymptote occurs when the degree of N x is greater than the degree of D x by one. Because the degree of a polynomial is constant, it is impossible to have both relationships at the same time. 88. (a) f x x 1 x3 8 (b) f x x 2 x3 1 (c) f x (d) f x 2 x2 9 x 2 x 1 2 x 2 x 3 x 1 x 2 89. No; Yes; 2 x 2 18 x2 x 2 Not every rational function is a polynomial because 1 3 and h x are rational functions, but g x x 2 x they are not polynomials. Every polynomial f x is a rational function because it can be written as f x 1 . 2 x 2 2 x 12 x2 x 2 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 252 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions Review Exercises for Chapter 2 1. (a) y 2×2 2. (a) y Vertical stretch x2 4 Vertical shift four units downward y y 4 3 3 2 1 2 x –4 –3 x −4 −3 −2 −1 −1 1 2 3 –1 4 1 3 4 –2 −2 −3 −4 (b) y –5 2 x 2 (b) y Vertical stretch and a reflection in the x-axis y 4 x2 Reflection in the x-axis and a vertical shift four units upward y 4 3 5 2 1 3 x −4 −3 −2 −1 1 2 3 2 4 1 x −4 −3 −3 −1 −1 −4 (c) y 1 3 4 −2 −3 x2 2 Vertical shift two units upward x 3 (c) y 2 Horizontal shift three units to the right y y 4 3 5 4 1 3 x −4 −3 −2 −1 −1 1 2 3 4 2 −2 1 −3 x −4 (d) y –2 –1 –1 2 3 4 5 –2 2 x 2 1 Horizontal shift two units to the left (d) y 1 x2 1 2 Vertical shrink each y -value is multiplied by 12 , y 4 and a vertical shift one unit downward y 1 −4 −3 −2 −1 −1 4 x 1 2 3 3 4 2 −2 1 −3 −4 x −4 −3 −2 2 3 4 −2 −3 −4 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises ffor Chapter 2 x2 2 x 3. g x 3 4x x2 6. h x x2 2 x 1 1 x 1 2 x2 4x 3 1 x2 4x 4 4 3 Vertex: 1, 1 ª x 2 ¬ Axis of symmetry: x x2 2x 0 1 2 x 2 x x 2 7º ¼ 2 y 7 10 8 Vertex: 2, 7 x-intercepts: 0, 0 , 2, 0 6 4 Axis of symmetry: x y 7 6 5 4 3 0 3 4x x 0 x2 4x 3 x 2 3 4 5 6 4 r 6x x2 x −2 4 28 2 2 4 6 8 10 4 1 3 2r 2 4. f x 2 21 4r −2 2 2 x −3 −2 −1 253 x-intercepts: 2 r 7 7, 0 x 6x 9 9 2 x 3 2 2t 2 4t 1 7. f t 9 2 t 2 2t 1 1 1 Vertex: 3, 9 3 2 ª t 1 ¬ x6 x 2 t 1 Axis of symmetry: x 6x x2 0 1º 1 ¼ 2 2 3 Vertex: 1, 3 x-intercepts: 0, 0 , 6, 0 Axis of symmetry: t y 10 0 2 t 1 1 2 3 8 2t 1 6 2 y 3 6 5 4 3 4 t 1 2 −2 3 2 r x −2 4 2 8 t x 2 8 x 10 5. f x 2 8. f x 2 x 4 x t −3 −2 −1 1 2 3 4 5 6 4 10 6 · , 0 ¸¸ 2 ¹ x 4 2 x-intercepts: 4 r 2 4 y Vertex: 4, 4 6 4 r x 4 y 6 8 6 2 r 6 x 2 8 x 12 x 2 8 x 16 16 12 4 Axis of symmetry: x x 4 1r 6 Vertex: 4, 6 0 6 2 § t-intercepts: ¨¨1 r © x 2 8 x 16 16 10 x 4 2 1 10 Axis of symmetry: x 4 4 x −8 6 6, 0 −4 −2 0 x 2 8 x 12 −4 0 x 2 x 6 2 −6 x-intercepts: 2, 0 , 6, 0 2 x −2 −2 8 −4 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 254 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Functions 4 x 2 4 x 13 9. h x 11. h x 4 x 2 x 13 x2 5x 4 x 2 x 14 14 13 2 2 2 5· 41 § ¨x ¸ 2 4 © ¹ 12 Vertex: 12 , 12 0 4 x 12 2 3 x 12 2 20 12 −3 −2 −1 Axis of symmetry: x 0 x2 5x 4 −10 x 1 2 3 x 6 r 6 3 12. f x 32 2 41 1 21 4 x2 4x 5 2 ª§ º 1· 4 «¨ x ¸ 1» 2¹ «¬© »¼ 3r 2 2 2 52 4 1 4 1 1 5· § 4¨ x 2 x ¸ 4 4 4¹ © 21 6r 5 r 5 r 41 2 x2 6x 1 x 8 2 2 § 1 · Vertex: ¨ , 4 ¸ © 2 ¹ y 2 Axis of symmetry: x −8 −6 −4 −2 −2 x −2 −4 2 4 8 10 0 −6 −8 y 12 1· § 4¨ x ¸ 4 2¹ © x-intercepts: 3 r 2 2, 0 −2 2 −2 § 5 r 41 · x-intercepts: ¨¨ , 0 ¸¸ 2 © ¹ 8 Vertex: 3, 8 0 −2 5 x2 6 x 9 9 1 2 −4 10 x 6x 1 x 3 −6 −4 2 10. f x x −8 Axis of symmetry: 5 x 2 15 No real zeros x-intercepts: none y § 5 41 · Vertex: ¨ , ¸ 4¹ © 2 y 12 Axis of symmetry: x 25 25 4 4 4 5· 25 16 § ¨x ¸ 2¹ 4 4 © 4 x 2 x 14 1 13 4 x 12 x2 5x 4 x x 2 4 6 1 2 4 x2 4 x 5 4 r 42 4 4 5 24 4 r 64 8 4 r 8i 8 1 ri 2 The equation has no real zeros. x-intercepts: none INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises ffor Chapter 2 1 2 x 5x 4 3 1§ 2 25 25 · 4¸ ¨ x 5x 3© 4 4 ¹ 13. f x 1 ª§ 5· 41º «¨ x ¸ » 3 «¬© 2¹ 4 »¼ 15. Vertex: 4, 1 Ÿ f x a x 4 Point: 2, 1 Ÿ 1 2 2 2 4a 12 a 2 1 12 x 4 f x a2 4 2 255 1 1 2 1§ 5· 41 ¨x ¸ 3© 2¹ 12 § 5 41 · Vertex: ¨ , ¸ © 2 12 ¹ Axis of symmetry: x 5 r x a x 2 Point: 0, 3 Ÿ 3 a0 2 2 3 4a 2 1 4a 1 4 a 5 2 x2 5x 4 0 16. Vertex: 2, 2 Ÿ f x 21 5 r 41 2 17. Vertex: 1, 4 Ÿ f x § 5 r 41 · , 0 ¸¸ x-intercepts: ¨¨ 2 © ¹ 2 4 2 a x 1 Point: 2, 3 Ÿ 3 a21 1 y 2 1 x 2 2 2 4 f x 52 4 1 4 2 x 1 f x 2 2 4 a 4 4 18. Vertex: 2, 3 Ÿ f x 2 x −8 −6 14. f x −4 −2 Point: 1, 6 Ÿ 6 2 6 9a 3 −6 3 9a 1 3 a 1 2 6 x 24 x 22 2 3 x 2 12 x 11 3x 2 2 3 4 11 3x 2 2 1 Axis of symmetry: x 12 2 r 3 3 x (b) x x y y 2x 2 y y 2 2 A 4 3 11 P 1000 500 x xy x 500 x 500 x x 2 y (c) A 14 500 x x 2 x 2 500 x 62,500 62,500 12 10 8 x 250 6 2 62,500 4 x-intercepts: § 3 · , 0 ¸¸ ¨¨ 2 r 3 © ¹ 3 y 23 12 r 12 6 3 19. (a) 3x 2 12 x 11 12 r 2 2 2 1 x 2 3 3 f x Vertex: 2, 1 x a 1 2 −4 3 x 2 4 x 4 4 11 0 a x 2 2 x –6 –4 –2 4 6 8 10 The maximum area occurs at the vertex when 500 250 250. x 250 and y The dimensions with the maximum area are x 250 meters and y 25 meters. 250 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 256 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Functions 10 p 2 800 p 20. R (a) R 20 $12,000 R 25 $13,750 R 30 $15,000 4 x3 x3 , f x 24. y y 3 2 1 (b) The maximum revenue occurs at the vertex of the parabola. 800 2 10 b 2a R 40 −3 −2 −1 x 2 3 −2 −3 $40 Transformation: Reflection in the x-axis and a vertical stretch $16,000 The revenue is maximum when the price is $40 per unit. 6 x4 x4 , f x 25. y The maximum revenue is $16,000. y Any price greater or less than $40 per unit will not yield as much revenue. 7 5 4 70,000 120 x 0.055 x 2 21. C 1 −1 3 2 The minimum cost occurs at the vertex of the parabola. 120 | 1091 units 2 0.055 b Vertex: 2a About 1091 units should be produced each day to yield a minimum cost. 1 0.107 x 5.68 x 48.5 0 0.107 x 2 5.68 x 74.5 1 2 3 4 Transformation: Reflection in the x-axis and a vertical shift six units upward 2 x 8 x4 , f x 26. y 22. 26 x −4 −3 − 2 4 2 5.68 r x 5.68 2 y 10 4 0.107 74.5 8 6 2 0.107 4 x | 23.7, 29.4 The age of the bride is about 24 years when the age of the groom is 26 years. x −2 −2 2 4 6 8 10 Transformation: Horizontal shift eight units to the right and a vertical stretch y Age of groom 27 26 24 5 y 23 4 22 3 x 2 20 21 22 23 24 25 Age of bride 1 x −2 23. y x 5 x5 , f x 27. y 25 x 2 x3 , f x 1 2 3 5 6 3 −3 −4 y 4 Transformation: Horizontal shift five units to the right 3 2 1 x − 4 −3 − 2 1 2 4 −2 −3 −4 INSTRUCTOR USE ONLY Transformation: Reflection in the x-axis and a horizontal shift two units to the right © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises ffor Chapter 2 1 x5 3 2 x5 , f x 28. y 36. f x x3 8 x 2 0 x3 8 x 2 0 x2 x 8 y 8 257 10 − 10 10 6 −6 −4 − 80 Zeros: x 0 of multiplicity 2 (even multiplicity) 4 x −2 2 4 x 8 of multiplicity 1 (odd multiplicity) 6 Turning points: 2 Transformation: Vertical shrink and a vertical shift three units upward 37. f x 18 x3 12 x 2 0 18 x3 12 x 2 0 6 x 2 3 x 2 2 x 2 5 x 12 29. f x The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. 1 x3 2 x 2 30. f x x The degree is odd and the leading coefficient is positive. The graph falls to the left and rises to the right. 3 4 x 3x 2 2 4 31. g x 2 3 Zeros: x 0 of multiplicity 2 (even multiplicity) Turning points: 2 2 −3 3 The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right. x7 8 x 2 8 x 32. f x −2 38. g x x 4 x3 12 x 2 0 x 4 x3 12 x 2 0 x 2 x 2 x 12 0 x2 x 4 x 3 The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. 33. f x 3x 2 20 x 32 0 3x 2 20 x 32 0 3x 4 x 8 4 3 Zeros: x and x 40 −12 4 8, 0 of multiplicity 2 (even multiplicity) x 4 of multiplicity 1 (odd multiplicity) x 3 of multiplicity 1 (odd multiplicity) Turning points: 3 both of multiplicity 1 (odd multiplicity) 0 Zeros: x − 80 Turning points: 1 34. f x x3 x 2 2 39. f x x x 3 2 x x 3 2 of multiplicity 1 (odd multiplicity) 3 −6 Zeros: x 0 of multiplicity 1 (odd multiplicity) 6 (a) The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. 1 (b) Zero: x −5 (c) x 3 of multiplicity 2 (even multiplicity) x –3 –2 1 0 1 2 f x 34 10 0 –2 –2 –6 Turning points: 2 35. f t t 3 3t 0 t 3 3t 0 (d) 3 y 4 3 −5 tt 3 2 4 2 (− 1, 0) 1 x −4 −3 −2 Zeros: t 0, r 3, all of multiplicity 1 (odd multiplicity) 1 2 3 4 −3 −3 −4 INSTRUCTOR USE ONLY Turning points: 2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 258 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Functions 2 x3 4 x 2 40. g x (c) (a) The degree is odd and the leading coefficient, 2, is positive. The graph falls to the left and rises to the right. (b) g x 2 x3 4 x 2 0 2 x3 4 x 2 0 2×2 x 2 –2 1 0 1 2 h x –4 2 0 2 –4 (d) y 4 2 ( 3, 0) (− 3, 0) −4 −3 x x 2 x −1 −1 1 3 4 −2 −3 2, 0 Zeros: x (0, 0) 3 2 0 (c) x −4 x –3 2 1 0 1 g x –18 0 2 0 6 6x 3 43. 5 x 3 30 x 2 3x 8 30 x 2 18 x y (d) 15 x 8 4 15 x 9 3 2 −4 −3 17 (0, 0) (− 2, 0) x −1 −1 1 2 3 30 x 2 3 x 8 5x 3 4 −2 −3 6x 3 17 5x 3 −4 4 3 44. 3x 2 4 x 7 x x x 5x 3 3 41. f x 2 (a) The degree is even and the leading coefficient is positive. The graph rises to the left and rises to the right. x –4 3 2 –1 0 1 2 3 f x 100 0 –18 –8 0 0 10 72 (d) (− 3, 0) 4x 7 3x 2 4 29 3 3 3x 2 5x 4 45. x 2 5 x 1 5 x3 21x 2 25 x 4 y −4 8 3 29 3 0, 1, 3 (b) Zeros: x (c) 4x 3 (1, 0) 5 x 3 25 x 2 5 x x −2 −1 1 2 3 4 (0, 0) 4 x 2 20 x 4 4 x 2 20 x 4 −15 −18 0 −21 3x x 2 42. h x 5 x3 21x 2 25 x 4 x2 5x 1 4 (a) The degree is even and the leading coefficient, 1, is negative. The graph falls to the left and falls to the right. (b) g x 3x 2 x 4 0 3x 2 x 4 0 x2 3 x2 Zeros: x 0, r 5 x 4, x z 3x 2 5 r 2 29 2 3 46. x 1 3 x 0 x 0 x 2 0 x 0 2 4 3x 4 3 3x 2 3x 2 0 3x 2 3 3 3 4 3x x2 1 3x 2 3 3 x2 1 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises ffor Chapter 2 259 x 2 3x 2 47. x 0 x 2 x 3 x 4 x 2 6 x 3 2 4 3 x 4 0 x3 2 x 2 3×3 2 x 2 6 x 3×3 0 x 2 6 x 2×2 0 x 3 2×2 0 x 4 1 x 3x 4 x 6 x 3 x2 2 4 3 2 x 2 3x 2 1 x2 2 3x 2 5 x 8 48. 2 x 0 x 1 6 x 10 x 13 x 2 5 x 2 2 4 3 6 x 4 0 x3 3x 2 10 x3 16 x 2 5 x 10 x3 0 x 2 5 x 16 x 2 0 x 2 16 x 2 0 x 8 10 6 x 4 10 x3 13 x 2 5 x 2 2 x2 1 49. 2 6 –4 6 – 27 18 12 16 – 22 –8 8 –11 –4 –8 0.1 0.1 6 x3 8 x 2 11x 4 0 0.5 53. f x 0.5 4 20 (a) 0.8 4 19.5 0.1x 2 0.8 x 4 2 – 25 16 66 – 72 48 – 48 2 –9 –6 0 2 x 25 x 66 x 48 x 8 3 52. –4 2 5 5 8 x 2 0.3 0.1×3 0.3 x 2 0.5 x 5 51. 8 10 2 x2 1 0 6 x 4 4 x3 27 x 2 18 x x 2 50. 5 3x 2 5 x 8 –8 –20 – 52 8 13 –2 0 20 9 – 20 14 11 –3 3 0 0 20 –11 –3 0 0 1 is a zero of f. 3 4 Yes, x x z 8 (d) 5 x 2 13 x 2, x z 4 0 9 14 –3 0 15 18 3 0 24 4 0 0 20 20 (c) 50 –1 Yes, x (b) 2 x 2 9 x 6, 33 5 x3 33 x 2 50 x 8 x 4 19.5 x 5 20 x 4 9 x3 14 x 2 3 x 3 4 is a zero of f. 20 9 0 14 0 –3 0 0 0 20 9 –14 –3 0 Yes, x 0 is a zero of f. 1 20 9 20 14 29 –3 15 0 12 20 29 15 12 12 INSTRUCTOR USE ONLY No, x 1 is not a zero off ff. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 260 NOT FOR SALE Chapter 2 3 x3 8 x 2 20 x 16 54. f x (a) Polynomial ynomial and Rational Functions 4 3 –8 – 20 16 12 16 –16 4 –4 0 3 Yes, x (b) –4 (c) 2 3 16 –12 80 – 240 3 – 20 60 – 224 –3 4 is not a zero of f. –8 – 20 16 2 –4 –16 –6 – 24 0 2 3 10 3 3 24 3 2 10 – 24 20 44 1 –3 7 – 21 – 45 135 155 – 465 – 421 So, f 3 20 3 44 421. (b) Remainder Theorem: f 1 1 4 10 1 3 24 1 2 20 1 44 9 is a zero of f. – 20 1 Synthetic Division: 3 –8 –3 11 9 3 –11 –9 25 1 1 is not a zero of f. So, f 1 No, x 4 Synthetic Division: – 20 Yes, x 3 421 –8 3 –1 f 3 3 3 (d) (a) Remainder Theorem: 4 is a zero of f. No, x x 4 10 x3 24 x 2 20 x 44 55. f x 16 –1 1 10 – 24 20 44 –1 –9 33 – 53 9 –33 53 –9 9. 2t 5 5t 4 8t 20 56. g t (a) Remainder Theorem: g 4 2 4 5 4 5 4 8 4 20 3276 Synthetic Division: –4 2 –5 0 0 –8 20 2 –8 –13 52 52 –208 –208 832 824 – 3296 – 3276 So, g 4 3276. (b) Remainder Theorem: g 2 2 2 5 5 2 4 8 2 20 0 Synthetic Division: 2 So, g 2 –5 0 0 –8 20 5 2 4 5 2 4 10 4 2 10 4 2 10 2 8 10 2 –20 2 2 2 5 2 2 2 0. 0 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises ffor Chapter 2 x 3 4 x 2 25 x 28; Factor: x 4 57. f x (a) 4 1 1 (b) x 2 3 x 4 x 1 x 4 The remaining factors are x 1 and x 4 . 4 25 28 4 32 28 (c) f x 8 7 0 (d) Zeros: 2, 1, 3, 4 Yes, x 4 is a factor of f x . (b) x 8 x 7 261 x 1 x 4 x 2 x 3 (e) 40 x 7 x 1 2 The remaining factors are x 7 and x 1 . −3 5 x3 4 x 2 25 x 28 (c) f x −10 x 7 x 1 x 4 (a) (e) x 4 11x 3 41x 2 61x 30 60. f x (d) Zeros: 7, 1, 4 80 −8 2 1 –11 41 – 61 30 2 –18 – 30 1 –9 23 46 –15 1 –9 23 –15 5 –20 15 –4 3 0 0 5 5 −60 1 2 x 3 11x 2 21x 90; Factor: x 6 58. f x (a) –6 11 21 – 90 –12 6 90 –1 –15 0 2 2 Yes, x 2 and x 5 are both factors of f x . (b) x 2 4 x 3 The remaining factors are x 1 and x 3 . Yes, x 6 is a factor of f x . (b) 2 x x 15 2 2x 5 x 3 (d) Zeros: x (e) 2x 5 x 3 x 6 (d) Zeros: x 52 , 3, 6 (e) 50 x 1 x 3 x 2 x 5 (c) f x The remaining factors are 2 x 5 and x 3 . (c) f x x 1 x 3 1, 2, 3, 5 4 −6 12 −8 −7 61. A | 0.0022t 3 0.044t 2 0.17t 2.3 5 62. 8 −100 x 4 4 x 3 7 x 2 22 x 24 59. f x 0 Factors: x 2 , x 3 (a) –2 3 1 1 12 0 1 –4 7 22 24 –2 12 –10 –24 1 –6 5 12 0 –6 5 12 3 –9 –12 –3 –4 0 The model is a good fit to the actual data. INSTRUCTOR USE ONLY Yes, x 2 and x 3 are both factors of f x . © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 262 63. Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Functions t A, actual A, cubic model 0 2.3 2.3 1 2.4 2.5 2 2.9 2.8 3 3.2 3.1 4 3.6 3.5 5 4.0 4.0 6 4.2 7 64. 18 –0.0022 0.044 0.17 2.3 –0.0396 0.0792 4.4856 0.0044 0.2492 6.7856 –0.0022 A 18 | 6.8 million No, the model falls to the right as t increases since the degree is odd and the leading coefficient is negative. 65. 8 100 8 10i 4.4 66. 5 49 5 7i 4.9 4.9 67. i 2 3i 1 3i 8 5.4 5.3 68. 5i i 2 1 5i 9 5.8 5.8 10 6.4 6.2 11 6.5 6.6 12 6.9 6.9 § 2 70. ¨¨ © 2 2 · § 2 i¸ ¨ 2 ¸¹ ¨© 2 71. 7i 11 9i 2 · i¸ 2 ¸¹ 77i 63i 2 72. 1 6i 5 2i 69. 7 5i 4 2i 2 2 i 2 2 2 2 2 i 2 § 2 · 2¨¨ i ¸¸ © 2 ¹ i 20 9i 5 28i 12 20i 9i 2 17 28i 9 20i 20 30i 16i 24i 2 20 46i 24 75. 8 5i 2 4 7i 2 2 64 80i 25i 2 64 80i 25 4 46i 76. 4 7i i 18 12i 3i 2i 2 i 18 9i 2 5 2i 30i 12i 2 73. 10 8i 2 3i 3 7i 2i 74. i 6 i 3 2i 63 77i 7 4 5i 2i 39 80i 16 56i 49i 2 16 56i 49i 2 32 98i 2 66 77. 6i 4 i 6i 4i ˜ 4i 4i 79. 4 2 2 3i 1 i 24 10i i 2 16 1 23 10i 17 23 10 i 17 17 8 5i i 78. ˜ i i 8i 5i 2 i 2 5 8i 1 5 8i 4 2 3i 2 1i ˜ ˜ 2 3i 2 3i 1 i 1 i 8 12i 2 2i 4 9 11 8 12 i 1i 13 13 §8 · § 12 · ¨ 1¸ ¨ i i ¸ © 13 ¹ © 13 ¹ 21 1 i 13 13 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises ffor Chapter 2 80. 1 4i 5 2 i 1 5 2 i 1 4i 18 81i 2i 9i 2 4 81i 2 9 83i 9 83 i 85 85 85 0 5×2 2 x2 x r 2 i 5 x r 10 i 5 2 5 2 x 2 14 x r 12 i Zeros: x 4, 6, 2i, 2i 2 x 3i x 3i 5, 8, 3 r i 4 x 3 8 x 2 3 x 15 Possible rational zeros: r1, r 3, r 5, r15, r 12 , r 32 , r 52 , r 15 , r 14 , r 34 , r 54 , r 15 2 4 3x 4 4 x3 5 x 2 8 Possible rational zeros: r1, r 2, r 4, r 8, r 13 , r 23 , r 43 , r 83 x 3 3 x 2 28 x 60 93. f x Possible rational zeros: r1, r 2, r 3, r 4, r 5, r 6, r10, r12, r15, r 20, r 30, r 60 10 1 2 9 x 1 r 9 x 1 r 3i –2 60 –2 –2 60 1 –30 0 x 2 x 2 x 30 x 2 x 6 x 5 32 4 6 27 2, x The zeros of f x are x 26 639 6, and x 5. 4 x 3 27 x 2 11x 42 94. f x Possible rational zeros: r 14 , r 12 , r 34 , r1, r 32 , r 74 , 12 3 r 3i 71 12 28 3 x3 3 x 2 28 x 60 0 3 r 4x x 3 1 1 b 2 4ac 2a 1 r 4 71 i 4 r 2, r 3, r 72 , r 21 , r 6, r 7, r 21 , r14, r 21, r 42 4 2 –1 4 2 4 –27 11 42 –4 31 – 42 –31 42 0 0, 3 4 x3 27 x 2 11x 42 86. f x 10i 0, r 10i x 8 x 5 91. f x b r Zeros: x 10i x x 4 x 6 x 2i x 2i 92. f x 84. 6 x 2 3x 27 85. f x x x 0 x2 2 x 1 3 r x x 2 10 Zeros: x 0 8x 2, 9 x3 10 x 88. f x 90. f x 83. x 2 2 x 10 x Zeros: x Zeros: x 2 x 1 x 2 x 9 89. f x 81. 5 x 2 2 82. 2 8 x 2 x 2 11x 18 87. f x 2 i 1 4i 1 4i 10 5i 2 8i i 4i 2 2 9i 9 i ˜ 2 9i 2 9i 263 x 4 x 9 9, 4 2 x 1 4 x 2 31x 42 x 1 x 6 4x 7 INSTRUCTOR USE ONLY Zeros: x The zeros of f x are x 1, x 7 , and x 4 6. © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 264 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions x 3 10 x 2 17 x 8 95. f x Possible rational zeros: r1, r 2, r 4, r 8 1 1 1 –10 17 –8 1 –9 8 –9 8 0 x3 10 x 2 17 x 8 x 1 x2 9x 8 The zeros of f x are x 1 and x x 1 x 1 x 8 2 x 8 8. x 3 9 x 2 24 x 20 96. f x x 1 x 4 x3 11x 2 x 12 97. f x Possible rational zeros: r1, r 2, r 4, r 5, r10, r 20 Possible rational zeros: r1, r 2, r 3, r 4, r 6, r12 –5 3 1 9 1 24 20 –5 –20 –20 4 4 0 x3 9 x 2 24 x 20 1 1 x 5 x2 4 x 4 –4 1 –11 1 –12 3 12 3 12 4 1 4 0 1 4 1 4 –4 0 –4 0 1 0 2 x 5 x 2 . 5 and x The zeros of f x are x 2. 1 x 4 x3 11x 2 x 12 The zeros of f x are x x 3 x 4 x2 1 3 and x 4. 25 x 4 25 x3 154 x 2 4 x 24 98. f x Possible rational zeros: r1, r 2, r 3, r 4, r 6, r 8, r12, r 24, r 15 , r 52 , r 53 , r 54 , r 56 , r 85 , r 12 , 5 1 , r 2 , r 3 , r 4 , r 6 , r 8 , r 12 , r 24 r 24 , r 25 5 25 25 25 25 25 25 25 –3 25 25 2 25 25 25 –154 –4 24 –75 150 12 –24 – 50 –4 8 0 – 50 –4 8 50 0 –8 0 –4 0 25 x 4 25 x3 154 x 2 4 x 24 x 3 x 2 25 x 2 4 3, x The zeros of f x are x 2, and x x3 4 x 2 x 4, Zero: i 99. f x Because i is a zero, so is i. i i –4 1 –4 i 1 4i 4 1 4 i 4i 0 4 i 4i i 4i –4 0 1 f x 2 r . 5 100. h x x3 2 x 2 16 x 32 Because 4i is a zero, so is 4i. 1 1 x 3 x 2 5x 2 5x 2 . x i x i x 4 –1 2 16 32 4i 16 8i –32 –1 2 4i 8i 0 –1 2 4i 8i 4i 8i 2 0 4i 4i –1 h x x 4i x 4i x 2 INSTRUCTOR USE ONLY Zeros: Zeros x ri, 4 Zeros: x r4i, 2 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE 2 x 4 3 x3 13 x 2 37 x 15, Zero: 2 i 101. g x 105. g x Because 2 i is a zero, so is 2 i. 2i 2 2 –4 3 13 37 –15 4 2i 5i 15 1 2i 13 5i 31 3i 6 3i 2 1 2i 13 5i 6 3i 4 2i 10 5i 3 6 3i 2 5 2 r i, 12 , 3 12 –208 52 0 1 0 3 52 –4 16 –52 –4 13 0 x 4 2 2 r 3i. 2 x 4 ª¬ x 2 3i ºª ¼¬ x 2 3i º¼ 1i 11 14 –6 11 3i 4 7 4i 3 3i 6 0 4 7 4i 3 3i 4 4i 3 3i 0 3 4 1 1 8 8 –72 –153 3 33 123 153 11 41 51 0 1 1 x x 1 i x 1 i 4x 3 Zeros: 0, 34 , 1 i, 1 i 103. f x 3 x –2 1 1 41 51 –3 –24 –51 8 17 8 r 8 2 4 1 17 8 r 21 4 2 x3 4 x 2 5 x The zeros of f x are 3, 3, 4 i, 4 i. x x2 4x 5 f x 4 r i. x 3 x 3 x 4i x 4 i 3i is a zero, so is 3i. 107. Because 0, 5, 1 Multiply by 3 to clear the fraction. x3 7 x 2 36 104. g x 11 By the Quadratic Formula, the zeros of x 2 8 x 17 are x x 5 x 1 Zeros: x x 2 3i x 2 3i x 4 8 x3 8 x 2 72 x 153 106. f x –3 x ª¬ x 1 i ºª ¼¬ x 1 i º¼ 4 x 3 f x 2 x 4 2 4 4i 4 x 2 4 x 13 By the Quadratic Formula the zeros of x 2 4 x 13 4 and are x 2 r 3i. The zeros of g x are x x 4 x 11x 14 x 6 1i 208 0 1 x 0. Because 1 i is a zero, so is 1 i. 40 –3 4 x 4 11×3 14 x 2 6 x One zero is x 3 0 g x 3 4 –4 0 x 2 i x 2 i 2x 1 x 3 102. f x 4 0 2 ª¬ x 2 i ºª ¼¬ x 2 i º¼ 2 x 5 x 3 Zeros: x x 4 4 x3 3 x 2 40 x 208, Zero: x 1 g x g x 265 1 –4 2i Review Exercises ffor Chapter 2 f x 3 x 23 x 4 x 3i x –7 0 36 3x 2 x 4 x 2 3 –2 18 –36 3 x 2 14 x 8 x 2 3 –9 18 0 The zeros of x 2 9 x 18 3i 3x 4 14 x3 17 x 2 42 x 24 x 3 x 6 are x 3, 6. The zeros of g x are x g x x 2 x 3 x 6 2, 3, 6. Note: f x a 3 x 4 14 x3 17 x 2 42 x 24 , where a is any real nonzero number, has zeros 23 , 4, and r 3i. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 266 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions 108. Because 1 2i is a zero, so is 1 2i. f x x 2 x 3 x 1 2i x 1 2i x2 x 6 ª x 1 ¬ 2 4º ¼ 1 (a) Domain: all real numbers x except x 0 (b) No intercepts 3x x 10 (c) Vertical asymptote: x Domain: all real numbers x except x 4 x3 2 5x 2, x 3 2×2 117. f x x 4 x3 3 x 2 17 x 30 110. f x x 2 x 1 2 Vertical asymptotes: x x2 x 6 x2 2x 5 109. f x x2 x 4 x3 4 x 2 x 3x 2 116. h x 10 Horizontal asymptote: y (d) 4 x3 5x 2 Domain: all real numbers x except x 0 2 5 0 x –1 1 2 1 2 1 y 3 2 –6 6 3 2 y 111. f x 8 x 2 10 x 24 8 x 4 x 6 1 Domain: all real numbers x except x 112. f x x −4 −3 x2 x 2 x2 4 4 and x 1 −1 3 4 6 4 x 118. f x Domain: all real numbers x (a) Domain: all real numbers x except x 113. f x 4 x 3 Vertical asymptote: x (b) No intercepts (c) Vertical asymptote: x 3 Horizontal asymptote: y (d) 2 x 5x 3 x2 2 x y Horizontal asymptote: y 0 Horizontal asymptote: y 0 2 114. f x 0 0 –3 –2 –1 1 2 3 4 3 –2 –4 4 2 4 3 2 y 115. f x 5 x 20 x 2 2 x 24 5x 4 4 3 2 1 x x 6 x 4 –3 –2 –1 1 2 3 4 –2 5 ; x z 4 x 6 Vertical asymptote: x Horizontal asymptote: y –3 6 0 INSTRUCTOR USE ONLY © 2012 Cengage Learning. 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NOT FOR SALE Review Exercises ffor Chapter 2 2 x 1 x 119. g x x 2 x 1 (d) (a) Domain: all real numbers x except x 1 x –2 –1 0 1 2 y 20 17 1 0 1 20 17 (b) x-intercept: 2, 0 y y-intercept: 0, 2 2 (c) Vertical asymptote: x 1 x –1 y 1 2 0 1 1 Horizontal asymptote: y (d) 267 2 3 –4 5 2 −2 x (0, 0)1 −1 2 −1 −2 2 2x x2 4 122. f x y 6 (a) Domain: all real numbers x 4 (0, 2) (−2, 0) 2 (b) Intercept: 0, 0 x (c) Horizontal asymptote: y 0 –4 (d) –6 x –2 –1 0 1 2 2 5 0 2 5 1 2 –8 y x 4 x 7 120. f x 1 2 y (a) Domain: all real numbers x except x 7 2 (b) x-intercept: 4, 0 1 (0, 0) § 4· y-intercept: ¨ 0, ¸ © 7¹ x 1 2 –1 (c) Vertical asymptote: x –2 7 Horizontal asymptote: y (d) 1 x –2 –1 0 1 2 y 2 3 5 8 4 7 1 2 2 5 y x x2 1 123. f x (a) Domain: all real numbers x (b) Intercept: 0, 0 (c) Horizontal asymptote: y 0 8 (d) 6 (0, 47) 4 x −4 − 2 x –2 –1 0 1 2 y 2 5 1 2 0 1 2 2 5 (4, 0) 2 4 10 12 −4 −6 y −8 2 2 121. f x 5x 4×2 1 1 (0, 0) x 1 (a) Domain: all real numbers x –1 (b) Intercept: 0, 0 –2 (c) Horizontal asymptote: y 2 5 4 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 268 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 9 124. h x x 3 (a) Domain: all real numbers x except x 3 2 (a) Domain: all real numbers x except x r2 (b) Intercept: 0, 0 (b) y-intercept: 0, 1 (c) Vertical asymptote: x (c) Vertical asymptotes: x 3 Horizontal asymptote: y (d) 2×2 x 4 126. y 2 2, x Horizontal asymptote: y 0 x –3 –2 –1 0 1 2 y 1 4 9 25 9 16 1 9 4 9 (d) 2 2 x r5 r4 r3 r1 0 y 50 21 8 3 18 5 2 3 0 y y 8 6 6 4 4 (0, 0) x 2 –6 –4 (0, 1) −2 4 6 x 2 4 6 8 −2 6 x 2 x2 1 125. f x 6 x 2 11x 3 3x 2 x 3x 1 2 x 3 127. f x (a) Domain: all real numbers x (c) Horizontal asymptote: y (d) x r3 y r2 27 5 2x 3 1 ,x z 3 x x 3x 1 (b) Intercept: 0, 0 24 5 6 r1 0 –3 0 (a) Domain: all real numbers x except x 1 x 3 0 and §3 · (b) x-intercept: ¨ , 0 ¸ ©2 ¹ y (c) Vertical asymptote: x 4 2 0 Horizontal asymptote: y 2 x –2 –1 1 2 3 4 y 7 2 5 –1 1 2 1 5 4 (0, 0) −6 −4 x −2 2 −8 4 6 (d) y 2 −8 −6 −4 x −2 −4 4 6 3 ,0 2 ( ( 8 −6 −8 INSTRUCTOR USE ONLY © 2012 Cengage Learning. 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NOT FOR SALE Review Exercises ffor Chapter 2 6 x2 7 x 2 4×2 1 2 x 1 3x 2 128. f x 3x 2 1 ,x z 2x 1 2 2x 1 2x 1 (a) Domain: all real numbers x except x (c) Vertical asymptote: x 1 Slant asymptote: y x 1 (d) r 1 2 x –6 y 3 2 13 2 –2 37 5 §2 · (b) x-intercept: ¨ , 0 ¸ ©3 ¹ –5 1 2 5 2 0 4 1 17 5 269 y 4 (0, 1) y-intercept: 0, 2 x −6 −4 1 2 (c) Vertical asymptote: x (d) x –3 –2 –1 0 2 3 1 2 4 6 y 11 5 8 3 5 –2 0 1 3 4 5 3×3 2 x 2 3x 2 3x 2 x 4 3x 2 x 1 x 1 131. f x 3x 4 x 1 y 3x 2 x 1 3x 4 1 23 x , x z 1 3 3x 4 2 x −3 −2 −1 2 (23, 0) (a) Domain: all real numbers x except x 4 x 3 3 (0, −2) 2 x3 x2 1 1· § y-intercept: ¨ 0, ¸ 2¹ © (b) Intercept: 0, 0 (c) Slant asymptote: y 1 and §2 · (b) x-intercepts: 1, 0 , ¨ , 0 ¸ ©3 ¹ 2x 2x 2 x 1 (a) Domain: all real numbers x (d) 2 3 2 Horizontal asymptote: y 129. f x −2 2x x –2 –1 0 1 2 y 16 5 –1 0 1 16 5 (c) Vertical asymptote: x 4 3 Slant asymptote: y x (d) 1 3 x –3 –2 0 1 2 3 y 44 13 12 5 1 2 0 2 14 5 y 3 2 y 1 (0, 0) −3 130. f x −2 −1 4 x 1 2 3 3 −2 2 −3 (0, − 12 ( 1 (1, 0) −2 −1 2 x2 1 x 1 x 1 (23 , 0( 3 x 4 −2 2 x 1 (a) Domain: all real numbers x except x 1 (b) y-intercept: 0, 1 INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 270 NOT FOR SALE Chapter 2 Polynomial ynomial and Rational Function Functions 3x 3 4 x 2 12 x 16 3x 2 5 x 2 x 2 x 2 3x 4 132. f x 0.5 x 500 ,0 x x 0.5 1 Horizontal asymptote: C x 2 3x 1 x 2 3x 4 5 , x z 2 3x 1 528 p , 0 d p 100 100 p 134. C (a) Domain: all real numbers x except x 1 x 3 2 and (a) §4 · (b) x-intercepts: ¨ , 0 ¸, 2, 0 ©3 ¹ 4000 0 100 0 y-intercept: 0, 8 528 25 (c) Vertical asymptote: x 1 3 (b) When p 25, C 100 25 Slant asymptote: y x 3 When p 50, C 100 50 x 0 When p 75, C 100 75 (d) y –4 –1 96 13 0.5 As x increases, the average cost per unit approaches the 0.5 $0.50. horizontal asymptote, C 3x 1 x 3 C x 133. C 21 4 –8 1 1 2 2 0 4 16 11 528 50 528 75 $176 million. $528 million. $1584 million. (c) As p o 100, C o f. No, it is not possible. y 4 (43, 0) 2 x − 6 − 4 −2 −2 4 6 (2, 0) −6 (0, − 8) −8 Chapter Test for Chapter 2 x2 1. f x (a) g x 3. (a) y 1 x 2 60 x 900 900 5 20 2 x2 Reflection in the x-axis followed by a vertical shift two units upward x 32 (b) g x 2 2. Vertex: 3, 6 a x 3 2 6 1 ª x 30 20 ¬ 1 x 30 20 2 2 900º 5 ¼ 50 Vertex: 30, 50 Horizontal shift 32 units to the right y 1 x 2 3x 5 20 The maximum height is 50 feet. (b) The constant term, c 5, determines the height at which the ball was thrown. Changing this constant results in a vertical translation of the graph, and, therefore, changes the maximum height. Point on the graph: 0, 3 3 a03 9 9a Ÿ a 2 6 x 3 1 2 6. INSTRUCTOR USE ONLY So, y © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Chapter Test ffor Chapter 2 34 t 5 2t 2 4. h t 10. Because 2 i is a zero, so is 2 i. y 5 4 3 The degree is odd and the leading coefficient is negative. The graph rises to the left and falls to the right. −4 − 3 − 2 −1 x x 3 t x 4 7 x3 17 x 2 15 x 6. 2 2 2 So, any non-zero real number, has the zeros 0, 3, and 2 i. 11. Because 1 5 2 f x x 1 . x2 1 2 x 1 3i x 1 3i x 2 x 2 8 6 12 x 4 6 x3 16 x 2 24 x 16 4 3 6 9 9 . x 2 –5 –6 15 5 0 –15 0 –6 0 8. (a) 10i 3 25 a is any non-zero real number, has the zeros 1 3×3 14 x 2 7 x 10 Possible rational zeros: r 13 , r 23 , r1, r 53 , r 2, r 10 , r 5, r10 3 1 3 x 52 2 x 2 6 3 3 3 x 3 –7 10 3 17 10 17 10 0 x 1 3 x 2 17 x 10 x 1 3x 2 x 5 1, 23 , 5 Zeros: x x 4 9 x 2 22 x 24 13. f x 4 3i 2 Possible rational zeros: r1, r 2, r 3, r 4, r 6, r 8, r12, r 24 43 –2 0 9 22 –24 –2 4 10 1 –2 –5 –12 24 0 1 –2 –5 12 4 8 12 2 3 0 1 7 5 2 i ˜ 2i 2i 52 i 14 3×3 14 x 2 7 x 10 10i 3 5i 3i 3i, 2, and 2. 12. f x 3 5i 3i 2 x2 x2 a x 4 6 x3 16 x 2 24 x 16 , where Note: f x 2 x3 4 x 2 3 x 6 r 5 2 i 3i 4 5 ,x 2 9. x 1 x2 2x 4 x2 4x 4 2 x x 2 3i –3 2 x 52 x 2 3 (b) x 1 0 2 x 3 5 x 2 6 x 15 Zeros: x 3i. –5 2 x3 5 x 2 6 x 15 2 3i is a zero, so is 1 0 2 x4 5×2 3 x 2 7. f x a x 4 7 x3 17 x 2 15 x , where a is Note: f x x 1 3x x 2 i x x3 7 x 2 17 x 15 2 3×3 4 x 1 x2 1 x 2 i x 2i x x 3 x2 4x 5 2 3 4 5 −2 −3 −4 −5 3x 0 x 3x So, x 0 x 3 x 2i f x x 1 3x 2 x 1 5. x 2 0 x 1 3 x3 0 x 2 4 x 1 3 271 4 41 2i 1 f x x 2 x 4 x2 2 x 3 By the Quadratic Formula, the zeros of x 2 2 x 3 are x 1 r 1 r 2i. 2i. The zeros of f are: x 2, 4, INSTRUCTOR U USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 272 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions 4 1 x2 14. h x x2 2 x 1 y 16. g x 4 x2 x2 2 x 2 x x 1 3 x 1 y-intercept: 0, 2 4 3 2 1 (− 2, 0) x2 (2, 0) −1 1 x 2 Vertical asymptote: x 1 Slant asymptote: y x 1 −2 x-intercepts: r 2, 0 y 10 Vertical asymptote: x 0 8 6 1 Horizontal asymptote: y 4 2 2 x 5 x 12 x 2 16 2x 3 x 4 2 15. f x x −8 −6 −4 y 2 −4 8 4 6 8 (0, −2) −6 6 (− 32, 0( (0, 34 ( x 4 x 4 2x 3 ,x z 4 x 4 −8 −6 − 4 x −2 2 −4 § 3 · x-intercept: ¨ , 0 ¸ © 2 ¹ 18.47 x 2.96 ,0 x 0.23 x 1 17. y 2 4 The limiting amount of CO2 uptake is determined by the horizontal asymptote. 18.47 | 80.3 mg dm 2 hr. 0.23 y § 3· y-intercept: ¨ 0, ¸ © 4¹ 90 4 Vertical asymptote: x Horizontal asymptote: y 2 0 100 0 Problem Solving for Chapter 2 1. V l ˜w˜h x2 x 3 x2 x 3 20 x 3 x 20 3 2 f x d x 0 Possible rational zeros: r1, r 2, r 4, r 5, r10, r 20 2 1 1 3 0 20 2 10 20 5 10 0 x 2 or x x+3 0 x 5 r q x r x d x . The statement should be corrected to read f 1 because x x 2 x 2 5 x 10 d x q x r x , you have 2. False. Because f x f x x 1 q x f 1 x 1 2 . 3. If h 0 and k 0, then a 1 produces a stretch that is reflected in the x-axis, and 1 a 0 produces a shrink that is reflected in the x-axis. 15i 2 Choosing the real positive value for x we have: x 2 and x 3 5. The dimensions of the mold are 2 inches u 2 inches u 5 inches. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Problem Solving ffor Chapter 2 4. (a) y1 y2 13 x 2 3 5 x 2 5 5 273 1 is decreasing. 3 is increasing. 8 −12 12 y1 y2 −8 (b) The graph is either always increasing or always decreasing. The behavior is determined by a. If a ! 0, g x will always be increasing. If a 0, g x will always be decreasing. x5 3×3 2 x 1 (c) H x Since H x is not always increasing or always decreasing, H x z a x h 5 k. 6 −9 9 −6 ax 5. f x x b 2 (a) b z 0 Ÿ x b is a vertical asymptote. a causes a vertical stretch if a ! 1 and a vertical shrink if 0 a 1. For a ! 1, the graph becomes wider as a increases. When a is negative, the graph is reflected about the x-axis. (b) a z 0. Varying the value of b varies the vertical asymptote of the graph of f. For b ! 0, the graph is translated to the right. For b 0, the graph is reflected in the x-axis and is translated to the left 6. G 0.003t 3 0.137t 2 0.458t 0.839, 2 d t d 34 7. f x 2 x2 x 1 x 1 (a) 6 60 (a) −9 −10 9 45 −5 (b) The tree is growing most rapidly when it is approximately 15.2 years old. The graph has a “hole” when x vertical asymptotes. 0.009t 2 0.274t 0.458 (c) y −6 b 2a 0.274 | 15.2222 2 0.009 y 15.2222 | 2.5434 (b) 2 x2 x 1 x 1 (c) As x o 1, Vertex: 15.2222, 2.5434 2x 1 x 1 x 1 1. There are no 2 x 1, x z 1 2×2 x 1 o 3 x 1 (d) In both (b) and (c) the point of diminishing returns occured when t | 15.2. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 274 NOT FOR SALE Chapter 2 8. Let x Polynomial ynomial and Rational Function Functions length of the wire used to form the square. Then 100 x (a) let s length of wire used to form the circle. the side of the square. Then 4 s Let r the radius of the circle. Then 2S r is S r 2 S¨ x Ÿ s x and the area of the square is s 2 4 100 x Ÿ r 2 § x· ¨ ¸ . ©4¹ 100 x and the area of the circle 2S 2 § 100 x · ¸ . © 2S ¹ The combined area is: 2 Ax § x· § 100 x · ¨ ¸ S¨ ¸ © 4¹ © 2S ¹ 2 § 10,000 200 x x 2 · x2 S¨ ¸ 16 4S 2 © ¹ 2500 50 x x2 x2 S S 16 4S 1 · 2 50 x 2500 §1 ¨ ¸x S S 4S ¹ © 16 50 2500 §S 4· 2 x ¨ ¸x S S © 16S ¹ (b) Domain: Since the wire is 100 cm, 0 d x d 100. (c) A x 50 2500 §S 4· 2 x ¨ ¸x 16 S S S © ¹ 800 · 2500 § S 4 ·§ 2 x ¨ ¸¨ x S 4 ¹¸ S © 16S ¹© 2 2 800 §S 4·ª 2 § 400 · § 400 · º 2500 x x « ¨ ¸ ¨ ¸ ¨ ¸ » S 4 S © 16S ¹ «¬ ©S 4¹ © S 4 ¹ »¼ 2 2500 §S 4·ª § 400 ·º § S 4 ·§ 400 · ¨ ¸ «x ¨ ¸» ¨ ¸¨ ¸ S © 16S ¹ ¬ © S 4 ¹¼ © 16S ¹© S 4 ¹ 2 2 10,000 2500 §S 4·ª § 400 ·º ¨ ¸ «x ¨ ¸» S S 4 S © 16S ¹ ¬ © S 4 ¹¼ 2 2500 §S 4·ª § 400 ·º ¨ ¸ «x ¨ ¸» S 4 © 16S ¹ ¬ © S 4 ¹¼ The minimum occurs at the vertex when x 400 S 4 | 56 cm and A x | 350 cm 2 . The maximum occurs at one of the endpoints of the domain. When x 0, A x | 796 cm 2 . When x 100, A x 625 cm 2 . Thus, the area is maximum when x 0 cm. (d) Answers will vary. Graph A x to see where the minimum and maximum values occur. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Problem Solving ffor Chapter 2 1 z 9. (a) zm 1 1i ˜ 1i 1i 1 1 i 2 2 1 1i 1i 2 10. y (a) 1 z (b) zm 1 3i 3i 10 275 1 z 1 2 8i 1 2 8i ˜ 2 8i 2 8i 1 2 2 8i i 68 34 17 (c) zm 1 3i ˜ 3i 3i 3 1 i 10 10 ax 2 bx c 0, 4 : 4 (b) c 1, 0 : 0 a b c Ÿ a b 2, 2 : 2 4a 2b c Ÿ 4a 2b L1 4 6 Solve the system of equations: 0 –4 1 0 4a 2b 6 Ÿ 2a b 3 2 2 a 4 Ÿ a b 4 4 0 a 1 1 b 4 6 – 10 b 5 b Thus, y x 2 5 x 4. Check: 4, 0 : 0 6, 10 : 10 11. (a) Slope 4 2 54 4 6 2 56 4 9 4 3 2 5. Slope of tangent line is less 4 1 2 1 3. Slope of tangent line is greater 12. (a) x 2 y than 5. (b) Slope Ax 100 x 2 x2 § 100 x · 50 x x¨ ¸ 2 2 © ¹ 100 Ÿ y xy Domain: 0 x 100 than 3. 4.41 4 (c) Slope 2.1 2 less than 4.1. (d) Slope Use the “Quad Reg” feature of your graphing utility to obtain y x2 5x 4 L2 (b) 4.1. Slope of tangent line is A 1400 1200 1000 f 2 h f 2 800 2 h 2 400 600 200 2 h 2 4 x 20 h Slope Ax (c) 4 h, h z 0 4 1 3 41 5 4 0.1 4.1 60 80 100 Maximum of 1250 m 2 at x 4h h 2 h 4 h, h z 0 (e) 40 A 50 The results are the same as in (a)–(c). x 50 m, y 25 m 1 2 x 100 x 2 1 x 2 100 x 2500 1250 2 1 2 x 50 1250 2 1250 m 2 is the maximum. 50 m, y 25 m (f ) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at 2, 4 is 4. INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 276 Chapter 2 NOT FOR SALE Polynomial ynomial and Rational Function Functions ax b cx d 13. f x f x has a vertical asymptote at x d and a horizontal asymptote at y c a . c (i) a ! 0 (ii) a ! 0 b 0 b ! 0 c ! 0 c 0 d 0 d 0 x d is positive. c x y a is positive. c y a is negative. c Both asymptotes are positive on graph (d). d is negative. c Both asymptotes are negative on graph (b). (iii) a 0 (iv) a ! 0 b ! 0 b 0 c ! 0 c ! 0 d 0 d ! 0 x d is positive. c x y a is negative. c y a is positive. c The vertical asymptote is positive and the horizontal asymptote is negative on graph (a). d is negative. c The vertical asymptote is negative and the horizontal asymptote is positive on graph (c). INSTRUCTOR USE ONLY © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved.