C H a P T E R 2 Polynomial and Rational Functions

Home , Exercise (mathematics)

CHAPTER 2 Polynomial and Rational Functions

Section 2.1 Quadratic Functions and Models ...... 136

Section 2.2 Polynomial Functions of Higher Degree ...... 151

Section 2.3 Polynomial and Synthetic Division ...... 168

Section 2.4 Complex Numbers ...... 180

Section 2.5 Zeros of Polynomial Functions ...... 187

Section 2.6 Rational Functions ...... 205

Section 2.7 Nonlinear Inequalities ...... 222

Review Exercises ...... 237

Problem Solving ...... 258

Practice Test ...... 262 CHAPTER 2 Polynomial and Rational Functions

Section 2.1 Quadratic Functions and Models

You should know the following facts about parabolas. ■ f x ax2 bx c, a 0, is a quadratic function, and its graph is a parabola. ■ If a > 0, the parabola opens upward and the vertex is the point with the minimum y-value. If a < 0, the parabola opens downward and the vertex is the point with the maximum y-value. ■ The vertex is b2a, f b2a. ■ To find the x-intercepts (if any), solve ax2 bx c 0.

■ The standard form of the equation of a parabola is f x ax h2 k where a 0. (a) The vertex is h, k. (b) The axis is the vertical line x h.

Vocabulary Check 1. nonnegative integer; real 2. quadratic; parabola 3. axis or axis of symmetry 4. positive; minimum 5. negative; maximum

1. f x x 22 opens upward and has vertex 2, 0. 2. f x x 42 opens upward and has vertex 4, 0. Matches graph (g). Matches graph (c).

3. f x x2 2 opens upward and has vertex 0, 2. 4. f x 3 x2 opens downward and has vertex 0, 3. Matches graph (b). Matches graph (h).

5. f x 4 x 22 x 22 4 opens downward 6. f x x 12 2 opens upward and has vertex and has vertex 2, 4. Matches graph (f). 1, 2. Matches graph (a).

7. f x x 32 2 opens downward and has 8. f x x 42 opens downward and has vertex 4, 0. vertex 3, 2. Matches graph (e). Matches graph (d).

136 Section 2.1 Quadratic Functions and Models 137

1 2 1 2 9. (a) y 2x (b) y 8x

y y

5 6

4 4

3 2

2 x −6 −4 46 1 −2

x −4 −3 −2 −1 123 −1 −6

Vertical shrink Vertical shrink and reflection in the x-axis

3 2 2 (c) y 2x (d) y 3x

y y

5 6

4 4

3 2

2 x −6 −4 −2 246 1

x −3 −2 −1 123 −1

Vertical stretch Vertical stretch and reflection in the x-axis

10. (a) y x2 1 (b) y x2 1

y y

5 4

4 3

3 2

2 1

x −3 −2 23 x −3 −2 −1 123 −1 −2

Vertical translation one unit upward Vertical translation one unit downward (c) y x2 3 (d) y x2 3

y y

10 8

8 6

6 4

x −6–4 46 x −6 −4 −2 246 −2 −4

Vertical translation three units upward Vertical translation three units downward 138 Chapter 2 Polynomial and Rational Functions

11. (a) y x 12 (b) y 3x2 1

y y

5 5

4 4

3 3

x x − − 2 1 1234 −3 −2 −1 1 23 − 1 −1

Horizontal translation one unit to the right Horizontal shrink and a vertical translation one unit upward 1 2 2 (c) y 3x 3 (d) y x 3

y y

8 10

6 8

x 2 −6 −226 −2 x −8 −6 −4 −224 −4 −2

Horizontal stretch and a vertical translation three units Horizontal translation three units to the left downward

1 2 1 2 12. (a) y 2 x 2 1 (b) y 2 x 1 3

y y

8 10 6 8 4 6 4 x −6 −4 −22 6810 x −8268−6 −4

−4 −6

Horizontal translation two units to the right, vertical Horizontal translation one unit to the right, horizontal 1 shrink each y-value is multiplied by 2 , reflection in stretch (each x-value is multiplied by 2), and vertical the x-axis, and vertical translation one unit upward translation three units downward

1 2 2 (c) y 2 x 2 1 (d) y 2x 1 4

y y

6 7 4 2 4 x −8 −6 −4 264 3 2 −4 1 −6 x −4234−3 −2 −1 1 − −8 1

Horizontal translation two units to the left, vertical Horizontal translation one unit to the left, horizontal 1 1 shrink each y-value is multiplied by 2 , reflection in shrink each x-value is multiplied by 2 , and vertical trans- x-axis, and vertical translation one unit downward lation four units upward Section 2.1 Quadratic Functions and Models 139

13. f x x2 5 14. hx 25 x2 Vertex: 0, 5 Vertex: 0, 25 Axis of symmetry:x 0 or the y -axis Axis of symmetry: x 0 Find x-intercepts: y Find x-intercepts: y 2 x2 5 0 2 30 1 25 x 0 x 2 2 x 5 −4 −3 −11 34 x 25 x ±5 −2 x ±5 −3 x-intercepts: x-intercepts: ±5, 0 x −20−10 10 20 5, 0, 5, 0 −6

1 2 1 2 1 2 1 2 15. f x 2x 4 2 x 0 4 16. f x 16 4x 4x 16 Vertex: 0, 4 Vertex: 0, 16 Axis of symmetry:x 0 or the y -axis Axis of symmetry: x 0 Find x-intercepts: y Find x-intercepts: y 3 1 x2 4 0 1 2 18 2 2 16 4x 0 2 1 x 8 2 12 x x 64 −4 −3 −1 1234 9 x ±8 ±22 x ±8 −2 6 x-intercepts: −3 x-intercepts: ±8, 0 3 x − − − 22, 0, 22, 0 −5 9 6 3 369 −3

17. f x x 52 6 y 18. f x x 62 3 y Vertex: 5, 6 20 Vertex: 6, 3 50 16 40 Axis of symmetry: x 5 12 Axis of symmetry: x 6 30 Find x-intercepts: Find x-intercepts: 20 2 x 2 x 5 6 0 −20 −12 4 8 x 6 3 0 10

2 x 5 6 −8 2 x x 6 3 −20−10 10 20 30 x 5 ±6 Not possible for real x x 5 ± 6 No x-intercepts x-intercepts: 5 6, 0, 5 6, 0

19. hx x2 8x 16 x 42 20. gx x2 2x 1 x 12 Vertex: 4, 0 y Vertex: 1, 0 y Axis of symmetry: x 4 20 Axis of symmetry: x 1 6 5 16 x-intercept: 4, 0 x-intercept: 1, 0 4 12 3 8 2 4 1

x x −4481216 −4 −3 −2 −112 140 Chapter 2 Polynomial and Rational Functions

5 1 21. fx x2 x 22. f x x2 3x 4 4 1 1 5 9 9 1 x2 x x2 3x 4 4 4 4 4 4 1 2 3 2 x 1 x 2 2 2 1 3 Vertex: , 1 Vertex: , 2 y 2 2 4 1 3 3 Axis of symmetry: x y Axis of symmetry: x 2 2 2 5 1 Find x-intercepts: Find x-intercepts: x 4 −5 −4 −3 −2 −112

5 3 1 x2 x 0 x2 3x 0 −2 4 4 −3

1 ± 1 5 1 3 ± 9 1 x x 2 x 2 −2 −1 123 Not a real number 3 ± 2 2 No x-intercepts 3 x-intercepts: ± 2, 0 2

23. f x x2 2x 5 24. f x x2 4x 1 x2 4x 1 x2 2x 1 1 5 x2 4x 4 4 1 x 12 6 x 22 5 Vertex: 1, 6 Vertex: 2, 5 Axis of symmetry: x 1 Axis of symmetry: x 2 Find x-intercepts: Find x-intercepts: x2 4x 1 0 x2 2x 5 0 x2 4x 1 0 x2 2x 5 0 4 ± 16 4 x 2 2 ± 4 20 x 2 2 ± 5 1 ± 6 x-intercepts: 2 ± 5, 0 x-intercepts: 1 6, 0, 1 6, 0 y 5 y 4 6 2 1 x −6 −5 −3 −2 −112 x − −42 6 2 −2 −3

−4 Section 2.1 Quadratic Functions and Models 141

25. hx 4x2 4x 21 26. f x 2x2 x 1 1 1 1 4x2 x 4 21 2x2 x 1 4 4 2 1 2 1 2 1 4x 20 y 2x 2 1 2 4 16 1 1 2 7 Vertex: , 20 2x y 2 4 8 6

1 1 7 5 Axis of symmetry: x 20 Vertex: , 2 4 8 4 10 Find x-intercepts: 1 3 x Axis of symmetry: x − − 4 2 8484 4 x 4x 21 0 1 Find x-intercepts: 4 ± 16 336 x x −3 −2 −1 123 24 2 x2 x 1 0 Not a real number ⇒ No x-intercepts 1 ± 1 8 x 22 Not a real number No x-intercepts

1 2 1 2 27. f x 4x 2x 12 28. f x 3x 3x 6 1 2 1 1 2 4 x 8x 16 4 16 12 3 x 9x 6 1 2 1 2 81 181 4 x 4 16 3 x 9x 4 3 4 6 1 92 3 Vertex: 4, 16 3 x 2 4 9 3 Axis of symmetry: x 4 Vertex: 2, 4

y 9 y Find x-intercepts: Axis of symmetry: x 2 1 2 4 2 4 x 2x 12 0 Find x-intercepts: x x x2 x −8 48 16 1x2 3x 6 0 −2 46810 8 48 0 3 −2 x 4x 12 0 x2 9x 18 0 −4 − 12 −6 x 4 or x 12 x 3x 6 0 −16 x-intercepts: 4, 0, 12, 0 −20 x-intercepts: 3, 0, 6, 0

29. f x x2 2x 3 x 12 4 30. f x x2 x 30

Vertex: 1, 4 5 x2 x 30 2 1 1 Axis of symmetry: x 1 x x 4 4 30 − 87 1 2 121 35 x-intercepts: 3, 0, 1, 0 x 2 4 1 121 − −5 Vertex: 2, 4 10 10 1 Axis of symmetry: x 2 x-intercepts: 6, 0 , 5, 0 −80 142 Chapter 2 Polynomial and Rational Functions

31. gx x2 8x 11 x 42 5 32. f x x2 10x 14

Vertex: 4, 5 14 x2 10x 25 25 14

Axis of symmetry: x 4 x 52 11 5

− x-intercepts: 4 ± 5, 0 −18 12 Vertex: 5, 11 20 10 −6 Axis of symmetry: x 5 ± x-intercepts: 5 11, 0 −15

2 2 33. f x 2x 16x 31 48 34. f x 4x 24x 41 2x 42 1 4x2 6x 41 Vertex: 4, 1 4x2 6x 9 36 41 −612 Axis of symmetry: x 4 4x 32 5 0 −12 0 6 ± 1 x-intercepts: 4 2 2, 0 Vertex: 3, 5 Axis of symmetry: x 3

No x-intercepts −20

1 2 1 2 3 2 35. g x 2 x 4x 2 2 x 2 3 36. f x 5 x 6x 5 3 27 4 2 Vertex: 2, 3 5 x 6x 9 5 3 3 42 2 6 Axis of symmetry: x 2 5 x 3 5 −84 x ± Vertex: 3, 42 -intercepts: 2 6, 0 5 −14 10 −4 Axis of symmetry: x 3 ± x-intercepts: 3 14, 0 −10

37. 1, 0 is the vertex. 38. 0, 1 is the vertex. y ax 12 0 ax 12 f x ax 02 1 ax2 1 Since the graph passes through the point 0, 1, we have: Since the graph passes through 1, 0, 1 a0 12 0 a12 1 1 a 1 a. y 1x 12 x 12 So, y x2 1.

39. 1, 4 is the vertex. 40. 2, 1 is the vertex. y ax 12 4 f x ax 22 1 Since the graph passes through the point 1, 0, we have: Since the graph passes through 0, 3, 0 a1 12 4 3 a0 22 1 4 4a 3 4a 1 1 a 4 4a y 1x 12 4 x 12 4 1 a. So, y x 22 1. Section 2.1 Quadratic Functions and Models 143

41. 2, 2 is the vertex. 42. 2, 0 is the vertex. y ax 22 2 f x ax 22 0 ax 22 Since the graph passes through the point 1, 0, we have: Since the graph passes through 3, 2, 0 a1 22 2 2 a3 22 2 a 2 a. y 2x 22 2 So, y 2x 22.

43. 2, 5 is the vertex. 44. 4, 1 is the vertex. f x ax 22 5 f x ax 42 1 Since the graph passes through the point 0, 9, we have: Since the graph passes through 2, 3, 9 a0 22 5 3 a2 42 1 4 4a 3 4a 1 1 a 4 4a fx 1x 22 5 x 22 5 1 a. So, f x x 42 1.

45. 3, 4 is the vertex. 46. 2, 3 is the vertex. fx ax 32 4 f x ax 22 3 Since the graph passes through the point 1, 2, we have: Since the graph passes through 0, 2, 2 a1 32 4 2 a0 22 3 2 4a 2 4a 3 1 2 a 1 4a 1 2 1 f x 2 x 3 4 4 a. 1 2 So, f x 4 x 2 3.

47. 5, 12 is the vertex. 48. 2, 2 is the vertex. fx ax 52 12 f x ax 22 2 Since the graph passes through the point 7, 15, we have: Since the graph passes through 1, 0, 15 a7 52 12 0 a1 22 2 ⇒ 3 3 4a a 4 0 a 2 3 2 f x 4 x 5 12 2 a. So, f x 2x 22 2.

1 3 5 3 49. 4, 2 is the vertex. 50. 2, 4 is the vertex. 12 3 52 3 f x a x 4 2 f x a x 2 4 Since the graph passes through the point 2, 0, Since the graph passes through 2, 4, we have: 52 3 4 a 2 2 4 12 3 0 a 2 4 2 81 3 4 4 a 4 3 49 ⇒ 24 2 16a a 49 19 81 4 4 a 24 12 3 f x 49 x 4 2 19 81 a. 19 52 3 So,f x 81 x 2 4 . 144 Chapter 2 Polynomial and Rational Functions

5 51. 2, 0 is the vertex. 52. 6, 6 is the vertex. 52 2 f x a x 2 f x a x 6 6 7 16 61 3 Since the graph passes through the point 2, 3 , Since the graph passes through 10, 2 , we have: 3 61 2 2 a 10 6 6 16 7 52 3 a 2 2 3 1 2 100a 6 16 3 a 9 1 2 100a 2 f x 16x 5 3 2 450 a. So,f x 450x 62 6 .

53. y x2 16 54. y x2 6x 9 x-intercepts: ±4, 0 x-intercept: 3, 0 0 x2 16 0 x2 6x 9 x2 16 0 x 32 x ±4 x 3 0 ⇒ x 3

55. y x2 4x 5 56. y 2x2 5x 3 1 x-intercepts: 5, 0 , 1, 0 x-intercepts: 2, 0 , 3, 0 0 x2 4x 5 0 2x2 5x 3 0 x 5x 1 0 2x 1x 3 ⇒ 1 x 5 or x 1 2 x 1 0 x 2 x 3 0 ⇒ x 3

57. fx x2 4x 4 58. f x 2x2 10x 14 x-intercepts: 0, 0, (4,0 x-intercepts: 0, 0, 5, 0 −48 2 2 0 x 4x 0 2x 10x −16 0 x x 4) −4 0 2x x 5 −6 x 0 or x 4 2x 0 ⇒ x 0 The x-intercepts and the solutions of f x 0 are the same. x 5 0 ⇒ x 5 The x-intercepts and the solutions of f x 0 are the same.

59. fx x2 9x 18 12 60. f x x2 8x 20 10 x-intercepts: 3, 0, 6, 0 x-intercepts: 2, 0, 10, 0 −412

2 2 0 x 9x 18 −8 16 0 x 8x 20 0 x 3) x 6 −4 0 x 2 x 10 −40 x 3 or x 6 x 2 0 ⇒ x 2 The x-intercepts and the solutions of f x 0 are the same. x 10 0 ⇒ x 10 The x-intercepts and the solutions of f x 0 are the same. Section 2.1 Quadratic Functions and Models 145

61. fx 2x2 7x 30 10 62. f x 4x2 25x 21 10 −92 5 −5 10 3 x-intercepts: 2, 0 , 6, 0 x-intercepts: 7, 0 , 4, 0 0 2x2 7x 30 0 4x2 25x 21 0 2x 5) x 6 −40 0 x 7 4x 3 −70 5 ⇒ x 2 or x 6 x 7 0 x 7 ⇒ 3 The x-intercepts and the solutions of f x 0 are the same. 4 x 3 0 x 4 The x-intercepts and the solutions of f x 0 are the same.

1 7 2 10 2 10 63. f x 2 x 6x 7 64. f x 10 x 12x 45 −18 4 x-intercepts: 1, 0, 7, 0 x-intercepts: 15, 0, 3, 0

1 2 −10 14 7 2 0 2 x 6x 7 0 10 x 12x 45 2 0 x 6x 7 −6 0 x 15 x 3 −60 0 x 1x 7 x 15 0 ⇒ x 15 x 1 or x 7 x 3 0 ⇒ x 3 The x-intercepts and the solutions of f x 0 are the same. The x-intercepts and the solutions of f x 0 are the same.

65. f x x 1x 3 opens upward 66. f x x 5x 5 x 1x 3 x 5x 5 x2 2x 3 x2 25, opens upward gx x 1x 3 opens downward gx f x, opens downward x 1x 3 gx x2 25 x2 2x 3 Note: f x ax2 25 has x-intercepts 5, 0 and 5, 0 for all real numbers a 0. x2 2x 3 Note: f x ax 1x 3 has x-intercepts 1, 0 and 3, 0 for all real numbers a 0.

67. f x x 0x 10 opens upward 68. f x x 4x 8 x2 10x x2 12x 32, opens upward gx x 0x10 opens downward gx f x, opens downward x2 10x gx x2 12x 32 Note: f x ax 0x 10 axx 10 has Note: f x ax 4x 8 has x-intercepts 4, 0 and x-intercepts 0, 0 and 10, 0 for all real numbers 8, 0 for all real numbers a 0. a 0.

1 5 69. f x x 3 x 2 2 opens upward 70. f x 2 x 2 x 2 1 5 x 3 x 2 2 2 x 2 x 2 2 1 x 3 2x 1 2 x 2x 5 2x2 7x 3 2x2 x 10, opens upward gx 2x2 7x 3 opens downward gx f x, opens downward 2x2 7x 3 gx 2x2 x 10 5 5 Note: f x a x 3 2x 1 has x-intercepts Note: f x a x 2 x 2 has x-intercepts 2, 0 1 3, 0 and 2, 0 for all real numbers a 0. and 2, 0 for all real numbers a 0. 146 Chapter 2 Polynomial and Rational Functions

71. Let x the first number and y the second number. 72. Let x first number and y second number. Then, Then the sum is x y S, y S x. The product is x y 110 ⇒ y 110 x. Px xy xS x. The product is Px xy x110 x 110x x2. Px Sx x2 Px x2 110x x2 Sx x2 110x 3025 3025 S2 S2 x2 Sx 4 4 x 552 3025 S 2 S2 x 552 3025 x 2 4 The maximum value of the product occurs at the vertex of The maximum value of the product occurs at the vertex Px and is 3025. This happens when x y 55. of Px and is S24. This happens when x y S2.

73. Let x the first number and y the second number. 74. Let x the first number and y the second number. Then the sum is 42 x Then the sum is x 3y 42 ⇒ y . 24 x 3 x 2y 24 ⇒ y . 2 42 x The product is Px xy x . 24 x 3 The product is Px xy x . 2 1 2 1 P x x 42x Px x2 24x 3 2 1 2 1 x 42x 441 441 x2 24x 144 144 3 2 1 1 2 2 1 1 x 21 441 x 21 147 x 122 144 x 122 72 3 3 2 2 The maximum value of the product occurs at the vertex The maximum value of the product occurs at the vertex of Px and is 147. This happens when x 21 and of P x and is 72. This happens when x 12 and 42 21 y 7. Thus, the numbers are 21 and 7. y 24 12 2 6. Thus, the numbers are 12 and 6. 3

75. (a) (b) This area is maximum when xA 100 1 x 25 feet and y 3 333 feet. y 5 600

0 60 0

This area is maximum when x 25 feet and 100 1 y 3 333 feet.

—CONTINUED— Section 2.1 Quadratic Functions and Models 147

75. —CONTINUED— 8 (d) A x50 x (e) They are all identical. 3 1 8 x 25 feet and y 333 feet x2 50x 3 8 x2 50x 625 625 3 8 x 252 625 3 8 5000 x 252 3 3 The maximum area occurs at the vertex and is 50003 square feet. This happens when x 25 feet and y 200 4253 1003 feet. The dimensions 1 are 2x 50 feet by 333 feet.

1 76. (a) Radius of semicircular ends of track: r y (c) Area of rectangular region: 2 Distance around two semicircular parts of track: 200 2x A xy x 1 d 2r 2 y y 2 1 2 200x 2x (b) Distance traveled around track in one lap: 2 d y 2x 200 2 x 100x y 200 2x 2 2 x 100x 2500 2500 200 2x y 2 5000 2 x 50

The area is maximum when x 50 and 200 2 50 100 y .

4 24 77. y x2 x 12 9 9 b 249 4 24 The vertex occurs at 3. The maximum height is y3 32 3 12 16 feet. 2a 249 9 9

16 9 78. y x2 x 1.5 2025 5 (a) The ball height when it is punted is the y -intercept. 16 9 y 02 0 1.5 1.5 feet 2025 5 b 95 3645 (b) The vertex occurs at x . 2a 2162025 32 3645 16 3645 2 9 3645 The maximum height is f 1.5 32 2025 32 5 32 6561 6561 6561 13,122 96 6657 1.5 feet 104.02 feet. 64 32 64 64 64 64 —CONTINUED— 148 Chapter 2 Polynomial and Rational Functions

78. —CONTINUED— (c) The length of the punt is the positive x -intercept. 16 9 0 x 2 x 1.5 2025 5 95 ± 952 41.5162025 1.8 ± 1.81312 x 322025 0.01580247 x 0.83031 or x 228.64 The punt is approximately 228.64 ft.

79. C 800 10x 0.25x2 0.25x2 10x 800 80. C 100,000 110x 0.045x2 b 10 110 The vertex occurs at x 20. The vertex occurs at x 1222. 2a 20.25 20.045 The cost is minimum when x 20 fixtures. The cost is minimum when x 1222 units.

81. P 0.0002x2 140x 250,000 82. P 230 20x 0.5x2 b 140 b 20 The vertex occurs at x 350,000. The vertex occurs at x 20. 2a 20.0002 2a 20.5 The profit is maximum when x 350,000 units. Because x is in hundreds of dollars, 20 100 2000 dollars is the amount spent on advertising that gives maximum profit.

83. Rp 25p2 1200p 84. R p 12p2 150p (a) R20 $14,000 thousand (a) R$4 12$42 150$4 $408 R25 $14,375 thousand R$6 12$62 150$6 $468 R30 $13,500 thousand R$8 12$82 150$8 $432 (b) The revenue is a maximum at the vertex. (b) The vertex occurs at b 1200 b 150 24 p $6.25 2a 225 2a 212 R24 14,400 Revenue is maximum when price $6.25 per pet. The unit price that will yield a maximum revenue of The maximum revenue is $14,400 thousand is $24. f$6.25 12$6.252 150$6.25 $468.75.

85. C 4299 1.8t 1.36t2, 0 ≤ t ≤ 43

(a) 5000 (b) Vertex 0, 4299 The vertex occurs when y 4299 which is the maximum average annual consumption. The warnings may not have had an immediate effect, but over time 0 43 0 they and other findings about the health risks and the increased cost of cigarettes have had an effect. (c) C40 2051 209,128,0942051 Annually: 8879 cigarettes 48,308,590 8879 Daily: 24 cigarettes 366 Section 2.1 Quadratic Functions and Models 149

86. (a) and (c) 87. (a) 25

950

0 100

−5 4 12 650 (b) 0.002s2 0.005s 0.029 10 (b) y 4.303x2 49.948x 886.28 2 s2 5s 29 10,000 (d) 1996 2 s2 5s 10,029 0 (e) Vertex occurs at a 2, b 5, c 10,029 b 49.948 x 5.8 5 ± 52 4210,029 2a 24.303 s 22 Minimum occurs at year 1996. 5 ± 80,257 s (f) x 18 4 y 4.303182 49.94818 886.28 1381.388 s 72.1, 69.6 There will be approximately 1,381,000 hairdressers The maximum speed if power is not to exceed and cosmetologists in 2008. 10 horsepower is 69.6 miles per hour.

88. (a) and (c) (b) y 0.0082x2 0.746x 13.47

31 (d) The maximum of the graph is at x 45.5, or about 45.5 mi/h. Algebraically, the maximum occurs at b 0.746 x 45.5 mi/h. 2a 20.0082 10 80 20

2 5 53 89. True. The equation 12x 1 0 has no real solution, 90. True. The vertex of f x is 4, 4 and the vertex of g x 5 71 so the graph has no x-intercepts. is 4, 4 .

91. f x ax2 bx c b ax2 x c a b b2 b2 ax2 x c a 4a2 4a2 b 2 b2 ax c 2a 4a b 2 4ac b2 ax 2a 4a b b2 b f a b c 2a 4a2 2a b2 b2 c 4a 2a b2 2b2 4ac 4ac b2 4a 4a b 4ac b2 b b So, the vertex occurs at , , f . 2a 4a 2a 2a 150 Chapter 2 Polynomial and Rational Functions

92. Conditions (a) and (d) are preferable because profits 93. Yes. A graph of a quadratic equation whose vertex is would be increasing. 0, 0 has only one x-intercept.

94. If f x ax2 bx c has two real zeros, then by the Quadratic Formula they are b ± b2 4ac x . 2a The average of the zeros of f is b b2 4ac b b2 4ac 2b 2a 2a 2a b . 2 2 2a This is the x -coordinate of the vertex of the graph.

7 3 95. 4, 3 and 2, 1 96. , 2, m 2 2 1 3 2 1 m 3 7 2 4 6 3 y 2 x 2 2 1 y 1 x 2 3 21 3 y 2 x 2 4 1 2 y 1 x 3 13 3 3 y x 2 4 1 5 y x 3 3

⇒ 4 4 97. 4x 5y 10 y 5x 2 and m 5 98. y 3x 2 The slope of the perpendicular line through 0, 3 is m 3 m 5 and the y-intercept is b 3. 4 For a parallel line,m 3. So, for 8, 4, the line is y 5x 3 4 y 4 3x 8 y 4 3x 24 y 3x 20.

For Exercises 99–104, let f x 14x 3, and gx 8x2.

99. f g3 f 3 g3 100. g f 2 822 142 3 32 28 3 7 143 3 832 27

4 4 4 f 141.5 3 24 4 101. fg f g 102. 1.5 7 7 7 g 81.52 18 3

4 4 2 14 38 7 7 128 1408 11 49 49

103. f g1 fg1 f 8 148 3 109 104. g f 0 g f 0 g140 3 g3 832 72

105. Answers will vary. Section 2.2 Polynomial Functions of Higher Degree 151

Section 2.2 Polynomial Functions of Higher Degree

You should know the following basic principles about polynomials. ■ n n1 . . . 2 f x anx an1x a2x a1x a0, an 0, is a polynomial function of degree n. ■ If f is of odd degree and

(a)an > 0, then (b)an < 0, then 1.fx → as x → . 1. fx → as x → . 2.fx → as x → . 2. fx → as x → . ■ If f is of even degree and

(a)an > 0, then (b)an < 0, then 1.fx → as x → . 1. fx → as x → . 2.fx → as x → . 2. fx → as x → . ■ The following are equivalent for a polynomial function. (a)x a is a zero of a function. (b)x a is a solution of the polynomial equation fx 0. (c)x a is a factor of the polynomial. (d)a, 0 is an x-intercept of the graph of f. ■ A polynomial of degree n has at most n distinct zeros and at most n 1 turning points. ■ A factor x ak, k > 1, yields a repeated zero of x a of multiplicity k. (a) If k is odd, the graph crosses the x-axis at x a. (b) If k is even, the graph just touches the x-axis at x a. ■ If f is a polynomial function such that a < b and fa fb, then f takes on every value between fa and fb in the interval a, b. ■ If you can find a value where a polynomial is positive and another value where it is negative, then there is at least one real zero between the values.

Vocabulary Check 1. continuous 2. Leading Coefficient Test 3. n; n 1 4. solution; x a; x-intercept 5. touches; crosses 6. standard 7. Intermediate Value

1. fx 2x 3 is a line with y-intercept 0, 3. 2. fx x2 4x is a parabola with intercepts 0, 0 and Matches graph (c). 4, 0 and opens upward. Matches graph (g).

3. fx 2x2 5x is a parabola with x-intercepts 4. fx 2x3 3x 1 has intercepts 0, 1, 1, 0, 5 1 1 1 1 0, 0 and 2, 0 and opens downward. Matches 2 2 3, 0 and 2 2 3, 0 . Matches graph (f). graph (h).

1 4 2 ± 1 3 2 4 4 5. f x 4x 3x has intercepts 0, 0 and 2 3, 0 . 6. f x 3x x 3 has y-intercept 0, 3 . Matches graph (a). Matches graph (e).

4 3 1 5 3 9 7. f x x 2x has intercepts 0, 0 and 2, 0 . 8. f x 5x 2x 5x has intercepts 0, 0 , 1, 0 , Matches graph (d). 1, 0, 3, 0, 3, 0. Matches graph (b). 152 Chapter 2 Polynomial and Rational Functions

9. y x3 (a)fx x 23 (b) fx x3 2

y y

4 3 3 2 2 1 1 x −4 −3 −2234 x −3 −2 2345

−2 −3 −4 −4 −5

Horizontal shift two units to the right Vertical shift two units downward

1 3 3 (c)f x 2x (d) f x x 2 2

y y

4 3 3 2 2 1 1 x −3 −21245 x −4 −3 −2234 −2 −2 −3 −3 −4 −4 −5

Reflection in the x-axis and a vertical shrink Horizontal shift two units to the right and a vertical shift two units downward

10. y x5 (a) fx x 15 (b) fx x5 1

y y

4 4 3 3 2 2 1 x x −4 −3 1234 −4 −3 −2 1234

−3 −3 −4 −4

Horizontal shift one unit to the left Vertical shift one unit upward

1 5 1 5 (c) f x 1 2x (d) f x 2 x 1

y y

4 4 3 3 2 2 1 x x −4 −3 −2 234 −5 −4 −3 −2 123

−3 −3 −4 −4

Reflection in the x-axis, vertical shrink each y-value Reflection in the x-axis, vertical shrink each y-value is 1 1 is multiplied by 2 , and vertical shift one unit upward multiplied by 2 , and horizontal shift one unit to the left Section 2.2 Polynomial Functions of Higher Degree 153

11. y x4 (a)fx x 34 (b) fx x4 3

y y

6 4 5 3 4 2 3 1 2 x −4 −3 −2234 1 x −5 −4 −3 −2 −1123

−2 −4

Horizontal shift three units to the left Vertical shift three units downward

4 1 4 (c)f x 4 x (d) f x 2 x 1

y y

6 5

3 2 1 x x −4 −3 −2 1234 −4 −3 −2 −11234 −1 −2 −2

Reflection in the x-axis and then a vertical Horizontal shift one unit to the right and a vertical shrink 1 shift four units upward each y-value is multiplied by 2 4 1 4 (e)f x 2x 1 (f) f x 2x 2

y y

6 6 5 5 4 3 2 1 x x −4 −3 −2 −1 1 234 −4 −3 −1134 −1 −1 −2

Vertical shift one unit upward and a horizontal shrink Vertical shift two units downward and a horizontal stretch 1 1 each y-value is multiplied by 2 each y-value is multiplied by 2

12. y x6

1 6 6 (a) f x 8 x (b) f x x 2 4

y y

4 3 2 1 x x −4 −3 −2 234 −5 −4 −2 123 −1 −2 −3 −4 −4 1 Vertical shrink each y-value is multiplied by 8 and Horizontal shift two units to the left and vertical shift four reflection in the x-axis units downward

—CONTINUED— 154 Chapter 2 Polynomial and Rational Functions

12. —CONTINUED—

6 1 6 (c) f x x 4 (d) f x 4 x 1

y y

4 4 3 3 2 2 1 x x −4 −3 −2 234 −4 −3 −2 234 −1 −2 −3 −4

Vertical shift four units downward Reflection in the x-axis, vertical shrink each y-value is 1 1 6 multiplied by 4 , and vertical shift one unit upward (e) f x 4 x 2 6 y (f) f x 2x 1

x −868−6 −2 2 x − −4 −3 −2 −11234 4 −1 −2 Horizontal stretch (each x-value is multiplied by 4), 1 and vertical shift two units downward Horizontal shrink each x-value is multiplied by 2 , and vertical shift one unit downward

1 3 2 13. f x 3x 5x 14. f x 2x 3x 1 Degree: 3 Degree: 2 1 Leading coefficient: 3 Leading coefficient: 2 The degree is odd and the leading coefficient is positive. The degree is even and the leading coefficient is positive. The graph falls to the left and rises to the right. The graph rises to the left and rises to the right.

7 2 6 15. g x 5 2x 3x 16. h x 1 x Degree: 2 Degree: 6 Leading coefficient: 3 Leading coefficient: 1 The degree is even and the leading coefficient is negative. The degree is even and the leading coefficient is negative. The graph falls to the left and falls to the right. The graph falls to the left and falls to the right.

17. fx 2.1x5 4x3 2 18. fx 2x5 5x 7.5 Degree: 5 Degree: 5 Leading coefficient: 2.1 Leading coefficient: 2 The degree is odd and the leading coefficient is negative. The degree is odd and the leading coefficient is positive. The graph rises to the left and falls to the right. The graph falls to the left and rises to the right.

3x4 2x 5 19. fx 6 2x 4x2 5x3 20. fx 4 Degree: 3 Degree: 4 Leading coefficient: 5 3 Leading coefficient: 4 The degree is odd and the leading coefficient is negative. The degree is even and the leading coefficient is positive. The graph rises to the left and falls to the right. The graph rises to the left and rises to the right. Section 2.2 Polynomial Functions of Higher Degree 155

2 2 7 3 2 21. h t 3 t 5t 3 22. f s 8 s 5s 7s 1 Degree: 2 Degree: 3 2 7 Leading coefficient: 3 Leading coefficient: 8 The degree is even and the leading coefficient is The degree is odd and the leading coefficient is negative. negative. The graph falls to the left and falls to the right. The graph rises to the left and falls to the right.

3 3 1 3 1 3 23. f x 3x 9x 1; g x 3x 24. f x 3 x 3x 2 , g x 3 x

8 6 g f g f −44 −99

−8 −6

25. fx x4 4x3 16x; gx x4 26. fx 3x 4 6x 2, gx 3x 4

12 5

f − 8 8 g g −66

f −20 −3

27. fx x2 25 28. (a) fx 49 x2 (a) 0 x2 25 x 5x 5 0 7 x7 x Zeros: x ±5 x ±7, both with multiplicity 1 (b) Each zero has a multiplicity of 1 (odd multiplicity). (b) Multiplicity of x 7 is 1. Turning point: 1 (the vertex of the parabola) Multiplicity of x 7 is 1.

− 30 30 (c) 55

−30 −30 30 −5

29. ht t 2 6t 9 30. (a) fx x2 10x 25 (a) 0 t2 6t 9 t 32 0 x 5 2 Zero: t 3 x 5, with multiplicity 2 (b)t 3 has a multiplicity of 2 (even multiplicity). (b) The multiplicity of x 5 is 2. Turning point: 1 (the vertex of the parabola) There is one turning point.

−25 15

−20 −5 156 Chapter 2 Polynomial and Rational Functions

1 2 1 2 31. f x 3x 3x 3 1 1 2 2 4 (a) 0 3 x 3 x 3 (c) 1 2 3 x x 2 −6 6 1 3 x 2 x 1

Zeros: x 2, x 1 −4 (b) Each zero has a multiplicity of 1 (odd multiplicity). Turning point: 1 (the vertex of the parabola)

1 5 3 5 37 32. (a) fx x2 x (b) The multiplicity of is 1. 2 2 2 2 1 5 3 5 37 a , b , c The multiplicity of is 1. 2 2 2 2 5 ± 52 41 3 There is one turning point. x 2 2 2 2 1 (c) 3 5 37 ± −8 4 2 4 5 ± 37 , both with multiplicity 1 2 −5

33. f x 3x3 12x2 3x

(a) 0 3x3 12x2 3x 3xx2 4x 1 (c) 8 Zeros:x 0, x 2 ± 3 (by the Quadratic −6 6 Formula) (b) Each zero has a multiplicity of 1 (odd multiplicity). −24 Turning points: 2

34. (a) gx 5xx 2 2x 1 (b) The multiplicity of x 0 is 1. 0 5xx 2 2x 1 The multiplicity of x 1 2 is 1. 0 xx 2 2x 1 The multiplicity of x 1 2 is 1. For x2 2x 1, a 1, b 2, c 1. There are two turning points.

2 ± 2 2 411 (c) 12 x 21 −1 3 2 ± 8 2 −16 1 ± 2 The zeros are 0, 1 2, and 1 2, all with multiplicity 1.

35. ft t3 4t2 4t

(a) 0 t3 4t2 4t tt2 4t 4 tt 22 (c) 5 Zeros: t 0, t 2 −7 8 (b)t 0 has a multiplicity of 1 (odd multiplicity). t 2 has a multiplicity of 2 (even multiplicity). −5 Turning points: 2 Section 2.2 Polynomial Functions of Higher Degree 157

36. (a) fx x 4 x 3 20x 2 37. gt t5 6t3 9t 0 x 2x 2 x 20 (a) 0 t5 6t3 9t tt4 6t2 9 tt2 32 2 2 0 x 2x 4x 5 tt 3 t 3 x 0, 4, 5 Zeros: t 0, t ±3 0 with multiplicity 2,4 and 5 with multiplicity 1. (b)t 0 has a multiplicity of 1 (odd multiplicity). (b) The multiplicity of x 0 is 2. t ±3 each have a multiplicity of 2 (even multiplicity). The multiplicity of x 5 is 1. Turning points: 4 The multiplicity of x 4 is 1. (c) 6 There are three turning points.

25 (c) −9 9 −6 6

−6

−150

38. (a) f x x 5 x 3 6x 39. fx 5x4 15x2 10 0 xx 4 x 2 6 (a) 0 5x 4 15x2 10 0 xx 2 3x 2 2 5x 4 3x2 2 x 0, ±2, all with multiplicity 1 5x2 1x2 2 (b) The multiplicity of x 0 is 1. No real zeros The multiplicity of x 2 is 1. (b) Turning point: 1

The multiplicity of x 2 is 1. (c) 40 There are two turning points.

−5 −9 9

−6

40. (a) f x 2x4 2x2 40 41. gx x3 3x2 4x 12 0 2x4 2x2 40 (a) 0 x3 3x2 4x 12 x2x 3 4x 3 0 2x2 4x 5x 5 x2 4x 3 x 2x 2x 3 x ±5, both with multiplicity 1 Zeros: x ± 2, x 3 (b) The multiplicity of x 5 is 1. (b) Each zero has a multiplicity of 1 (odd multiplicity). The multiplicity of x 5 is 1. Turning points: 2

There is one turning point. (c) 4

−8 7 (c) 20

−6 6

−16

−60 158 Chapter 2 Polynomial and Rational Functions

42. (a) f x x 3 4x 2 25x 100 43. y 4x3 20x2 25x

0 x 2x 4 25x 4 (a) 12 0 x 2 25x 4 0 x 5 x 5 x 4 −2 6

x ±5, 4, all with multiplicity 1 −4 (b) The multiplicity of x 5 is 1. 5 (b) x-intercepts: 0, 0 , 2, 0 The multiplicity of x 5 is 1. (c) 0 4x3 20x2 25x The multiplicity of x 4 is 1. 0 x2x 52 There are two turning points. 5 x 0 or x 2 (c) 140 (d) The solutions are the same as the x-coordinates of the x-intercepts.

−9 9

−20

44. y 4x 3 4x 2 8x 8 45. y x5 5x3 4x

(a) 2 (a) 4

−3 3

−66

−11 −4 (b) (1, 0, 1.414214, 0, 1.414214, 0 (b) x-intercepts: 0, 0, ±1, 0, ±2, 0 (c) 0 4x3 4x2 8x 8 (c) 0 x5 5x3 4x 0 4x2x 1 8x 1 0 xx2 1x2 4 0 4x2 8x 1 0 xx 1x 1x 2x 2 0 4x2 2x 1 x 0, ±1, ±2 x ±2, 1 (d) The solutions are the same as the x-coordinates of the x-intercepts. (d) The intercepts match part (b).

1 3 2 46. y 4 x x 9 1 12 3 2 (a) (c) 0 4 x x 9 x 0, ±3 −18 18 x-intercepts: 0, 0, ±3, 0

−12 (d) The intercepts match part (b). (b) 0, 0, 3, 0, 3, 0

47. fx x 0x 10 48. f x x 0x 3 fx x2 10x xx 3 Note: fx ax 0x 10 axx 10 x 2 3x has zeros 0 and 10 for all real numbers a 0. Note: f x axx 3 has zeros 0 and 3 for all real numbers a. Section 2.2 Polynomial Functions of Higher Degree 159

49. fx x 2x 6 50. f x x 4x 5 fx x 2x 6 x 4x 5 fx x2 4x 12 x 2 x 20 Note: fx ax 2x 6 has zeros 2 and 6 Note: f x ax 4x 5 has zeros 4 and 5 for all for all real numbers a 0. real numbers a.

51. fx x 0x 2x 3 52. f x x 0x 2x 5 xx 2x 3 xx 2x 5 x3 5x2 6x xx 2 7x 10 Note: fx axx 2x 3 has zeros 0,2, 3 for x 3 7x 2 10x all real numbers a 0. Note: f x axx 2x 5 has zeros 0, 2, 5 for all real numbers a.

53. fx x 4x 3x 3x 0 54. f x x 2x 1x 0x 1x 2 x 4x2 9x xx 2x 1x 1x 2 x4 4x3 9x2 36x xx 2 4x 2 1 Note: fx ax4 4x3 9x2 36x has these zeros xx 4 5x 2 4 for all real numbers a 0. x 5 5x 3 4x Note: f x axx 2x 1x 1x 2 has zeros 2, 1, 0, 1, 2 for all real numbers a.

55. fx x 1 3 x 1 3 56. f x x 2x 4 5 x 4 5 x 1 3 x 1 3 x 2x 4 5x 4 5 2 x 12 3 x 2x 42 5 x2 2x 1 3 xx 42 5x 2x 42 10 x2 2x 2 x 3 8x 2 16x 5x 2x 2 16x 32 10 2 Note: f x a x 2x 2 has these zeros for all real x 3 10x 2 27x 22 numbers a 0. Note: f x ax 3 10x 2 27x 22 has these zeros for all real numbers a.

57. fx x 2x 2 58. f x x 8x 4 x 22 x2 4x 4 x 8x 4 x2 12x 32 Note: f x ax2 4x 4, a 0, has degree 2 and Note: f x ax2 12x 32, a 0, has degree 2 zero x 2. and zeros x 8 and 4.

59. f x x 3x 0x 1 60. f x x 2x 4x 7 xx 3x 1 x3 2x2 3x x 2x2 11x 28 x3 9x2 6x 56 Note: f x ax3 2x2 3x, a 0, has degree 3 and Note: f x ax3 9x2 6x 56, a 0, has degree zeros x 3, 0, 1. 3 and zeros x 2, 4, and 7.

61. f x x 0x 3x 3 62. f x x 93 x3 27x2 243x 729 xx 3x 3 x3 3x Note: f x ax3 27x2 243x 729, a 0, has degree 3 and zero x 9. Note: f x ax3 3x, a 0, has degree 3 and zeros x 0, 3, 3. 160 Chapter 2 Polynomial and Rational Functions

63. f x x 52x 1x 2 x4 7x3 3x2 55x 50 or f x x 5x 12x 2 x4 x3 15x2 23x 10 or f x x 5x 1x 22 x4 17x2 36x 20 Note: Any nonzero scalar multiple of these functions would also have degree 4 and zeros x 5, 1, 2.

64. f x x 4x 1x 3x 6 x 4 4x3 23x2 54x 72 Note: f x ax 4 4x3 23x2 54x 72, a 0, has degree 4 and zeros x 4, 1, 3, and 6.

65. f x x4x 4 x5 4x4 or f x x3x 42 x5 8x4 16x3 or f x x2x 43 x5 12x4 48x3 64x2 or f x xx 44 x5 16x4 96x3 256x2 256x Note: Any nonzero scalar multiple of these functions would also have degree 5 and zeros x 0 and 4.

66. f x x 32x 1x 5x 6 x5 6x4 22x3 108x2 189x 270 or f x x 3x 12x 5x 6 x5 10x 4 14x3 88x2 183x 90 or f x x 3x 1x 52x 6 x5 14x 4 50x3 68x2 555x 450 or f x x 3x 1x 5x 62 x5 15x 4 59x3 63x2 648x 540 Note: Any nonzero multiple of these functions would also have degree 5 and zeros x 3, 1, 5, and 6.

67. f x x3 9x xx2 9 xx 3x 3 68. gx x 4 4x2 x2x 2x 2 (a) Falls to the left; rises to the right (a) Rises to the left; rises to the right (b) Zeros: 0, 3, 3 (b) Zeros: 2, 0, 2 (c) (c) x 3 2 1 01 2 3 x 0.5 1 1.5 2.5 f x 010808 10 0 gx 0.94 3 3.94 14.1

y (d) (d) y 12 4 3 (0, 0) 2 4 − 1 ( 3, 0) (3, 0) (−2, 0) (0, 0) (2, 0) x x −12 −8 −4 4 812 −4 −3 −1 1 3 4 −4

−8

−4

1 2 1 2 7 69. f t 4 t 2t 15 4 t 1 2 7 (a) Rises to the left; rises to the right (d) The graph is a parabola with vertex 1, 2 . (b) No real zero (no x-intercepts) y

t −4 −224 Section 2.2 Polynomial Functions of Higher Degree 161

70. gx x2 10x 16 x 2x 8 71. f x x3 3x2 x2x 3 (a) Falls to the left; falls to the right (a) Falls to the left; rises to the right (b) Zeros: 2, 8 (b) Zeros: 0, 3 (c) (c) x 13579 x 1 01 2 3

gx 75957 f x 4 0 2 4 0

(d) y (d) y

10 1 (0, 0) (3, 0) 8 x −112 4 6

2 −3 (2, 0) (8, 0) x −4 46 10

72. f x 1 x3 73. fx 3x3 15x2 18x 3xx 2x 3 (a) Rises to the left; falls to the right (a) Falls to the left; rises to the right (b) Zero: 1 (b) Zeros: 0, 2, 3 (c) (c) x 2 1012 x 0122.5 3 3.5 f x 92107 f x 0601.875 0 7.875

(d) y (d) y 7 3 6 5 2 4 3 2 (1, 0) 1 x (0, 0) (2, 0) (3, 0) −2 −12 x −3 −2 −1 1 4 56 −1 −1 −2

74. f x 4x3 4x2 15x 75. f x 5x2 x3 x25 x x4x2 4x 15 (a) Rises to the left; falls to the right x2x 52x 3 (b) Zeros: 0, 5 (a) Rises to the left; falls to the right (c) x 5 4 3 2 1 01 3 5 (b) Zeros: 2, 0, 2 f x 0 16 18 12 4 0 6 (c) x 3 2 10123 (d) y f x 99 18 701514 27 5 (−5, 0) (0, 0) x (d) y −15 −10 510 20

8 −20 −3, 0 4 5, 0 ( 2 ( (0, 0) ( 2 ( x −4 −3 −2 1 2 3 4 162 Chapter 2 Polynomial and Rational Functions

76. f x 48x2 3x 4 3x2x2 16

(a) Rises to the left; rises to the right (d) (−4, 0) y (b) Zeros: 0, ±4 100 (0, 0) (4, 0) (c) x x 5 4 3 2 101 2 3 45 −6 −2 2 6 f x 675 0 189 144 45 0 45 144 189 0 675 −200

−300

2 1 3 2 77. f x x x 4 78. h x 3x x 4 (a) Falls to the left; rises to the right (a) Falls to the left; rises to the right (b) Zeros: 0, 4 (b) Zeros: 0, 4 (c) (c) x 1 0 1 23 45 x 1012345 25 32 125 f x 5 00253 8 9 h x 3 033 90 3

(d) y (d) y

2 14 (0, 0) (4, 0) 12 x −4 −2268 10 8 6 4

(0, 0) (4, 0) x −4 −2 24681012

1 2 2 1 2 3 79. g t 4 t 2 t 2 80. g x 10 x 1 x 3 (a) Falls to the left; falls to the right (a) Falls to the left; rises to the right (b) Zeros: 2, 2 (b) Zeros: 1, 3 (c) (c) t 3 2 1 0123 x 2 1 0124 25 9 9 25 g t 4 004 4 4 4 g x 12.5 0 2.7 3.2 0.9 2.5

(d) y (d) y (−2, 0) (2, 0) t 6 −3 −1123 −1 4 − 2 2 (−1, 0) (3, 0) x −6 −4 −2 468

−5

−6 Section 2.2 Polynomial Functions of Higher Degree 163

3 1 4 2 81. f x x 4x x x 2 x 2 82. f x 4x 2x

6 6

−99 −99

−6 −6

Zeros:0, 2, 2 all of multiplicity 1 Zeros:2.828 and 2.828 with multiplicity 1; 0, with multiplicity 2

1 2 1 2 2 83. g x 5 x 1 x 3 2x 9 84. h x 5 x 2 3x 5

14 21

−12 18 −12 12

−6 −3

5 Zeros:1 of multiplicity 2; 3 of multiplicity 1; Zeros:2, 3, both with multiplicity 2 9 2 of multiplicity 1

85. fx x3 3x2 3 86. f x 0.11x 3 2.07x 2 9.81x 6.88 x y1 10 The function has three zeros. They are in the intervals 3 51 0, 1, 6, 7, and 11, 12. They are approximately 0.845, 2 17 6.385, and 11.588. −5 5 x y x y 1 1 10 0 6.88 7 1.91 −10 03 −4 16 1 0.97 8 4.56 The function has three zeros. 11 They are in the intervals 2 5.34 9 6.07 − 1, 0, 1, 2 and 2, 3. They 2 1 10 3 6.89 10 5.78 are x 0.879, 1.347, 2.532. 33 4 6.28 11 3.03 419 5 4.17 12 2.84 6 1.12

87. gx 3x4 4x3 3 88. hx x 4 10x 2 3 x y1 x y 10 The function has four zeros. They are 4 509 4 99 in the intervals 4, 3, 1, 0, 0, 1, and 3, 4. They are approximately −5 5 3 132 3 6 ±3.113 and ±0.556. 2 13 2 21 10 −10 1 4 1 6 −4 4 The function has two zeros. 0 3 0 3 They are in the intervals 2, 1 and 0, 1. They 14 1 6 are x 1.585, 0.779. −30 277 2 21 3 348 3 6 499 164 Chapter 2 Polynomial and Rational Functions

89. (a) Volume l w h (c) Box Box Box height x Height Width Volume, V length width 36 2x 1 36 21 136 212 1156 Thus, Vx 36 2x36 2xx x36 2x2. 2 36 22 236 222 2048 (b) Domain: 0 < x < 18 3 36 23 336 232 2700 The length and width must be positive. 4 36 24 436 242 3136 (d) 3600 5 36 25 536 252 3380 6 36 26 636 262 3456 7736 27 36 272 3388 018 0 The volume is a maximum of 3456 cubic inches when the The maximum point on the graph occurs at x 6. height is 6 inches and the length and width are each This agrees with the maximum found in part (c). 24 inches. So the dimensions are 6 24 24 inches.

90. (a) Volume l w h 24 2x24 4xx (c) V 212 x 46 xx 720 600 8x12 x6 x 480 (b) x > 0, 12 x > 0, 6 x > 0 360 240 x < 12 x < 6 120 Domain: 0 < x < 6 x 123456

x 2.6 corresponds to a maximum of about 665 cubic inches.

91. (a)A l w 12 2xx 2x2 12x square inches (e) 4000 (b) 16 feet 192 inches V l w h 0 6 12 2xx192 0

384x2 2304x cubic inches Maximum: 3, 3456 (c) Since x and 12 2x cannot be negative, we have The maximum value is the same. 0 < x < 6 inches for the domain. (d) When x 3, the volume is a (f) No. The volume is a product of the constant length and x V maximum with V 3456 in.3 . The the cross-sectional area. The value of x would remain the 00 dimensions of the gutter cross-section same; only the value of V would change if the length was are 3 inches 6 inches 3 inches. changed. 1 1920 2 3072 3 3456 4 3072 5 1920 60 Section 2.2 Polynomial Functions of Higher Degree 165

4 3 2 ≥ 92. (a) V 3 r r 4r (b) r 0 4 3 3 3 16 3 V 3r 4r (d) V 120 ft 3 r 16 3 3 r r 1.93 ft (c) 150 length 4r 7.72 ft

0 2 0

3 2 3 2 93. y1 0.139t 4.42t 51.1t 39 94. y 0.056t 1.73t 23.8t 29

200 180

7 13 7 13 140 120 The model is a good fit to the actual data. The data fit the model closely.

95. Midwest: y1 18 $259.368 thousand $259,368 96. Answers will vary. South: y2 18 $223.472 thousand $223,472 Example: The median price of homes in the South are all lower than those in the Midwest. The curves do Since the models are both cubic functions with positive not intersect. leading coefficients, both will increase without bound as t increases, thus should only be used for short term projections.

97. G 0.003t 3 0.137t 2 0.458t 0.839, 2 ≤ t ≤ 34

(a) 60 (c) y 0.009t2 0.274t 0.458 b 0.274 15.222 2a 20.009 −10 45 y15.222 2.543 −5 Vertex 15.22, 2.54 (b) The tree is growing most rapidly at t 15. (d) The x-value of the vertex in part (c) is approximately equal to the value found in part (b).

1 3 2 98. R 100,000 x 600x 99. False. A fifth degree polynomial can have at most four turning points. 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 corresponds to spending $2,000,000 on advertising to obtain a revenue of $160 million.

101. True. A polynomial of degree 7 with a negative leading 100. True. f x x 16 has one repeated solution. coefficient rises to the left and falls to the right.

102. (a) Degree: 3 (c) Degree: 4 Leading coefficient: Positive Leading coefficient: Positive (b) Degree: 2 (d) Degree: 5 Leading coefficient: Positive Leading coefficient: Positive 166 Chapter 2 Polynomial and Rational Functions

103. fx x4; fx is even. y (a) gx fx 2 5 4

Vertical shift two units upward 3 gx f x 2 2 1 f x 2 x −3 −2 −1 123 gx −1 Even (b)gx f x 2 (c) gx f x x4 x 4 Horizontal shift two units to the left Reflection in the y- axis. The graph looks the same. Neither odd nor even Even

4 1 1 4 (d) g x f x x (e) g x f 2x 16x Reflection in the x-axis Horizontal stretch Even Even

1 1 4 34 344 3 ≥ (f) g x 2 f x 2x (g) g x f x x x , x 0 Vertical shrink Neither odd nor even Even

(h) gx f f x f f x f x 4 x 44 x16 Even

1 5 5 3 104. (a)y1 3 x 2 1 is decreasing. (c) H x x 3x 2x 1 3 5 y2 5 x 2 3 is increasing. Since H x is not always increasing or always decreasing, Hx cannot be written in the form ax h5 k. 8

−12 12 y y 2 1 −99

−8

−6 (b) The graph is either always increasing or always decreasing. The behavior is determined by a. If a > 0, gx will always be increasing. If a < 0, gx will always be decreasing.

105. 5x2 7x 24 5x 8x 3 106. 6 x3 61x2 10x x6x2 61x 10 x6x 1x 10

107. 4 x4 7x3 15x2 x24x2 7x 15 108. y3 216 y3 63 x24x 5x 3 y 6y2 6y 36

109. 2 x2 x 28 0 110. 3 x2 22x 16 0 2x 7x 4 0 3x 2x 8 0 ⇒ 7 2 x 7 0 x 2 3 x 2 0 or x 8 0 ⇒ 2 x 4 0 x 4 x 3 orx 8 Section 2.2 Polynomial Functions of Higher Degree 167

111. 12x2 11x 5 0 112. x2 24x 144 0 3x 14x 5 0 x 122 0 ⇒ 1 3 x 1 0 x 3 x 12 0 ⇒ 5 4 x 5 0 x 4 x 12

113. x2 2x 21 0 114. x2 8x 2 0 x2 2x 12 21 1 0 x2 8x 2 x 12 22 0 x2 8x 16 2 16 x 12 22 x 42 14 x 1 ±22 x 4 ±14 x 1 ± 22 x 4 ± 14

115. 2 x2 5x 20 0 116. 3 x2 4x 9 0 5 4 2 x2 x 20 0 x2 x 3 0 2 3 5 5 2 25 4 2 x2 x 20 0 x2 x 3 2 4 8 3 5 2 185 4 4 4 2 x 0 x2 x 3 4 8 3 9 9 5 2 185 2 2 31 x x 4 16 3 9 5 185 2 31 x ± x ± 4 4 3 9 5 ± 185 2 31 x x ± 4 3 3 2 ± 31 x 3

117. fx x 42 y 118. f x 3 x2 y 7 Common function: y x2 4 6 Reflection in the x-axis

5 and vertical shift of three 2 Transformation: Horizontal 2 4 units upward of y x shift four units to the left 1 3 x −4 −3 −1 134 2 −1 1 −2 x −3 −7 −6 −5 −4 −3 −2 −1 1 − 1 −4

119. fx x 1 5 y 120. f x 7 x 6 y 1 Common function: y x Horizontal shift of six units 15 x 12 −3 −2 −1 123 to the right, reflection in the Transformation: Horizontal − 1 x-axis, and vertical shift of 9 shift one unit to the left and − 2 seven units upward 6 a vertical shift five units −3 of y x downward 3 x −36912153 − 5 −3 168 Chapter 2 Polynomial and Rational Functions

y 1 y 121. f x 2 x 9 122. f x 10 3 x 3 9 Common function: y x 6 5 Horizontal shift of three units 8 7 4 to the left, vertical shrink Transformation: Vertical 6 3 each y-value is multiplied 5 stretch each y-value is 1 2 by , reflection in the x-axis 4 multiplied by 2 , then a 3 1 and vertical shift of ten 3 vertical shift nine units x 2 −6 −3 −2 −121 units upward of y x 1 upward x −2 −121 3456789

Section 2.3 Polynomial and Synthetic Division

You should know the following basic techniques and principles of polynomial division. ■ The Division Algorithm (Long Division of Polynomials) ■ Synthetic Division ■ f k is equal to the remainder of fx divided by x k (the Remainder Theorem). ■ fk 0 if and only if x k is a factor of fx.

Vocabulary Check 1. f x is the dividend; dx is the divisor; gx is the quotient; rx is the remainder 2. improper; proper 3. synthetic division 4. factor 5. remainder

x2 4 x4 3x2 1 39 1. y and y x 2 2. y and y x2 8 1 x 2 2 x 2 1 x2 5 2 x2 5 x 2 x2 8 x 2 ) x2 0x 0 x2 5 ) x4 3x2 1 x2 2x x4 5x2 2x 0 8x2 1 2x 4 8x2 40 4 39 x2 4 x4 3x2 1 39 Thus, x 2 and y y . Thus, x2 8 and y y . x 2 x 2 1 2 x2 5 x2 5 1 2

x5 3x3 4x 3. y and y x3 4x 1 x2 1 2 x2 1 x3 4x (a) and (b) (c) x2 0x 1 ) x5 0x4 3x3 0x2 0x 0 6 x5 0x4 x3 4x3 0x2 0x −99 4x3 0x2 4x 4x 0

−6 x5 3x3 4x Thus, x3 4x and y y . x2 1 x2 1 1 2 Section 2.3 Polynomial and Synthetic Division 169

x3 2x2 5 2x 4 4. y and y x 3 1 2 2 2 x x 1 x x 1 x 3 (a) and (b) (c) x2 x 1 ) x3 2x2 0x 5 3 2 8 x x x 3x2 x 5 2 −12 12 3x 3x 3 2x 8 −8 x3 2x2 5 2x 4 Thus, x 3 and y y . x2 x 1 x2 x 1 1 2

5. 2x 4 6. 5 x 3 x 3 ) 2x2 10x 12 x 4 ) 5x2 17x 12 2x2 6x 5 x2 20x 4x 12 3 x 12 4x 12 3 x 12 0 0 2x2 10x 12 5x2 17x 12 2x 4 5x 3 x 3 x 4

7. x2 3x 1 8. 2 x2 4x 3 4x 5 ) 4x3 7x2 11x 5 3 x 2 ) 6x3 16x2 17 x 6 4x3 5x2 6 x3 4x2 12x2 11x 12x2 17 x 12x2 15x 12x2 8 x 4x 5 9 x 6 4x 5 9 x 6 0 0 4x3 7x2 11x 5 6x3 16x2 17x 6 x2 3x 1 2x2 4x 3 4x 5 3x 2

9. x3 3x2 1 10. x2 7x 18 x 2 ) x4 5x3 6x2 x 2 x 3 ) x3 4x2 3x 12 x4 2x3 x3 3x2 3x3 6x2 7 x2 3x 3x3 6x2 7 x2 21x x 2 1 8 x 12 x 2 18x 54 0 4 2 x4 5x3 6x2 x 2 x3 4x2 3x 12 42 x3 3x2 1 x2 7x 18 x 2 x 3 x 3 170 Chapter 2 Polynomial and Rational Functions

11. 7 12. 4 x 2 ) 7x 3 2 x 1 ) 8x 5 7x 14 8 x 4 11 9 7x 3 11 8x 5 9 7 4 x 2 x 2 2x 1 2x 1

13. 3x 5 14. x 2x2 0x 1 ) 6x3 10x2 x 8 x2 0x1 ) x3 0x2 0x 9 6x3 0x2 3x x3 0 x2 x 10x2 2x 8 x 9 2 10x 0x 5 x3 9 x 9 x 2x 3 x2 1 x2 1 6x3 10x2 x 8 2x 3 3x 5 2x2 1 2x2 1

15. x2 2x 4 x4 3x2 1 2x 11 x2 2x 3 ) x4 0x3 3x2 0x 1 ⇒ x2 2x 4 x2 2x 3 x2 2x 3 x4 2x3 3x2 2x3 0x2 0x 2x3 4x2 6x 4x2 6x 1 4x2 8x 12 2x 11

16. x2 17. x 3 x3 0x2 0x 1 ) x5 0x4 0x3 0x2 0x 7 x3 3x2 3x 1 ) x4 0x3 0x2 0x 0 4 3 2 x5 0x4 0x3 x2 x 3x 3x x 3x3 3x2 x 0 x2 7 3x3 9x2 9x 3 x5 7 x2 7 x2 6x2 8x 3 x3 1 x3 1 x4 6x2 8x 3 x 3 x 13 x 13

18. 2 x 19. 5317 15 25 x2 2x 1 ) 2x3 4x2 15x 5 15 10 25 3 2 5 0 2 x3 4x2 2x 3x3 17x2 15x 25 3x2 2x 5 17x 5 x 5 2x 3 4x 2 15x 5 17x 5 2x x 12 x 2 2x 1 Section 2.3 Polynomial and Synthetic Division 171

20. 3518 7 6 21. 24 8 9 18 15 9 6 8 0 18 5 3 2 0 4 0 9 0 5x 3 18x 2 7x 6 4x3 8x2 9x 18 5x 2 3x 2 4x2 9 x 3 x 2

22. 2918 16 32 23. 10 1 0 75 250 18 0 32 10 100 250 9 0 16 0 1 10 25 0 9x 3 18x 2 16x 32 x3 75x 250 9x 2 16 x2 10x 25 x 2 x 10

24. 6316 0 72 25. 456 0 8 18 12 72 20 56 224 3 2 12 0 5 14 56 232 3x 3 16x 2 72 5x3 6x2 8 232 3x 2 2x 12 5x2 14x 56 x 6 x 4 x 4

26. 250 6 8 27. 61050 0 0 800 10 20 52 60 60 360 2160 5 10 26 44 10 10 60 360 1360 5x 3 6x 8 44 10x4 50x3 800 1360 5x 2 10x 26 10x3 10x2 60x 360 x 2 x 2 x 6 x 6

28. 3113 0 0 120 80 3 48 144 432 936 1 16 48 144 312 856 x 5 13x 4 120x 80 856 x 4 16x 3 48x 2 144x 312 x 3 x 3

29. 81 0 0 512 30. 910 0 729 8 64 512 9 81 729 1 8 64 0 1 9 81 0 x3 512 x3 729 x2 8x 64 x2 9x 81 x 8 x 9

31. 2 3 0 0 0 0 32. 2 3 0 0 0 0 6 12 24 48 6 12 24 48 3 6 12 24 48 3 6 12 24 48 3x4 48 3x 4 48 3x3 6x2 12x 24 3x 3 6x 2 12x 24 x 2 x 2 x 2 x 2

33. 6 1 0 0 180 0 34. 1 1 2 3 5 6 36 216 216 1 3 6 1 6 36 36 216 1 3 6 11 180x x4 216 5 3x 2x2 x3 11 x3 6x2 36x 36 x2 3x 6 x 6 x 6 x 1 x 1 172 Chapter 2 Polynomial and Rational Functions

1 3 35. 2 4 16 23 15 36. 2 3 4 0 5 9 3 9 2 7 15 2 4 8 4 14 30 0 1 3 49 3 2 4 8 4x3 16x2 23x 15 4x2 14x 30 3x 3 4x 2 5 1 3 49 x 12 3x 2 x x 32 2 4 8x 12

37. fx x3 x2 14x 11, k 4 38. f x x3 5x2 11x 8, k 2 411 14 11 215 11 8 4 12 8 2 14 6 1 3 2 3 1 7 3 2 fx x 4x2 3x 2 3 f x x 2x2 7x 3 2 f4 43 42 144 11 3 f 2 23 522 112 8 8 20 22 8 2

4 3 2 2 3 2 1 39. f x 15x 10x 6x 14, k 3 40. f x 10x 22x 3x 4, k 5 2 15 10 6 0 14 1 10 22 3 4 3 5 8 7 10 0 4 3 2 4 5 34 13 15 0 6 4 3 10 20 7 5 2 3 34 1 2 13 f x x 3 15x 6x 4 3 f x x 5 10x 20x 7 5 2 24 23 22 34 1 13 12 1 f 3 15 3 10 3 6 3 14 3 f 5 10 5 22 5 3 5 4 2 22 3 65 13 25 25 5 4 25 5

41. f x x3 3x2 2x 14, k 2 42. f x x3 2x2 5x 4, k 5 2 1 3 2 14 5 1 2 5 4 2 2 32 6 5 25 5 10 1 3 2 32 8 1 2 5 25 6

f x x 2x2 3 2x 32 8 f x x 5 x2 2 5 x 25 6 3 2 f 2 2 32 22 14 8 f 5 5 3 252 55 4 55 10 55 4 6

43. f x 4x3 6x2 12x 4, k 1 3 1 3 4 6 12 4 4 43 10 23 4 4 2 43 2 23 0 f x x 1 34x2 2 43x 2 23 0 3 2 f 1 3 41 3 61 3 121 3 4 0

44. f x 3x3 8x2 10x 8, k 2 2 f x x 2 2 3x2 2 32 x 8 42 0 2 2 3 8 10 8 f 2 2 32 2 3 82 2 2 102 2 8 6 32 2 42 8 320 142 86 42 102 2 8 3 2 3 2 8 4 2 0 60 422 48 322 20 102 8 0 Section 2.3 Polynomial and Synthetic Division 173

45. fx 4x3 13x 10 46. gx x 6 4x 4 3x 2 2 (a) 140 13 10 (a) 210 4 0 3 0 2 4 4 9 2 4 0 0 6 12 4 4 9 1 1 2 0 0 3 6 14 f 1 1 g2 14 (b) 24 0 13 10 (b) 410 4 0 3 0 2 8 16 6 4 16 48 192 780 3120 4 8 3 4 1 4 12 48 195 780 3122 f 2 4 g4 3122 1 (c) 2 4 0 13 10 (c) 310 4 0 3 0 2 2 1 6 3 9 15 45 144 432 4 2 12 4 1 3 5 15 48 144 434 f 1 4 2 g3 434 (d) 84 0 13 10 (d) 110 4 0 3 0 2 32 256 1944 1 1 3 3 0 0 4 32 243 1954 1 1 3 3 0 0 2 f 8 1954 g1 2

47. hx 3x3 5x2 10x 1 48. f x 0.4x 4 1.6x 3 0.7x 2 2 (a) 33 5 10 1 (a) 1 0.4 1.6 0.7 0 2 9 42 96 0.4 1.2 0.5 0.5 3 14 32 97 0.4 1.2 0.5 0.5 2.5 h3 97 f 1 2.5 1 (b) 3 3 5 10 1 (b) 2 0.4 1.6 0.7 0 2 8 1 2 3 0.8 4.8 11 22 5 3 6 8 3 0.4 2.4 5.5 11 20 1 5 f 2 20 h 3 3 (c) 23 5 10 1 (c) 5 0.4 1.6 0.7 0 2 6 2 16 2.0 2.0 13.5 67.5 3 1 8 17 0.4 0.4 2.7 13.5 65.5 h2 17 f 5 65.5 (d) 53 5 10 1 (d) 10 0.4 1.6 0.7 0 2 15 50 200 4.0 56.0 567 5670 3 10 40 199 0.4 5.6 56.7 567 5668 h5 199 f 10 5668

49. 210 7 6 50. 410 28 48 2 4 6 4 16 48 1 2 3 0 1 4 12 0 x3 7x 6 x 2x2 2x 3 x 3 28x 48 x 4x 2 4x 12 x 2x 3x 1 x 4x 6x 2 Zeros: 2, 3, 1 Zeros: 4, 2, 6 174 Chapter 2 Polynomial and Rational Functions

1 2 51. 2 2 15 27 10 52. 3 48 80 41 6 1 7 10 32 32 6 2 14 20 0 48 48 9 0 2 x3 15x2 27x 10 3 2 2 2 48x 80x 41x 6 x 3 48x 48x 9 1 2 x 2 2x 14x 20 2 x 3 4x 3 12x 3 2x 1x 2x 5 3x 24x 34x 1 1 Zeros: 2, 2, 5 2 3 1 Zeros: 3, 4, 4

53. 3 1 2 3 6 54. 2 1 2 2 4 3 3 23 6 2 22 2 4 1 2 3 2 3 0 1 2 2 22 0 3 1 2 3 23 2 1 2 2 22 3 2 3 2 22 1 2 0 1 2 0 x3 2x2 3x 6 x 3 x 3 x 2 x 3 2x2 2x 4 x 2x 2x 2 Zeros: ±3, 2 Zeros: 2, 2, 2

55. 1 3 1 3 0 2 1 3 1 3 2 1 2 3 1 3 0 1 3 1 2 3 1 3 1 3 1 3 1 1 0 x3 3x2 2 x 1 3 x 1 3 x 1 x 1x 1 3 x 1 3 Zeros: 1, 1 ± 3

56. 2 5 1 1 13 3 2 5 7 35 3 1 1 5 6 35 0 2 5 1 1 5 6 35 2 5 6 35 1 3 0 x 3 x 2 13x 3 x 2 5x 2 5x 3 Zeros: 2 5, 2 5, 3

57. f x 2x3 x2 5x 2; Factors: x 2, x 1 (a) 22 1 5 2 (b) The remaining factor of f x is 2x 1. 4 6 2 (c) f x 2x 1x 2x 1 2 3 1 0 1 (d) Zeros: 2, 2, 1 123 1 (e) 7 2 1 2 1 0 Both are factors of f x since the remainders are zero. −66

−1 Section 2.3 Polynomial and Synthetic Division 175

58. f x 3x3 2x2 19x 6; Factors: x 3, x 2 (a) 33 2 19 6 (c) f x 3x3 2x2 19x 6 9 21 6 3x 1x 3x 2 3 7 2 0 1 (d) Zeros: 3, 3, 2 237 2 6 2 (e) 35 3 1 0

(b) The remaining factor is 3x 1. −4 3

−10

59. f x x4 4x3 15x2 58x 40; Factors: x 5, x 4 (a) 514 15 58 40 (c) f x x 1x 2x 5x 4 5 5 50 40 (d) Zeros: 1, 2, 5, 4 1 1 10 8 0 (e) 20 41 1 10 8 −66 4 12 8 1 3 2 0 Both are factors of f x since the remainders are zero. −180 (b) x2 3x 2 x 1x 2 The remaining factors are x 1 and x 2.

60. f x 8x 4 14x3 71x2 10x 24; Factors: x 2, x 4 (a) 2814 71 10 24 (c) f x 4x 32x 1x 2x 4 16 60 22 24 3 1 (d) Zeros: 4, 2, 2, 4 8 30 11 12 0 (e) 40 −35 4830 11 12 32 8 12 8 2 3 0 −380 (b) 8x2 2x 3 4x 32x 1 The remaining factors are 4x 3 and 2x 1.

61. f x 6x3 41x2 9x 14; Factors: 2x 1, 3x 2 1 (a) 2 6 41 9 14 (b) 6x 42 6 x 7 3 19 14 f x This shows that 6x 7, 6 38 28 0 1 2 x 2 x 3 2 6 38 28 3 f x 4 28 so x 7. 2x 13x 2 6 42 0 The remaining factor is x 7. Both are factors since the remainders are zero. 1 2 (c) f x x 72x 13x 2 (d) Zeros: 7, , 2 3

(e) 320

−93

−40 176 Chapter 2 Polynomial and Rational Functions

62. f x 10x3 11x2 72x 45; 63. f x 2x3 x2 10x 5; Factors: 2x 5, 5x 3 Factors: 2x 1, x 5 5 (a) 1 2 1 10 5 (a) 2 10 11 72 45 2 25 90 45 1 0 5 2 0 10 0 10 36 18 0 10 3 10 36 18 5 2 0 5 6 18 2 5 10 2 25 0 10 30 0 (b) 10x 30 10x 3 Both are factors since the remainders are zero. fx (b) 2x 25 2x 5 This shows that 10x 3, 5 3 x 2 x 5 f x This shows that 2x 5, f x x 1x 5 2 so x 3. 2x 5 5x 3 f x so x 5. The remaining factor is x 3. 2x 1x 5 (c) f x x 32x 55x 3 The remaining factor is x 5. 5 3 (c) f x x 5x 5 2x 1 (d) Zeros: 3, , 2 5 1 (d) Zeros: 5, 5, (e) 100 2 (e) 14

−44

−66 −80

−6

64. f x x3 3x2 48x 144; Factors: x 43 , x 3 (a) 31 3 48 144 (c) f x x 43 x 43 x 3 3 0 144 (d) Zeros: ±43, 3 1 0 48 0 (e) 60 4 3 1 0 48 −88 43 48 1 43 0

−240 (b) The remaining factor is x 43 .

65. f x x3 2x2 5x 10 66. gx x 3 4x 2 2x 8 (a) The zeros of f are 2 and ±2.236. (a) The zeros of g are x 4, x 1.414, x 1.414. (b) An exact zero is x 2. (b)x 4 is an exact zero. (c) 212 5 10 (c) 414 2 8 2 0 10 4 0 8 1 0 5 0 1 0 2 0 2 f x x 2 x 5 fx x 4x 2 2 x 2 x 5x 5 x 4x 2x 2 Section 2.3 Polynomial and Synthetic Division 177

67. ht t3 2t2 7t 2 68. f s s 3 12s 2 40s 24 (a) The zeros of h are t 2, t 3.732, t 0.268. (a) The zeros of f are s 6, s 0.764, s 5.236 (b) An exact zero is t 2. (b)s 6 is an exact zero. (c) 212 7 2 (c) 6112 40 24 2 8 2 6 36 24 1 4 1 0 1 6 4 0 ht t 2t2 4t 1 f s s 6s 2 6s 4 By the Quadratic Formula, the zeros of t2 4t 1 s 6s 3 5s 3 5 are 2 ± 3. Thus, ht t 2t 2 3t 2 3 t 2t 2 3 t 2 3 .

4x3 8x2 x 3 x3 x2 64x 64 69. 70. 2x 3 x 8 3 2 4 8 1 3 811 64 64 6 3 3 8 56 64 4 2 2 0 1 7 8 0 4x3 8x2 x 3 x 3 x 2 64x 64 4x2 2x 2 22x2 x 1 x 2 7x 8, x 8 3 x 2 x 8 4x3 8x2 x 3 3 Thus, 2x2 x 1, x . 2x 3 2

x4 6x3 11x2 6x x4 6x3 11x2 6x x 4 9x 3 5x 2 36x 4 x 4 9x 3 5x 2 36x 4 71. 72. x2 3x 2 x 1x 2 x2 4 x 2x 2 11 6 11 6 0 219 5 36 4 1 5 6 0 2 22 34 4 1 5 6 0 0 1 11 17 2 0 21 5 6 0 2111 17 2 2 6 0 2 18 2 1 3 0 0 1 9 1 0 x4 6x3 11x2 6x x2 3x, x 2, 1 x 4 9x 3 5x 2 36x 4 x x x 2 9x 1, x ±2 1 2 x 2 4

73. (a) and (b)

1800

3 13 1200

—CONTINUED— 178 Chapter 2 Polynomial and Rational Functions

73. —CONTINUED—

(c) M 0.242t3 12.43t2 173.4t 2118 (d) 18 0.242 12.43 173.4 2118 4.356 145.332 505.224 Year, t Military Personnel M 0.242 8.074 28.068 1612.776 3 1705 1703 M18 1613 thousand 4 1611 1608 No, this model should not be used to predict the 5 1518 1532 number of military personnel in the future. It predicts an increase in military personnel until 2024 6 1472 1473 and then it decreases and will approach negative 7 1439 1430 infinity quickly. 8 1407 1402 9 1386 1388 10 1384 1385 11 1385 1393

12 1412 1409 13 1434 1433 The model is a good fit to the actual data.

4 74. (a) and (b) 75. False. If 7x 4 is a factor of f, then 7 is a zero of f.

2 12 0

(b) R 0.0026t3 0.0292t2 1.558t 15.632 (c) 18 0.0026 0.0292 1.558 15.632 0.0468 0.3168 33.7464 0.0026 0.0176 1.8748 49.3784 For the year 2008, the model predicts a monthly rate of about $49.38.

76. True. 77. True. The degree of the numerator is greater than the 1 degree of the denominator. 2 6 1 92 45 184 4 48 3 2 45 0 92 48 6 4 90 0 184 96 0 fx 2x 1x 1x 2x 33x 2x 4

78. fx x kqx r (a)k 2, r 5, qx any quadratic ax2 bx c (b)k 3, r 1, qx any quadratic ax2 bx c where a > 0. One example: where a < 0. One example: fx x 2x2 5 x3 2x2 5 fx x 3x2 1 x3 3x2 1 Section 2.3 Polynomial and Synthetic Division 179

79. x2n 6xn 9 80. x2n xn 3 xn 3 ) x3n 9x2n 27xn 27 xn 2 ) x3n 3x2n 5xn 6 x3n 3x2n x3n 2x2n 6x2n 27xn x2n 5xn 6x2n 18xn x2n 2xn 9xn 27 3 xn 6 9xn 27 3 xn 6 0 0 x3n 9x2n 27xn 27 x3n 3x2n 5xn 6 x2n 6xn 9 x2n xn 3 xn 3 xn 2

81. A divisor divides evenly into a dividend if the remainder 82. You can check polynomial division by multiplying the is zero. quotient by the divisor. This should yield the original dividend if the multiplication was performed correctly.

83. 514 3 c 84. 21 0 0 2 1 c 5 45 210 2 4 8 20 42 1 9 42 c 210 1 2 4 10 21 c 42 To divide evenly,c 210 must equal zero. Thus, c must To divide evenly,c 42 must equal zero. Thus, c must equal 210. equal 42.

85. fx x 32x 3x 13 86. In this case it is easier to evaluate f2 directly because f x is in factored form. To evaluate using synthetic The remainder when k 3 is zero since x 3 division you would have to expand each factor and then is a factor of fx. multiply it all out.

87. 9 x2 25 0 88. 16x2 21 0 3x 53x 5 0 16x2 21 5 21 3 x 5 0 ⇒ x x2 3 16 5 21 3 x 5 0 ⇒ x x ± 3 16 21 x ± 4

89. 5 x2 3x 14 0 90. 8 x2 22x 15 0 5x 7x 2 0 4x 52x 3 0 ⇒ 7 5 x 7 0 x 5 4 x 5 0 or 2 x 3 0 ⇒ 5 3 x 2 0 x 2 x 4 orx 2

91. 2 x2 6x 3 0 b ± b2 4ac 6 ± 62 423 6 ± 12 x 2a 22 4 3 ± 3 2 180 Chapter 2 Polynomial and Rational Functions

92. x2 3x 3 0 3 ± 32 413 3 ± 21 x 21 2

93. f x x 0x 3x 4 94. f x x 6x 1 xx 3x 4 xx2 7x 12 x 6x 1 x3 7x2 12x x2 5x 6 Note: Any nonzero scalar multiple of f x would also Note: Any nonzero scalar multiple of f x would also have these zeros. have these zeros.

95. f x x 3x 1 2 x 1 2 96. f x x 1x 2x 2 3x 2 3 x 3x 1 2x 1 2 x 1x 2x 2 3x 2 3 2 x 3x 12 2 2 x2 x 2x 22 3 x 3x2 2x 1 x2 x 2x2 4x 1 x3 x2 7x 3 x4 3x3 5x2 9x 2 Note: Any nonzero scalar multiple of f x would also Note: Any nonzero scalar multiple of f x would also have have these zeros. these zeros.

Section 2.4 Complex Numbers

■ Standard form:a bi . If b 0, then a bi is a real number. If a 0 and b 0, then a bi is a pure imaginary number. ■ Equality of Complex Numbers:a bi c di if and only if a c and b d ■ Operations on complex numbers (a) Addition: a bi c di a c b di (b) Subtraction: a bi c di a c b di (c) Multiplication: a bic di ac bd ad bci a bi a bi c di ac bd bc ad (d) Division: i c di c di c di c2 d 2 c2 d 2 ■ The complex conjugate of a bi is a bi: a bia bi a2 b2 ■ The additive inverse of a bi is a bi. ■ a ai for a > 0.

Vocabulary Check 1. (a) iii (b) i (c) ii 2. 1; 1 3. principal square 4. complex conjugates Section 2.4 Complex Numbers 181

1. a bi 10 6i 2. a bi 13 4i 3. a 1 b 3i 5 8i a 10 a 13 a 1 5 ⇒ a 6 b 6 b 4 b 3 8 ⇒ b 5

4. a 6 2bi 6 5i 5. 4 9 4 3i 6. 3 16 3 4i 2 b 5 5 b 2 a 6 6 a 0

7. 2 27 2 27i 8. 1 8 1 22i 9. 75 75i 53i 2 33i

10. 4 2i 11. 8 8 0i 8 12. 45

13. 6i i2 6i 1 14. 4i 2 2i 41 2i 15. 0.09 0.09 i 1 6i 4 2i 0.3i

16. 0.0004 0.02i 17. 5 i 6 2i 11 i 18. 13 2i 5 6i 8 4i

19. 8 i 4 i 8 i 4 i 20. 3 2i 6 13i 3 2i 6 13i 4 3 11i

21. 2 8 5 50 2 22i 5 52i 3 32i

22. 8 18 4 32i 8 32i 4 32i 23. 13i 14 7i 13i 14 7i 4 14 20i

3 5 5 11 3 5 5 11 24. 22 5 8i 10i 17 18i 25. 2 2i 3 3 i 2 2i 3 3 i 9 15 10 22 6 6 i 6 6 i 1 7 6 6i

26. 1.6 3.2i 5.8 4.3i 4.2 7.5i 27. 1 i3 2i 3 2i 3i 2i2 3 i 2 5 i

28. 6 2i2 3i 12 18i 4i 6i 2 29. 6 i5 2i 30i 12i2 30i 12 12 22i 6 6 22i 12 30i

30. 8i9 4i 72i 32i 2 31. 14 10 i14 10 i 14 10i2 32 72i 14 10 24 182 Chapter 2 Polynomial and Rational Functions

32. 3 15i3 15i 3 15i2 33. 4 5i2 16 40i 25i 2 3 151 16 40i 25 3 15 18 9 40i

34. 2 3i2 4 12i 9i 2 35. 2 3i2 2 3i2 4 12i 9i2 4 12i 9i2 4 9 12i 4 12i 9 4 12i 9 5 12i 10

36. 1 2i2 1 2i2 1 4i 4i 2 1 4i 4i 2 37. The complex conjugate of 6 3i is 6 3i. 1 4i 4i 2 1 4i 4i 2 6 3i6 3i 36 3i2 36 9 45 8i

38. The complex conjugate of 7 12i is 7 12i. 39. The complex conjugate of 1 5i is 1 5i. 7 12i7 12i 49 144i 2 1 5i1 5i 12 5i2 49 144 1 5 6 193

40. The complex conjugate of 3 2i is 3 2i. 41. The complex conjugate of 20 25i is 25i. 3 2i3 2i 9 2i 2 25i25i 20i2 20 9 2 11

42. The complex conjugate of 15 15i is 15i. 43. The complex conjugate of 8 is 8. 15i15i 15i 2 15 15 88 8

5 5 i 5i 44. The complex conjugate of 1 8 is 1 8. 45. 5i i i i 1 1 81 8 1 28 8 9 42

14 2i 28i 28i 2 2 4 5i 46. 7i 47. 2i 2i 4i 2 4 4 5i 4 5i 4 5i 24 5i 8 10i 8 10 i 16 25 41 41 41

5 1 i 5 5i 5 5i 5 5 3 i 3 i 3 i 48. i 49. 1 i 1 i 1 i 2 2 2 2 3 i 3 i 3 i 9 6i i 2 8 6i 4 3 i 9 1 10 5 5

6 7i 1 2i 6 12i 7i 14i 2 6 5i 6 5i i 50. 51. 1 2i 1 2i 1 4i 2 i i i 20 5i 20 5 6i 5i 2 i 4 i 5 6i 5 5 5 1 Section 2.4 Complex Numbers 183

8 16i 2i 16i 32i2 52. 8 4i 2i 2i 4i2

3i 3i 3i 9 40i 5i 5i 53. 54. 4 5i2 16 40i 25i2 9 40i 9 40i 2 3i2 4 12i 9i2 27i 120i2 120 27i 5i 5 12i 81 1600 1681 5 12i 5 12i 120 27 25i 60i2 i 1681 1681 25 144i2 60 25i 60 25 i 169 169 169

2 3 21 i 31 i 2i 5 2i2 i 52 i 55. 56. 1 i 1 i 1 i1 i 2 i 2 i 2 i2 i 2 i2 i 2 2i 3 3i 4i 2i 2 10 5i 1 1 4 i 2 1 5i 12 9i 2 5 1 5 12 9 i i 2 2 5 5

i 2i i3 8i 2i3 2i 1 i 3 1 i4 i 3i 57. 58. 3 2i 3 8i 3 2i3 8i i 4 i i4 i 3i 8i2 6i 4i2 4 i 4i i 2 3i 9 24i 6i 16i2 4i i 2 4i2 9i 5 1 4i 9 18i 16 1 4i 1 4i 4 9i 25 18i 5 20i 25 18i 25 18i 1 16i 2

100 72i 225i 162i2 5 20 i 625 324 17 17 100 297i 162 949 62 297i 62 297 i 949 949 949

59. 6 2 6i2i 12i 2 23 1 60. 5 10 5i10i 23 50i 2 521 52

2 2 2 2 61. 10 10i 10i2 10 62. 75 75i 75i 2 75

63. 3 5 7 10 3 5i7 10i 21 310i 75i 50i2 21 50 75 310 i 21 52 75 310 i 184 Chapter 2 Polynomial and Rational Functions

2 64. 2 6 2 6i2 6i 65. x2 2x 2 0; a 1, b 2, c 2 4 26i 26i 6i2 ± 2 2 2 4 1 2 x 4 26i 26i 61 2 1 2 ± 4 4 6 46i 2 2 46i 2 ± 2i 2 1 ± i

66. x2 6x 10 0; a 1, b 6, c 10 67. 4x2 16x 17 0; a 4, b 16, c 17 6 ± 62 4110 16 ± 162 4417 x x 21 24 6 ± 4 16 ± 16 2 8 3 ± i 16 ± 4i 1 2 ± i 8 2

68. 9x2 6x 37 0; a 9, b 6, c 37 69. 4x2 16x 15 0; a 4, b 16, c 15 6 ± 62 4937 16 ± 162 4415 x x 29 24 6 ± 1296 16 ± 16 16 ± 4 18 8 8 1 36i 1 12 3 20 5 ± ± 2i x or x 3 18 3 8 2 8 2

3 70. 16t2 4t 3 0; a 16, b 4, c 3 71. x2 6x 9 0 Multiply both sides by 2. 2 4 ± 42 4163 t 3 x2 12x 18 0 216 12 ± 122 4318 4 ± 176 x 23 32 12 ± 72 4 ± 411i 6 32 12 ± 62i 1 11 2 ± 2i ± i 6 8 8

7 3 5 72. x2 x 0 73. 1.4x2 2x 10 0 Multiply both sides by 5. 8 4 16 2 14x2 12x 5 0; a 14, b 12, c 5 7 x 10x 50 0 ± 2 12 ± 122 4145 10 10 4 7 50 x x 214 2 7 ± ± 12 ± 136 10 1500 10 10 15 28 14 14 12 ± 2i34 5 ± 515 5 515 ± 28 7 7 7 3 34 ± i 7 14 Section 2.4 Complex Numbers 185

74. 4.5x2 3x 12 0; a 4.5, b 3, c 12 75. 6i 3 i 2 6i2i i2 3 ± 32 44.512 61i 1 x 24.5 6i 1 3 ± 207 3 ± 3i23 1 23 ± i 1 6i 9 9 3 3

76. 4i 2 2i 3 4 2i 77. 5i5 5i2i2i 78. i3 1i 3 1i i 511i 5i

79. 753 53i 3 80. 26 2i6 8i 6 8i 4i 2 8 533 3i 3 12533 1i 3753i

1 1 1 i i i 1 1 1 8i 8i 1 81. i 82. i i3 i i i i2 1 2i3 8i 3 8i 8i 64i 2 8

83. (a) z1 9 16i, z2 20 10i 1 1 1 1 1 20 10i 9 16i 29 6i (b) z z1 z2 9 16i 20 10i 9 16i 20 10i 340 230i 340 230i 29 6i 11,240 4630i 11,240 4630 z i 29 6i 29 6i 877 877 877

84. (a) 23 8 (b) 1 3i3 1 3 31 23i 313i2 3i3 1 33i 9i2 33i 3 1 33i 9 33i 8 3 2 3 (c) 1 3i 13 312 3i 313i 3i 1 33i 9i2 33i3 1 33 i 9 33i 8

4 85. (a) 2 16 86. (a) i 40 i 410 110 1 4 (b) 2 16 (b) i 25 i 46 i 16i i 4 4 4 (c) 2i 2 i 16 1 16 (c) i50 i 412i 2 11 1 4 4 4 (d) 2i 2 i 16 1 16 (d)i 67 i 416i 3 1i i

87. False, if b 0 then a bi a bi a. 88. True That is, if the complex number is real, the number equals x 4 x2 14 56 its conjugate. 4 2 ? i6 i6 14 56 ? 36 6 14 56 56 56 186 Chapter 2 Polynomial and Rational Functions

89. False i44 i150 i 74 i109 i61 i411 i437i2 i418i2 i427i i415i 111 1371 1181 127i 115i 1 1 1 i i 1

90. 66 6i6i 6i2 6

2 91. a1 b1i a2 b2i a1a2 a1b2i a2b1i b1b2i a1a2 b1b2 a1b2 a2b1 i The complex conjugate of this product is a1a2 b1b2 a1b2 a2b1 i. The product of the complex conjugates is:

2 a1 b1i a2 b2i a1a2 a1b2i a2b1i b1b2i a1a2 b1b2 a1b2 a2b1 i Thus, the complex conjugate of the product of two complex numbers is the product of their complex conjugates.

92. a1 b1i a2 b2i 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. Thus, the complex conjugate of the sum of two complex numbers is the sum of their complex conjugates.

93. 4 3x 8 6x x2 x2 3x 12

3 2 2 3 2 2 1 2 1 94. x 3x 6 2x 4x x 3x 6 2x 4x 95. 3x 2 x 4 3x 12x 2x 2 3 2 2 23 x x 2x 6 3x 2 x 2

96. 2x 52 2x2 22x5 52 97. x 12 19 98. 8 3x 34 4x2 20x 25 x 31 3x 42 x 31 x 14

4 99. 4 5x 6 36x 1 0 100. 5 x 3x 11 20x 15 101. V a2b 3 20x 24 18x 3 0 5 x 15x 55 20x 15 3 V 4a2b 2 x 27 0 30x 40 3V a2 2 x 27 40 4 4b x 30 3 27 3V x a 2 4b 1 3V 3Vb a 2 b 2b

m m 102. F 1 2 103. Let x # liters withdrawn and replaced. r2 m m 0.505 x 1.00x 0.605 r 2 1 2 F 2.50 0.50x 1.00x 3.00 m m m m F m m F 0.50x 0.50 r 1 2 1 2 1 2 F F F F x 1 liter Section 2.5 Zeros of Polynomial Functions 187

Section 2.5 Zeros of Polynomial Functions

■ You should know that if f is a polynomial of degree n > 0, then f has at least one zero in the complex number system. ■ You should know the Linear Factorization Theorem. ■ You should know the Rational Zero Test. factors of constant term ■ You should know shortcuts for the Rational Zero Test. Possible rational zeros factors of leading coefficient (a) Use a graphing or programmable calculator. (b) Sketch a graph. (c) After finding a root, use synthetic division to reduce the degree of the polynomial. ■ You should know that if a bi is a complex zero of a polynomial f, with real coefficients, then a bi is also a complex zero of f. ■ You should know the difference between a factor that is irreducible over the rationals (such as x2 7 ) and a factor that is irreducible over the reals (such as x2 9 ). ■ You should know Descartes’s Rule of Signs. (For a polynomial with real coefficients and a non-zero constant term.) (a) The number of positive real zeros of f is either equal to the number of variations of sign of f or is less than that number by an even integer. (b) The number of negative real zeros of f is either equal to the number of variations in sign of fx or is less than that number by an even integer. (c) When there is only one variation in sign, there is exactly one positive (or negative) real zero. ■ You should be able to observe the last row obtained from synthetic division in order to determine upper or lower bounds. (a) If the test value is positive and all of the entries in the last row are positive or zero, then the test value is an upper bound. (b) If the test value is negative and the entries in the last row alternate from positive to negative, then the test value is a lower bound. (Zero entries count as positive or negative.)

Vocabulary Check 1. Fundamental Theorem of Algebra 2. Linear Factorization Theorem 3. Rational Zero 4. conjugate 5. irreducible; reals 6. Descarte’s Rule of Signs 7. lower; upper

1. f x xx 62 2. f x x2x 3x2 1 x2x 3x 1x 1 The zeros are: x 0, x 6 The five zeros are: 0, 0, 3, ±1

3. gx x 2x 43 4. f x x 5x 82 5. f x x 6x ix i The zeros are: x 2, x 4 The three zeros are: 5, 8, 8 The three zeros are: x 6, x i, x i

6. ht t 3t 2t 3it 3i 7. fx x3 3x2 x 3 The four zeros are: 3, 2, ±3i Possible rational zeros: ±1, ±3 Zeros shown on graph: 3, 1, 1 188 Chapter 2 Polynomial and Rational Functions

8. f x x 3 4x 2 4x 16 9. fx 2x4 17x3 35x2 9x 45 Possible rational zeros: ±1, ±2, ±4, ±8, ±16 Possible rational zeros: ±1, ±3, ±5, ±9, ±15, ±45, ± 1 ± 3 ± 5 ± 9 ± 15 ± 45 Zeros shown on graph: 2, 2, 4 2, 2, 2, 2, 2 , 2 3 Zeros shown on graph: 1, 2, 3, 5

10. f x 4x 5 8x 4 5x 3 10x 2 x 2 11. fx x3 6x2 11x 6 ± ± ± 1 ± 1 ± ± ± ± Possible rational zeros: 1, 2, 2, 4 Possible rational zeros: 1, 2, 3, 6 1 1 Zeros shown on graph: 1, 2, 2, 1, 2 11 6 11 6 1 5 6 1 5 6 0 x3 6x2 11x 6 x 1x2 5x 6 x 1x 2x 3 Thus, the rational zeros are 1, 2, and 3.

12. f x x 3 7x 6 13. gx x3 4x2 x 4 x2x 4 1x 4 Possible rational zeros: ±1, ±2, ±3, ±6 x 4x2 1 310 7 6 x 4x 1x 1 3 9 6 Thus, the rational zeros of gx are 4 and ±1. 1 3 2 0 f x x 3x 2 3x 2 x 3x 2x 1 Thus, the rational zeros are 2, 1, 3.

14. hx x 3 9x 2 20x 12 15. ht t3 12t2 21t 10 Possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12 Possible rational zeros: ±1, ±2, ±5, ±10 119 20 12 11 12 21 10 1 8 12 1 11 10 1 8 12 0 1 11 10 0 hx x 1x 2 8x 12 t3 12t2 21t 10 t 1t2 11t 10 x 1x 2x 6 t 1t 1t 10 Thus, the rational zeros are 1, 2, 6. t 12t 10 Thus, the rational zeros are 1 and 10.

16. px x 3 9x 2 27x 27 17. Cx 2x3 3x2 1 ± ± ± ± ± ± 1 Possible rational zeros: 1, 3, 9, 27 Possible rational zeros: 1, 2 319 27 27 12 3 0 1 3 18 27 2 1 1 2 1 1 0 1 6 9 0 f x x 3x 2 6x 9 2x3 3x2 1 x 12x2 x 1 x 3x 3x 3 x 1x 12x 1 Thus, the rational zero is 3. x 122x 1 1 Thus, the rational zeros are 1 and 2. Section 2.5 Zeros of Polynomial Functions 189

18. f x 3x3 19x 2 33x 9 ± ± ± ± 1 Possible rational zeros: 1, 3, 9, 3 3319 33 9 9 30 9 3 10 3 0 f x x 33x 2 10x 3 x 33x 1x 3 1 Thus, the rational zeros are 3, 3.

19. fx 9x4 9x3 58x2 4x 24 20. f x 2x 4 15x 3 23x 2 15x 25 ± ± ± ± ± ± ± ± ± ± ± ± 1 ± 5 ± 25 Possible rational zeros: 1, 2, 3, 4, 6, 8, 12, 24, Possible rational zeros: 1, 5, 25, 2, 2, 2 ± 1, ± 2, ± 4, ± 8, ± 1, ± 2, ± 4, ± 8 3 3 3 3 9 9 9 9 5215 23 15 25 29 9 58 4 24 10 25 10 25 18 54 8 24 2 5 2 5 0 9 27 4 12 0 125 2 5 39 27 4 12 2 3 5 27 0 12 2 3 5 0 9 0 4 0 123 5 9 x4 9x3 58x2 4x 24 2 5 2 x 2 x 3 9x 4 2 5 0 x 2x 33x 23x 2 f x x 5x 1x 12x 5 ± 2 5 Thus, the rational zeros are 2, 3, and 3. Thus, the rational zeros are 5, 1, 1, 2.

21. z4 z3 2z 4 0 22. x 4 13x 2 12x 0 Possible rational zeros: ±1, ±2, ±4 xx 3 13x 12 0 3 111 0 2 4 Possible rational zeros of x 13x 12: 1 2 2 4 ±1, ±2, ±3, ±4, ±6, ±12 1 2 2 4 0 110 13 12 1 1 12 212 2 4 2 0 4 1 1 12 0 1 0 2 0 xx 1x 2 x 12 0 4 3 2 z z 2z 4 z 1 z 2 z 2 xx 1x 4x 3 0 The only real zeros are 1 and 2. The real zeros are 0, 1, 4, 3.

23. 2y4 7y3 26y2 23y 6 0 ± ± ± ± ± 1 ± 3 Possible rational zeros: 1, 2, 3, 6, 2, 2 127 26 23 6 2 9 17 6 2 9 17 6 0 62 9 17 6 12 18 6 2 3 1 0 2y4 7y3 26y2 23y 6 y 1y 62y2 3y 1 y 1y 62y 1y 1 y 12y 62y 1 1 The only real zeros are 1, 6, and 2. 190 Chapter 2 Polynomial and Rational Functions

24. x 5 x 4 3x 3 5x 2 2x 0 25. fx x3 x2 4x 4 xx 4 x 3 3x 2 5x 2 0 (a) Possible rational zeros: ±1, ±2, ±4 Possible rational zeros of x4 x3 3x2 5x 2: (b) y ±1, ±2 4 111 3 5 2 2 x 1 0 3 2 −6 −4 46 1 0 3 2 0 −4 210 3 2 −6 2 4 2 −8 1 2 1 0 xx 1x 2x 2 2x 1 0 (c) The zeros are: 2, 1, 2 xx 1x 2x 1x 1 0 The real zeros are 2, 0, 1.

26. f x 3x 3 20x 2 36x 16 27. fx 4x3 15x2 8x 3 ± ± ± ± ± ± 1 ± ± ± 1 ± 3 ± 1 ± 3 (a) Possible rational zeros: 1, 2, 4, 8, 16, 3, (a) Possible rational zeros: 1, 3, 2, 2, 4, 4 ± 2 ± 4 ± 8 ± 16 3, 3, 3, 3 (b) y (b) y 10 8 6 4 4 2 2 x x −6 −4 −2246810 −4 −2 6810 12 −4 −4 −6 −6 2 1 (c) Real zeros: 3, 2, 4 (c) The zeros are: 4, 1, 3

28. f x 4x 3 12x 2 x 15 29. fx 2x4 13x3 21x2 2x 8 ± ± ± ± ± 1 ± 3 ± ± ± ± ± 1 (a) Possible rational zeros: 1, 3, 5, 15, 2, 2, (a) Possible rational zeros: 1, 2, 4, 8, 2 ± 5, ± 15, ± 1, ± 3, ± 5, ± 15 2 2 4 4 4 4 (b) 16 (b) y 15

12 −48

−8 1 x (c) The zeros are: 2, 1, 2, 4 −9 −6 −3 6912

3 5 (c) Real zeros: 1, 2, 2

30. f x 4x 4 17x 2 4 31. fx 32x3 52x2 17x 3 1 3 1 3 ± ± ± ± 1 ± 1 (a) Possible rational zeros: ±1, ±3, ± , ± , ± , ± , (a) Possible rational zeros: 1, 2, 4, 2, 4 2 2 4 4 ± 1, ± 3, ± 1 , ± 3 , ± 1 , ± 3 (b) 9 6 8 16 16 32 32 (b) 6

−88

−13 −15 −2 ± ± 1 (c) Real zeros: 2, 2 1 3 (c) The zeros are: 8, 4, 1 Section 2.5 Zeros of Polynomial Functions 191

32. f x 4x 3 7x 2 11x 18 33. fx x4 3x2 2

(a) Possible rational zeros: ±1, ±2, ±3, ±6, ±9, ±18, (a) From the calculator we have x ±1 and x ±1.414. ± 1 ± 3 ± 9 ± 1 ± 3 ± 9 2, 2, 2, 4, 4, 4 (b) An exact zero is x 1.

8 (b) 110 3 0 2 −88 1 1 2 2 1 1 2 2 0 (c) 11 1 2 2 −24 1 0 2 1 0 2 0 1 145 (c) Real zeros: 2, ± 8 8 fx x 1x 1x2 2 x 1x 1x 2 x 2

34. Pt t 4 7t 2 12 35. hx x5 7x4 10x3 14x2 24x (a) t ±2, ±1.732 (a) hx xx4 7x3 10x2 14x 24 (b) 210 7 0 12 From the calculator we have x 0, 3, 4 and x ±1.414. 2 4 6 12 (b) An exact zero is x 3. 1 2 3 6 0 317 10 14 24 21 2 3 6 3 12 6 24 2 0 6 1 4 2 8 0 1 0 3 0 (c) 414 2 8 (c) Pt t 2t 2t2 3 4 0 8 t 2t 2t 3 t 3 1 0 2 0 hx xx 3x 4x2 2 xx 3x 4x 2 x 2

36. gx 6x 4 11x 3 51x 2 99x 27 37. f x x 1x 5ix 5i (a) x ±3, 1.5, 0.333 x 1x2 25 (b) 3611 51 99 27 x3 x2 25x 25 18 21 90 27 Note: f x ax3 x2 25x 25, where a is any 6 7 30 9 0 nonzero real number, has the zeros 1 and ±5i. 367 30 9 18 33 9 6 11 3 0 (c) gx x 3x 36x 2 11x 3 x 3x 33x 12x 3

38. f x x 4x 3ix 3i 39. f x x 6x 5 2ix 5 2i x 4x 2 9 x 6x 5 2ix 5 2i x 3 4x 2 9x 36 x 6x 52 2i2 Note: f x ax 3 4x 2 9x 36, where a is x 6x2 10x 25 4 any real number, has the zeros 4,3i and 3i. x 6x2 10x 29 x3 4x2 31x 174 Note: f x ax3 4x2 31x 174, where a is any nonzero real number, has the zeros 6, and 5 ± 2i. 192 Chapter 2 Polynomial and Rational Functions

40. f x x 2x 4 ix 4 i 41. If 3 2i is a zero, so is its conjugate, 3 2i. x 2x 2 8x 17 f x 3x 2x 1x 3 2ix 3 2i x 3 10x 2 33x 34 3x 2 x 1 x 3 2i x 3 2i 3x2 x 2x 32 2i2 Note: f x ax 3 10x 2 33x 34 where a is any real number, has the zeros 2, 4 ± i. 3x2 x 2x2 6x 9 2 3x2 x 2x2 6x 11 3x4 17x3 25x2 23x 22 Note: f x a3x4 17x3 25x2 23x 22, where a is 2 ± any nonzero real number, has the zeros 3, 1, and 3 2i.

42. If1 3i is a zero, so is its conjugate, 1 3i. 43. f x x4 6x2 27 f x x 52x 1 3ix 1 3i (a) f x x2 9x2 3 x 2 10x 25x 2 2x 4 (b) f x x2 9x 3x 3 x 4 8x 3 9x 2 10x 100 (c) f x x 3ix 3ix 3x 3 Note: f x ax 4 8x 3 9x 2 10x 100, where a is any real number, has the zeros 5, 5, 1 ± 3i.

44. f x x 4 2x 3 3x 2 12x 18 x2 2x 3 x 6 ) x4 2x3 3x2 12x 18 (a) f x x 2 6x 2 2x 3 x4 6x2 (b) f x x 6x 6x 2 2x 3 x4 2x3 3x2 12x 6 (c) f x x 6x 6x 1 2ix 1 2i 2x3 12x 2x3 3x2 18 3x2 18 3x2 0

45. f x x 4 4x 3 5x 2 2x 6 x2 2x 3 x2 2x 2 ) x4 4x3 5x2 2x 6 (a) fx x2 2x 2x2 2x 3 4 3 2 x 2x 2x (b) fx x 1 3 x 1 3 x2 2x 3 x42x3 7x2 2x 6 (c) fx x 1 3 x 1 3 x 1 2 i x 1 2 i 2x3 4x2 4x Note: Use the Quadratic Formula for (b) and (c). 2x3 3x2 6x 6 3x2 6x 6 3x2 6x 0 fx x2 2x 2x2 2x 3

46. fx x4 3x3 x2 12x 20 x2 3 x 5 2 ) 4 3 2 2 2 x 4 x 3x x 12x 20 (a) f x x 4 x 3x 5 x4 4 x2 3 29 3 29 (b) f x x 2 4x x 3x3 5 x2 12 x 2 2 3 3 29 3 29 3x 12x (c) f x x 2ix 2ix x 2 2 5x2 20 5x2 20 0 Section 2.5 Zeros of Polynomial Functions 193

47. fx 2x3 3x2 50x 75 Alternate Solution Since 5i is a zero, so is 5i. Since x ±5i are zeros of fx, x 5ix 5i x2 25 is a factor of fx. By long division we have: 5i 2 3 50 75 10i 50 15i 75 02x 3 2 3 10i 15i 0 x2 0x 25 ) 2x3 3x2 50x 75 5i 2 3 10i 15i 2 x3 0x2 50x 10i 15i 2x3 3x2 50x 75 2 3 0 3 x2 50x 75 3 2 The zero of 2x 3 is x 2. The zeros of f x 3x 50x 70 3 ± are x 2 and x 5i. 2 ± 3 Thus,f x x 25 2x 3 and the zeros of f are x 5i and x 2.

48. f x x 3 x 2 9x 9 Since 3i is a zero, so is 3i. 3i 1 1 9 9 3i 9 3i 9 1 1 3i 3i 0 3i 1 1 3i 3i 3i 3i 1 1 0 The zero of x 1 is x 1. The zeros of f are x 1 and x ±3i.

49. fx 2x4 x3 7x2 4x 4 Alternate Solution Since 2i is a zero, so is 2i. Since x ±2i are zeros of fx, x 2ix 2i x2 4 is a factor 2i 2 1 7 4 4 of f x . By long division we have: 4i 8 2i 4 2i 4 2 x2 x 1 2 1 4i 1 2i 2i 0 x2 0x 4 ) 2x4 x 3 7 x2 4 x 4 2i 2 1 4i 1 2i 2i 2 x4 0 x3 8 x2 4i 2i 2i x3 x 2 4 x 2 1 1 0 x3 0 x2 4 x The zeros of 2x2 x 1 2x 1x 1 2 1 x 0x 4 are x 2 and x 1. The zeros of f x are ± 1 x2 0 x 4 x 2i, x 2, and x 1. 0 Thus, fx x2 42x2 x 1 fx x 2ix 2i2x 1x 1 ± 1 and the zeros of f x are x 2i, x 2, and x 1.

50. gx x 3 7x 2 x 87 Since 5 2i is a zero, so is 5 2i. 5 2i 1 7 1 87 5 2i 14 6i 87 1 2 2i 15 6i 0 5 2i 1 2 2i 15 6i 5 2i 15 6i 1 3 0 The zero of x 3 is x 3. The zeros of f are x 3, 5 ± 2i. 194 Chapter 2 Polynomial and Rational Functions

51. gx 4x3 23x2 34x 10 Alternate Solution Since 3 i is a zero, so is 3 i. Since 3 ± i are zeros of gx, 3 i 4 23 34 10 x 3 i x 3 i x 3 ix 3 i 2 2 12 4i 37 i 10 x 3 i 2 4 11 4i 3 i 0 x 6x 10 is a factor of gx. By long division we have: 3 i 4 11 4i 3 i 34x 1 12 4i 3 i x2 6x 10 ) 4x3 23x2 34x 10 4 1 0 4 x3 24x2 40x The zero of 4x 1 is x 1. The zeros of gx are 4 4x324 x2 36x 10 x 3 ± i and x 1. 4 x2 36x 10 0 Thus,gx x2 6x 104x 1 and the zeros of gx ± 1 are x 3 i and x 4.

52. hx 3x3 4x 2 8x 8 Since 1 3 i is a zero, so is 1 3i. 1 3i 3 4 8 8 3 33i 10 23i 8 3 1 33i 2 23i 0

1 3i 3 1 33i 2 23i 3 33i 2 23i 3 2 0 2 2 ± The zero of 3x 2 is x 3. The zeros of f are x 3, 1 3i.

53. fx x4 3x3 5x2 21x 22 Since 3 2 i is a zero, so is 3 2 i, and x 3 2 i x 3 2 i x 3 2 ix 3 2 i 2 x 32 2 i x2 6x 11 is a factor of fx. By long division, we have: x2 3x 2 x2 6x 11 ) x4 3x3 5x2 21x 22 x4 6x3 11x2 3x3 16x2 21x 3x3 18x2 33x 2x2 12x 22 2x2 12x 22 0 Thus, f x x2 6x 11x2 3x 2 x2 6x 11x 1x 2 and the zeros of f are x 3 ± 2 i, x 1, and x 2. Section 2.5 Zeros of Polynomial Functions 195

54. f x x 3 4x 2 14x 20 55. fx x2 25 Since 1 3i is a zero, so is 1 3i. x 5ix 5i 1 3i 1 4 14 20 The zeros of fx are x ±5i. 1 3i 12 6i 20 1 3 3i 2 6i 0 1 3i 1 3 3i 2 6i 1 3i 2 6i 1 2 0 The zero of x 2 is x 2. The zeros of f are x 2, 1 ± 3i.

56. f x x 2 x 56 57. hx x2 4x 1 By the Quadratic Formula, the zeros of f x are By the Quadratic Formula, the zeros of hx are 1 ± 1 224 1 ± 223i 4 ± 16 4 x . x 2 ± 3. 2 2 2 1 223i 1 223i hx x 2 3 x 2 3 f x x x 2 2 x 2 3 x 2 3

58. gx x 2 10x 23 59. fx x4 81 By the Quadratic Formula, the zeros of f x are x2 9x2 9 10 ± 100 92 10 ± 8 x 3x 3x 3ix 3i x 5 ± 2. 2 2 The zeros of fx are x ±3 and x ±3i. gx x 5 2x 5 2

60. f y y 4 625 61. fz z2 2z 2 y2 25y2 25 By the Quadratic Formula, the zeros of f z are Zeros: y ±5, ±5i 2 ± 4 8 z 1 ± i. f y y 5y 5y 5iy 5i 2 fz z 1 iz 1 i z 1 iz 1 i

62. hx x 3 3x 2 4x 2 63. gx x3 6x2 13x 10 Possible rational zeros: ±1, ±2 Possible rational zeros: ±1, ±2, ±5, ±10 113 4 2 216 13 10 1 2 2 2 8 10 1 4 5 0 1 2 2 0 By the Quadratic Formula, the zeros of x2 2x 2 By the Quadratic Formula, the zeros of x2 4x 5 are 2 ± 4 8 4 ± 16 20 are x 1 ± i. x 2 ± i. 2 2 Zeros: x 1, 1 ± i The zeros of gx are x 2 and x 2 ± i. hx x 1x 1 ix 1 i gx x 2x 2 ix 2 i x 2x 2 ix 2 i 196 Chapter 2 Polynomial and Rational Functions

64. f x x 3 2x 2 11x 52 Possible rational zeros: ±1, ±2, ±4, ±13, ±26 412 11 52 4 24 52 1 6 13 0

6 ± 36 52 By the Quadratic Formula, the zeros of x2 6x 13 are x 3 ± 2i. 2 Zeros: x 4, 3 ± 2i f x x 4x 3 2ix 3 2i

65. hx x3 x 6 66. hx x3 9x2 27x 35 Possible rational zeros: ±1, ±2, ±3, ±6 Possible rational zeros: ±1, ±5, ±7, ±35 21 0 1 6 51 9 27 35 2 4 6 5 20 35 1 2 3 0 1 4 7 0 By the Quadratic Formula, the zeros of x2 2x 3 are By the Quadratic Formula, the zeros of x2 4x 7 2 ± 4 12 4 ± 16 28 x 1 ± 2 i. are x 2 ± 3i. 2 2 The zeros of hx are x 2 and x 1 ± 2 i. Zeros: 5, 2 ± 3i hx x 2x 1 2 i x 1 2 i hx x 5x 2 3ix 2 3i x 2x 1 2 ix 1 2 i

67. fx 5x3 9x2 28x 6 68. gx 3x 3 4x 2 8x 8 Possible rational zeros: Possible rational zeros: ± ± ± ± ± 1 ± 2 ± 3 ± 6 ± ± ± ± ± 1 ± 2 ± 4 ± 8 1, 2, 3, 6, 5, 5, 5, 5 1, 2, 4, 8, 3, 3, 3, 3 1 2 5 5 9 28 6 3 3 4 8 8 1 2 6 2 4 8 5 10 30 0 3 6 12 0 By the Quadratic Formula, the zeros of By the Quadratic Formula, the zeros of 5x2 10x 30 5x2 2x 6 are 3x2 6x 12 3x2 2x 4 are 2 ± 4 24 2 ± 4 16 x 1 ± 5 i. x 1 ± 3i. 2 2 1 ± Zeros: x 2, 1 ± 3i The zeros of f x are x 5 and x 1 5 i. 3 1 gx 3x 2x 1 3ix 1 3i f x x 5 5 x 1 5 i x 1 5 i 5x 1x 1 5 ix 1 5 i Section 2.5 Zeros of Polynomial Functions 197

69. gx x4 4x3 8x2 16x 16 70. hx x 4 6x 3 10x 2 6x 9 Possible rational zeros: ±1, ±2, ±4, ±8, ±16 Possible rational zeros: ±1, ±3, ±9 214 8 16 16 31 6 10 6 9 2 4 8 16 3 9 3 9 1 2 4 8 0 1 3 1 3 0 212 4 8 31 3 1 3 2 0 8 1 0 4 0 3 0 3 gx x 2x 2x2 4 x 22x 2ix 2i 1 0 1 0 2 ± The zeros of gx are 2 and ±2i. The zeros of x 1 are x i. Zeros: x 3, ±i hx x 3 2x ix i

71. fx x4 10x2 9 72. f x x 4 29x 2 100 x2 1x2 9 x 2 25x 2 4 x ix ix 3ix 3i Zeros: x ±2i, ±5i The zeros of fx are x ±i and x ±3i. f x x 2ix 2ix 5ix 5i

73. fx x3 24x2 214x 740 74. f s 2s 3 5s 2 12s 5 ± ± ± ± ± ± ± ± ± ± 1 ± 5 Possible rational zeros: 1, 2, 4, 5, 10, 20, 37, Possible rational zeros: 1, 5, 2, 2 ±74, ±148, ±185, ±370, ±740 10

2000

−10 10

−20 10 −10

−1000 1 Based on the graph, try s 2. Based on the graph, try x 10. 1 2 2 5 12 5 10 1 24 214 740 1 2 5 10 140 740 2 4 10 0 1 14 74 0 By the Quadratic Formula, the zeros of 2s2 2s 5 are By the Quadratic Formula, the zeros of x2 14x 74 are 2 ± 4 20 14 ± 196 296 s 1 ± 2i. x 7 ± 5i. 2 2 1 ± The zeros of fx are x 10 and x 7 ± 5i. The zeros of f s are s 2 and s 1 2i. 198 Chapter 2 Polynomial and Rational Functions

75. fx 16x3 20x2 4x 15 76. f x 9x 3 15x 2 11x 5 ± ± ± 1 ± 5 ± 1 ± 5 Possible rational zeros: Possible rational zeros: 1, 5, 3, 3, 9, 9 ± ± ± ± ± 1 ± 3 ± 5 ± 15 ± 1 ± 3 1, 3, 5, 15, 2, 2, 2, 2 , 4, 4, 5 ± 5 ± 15 ± 1 ± 3 ± 5 ± 15 ± 1 ± 3 ± 5 ± 15 4, 4 , 8, 8, 8, 8 , 16, 16, 16, 16

20 −55

−5

− 3 3 Based on the graph, try x 1. −5 1915 11 5 3 9 6 5 Based on the graph, try x 4. 3 9 6 5 0 4 16 20 4 15 12 24 15 By the Quadratic Formula, the zeros of 9x2 6x 5 are 16 32 20 0 6 ± 36 180 1 2 x ± i. By the Quadratic Formula, the zeros of 18 3 3 16x2 32x 20 44x2 8x 5 are 1 ± 2 The zeros of f x are x 1 and x 3 3i. 8 ± 64 80 1 x 1 ± i. 8 2 3 ± 1 The zeros of f x are x 4 and x 1 2i.

4 3 2 1 77. f x 2x 5x 4x 5x 2 Based on the graph, try x 2 and x 2. ± ± ± 1 Possible rational zeros: 1, 2, 2 22 5 4 5 2 4 2 4 2 20 2 1 2 1 0 1 2 2 1 2 1 1 0 1 −4 4 2 0 2 0 −5 The zeros of 2x2 2 2x2 1 are x ±i. 1 ± The zeros of f x are x 2, x 2, and x i.

78. gx x 5 8x 4 28x 3 56x 2 64x 32 218 28 56 64 32 2 12 32 48 32 Possible rational zeros: ±1, ±2, ±4, ±8, ±16, ±32 1 6 16 24 16 0 10 216 16 24 16 −10 10 2 8 16 16 1 4 8 8 0

− 10 214 8 8 Based on the graph, try x 2. 2 4 8 1 2 4 0

By the Quadratic Formula, the zeros of x2 2x 4 are 2 ± 4 16 x 1 ± 3i. 2 The zeros of gx are x 2 and x 1 ± 3i. Section 2.5 Zeros of Polynomial Functions 199

79. gx 5x5 10x 5xx4 2 80. hx 4x 2 8x 3 Let fx x4 2. Sign variations: 2, positive zeros: 2 or 0 Sign variations: 0, positive zeros: 0 hx 4x 2 8x 3 fx x4 2 Sign variations: 0, negative zeros: 0 Sign variations: 0, negative zeros: 0

81. hx 3x4 2x2 1 82. hx 2x 4 3x 2 Sign variations: 0, positive zeros: 0 Sign variations: 2, positive zeros: 2 or 0 hx 3x4 2x2 1 hx 2x 4 3x 2 Sign variations: 0, negative zeros: 0 Sign variations: 0, negative zeros: 0

83. gx 2x3 3x2 3 84. f x 4x 3 3x 2 2x 1 Sign variations: 1, positive zeros: 1 Sign variations: 3, positive zeros: 3 or 1 gx 2x3 3x2 3 f x 4x 3 3x 2 2x 1 Sign variations: 0, negative zeros: 0 Sign variations: 0, negative zeros: 0

85. fx 5x3 x2 x 5 86. f x 3x 3 2x 2 x 3 Sign variations: 3, positive zeros: 3 or 1 Sign variations: 0, positive zeros: 0 fx 5x3 x2 x 5 f x 3x 3 2x 2 x 3 Sign variations: 0, negative zeros: 0 Sign variations: 3, negative zeros: 3 or 1

87. fx x4 4x3 15 88. fx 2x3 3x 2 12x 8 (a) 414 0 0 15 (a) 423 12 8 4 0 0 0 8 20 32 1 0 0 0 15 2 5 8 40 4 is an upper bound. 4 is an upper bound. (b) 114 0 0 15 (b) 323 12 8 1 5 5 5 6 27 45 1 5 5 5 20 2 9 15 37 1 is a lower bound. 3 is a lower bound.

89. fx x4 4x3 16x 16 90. f x 2x 4 8x 3 (a) 514 0 16 16 (a) 320 0 8 3 5 5 25 205 6 18 54 138 1 1 5 41 189 2 6 18 46 141 5 is an upper bound. 3 is an upper bound. (b) 314 0 16 16 (b) 420 0 8 3 3 21 63 141 8 32 128 544 1 7 21 47 125 2 8 32 136 547 3 is a lower bound. 3 is a lower bound. 200 Chapter 2 Polynomial and Rational Functions

91. fx 4x3 3x 1 92. f z 12z 3 4z2 27z 9 ± ± 1 ± 1 ± ± ± ± 1 ± 3 ± 9 ± 1 Possible rational zeros: 1, 2, 4 Possible rational zeros: 1, 3, 9, 2, 2, 2, 3, ± 1 ± 3 ± 9 ± 1 ± 1 140 3 1 4, 4, 4, 6, 12 3 4 4 1 2 12 4 27 9 4 4 1 0 18 21 9 4 x3 3x 1 x 14x2 4x 1 12 14 6 0

x 12x 12 3 2 f z 2 z 2 6z 7z 3 1 Thus, the zeros are 1 and 2. 2z 33z 12z 3 3 1 3 Real zeros: 2, 3, 2

93. fy 4y3 3y2 8y 6 94. gx 3x3 2x 2 15x 10 Possible rational zeros: ±1, ±2, ±3, ±6, ± 1, ± 3, ± 1, ± 3 ± ± ± ± ± 1 ± 2 ± 5 ± 10 2 2 4 4 Possible rational zeros: 1, 2, 5, 10, 3, 3, 3, 3 3 4 3 8 6 2 4 3 3 2 15 10 3 0 6 2 0 10 4 0 8 0 3 0 15 0 3 2 3 2 4 y 3y 8y 6 y 4 4y 8 2 2 2 g x x 3 3x 15 3x 2 x 5 3 2 y 4 4 y 2 2 Thus, the only real zero is 3. 4y 3y2 2 3 Thus, the only real zero is 4.

4 25 2 1 3 2 95. P x x 4 x 9 96. f x 2 2x 3x 23x 12 1 4 2 ± ± ± ± ± ± ± 1 ± 3 4 4x 25x 36 Possible rational zeros: 1, 2, 3, 4, 6, 12, 2, 2 1 2 2 4 4x 9 x 4 423 23 12 1 8 20 12 4 2x 3 2x 3 x 2 x 2 2 5 3 0 ± 3 ± The rational zeros are 2 and 2. 1 2 1 f x 2 x 4 2x 5x 3 2 x 4 2x 1 x 3 1 The rational zeros are 3, 2, and 4.

1 1 97. f x x3 x2 x 1 3 2 4 4 98. f z 6 6z 11z 3z 2 1 4x3 x2 4x 1 ± ± ± 1 ± 1 ± 2 ± 1 4 Possible rational zeros: 1, 2, 2, 3, 3, 6 1 2 4 x 4x 1 1 4x 1 2611 3 2 1 2 12 2 2 4 4x 1 x 1 1 6 1 1 0 4 4x 1 x 1 x 1 f x 1x 26x 2 x 1 1 ± 6 The rational zeros are 4 and 1. 1 6 x 2 3x 1 2x 1 1 1 Rational zeros: 2, 3, 2 Section 2.5 Zeros of Polynomial Functions 201

99. fx x3 1 x 1x2 x 1 100. f x x 3 2 Rational zeros: 1 x 1 x 3 2x2 3 2x 3 4 Irrational zeros: 0 Rational zeros: 0 Matches (d). Irrational zeros: 1 x 3 2 Matches (a).

101. f x x3 x xx 1x 1 102. f x x 3 2x Rational zeros: 3 x 0, ±1 xx 2 2 Irrational zeros: 0 xx 2x 2 Matches (b). Rational zeros: 1 x 0 Irrational zeros: 2 x ±2 Matches (c).

103. (a) 15 (b) V l w h 15 2x9 2xx

x 9 x9 2x15 2x 9 − − 2x x 2x 15 Since length, width, and height must be positive, x 9 we have 0 < x < 2 for the domain. (c) V (d) 56 x9 2x15 2x 125 2 3 100 56 135x 48x 4x 75 50 4x3 48x2 135x 56 50

olume of box 1 7 V 25 The zeros of this polynomial are 2, 2, and 8. x x cannot equal 8 since it is not in the domain of V. 13425 Length of sides of [The length cannot equal 1 and the width cannot squares removed equal 7 . The product of 817 56 so it The volume is maximum when x 1.82. showed up as an extraneous solution.] The dimensions are: length 15 21.82 11.36 Thus, the volume is 56 cubic centimeters when x 1 centimeter or x 7 centimeters. width 9 21.82 5.36 2 2 height x 1.82 1.82 cm 5.36 cm 11.36 cm

104. (a) Combined length and width: (c) 13,500 4x 230 x 4x y 120 ⇒ y 120 4x 4 x 3 120x 2 13,500 0 Volume l w h x 2y x 3 30x 2 3375 0 x 2120 4x 15 1 30 0 3375 4x 230 x 15 225 3375 1 15 225 0 (b) 18,000 x 15x 2 15x 225 0 15 ± 155 Using the Quadratic Formula, x 15, . 2 030 0 15 155 The value of is not possible because Dimensions with maximum volume: 2 it is negative. 20 in. 20 in. 40 in. 202 Chapter 2 Polynomial and Rational Functions

105. P 76x3 4830x2 320,000, 0 ≤ x ≤ 60 2,500,000 76x3 4830x2 320,000 76x3 4830x2 2,820,000 0 The zeros of this equation are x 46.1, x 38.4, and x 21.0. Since 0 ≤ x ≤ 60, we disregard x 21.0. The smaller remaining solution is x 38.4. The advertising expense is $384,000.

106. P 45x 3 2500x 2 275,000 107. (a) Current bin:V 2 3 4 24 cubic feet 800,000 45x 3 2500x 2 275,000 New bin:V 524 120 cubic feet 0 45x 3 2500x 2 1,075,000 V 2 x3 x4 x 120 0 9x 3 500x 2 215,000 (b) x3 9x2 26x 24 120 The zeros of this equation are x 18.0, x 31.5, and x3 9x2 26x 96 0 x 42.0. Because 0 ≤ x ≤ 50, disregard x 18.02. The only real zero of this polynomial is x 2. All the The smaller remaining solution is x 31.5, or an advertising expense of $315,000. dimensions should be increased by 2 feet, so the new bin will have dimensions of 4 feet by 5 feet by 6 feet.

108. (a) A 250 x160 x 1.5160250 (c) A 250 2x160 x 60,000 60,000 2 x2 570x 20,000 0 (b) 60,000 x2 410x 40,000 570 ± 5702 4220,000 x 22 0 x2 410x 20,000 x must be positive, so 410 ± 4102 4120,000 x 21 570 484,900 x 31.6. 4 410 ± 248,100 2 The new length is 250 231.6 313.2 ft and the new width is 160 31.6 191.6 ft, so the new dimensions x must be positive, so are 191.6 ft 313.2 ft. 410 248,100 x 2 44.05. The new length is 250 44.05 294.05 ft and the new width is 160 44.05 204.05 ft, so the new dimensions are 204.05 ft 294.05 ft.

200 x 110. h(t 16t2 48t 6 109. C 100 , x ≥ 1 x2 x 30 Let h 64 and solve for t. C is minimum when 3x3 40x2 2400x 36000 0. 64 16t2 48t 6 The only real zero is x 40 or 4000 units. 16t2 48t 58 0 48 ± i1408 By the Quadratic Formula we have t . 32 Since the equation yields only imaginary zeros, it is not possible for the ball to have reached a height of 64 feet. Section 2.5 Zeros of Polynomial Functions 203

111. P R C xp C 112. (a) A 0.0167t3 0.508t2 5.60t 13.4

x140 0.0001x 80x 150,000 (b)12 The model is a good fit to the actual data. 0.0001x 2 60x 150,000 9,000,000 0.0001x2 60x 150,000

2 7 13 Thus, 0 0.0001x 60x 9,150,000. 0 60 ± 60 x 300,000 ± 10,00015i (c)A 8.5 when t 10 which corresponds to the 0.0002 year 2000. Since the solutions are both complex, it is not possible (d)A 9 when t 11 which corresponds to the to determine a price p that would yield a profit of year 2001. 9 million dollars. (e) Yes. The degree of A is odd and the leading coefficient is positive, so as x increases, A will increase. This implies that attendance will continue to grow.

113. False. The most nonreal complex zeros it can have is 114. False. f does not have real coefficients. two and the Linear Factorization Theorem guarantees that there are 3 linear factors, so one zero must be real.

115. gx fx. This function would have the same zeros 116. gx 3f x. This function has the same zeros as f as f x so r1, r2, and r3 are also zeros of g x . because it is a vertical stretch of f. The zeros of g are r1, r2, and r3.

117. gx fx 5. The graph of gx is a horizontal shift 118. gx f 2x. Note that x is a zero of g if and only if 2x r r r of the graph of fx five units to the right so the zeros is a zero of f. The zeros of g are 1, 2, and 3. 2 2 2 of g x are 5 r1, 5 r2, and 5 r3.

119. gx 3 fx. Since gx is a vertical shift of the graph 120. gx f x. Note that x is a zero of g if and only if of f x , the zeros of g x cannot be determined. x is a zero of f. The zeros of g are r1, r2, and r3.

121. fx x4 4x2 k 4 ± 42 41k 4 ± 24 k x2 2 ± 4 k 21 2 x ±2 ± 4 k (a) For there to be four distinct real roots, both 4 k and 2 ± 4 k must be positive. This occurs 1 when 0 < k < 4. Thus, some possible k-values are k 1, k 2, k 3, k 2, k 2, etc. (b) For there to be two real roots, each of multiplicity 2, 4 k must equal zero. Thus, k 4.

(c) For there to be two real zeros and two complex zeros,2 4 k must be positive and 2 4 k 1 must be negative. This occurs when k < 0. Thus, some possible k-values are k 1, k 2, k 2, etc. (d) For there to be four complex zeros,2 ± 4 k must be nonreal. This occurs when k > 4. Some possible k-values are k 5, k 6, k 7.4, etc.

122. (a) gx f x 2 (b) gx f 2x No. This function is a horizontal shift of f x. No. Since x is a zero of g if and only if 2x is a zero of f, Note that x is a zero of g if and only if x 2 is a the number of real and complex zeros of g is the same as zero of f; the number of real and complex zeros is the number of real and complex zeros of f. not affected by a horizontal shift. 204 Chapter 2 Polynomial and Rational Functions

1 y 123. Zeros: 2, 2, 3 124. 50 f x x 22x 1x 3 (−1, 0) 2x3 3x2 11x 6

y 10 (1, 0) (4, 0) x (3, 0) 45 8 1, 0 (2 ( 4 (−2, 0) (3, 0) x −8 −4 4 812

Any nonzero scalar multiple of f would have the same three zeros. Let gx af x, a > 0. There are infinitely many possible functions for f.

125. Answers will vary. Some of the factoring techniques are: 126. (a) Zeros of fx: 2, 1, 4 1. Factor out the greatest common factor. (b) The graph touches the x-axis at x 1 2. Use special product formulas. (c) The least possible degree of the function is 4 because a2 b2 a ba b there are at least 4 real zeros (1 is repeated) and a function can have at most the number of real zeros a2 2ab b2 a b2 equal to the degree of the function. The degree a2 2ab b2 a b2 cannot be odd by the definition of multiplicity. a3 b3 a ba2 ab b2 (d) The leading coefficient of f is positive. From the information in the table, you can conclude that the a3 b3 a ba2 ab b2 graph will eventually rise to the left and to the right. 3. Factor by grouping, if possible. (e) Answers may vary. One possibility is: 4. Factor general trinomials with binomial factors fx x 12x 2x 4 by “guess-and-test” or by the grouping method. x 12x 2x 4 5. Use the Rational Zero Test together with synthetic division to factor a polynomial. x2 2x 1x2 2x 8 6. Use Descartes’s Rule of Signs to determine the x 4 4x3 3x2 14x 8 number of real zeros. Then find any zeros and (f) y use them to factor the polynomial. (−2, 0) 2 (1, 0) (4, 0) 7. Find any upper and lower bounds for the real zeros x −32−1 35 to eliminate some of the possible rational zeros. −4 Then test the remaining candidates by synthetic −6 −8 division and use any zeros to factor the polynomial. −10

127. (a) fx x bix bi x2 b 128. (a)f x cannot have this graph since it also has a zero at x 0. (b) fx x a bix a bi (b)gx cannot have this graph since it is a quadratic x a bi x a bi function. Its graph is a parabola. x a2 bi2 (c)hx is the correct function. It has two real zeros, x 2 and x 3.5, and it has a degree of four, x2 2ax a2 b2 needed to yield three turning points. (d)k x cannot have this graph since it also has a zero at x 1. In addition, since it is only of degree three, it would have at most two turning points. Section 2.6 Rational Functions 205

129. 3 6i 8 3i 3 6i 8 3i 11 9i 130. 12 5i 16i 12 11i

131. 6 2i1 7i 6 42i 2i 14i2 20 40i 132. 9 5i9 5i 81 25i2 81 25 106

133. gx fx 2 134. gx f x 2 135. gx 2fx

y y y (6, 4) 4 3 10 (4, 2) (4, 8) 3 2 8 (2, 2) 2 6 (4, 2) (0, 0) (2, 0) 1 x (0, 4) −24−1 1 23 (2, 4) x (0, 0) 23456 −1 −2 x (−2, −2) (−2, 0) 2684 −2 −3 −2

Horizontal shift two units Vertical shift two units downward Vertical stretch (each y-value is to the right multiplied by 2)

1 136. g x f x 137. g x f 2x 138. g x f 2x

y y y (−4, 4) 4 (2, 4) 4 10 3 8 (0, 2) 3 6 (−2, 2) (8, 4) 1 (0, 2) 4 (2, 0) (1, 2) (4, 2) x (0, 2) −4 −3 −2 −1 1 2 −1 x − (−4, 0) 2684 −2 ( 1, 0) −2 x −212−1 Reflection in the y-axis Horizontal stretch each x-value Horizontal shrink each x-value is 1 is multiplied by 2 multiplied by 2

Section 2.6 Rational Functions

■ You should know the following basic facts about rational functions. (a) A function of the form f x NxDx, Dx 0, where Nx and Dx are polynomials, is called a rational function. (b) The domain of a rational function is the set of all real numbers except those which make the denominator zero. (c) If f x NxDx is in reduced form, and a is a value such that Da 0, then the line x a is a vertical asymptote of the graph of f. fx→ or fx→ as x→a. (d) The line y b is a horizontal asymptote of the graph of f if fx → b as x → or x → . Nx a xn a xn1 . . . a x a (e) Let fx n n1 1 0 where Nx and Dx have no common factors. m m1 . . . D x bmx bm1x b1x b0

1. If n < m, then the x-axis y 0 is a horizontal asymptote. a 2. If n m, then y n is a horizontal asymptote. bm 3. If n > m, then there are no horizontal asymptotes.

Vocabulary Check 1. rational functions 2. vertical asymptote 3. horizontal asymptote 4. slant asymptote 206 Chapter 2 Polynomial and Rational Functions

1 1. fx x 1 (a) (b) The zero of the denominator is x 1, so x 1 is a x f x x f x x f x vertical asymptote. The degree of the numerator is less 0.5 2 1.5 2 5 0.25 than the degree of the denominator so the x-axis, or y 0, is a horizontal asymptote. 0.9 10 1.1 10 10 0.1 (c) The domain is all real numbers except x 1. 0.99 100 1.01 100 100 0.01 0.999 1000 1.001 1000 1000 0.001

5x 2. f x x 1 (a) (b) The zero of the denominator is x 1, so x 1 is a x f x x f x x f x vertical asymptote. The degree of the numerator is equal 5 0.5 5 1.5 15 5 6.25 to the degree of the denominator, so the line y 1 5 is a horizontal asymptote. 0.9 45 1.1 55 10 5.55 (c) The domain is all real numbers except x 1. 0.99 495 1.01 505 100 5.05 0.999 4995 1.001 5005 1000 5.005

3x2 3. fx x2 1 (a) (b) The zeros of the denominator are x ±1 so both x f x x f x x f x x 1 and x 1 are vertical asymptotes. Since 0.5 1 1.5 5.4 5 3.125 the degree of the numerator equals the degree 3 of the denominator,y 1 3 is a horizontal 0.9 12.79 1.1 17.29 10 3.03 asymptote. 0.99 147.8 1.01 152.3 100 3.0003 (c) The domain is all real numbers except x ±1. 0.999 1498 1.001 1502 1000 3

4x 4. f x x2 1 (a) (b) The zeros of the denominator are x ±1 so both x f x x f x x f x x 1 and x 1 are vertical asymptotes. 0.5 2.66 1.5 4.8 5 0.833 Because the degree of the numerator is less than the degree of the denominator, the x-axis or y 0 0.9 18.95 1.1 20.95 10 0.40 is a horizontal asymptote. 0.99 199 1.01 201 100 0.04 (c) The domain is all real numbers except x ±1. 0.999 1999 1.001 2001 1000 0.004

1 4 5. fx 6. fx x2 x 23 Domain: all real numbers except x 0 Domain: all real numbers except x 2 Vertical asymptote: x 0 Vertical asymptote: x 2 Horizontal asymptote: y 0 Horizontal asymptote: y 0

Degree of Nx < degree of Dx Degree of Nx < degree of Dx Section 2.6 Rational Functions 207

2 x x 2 1 5x 5x 1 7. fx 8. fx 2 x x 2 1 2x 2x 1 Domain: all real numbers except x 2 1 Domain: all real numbers except x 2 Vertical asymptote: x 2 1 Horizontal asymptote: y 1 Vertical asymptote: x 2 Degree of Nx degree of Dx 5 Horizontal asymptote: y 2 Degree of Nx degree of Dx

x3 2x2 9. fx 10. fx x2 1 x 1 Domain: all real numbers except x ±1 Domain: all real numbers except x 1 Vertical asymptotes: x ±1 Vertical asymptote: x 1 Horizontal asymptote: None Horizontal asymptote: None

Degree of Nx > degree of Dx Degree of Nx > degree of Dx

3x2 1 3x2 x 5 11. fx 12. fx x2 x 9 x2 1 Domain: All real numbers. The denominator has no real Domain: All real numbers. The denominator has zeros. [Try the Quadratic Formula on the no real zeros. [Try the Quadratic Formula denominator.] on the denominator.] Vertical asymptote: None Vertical asymptote: None Horizontal asymptote: y 3 Horizontal asymptote: y 3 Degree of Nx degree of Dx Degree of Nx degree of Dx

2 1 x 1 13. fx 14. fx 15. fx x 3 x 5 x 4 Vertical asymptote: y 3 Vertical asymptote: x 5 Vertical asymptote: x 4 Horizontal asymptote: y 0 Horizontal asymptote: y 0 Horizontal asymptote: y 1 Matches graph (d). Matches graph (a). Matches graph (c).

x 2 x2 1 x 1x 1 5 16. f x 17. gx 18. hx 2 x 4 x 1 x 1 x2 2 Vertical asymptote: x 4 The only zero of gx is x 1. 5 0 2 x2 2 Horizontal asymptote: y 1 x 1 makes gx undefined. 5 Matches graph (b). 2 x2 2 2x2 2 5 5 x2 2 2 No real solution,hx has no real zeros. 208 Chapter 2 Polynomial and Rational Functions

3 x3 8 19. f x 1 20. gx x 3 x2 1 3 x3 8 1 0 0 x 3 x2 1

3 3 1 x 8 0 x 3 x3 8 x 3 3 x 2 x 6 is a zero of f x. x 2 is a real zero of gx.

x 4 1 x 3 x 3 1 21. f x , x 4 22. f x , x 3 x2 16 x 4 x2 9 x 3x 3 x 3 Domain: all real numbers x except x ±4 Domain: all real numbers x except x ±3 Horizontal asymptote: y 0 The degree of the numerator is less than the degree of the denominator, so the graph has the line y 0 as a Degree of N x < degree of D x horizontal asymptote. Vertical asymptote:x 4 Since x 4 is a common Vertical asymptote:x 3 Since x 3 is a common factor of Nx and Dx, x 4 is not a vertical asymptote factor of Nx and Dx, x 3 is not a vertical asymp- of f x. tote of f x.

x2 1 x 1x 1 x 1 x2 4 23. f x , x 1 24. f x x2 2x 3 x 1x 3 x 3 x2 3x 2 Domain: all real numbers x except x 1 and x 3 x 2x 2 x 2 , x 2 x 2x 1 x 1 Horizontal asymptote: y 1 Domain: all real numbers x except x 1 and x 2 Degree of Nx degree of Dx Horizontal asymptote: y 1 Vertical asymptote: x 3 Since x 1 is a common factor of Nx and Dx, x 1 Degree of Nx degree of Dx is not a vertical asymptote of f x. Vertical asymptote:x 1 Since x 2 is a common factor of Nx and Dx, x 2 is not a vertical asymptote of f x.

x2 3x 4 6x2 11x 3 25. f x 26. f x 2x2 x 1 6x2 7x 3 x 1x 4 x 4 2x 33x 1 3x 1 3 , x 1 , x 2x 1x 1 2x 1 2x 33x 1 3x 1 2 1 3 1 Domain: all real numbers x except x and x 1 Domain: all real numbers x except x or x 2 2 3 1 Horizontal asymptote: y Horizontal asymptote: y 1 2 Degree of Nx degree of Dx Degree of Nx degree of Dx 1 Vertical asymptote:x 3 Since 2x 3 is a common Vertical asymptote:x 1 Since x 1 is a common 3 2 factor of N x and D x , x 2 is not a vertical asymptote factor of Nx and Dx, x 1 is not a vertical asymp- of f x . tote of f x. Section 2.6 Rational Functions 209

1 1 27. fx 28. f x x 2 x 3 (a) Domain: all real numbers x except x 2 (a) Domain: all real numbers x except x 3 1 1 (b) y-intercept: 0, (b)y- intercept: 0, 2 3 (c) Vertical asymptote: x 2 (c) Vertical asymptote: x 3 Horizontal asymptote: y 0 Horizontal asymptote: y 0 (d) (d) x 4 3 1 01 x 012456 1 1 1 1 1 1 1 y 2 1 1 2 3 y 3 2 11 2 3

y y

2 3

2 1 1 (0, ( 2 1

x x − − 3 1 2 456 − −1 1 − 1 −2 (0, 3 ( −2 −3

1 1 1 29. hx 30. gx x 2 3 x x 3 (a) Domain: all real numbers x except x 2 (a) Domain: all real numbers x except x 3 1 1 (b) y-intercept: 0, (b) y-intercept: 0, 2 3 (c) Vertical asymptote: x 2 (c) Vertical asymptote: x 3 Horizontal asymptote: y 0 Horizontal asymptote: y 0 (d) (d) x 4 3 1 0 x 0124 5 6 y 1 1 1 1 1 1 1 1 2 2 y 3 2 1 1 2 3

y y

2 3 2 1 1 (0, 3 ( 0, − 1 1 ( 2 ( x x −4 −3 −1 12 4 −1 −1 −2 − 2 −3

1 1 Note: This is the graph of fx Note: This is the graph of fx x 2 x 3 (Exercise 27) reflected about the x-axis. (Exercise 28) reflected about the x-axis. 210 Chapter 2 Polynomial and Rational Functions

5 2x 2x 5 1 3x 3x 1 31. Cx 32. Px 1 x x 1 1 x x 1 (a) Domain: all real numbers x except x 1 (a) Domain: all real numbers x except x 1 5 1 (b) x-intercept: , 0 (b) x-intercept: , 0 2 3 y-intercept: 0, 5 y-intercept: 0, 1 (c) Vertical asymptote: x 1 (c) Vertical asymptote: x 1 Horizontal asymptote: y 2 Horizontal asymptote: y 3 (d) (d) x 4 3 2 012 x 1023 1 7 y 2154 C x 1532 1 2

y y

5 6 (0, 5) 4 − 5, 0 ( 2 (

x − − (0, 1) 1, 0 6 4 24 ()3 x −2 −1 234

x2 1 2t 2t 1 33. fx 34. ft x2 9 t t (a) Domain: all real numbers x (a) Domain: all real numbers t except t 0 (b) Intercept: 0, 0 1 (b) t-intercept: , 0 (c) Horizontal asymptote: y 1 2 (d) (c) Vertical asymptote: t 0 x ±1 ±2 ±3 Horizontal asymptote: y 2 y 1 4 1 (d) 10 13 2 1 t 2 1122 y 5 3 y 2 30 1 2

3 y 1 , 0 2 ( 2 ) t −2 −112 − (0, 0) 1 x −2 −112

−1 −3 Section 2.6 Rational Functions 211

s 1 35. gs 36. fx s2 1 x 22 (a) Domain: all real numbers s (a) Domain: all real numbers x except x 2 (b) Intercept: 0, 0 1 (b)y- intercept: 0, (c) Horizontal asymptote: y 0 4 (d) (c) Vertical asymptote: x 2 s 2 1 012 Horizontal asymptote: y 0 2 1 1 2 g s 5 2 0 2 5 (d)

1 3 5 7 y x 012 2 2 342 1 4 4 1 2 y 4 9 1 4 4 1 9 4

1 y 1 (0, 0) 0, − s ( 4 ( x 12 1 3 −1 −1

−2 −2

−3

−4

x2 5x 4 x 1x 4 x2 2x 8 x 4x 2 37. hx 38. gx x2 4 x 2x 2 x2 9 x 3x 3 (a) Domain: all real numbers x except x ±2 (a) Domain: all real numbers x except x ±3 (b)x- intercepts: 1, 0, 4, 0 8 (b)y- intercept: 0, y-intercept: 0, 1 9 x-intercepts: 4, 0, 2, 0 (c) Vertical asymptotes: x 2, x 2 Horizontal asymptote: y 1 (c) Vertical asymptotes: x ±3 Horizontal asymptote: y 1 (d) x 4 3 1 0134 (d) 10 28 10 2 x 5 4 2 0245 y 3 5 3 1 005 27 16 8 8 7 y 16 7 009 5 16 y

y 6 (0, 0.88) 4 6

4 2 (1, 0) x 2 (4, 0) −66−4 (4, 0) x −6246−4 (−2, 0) −2

−4

−6 212 Chapter 2 Polynomial and Rational Functions

2x2 5x 3 2x 1x 3 x2 x 2 x 1x 2 39. f x 40. f x x3 2x2 x 2 x 2x 1x 1 x3 2x2 5x 6 x 1x 2x 3 (a) Domain: all real numbers x except x 2, x ±1 (a) Domain: all real numbers x except x 1, x 2, or x 3 1 (b) x-intercepts: , 0, 3, 0 2 (b) x-intercepts: 1, 0, 2, 0 3 1 y-intercept: 0, y-intercept: 0, 2 3 (c) Vertical asymptotes:x 2 , x 1 and x 1 (c) Vertical asymptotes: x 2, x 1, x 3 Horizontal asymptotes: y 0 Horizontal asymptote: y 0 (d) (d) x 3 2 0 1.5 3 4 x 4 3 1 024 3 5 3 48 3 9 5 1 5 f x 4 4 2 5 0 10 y 35 12 003 9

y y

4 9 3 (2, 0) − 1, 0 6 2 ( 2 ( − 3 ( 1, 0) 1 (3, 0) x x −4 −343 245 0, − 1 ( 3 ( −2 3 0, − −3 ( 2 ( −4 −5

x2 3x xx 3 x 5x 4 5x 4 5 41. f x , x 3 42. f x , x 4 x2 x 6 x 3x 2 x 2 x2 x 12 x 4x 3 x 3 (a) Domain: all real numbers x except x 3 and x 2 (a) Domain: all real numbers x except x 4 or x 3 (b) Intercept: 0, 0 5 (b)y- intercept: 0, 3 (c) Vertical asymptote: x 2 x-intercept: none Horizontal asymptote: y 1 (c) Vertical asymptote: x 3 (d) x 1 01 34 Horizontal asymptote: y 0 1 (d) y 3 0321 x 2 0257

y 5 5 5 y 1 3 5 2 4 6 y 4

2 6

x 4 −646−4 −2 (0, 0) 2

−4 x 2468 −6 (0, −1.66) −4

−6 Section 2.6 Rational Functions 213

2x2 5x 2 3x2 8x 4 43. f x 44. f x 2x2 x 6 2x2 3x 2 2x 1x 2 2x 1 x 23x 2 3x 2 , x 2 , x 2 2x 3x 2 2x 3 x 22x 1 2x 1 (a) Domain: all real numbers x except x 2 and 1 (a) Domain: all real numbers x except x 2 or x 3 2 x 2 (b)y- intercept: 0, 2 1 2 (b)x- intercept: , 0 x-intercept: , 0 2 3 1 y-intercept: 0, 1 3 (c) Vertical asymptote: x 2 3 3 (c) Vertical asymptote: x Horizontal asymptote: y 2 2 Horizontal asymptote: y 1 (d) 2 (d) x 3 1 033 x 3 2 1 01 11 y 5 5012 7 1 1 y 3 5 3 3 5 y y

4 3 2 1 1 x x −523−4 −3 −2 1 −434−3 −2 −1 2 , 0 1 ) 2 ) ( , 0 ( 0, − 3 ) 3 ) (0, −2)

t 2 1 t 1t 1 x2 16 x 4x 4 45. f t t 1, t 1 46. f x x 4, x 4 t 1 t 1 x 4 x 4 (a) Domain: all real numbers t except t 1 (a) Domain: all real numbers x 4 (b)t- intercept: 1, 0 (b)y- intercept: 0, 4 y-intercept: 0, 1 x-intercept: 4, 0 (c) Vertical asymptote: none (c) Vertical asymptote: none Horizontal asymptote: none Horizontal asymptote: none (d) (d) t 3 2 01 x 6 4 05 y 4 3 1 0 y 2 049

y y

3 10

2 8

1 6 (1, 0) t 4 −3 −2 −1 123 (0, 4) − 1 (0, −1) 2 (−4, 0) −2 x −6246−2 −3 −2 214 Chapter 2 Polynomial and Rational Functions

x2 1 x2x 2 47. fx , gx x 1 48. fx , gx x x 1 x2 2x (a) Domain of f : all real numbers x except x 1 (a) Domain off : all real numbers x except 0 and 2 Domain of g: all real numbers x Domain of g: all real numbers x (b) Because x 1 is a factor of both the numerator (b) Since x2 2x is a factor of both the numerator and the denominator of f, x 1 is not a vertical and the denominator of f, neither x 0 nor x 2 asymptote. f has no vertical asymptotes. is a vertical asymptote of f. Thus,f has no (c) vertical asymptotes. x 3 2 1.5 1 0.5 01 (c) f x 4 3 2.5 Undef.1.5 1 0 x 10 11.5 2 2.5 3 gx 4 3 2.5 2 1.5 1 0 f x 1 Undef. 1 1.5 Undef. 2.5 3 g(x) 10 11.5 2 2.5 3 (d) 1

−4 2 (d) 2

−2 4

−3

(e) Because there are only a finite number of pixels, −2 the utility may not attempt to evaluate the function (e) Because there are only a finite number of pixels, where it does not exist. the utility may not attempt to evaluate the function where it does not exist.

x 2 1 2x 6 2 49. fx , gx 50. fx , gx x2 2x x x2 7x 12 x 4 (a) Domain of f : all real numbers x except x 0 and (a) Domain off : all real numbers x except 3 and 4 x 2 Domain of g: all real numbers x except 4 Domain of g: all real numbers x except x 0 (b) Since x 3 is a factor of both the numerator and the (b) Because x 2 is a factor of both the numerator and denominator of f, x 3 is not a vertical asymptote the denominator of f, x 2 is not a vertical asymptote. of f. Thus, f has x 4 as its only vertical asymptote. The only vertical asymptote of f is x 0. (c) x 0 123 4 56 (c) x 0.5 01230.5 1.5 1 2 f x 2 3 1 Undef. Undef. 2 1 2 1 f x 2 Undef. 2 1 3 Undef. 3 1 2 g(x) 2 3 1 2 Undef. 2 1 2 1 1 g x 2 Undef. 2 1 3 2 3 (d) 3 (d) 2

−1 8

−3 3

−3 −2 (e) Because there are only a finite number of pixels, the (e) Because there are only a finite number of pixels, utility may not attempt to evaluate the function where the utility may not attempt to evaluate the function it does not exist. where it does not exist. Section 2.6 Rational Functions 215

x2 4 4 x2 5 5 51. hx x 52. gx x x x x x (a) Domain: all real numbers x except x 0 (a) Domain: all real numbers x except x 0 (b) Intercepts: 2, 0, 2, 0 (b) No intercepts (c) Vertical asymptote: x 0 (c) Vertical asymptote: x 0 Slant asymptote: y x Slant asymptote: y x (d) (d) x 3 1 13 x 3 2 1 123 5 5 14 9 9 14 y 3 3 3 3 y 3 2 6 6 2 3

y y

y = x 6

2 2 y = x (−2, 0) x x −66−4 (2, 0) −6246−4 −2 −2 −2

−4 −4

−6

2x2 1 1 1 x2 1 53. fx 2x 54. fx x x x x x (a) Domain: all real numbers x except x 0 (a) Domain: all real numbers x except x 0 (b) No intercepts (b) x-intercepts: 1, 0, 1, 0 (c) Vertical asymptote: x 0 Slant asymptote: y 2x (c) Vertical asymptote: x 0 Slant asymptote: y x (d) x 4 2 24 6 (d) x 6 4 2 2 46 f x 33 9 9 33 73 4 2 2 4 6 35 15 3 3 15 35 f x 6 4 2 2 4 6 y y 6 y = −x 8 4 6 2 y = 2x 4 x − 2 −6 −4 −2 246 ( 1, 0) (1, 0) x −8 −6 −4 −2 468

−4 −6 −6 −8

x2 1 1 55. gx x x x (a) Domain: all real numbers x except x 0 (d) y x 4 2 24 6 (b) No intercepts 6 17 5 5 17 37 g x 4 2 2 4 6 4 (c) Vertical asymptote: x 0 2 y = x Slant asymptote: y x x −6 −4 −2 246

−6 216 Chapter 2 Polynomial and Rational Functions

x2 1 t 2 1 26 56. hx x 1 57. ft t 5 x 1 x 1 t 5 t 5 (a) Domain: all real numbers x except x 1 (a) Domain: all real numbers t except t 5 (b) Intercept: 0, 0 1 (b) Intercept: 0, 5 (c) Vertical asymptote: x 1 Slant asymptote: y x 1 (c) Vertical asymptote: t 5 (d) Slant asymptote: y t 5 x 4 2 2 46 (d) 16 4 16 36 t 7 6 4 3 0 h x 5 3 4 3 5 1 y 25 37 17 5 5 y

8 y

6 25 4 20 y = x + 1 2 15 (0, 0) − x y = 5 t −4 2468 5 −2 (0, −0.2) t −4 −20−15 −10 −5 10

x2 1 1 1 x3 x 58. fx x 59. fx x 3x 1 3 9 93x 1 x2 1 x2 1 ± 1 (a) Domain: all real numbers x except x 1 (a) Domain: all real numbers x except x 3 (b) Intercept: 0, 0 (b) Intercept: 0, 0 (c) Vertical asymptotes: x ±1 Slant asymptote: y x 1 (c) Vertical asymptote: x (d) 3 x 4 2 024 1 1 Slant asymptote: y x 64 8 8 64 3 9 f x 15 3 0 3 15

(d) y 1 x 3 2 1 2 02 9 4 1 1 4 y 8 5 2 2 0 7 y = x 2 y (0, 0) x −6 −4 −2 246 1 2 1 − 1 3 y = x 1 (0, 0) 3 9 3 x −1 1 2 1 4 3 3 3 Section 2.6 Rational Functions 217

x3 1 4x x2 x 1 1 60. gx x 61. fx x 2x2 8 2 2x2 8 x 1 x 1 (a) Domain: all real numbers x except x ±2 (a) Domain: all real numbers x except x 1 (b) Intercept: 0, 0 (b) y-intercept: 0, 1 (c) Vertical asymptotes: x ±2 (c) Vertical asymptote: x 1 Slant asymptote: y x 1 Slant asymptote: y x 2 (d) x 4 2 024 (d) 21 7 13 x 6 4 1 146 f x 5 3 1 3 3 27 8 1 1 8 27 g x 8 3 6 6 3 8 y

y 8 6 8 y = x 6 4 4 2 (0, −1) (0, 0) x x −4 −22468 −8 −6 −4 468 y = 1 x −4 2

2x2 5x 5 3 2x3 x2 2x 1 62. fx 2x 1 63. f x x 2 x 2 x2 3x 2 (a) Domain: all real numbers x 2 2x 1x 1x 1 x except x 1x 2 5 (b) y-intercept: 0, 2x 1x 1 2 , x 1 x 2 (c) Vertical asymptote: x 2 2x2 3x 1 Slant asymptote: y 2x 1 x 2 (d) x 6 3 1367 15 2x 7 , x 1 y 107 38 47 68 x 2 8 5 2 8 4 5 (a) Domain: all real numbers x except x 1 and x 2 y 1 (b)y- intercept: 0, 15 2 12 1 9 x-intercepts: , 0, 1, 0 6 2 y = 2x − 1 3 x (c) Vertical asymptote: x 2 −9 −6 −336912 15 Slant asymptote: y 2x 7 0, − 5 ( 2 ( (d) −9 3 x 4 3 2 01 y 45 1 2 28 202 0

18 12 (0, 0.5) 6 (1, 0) x −63−5 −4 −3 −1

−12 (0.5, 0) −18 y = 2x − 7 −24 −30 −36 218 Chapter 2 Polynomial and Rational Functions

2x3 x2 8x 4 x 2x 22x 1 x2 5x 8 2 64. f x 65. fx x 2 x2 3x 2 x 2x 1 x 3 x 3 9 2x 7 , x 2 Domain: all real numbers x except x 3 x 1 8 (a) Domain: all real numbers x except x 1 or x 2 y-intercept: 0, 3 (b)y- intercept: 0, 2 Vertical asymptote: x 3 1 x-intercepts: 2, 0, , 0 2 Slant asymptote: y x 2 (c) Vertical asymptote: x 1 Line: y x 2 Slant asymptote: y 2x 7 8 (d) x 3 2 1 0341 3 2 2 −14 10 y 5 1 35 4 028182 2 10 2

−8 y

12 y = 2x + 7 − ( 2, 0) (− 1, 0 ( 2 x −646−2 2 (0, −2)

2x2 x 1 1 3x2 x3 1 1 66. fx 2x 1 67. gx 3 x x 3 x 1 x 1 x2 x2 x2 Domain: all real numbers x except x 1 Domain: all real numbers x except x 0

Vertical asymptote: x 1 6 Vertical asymptote: x 0 12 Slant asymptote: y 2x 1 −12 12 Slant asymptote: y x 3 Line: y 2x 1 Line: y x 3 −12 12

−10 −4

12 2x x2 1 2 x 1 68. hx x 1 69. y 24 x 2 4 x x 3 Domain: all real numbers x except x 4 (a) x-intercept: 1, 0

Vertical asymptote: x 4 10 x 1 (b) 0 x 3 1 Slant asymptote: y x 1 2 −16 8 0 x 1 1 1 x Line: y x 1 −6 2

1 70. (a) x-intercept: 0, 0 71. y x x 2x 1 (b) 0 (a) x-intercepts:±1, 0 (b) 0 x x 3 x 1 x 0 2x x 0 x x2 1 x ±1 Section 2.6 Rational Functions 219

255p 72. (a) x-intercepts: 1, 0, 2, 0 73. C , 0 ≤ p < 100 100 p 2 (b) 0 x 3 (a) 2,000 x 0 x2 3x 2 0 x 1x 2 0 100 0 x 1, x 2 25510 (b) C10 28.33 million dollars 100 10 25540 C40 170 million dollars 100 40 25575 C75 765 million dollars 100 75 (c)C → as x → 100. No, it would not be possible to remove 100% of the pollutants.

25,000p 205 3t 74. C , 0 ≤ p < 100 75. N , t ≥ 0 100 p 1 0.04t (a) 300,000 (a) N5 333 deer N10 500 deer N25 800 deer 0 100 0 (b) The herd is limited by the horizontal asymptote:

25,00015 60 (b) C 4411.76 N 1500 deer 100 15 0.04 The cost would be $4411.76. 25,00050 C 25,000 100 50 The cost would be $25,000. 25,00090 C 225,000 100 90 The cost would be $225,000. (c)C → as x → 100. No. The model is undefined for p 100.

76. (a) 0.2550 0.75x C50 x (c) C

12.50 0.75x 4 1.0 C 50 x 4 0.8 50 3x 3x 50 0.6 C 450 x 4x 50 0.4 (b) Domain: x ≥ 0 and x ≤ 1000 50 0.2 x Thus,0 ≤ x ≤ 950. Using interval notation, the 200 400 600 800 1000 domain is 0, 950. (d) As the tank is filled, the concentration increases more slowly. It approaches the horizontal asymptote 3 of C 4 0.75. 220 Chapter 2 Polynomial and Rational Functions

77. (a)A xy and x 4y 2 30 30 y 2 x 4 30 2x 22 y 2 x 4 x 4 2x 22 2xx 11 Thus, A xy x . x 4 x 4

(b) Domain: Since the margins on the left and right are each 2 inches,x > 4. In interval notation, the domain is 4, .

y1 (Area) 160 102 84 76 72 70 69.143 69 69.333 70 70.909

4 40 0

The area is minimum when x 11.75 inches and y 5.87 inches. The area is minimum when x is approximately 12.

78. A xy and

x 3y 2 64 200 64 y 2 x 3

64 2x 58 3 39 y 2 x 3 x 3 0 2x 58 2xx 29 Thus, A xy x , x > 3. x 3 x 3 By graphing the area function, we see that A is minimum when x 12.8 inches and y 8.5 inches.

79. (a) Let t1 time from Akron to Columbus and t2 time (b) Vertical asymptote: x 25 from Columbus back to Akron. Horizontal asymptote: y 25 100 xt 100 ⇒ t (c) 200 1 1 x 100 yt 100 ⇒ t 2 2 y 25 65 50 t1 t2 200 0 t t 4 1 2 (d) x 30 35 40 45 50 55 60 100 100 4 x y y 15087.5 66.67 56.25 5045.83 42.86 100y 100x 4xy (e) Yes. You would expect the average speed for the round 25y 25x xy trip to be the average of the average speeds for the two parts of the trip. 25x xy 25y (f) No. At 20 miles per hour you would use more time in 25x yx 25 one direction than is required for the round trip at an 25x average speed of 50 miles per hour. Thus, y . x 25 Section 2.6 Rational Functions 221

80. (a) 600

8 13 0

5.816182 130.68 (b) S 763.81 0.004182 1.00 The sales in 2008 is estimated to be $763,810,000. 5.816 (c) Probably not. The graph has a horizontal asymptote at S 1454 million dollars. 0.004 Future sales may exceed this limiting value.

x 81. False. Polynomial functions do not have vertical 82. False. The graph of f x crosses y 0, which x2 1 asymptotes. is a horizontal asymptote.

83. Vertical asymptote: None ⇒ The denominator is not 84. Vertical asymptotes: x 2, x 1 ⇒ x 2x 1 zero for any value of x (unless the numerator is also zero are factors of the denominator. there). Horizontal asymptotes:None ⇒ The degree of the Horizontal asymptote:y 2 ⇒ The degree of the numerator is greater than the degree of the denominator. numerator equals the degree of the denominator. x3 f x is one possible function. There are 2x2 x 2x 1 fx is one possible function. There are many x2 1 many correct answers. correct answers.

85. x2 15x 56 x 8x 7 86. 3x2 23x 36 3x 4x 9

87. x3 5x2 4x 20 x 5x2 4 88. x3 6x2 2x 12 x2x 6 2x 6 x 5x 2ix 2i x 6x2 2 x 6x 2x 2

89. 10 3x ≤ 0 10 90. 5 2x > 5x 1 x 3 −323−2 −1 0 1 3 x ≥ 10 x 5 2x > 5x 5 0 123456 ≥ 10 7x > 0 x 3 x < 0

1 − ≥ −13 7 91. 4 x 2 < 20 3 7 92. 2 2x 3 5 2 2 x x − − 20 < 4x 8 < 20 4 2 0 2 468 2x 3 ≥ 10 −8 −6 −4 −2 0 2 4 12 < 4x < 28 2 x 3 ≤ 10 or 2 x 3 ≥ 10 3 < x < 7 2 x ≤ 13 2 x ≥ 7 ≤ 13 ≥ 7 x 2 x 2

93. Answers will vary. 222 Chapter 2 Polynomial and Rational Functions

Section 2.7 Nonlinear Inequalities

■ You should be able to solve inequalities. (a) Find the critical number. 1. Values that make the expression zero 2. Values that make the expression undefined (b) Test one value in each test interval on the real number line resulting from the critical numbers. (c) Determine the solution intervals.

Vocabulary Check 1. critical; test intervals 2. zeros; undefined values 3. P R C

1. x2 3 < 0 3 (a) x 3 (b) x 0 (c) x 2 (d) x 5 2 ? 2 ? 32 ? 2 ? 3 3 < 0 0 3 < 0 2 3 < 0 5 3 < 0 3 6 < 0 3 < 0 4 < 0 22 < 0 3 No, x 3 is not a Yes, x 0 is a solution. Yes, x 2 is a solution. No, x 5 is not a solution. solution.

2. x2 x 12 ≥ 0 (a) x 5 (b) x 0 (c) x 4 (d) x 3 ? ? ? ? 52 5 12 ≥ 0 02 0 12 ≥ 0 42 4 12 ≥ 0 32 3 12 ≥ 0 ? ? 8 ≥ 0 12 ≥ 0 16 4 12 ≥ 0 9 3 12 ≥ 0 ≥ Yes, x 5 is a No, x 0 is not a 8 ≥ 0 0 0 solution. solution. Yes, x 3 is a solution. Yes, x 4 is a solution.

x 2 3. ≥ 3 x 4 9 9 (a) x 5 (b) x 4 (c) x (d) x 2 2 5 2 ? 4 2 ? 9 ≥ 3 ≥ 3 2 ? 9 2 ? 5 4 4 4 2 ≥ 3 2 ≥ 3 9 9 2 4 2 4 7 ≥ 3 6 5 ≥ is undefined. ≥ 3 13 3 0 17 Yes, x 5 is a 9 Yes, x 2 is a solution. solution. No, x 4 is not a 9 No, x 2 is not a solution. solution. Section 2.7 Nonlinear Inequalities 223

3x2 4. < 1 x2 4 (a) x 2 (b) x 1 (c) x 0 (d) x 3 322 ? 312 ? 302 ? 332 ? < 1 < 1 < 1 < 1 22 4 12 4 02 4 32 4 12 3 0 1 27 < 1 < 1 < < 1 8 5 13 Yes, x 0 is a No, x 2 is not Yes, x 1 is a solution. No, x 3 is not a a solution. solution. solution.

3 2x 5 3 5. 2x2 x 6 2x 3x 2 6. 9 x3 25x2 0 7. 2 x 5 x 5 3 x29x 25 0 2 x 3 0 ⇒ x 2x 7 2 x2 0 ⇒ x 0 x 5 x 2 0 ⇒ x 2 25 7 9 x 25 0 ⇒ x 2 x 7 0 ⇒ x 3 9 2 Critical numbers: x , x 2 2 25 x 5 0 ⇒ x 5 The critical numbers are 0 and . 9 7 Critical numbers: x , x 5 2

x 2 xx 1 2x 2 8. 9. x2 ≤ 9 x 2 x 1 x 2x 1 x2 9 ≤ 0 x2 x 2x 4 x 2x 1 x 3x 3 ≤ 0 x 4x 1 Critical numbers: x ±3 x 2x 1 Test intervals: , 3, 3, 3, 3, x 4x 1 0 Test: Is x 3x 3 ≤ 0? x 4 0 ⇒ x 4 Interval x-Value Value of x2 9 Conclusion x 1 0 ⇒ x 1 , 3 x 4 16 9 7 Positive x 2x 1 0 3, 3 x 0 0 9 9 Negative x 2 0 ⇒ x 2 3, x 4 16 9 7 Positive x 1 0 ⇒ x 1 Solution set: 3, 3 The critical numbers are 2, 1, 1, 4. x −3 −2 −1 0123

10. x2 < 36 x2 36 < 0 x 6x 6 < 0 Critical numbers: x 6, x 6

Test intervals: , 6 ⇒ x 6x 6 > 0 6, 6 ⇒ x 6x 6 < 0 6, ⇒ x 6x 6 > 0 Solution interval: 6, 6

x −8 −6 −4 −202468 224 Chapter 2 Polynomial and Rational Functions

11. x 22 < 25 12. x 32 ≥ 1 x2 4x 4 < 25 x2 6x 8 ≥ 0 x2 4x 21 < 0 x 2x 4 ≥ 0 x 7x 3 < 0 Critical numbers: x 2, x 4 Critical numbers: x 7, x 3 Test intervals: , 2 ⇒ x 2x 4 > 0 Test intervals: , 7, 7, 3, 3, 2, 4 ⇒ x 2x 4 < 0 Test: Is x 7x 3 < 0? 4, ⇒ x 2x 4 > 0 Interval x-Value Value of Conclusion Solution intervals: , 2 4, x 7x 3 x , 7 x 10 313 39 Positive 123 4 5 7, 3 x 0 73 21 Negative 3, x 5 122 24 Positive

Solution set: 7, 3 −7 3 x −8 −6 −4 −2 024

13. x2 4x 4 ≥ 9 14. x2 6x 9 < 16 x2 4x 5 ≥ 0 x2 6x 7 < 0 x 5x 1 ≥ 0 x 1x 7 < 0 Critical numbers: x 5, x 1 Critical numbers: x 1, x 7

Test intervals: , 5, 5, 1, 1, Test intervals: , 1 ⇒ x 1x 7 > 0 Test: Is x 5x 1 ≥ 0? 1, 7 ⇒ x 1x 7 < 0 Interval x-Value Value of Conclusion 7, ⇒ x 1x 7 > 0 x 5x 1 Solution interval: 1, 7 , 5 x 6 17 7 Positive −1 7 5, 1 x 0 51 5 Negative x −2 0624 8 1, x 2 71 7 Positive Solution set: , 5 1,

x −6 −5 −4 −3 −2 −1 0 1 2

15. x2 x < 6 16. x2 2x > 3 x2 x 6 < 0 x2 2x 3 > 0 x 3x 2 < 0 x 3x 1 > 0 Critical numbers: x 3, x 2 Critical numbers: x 3, x 1

Test intervals: , 3, 3, 2, 2, Test intervals: , 3 ⇒ x 3x 1 > 0 Test: Is x 3x 2 < 0? 3, 1 ⇒ x 3x 1 < 0 Interval x-Value Value of Conclusion 1, ⇒ x 3x 1 > 0 x 3x 2 Solution intervals: , 3 1, , 3 x 4 16 6 Positive x 3, 2 x 0 32 6 Negative −4012−3 −2 −1 2, x 3 61 6 Positive

Solution set: 3, 2 x −312−2 −1 0 Section 2.7 Nonlinear Inequalities 225

17. x2 2x 3 < 0 18. x2 4x 1 > 0 x 3x 1 < 0 4 ± 16 4 x 2 ± 5 2 Critical numbers: x 3, x 1 Critical numbers: x 2 5, x 2 5 Test intervals: , 3, 3, 1, 1, Test intervals: , 2 5 ⇒ x2 4x 1 0 Test: Is x 3x 1 < 0? > 2 5, 2 5 ⇒ x2 4x 1 0 Interval x-Value Value of Conclusion < x 3x 1 2 5, ⇒ x2 4x 1 > 0 , 3 x 4 15 5 Positive Solution intervals: , 2 5 2 5, 3, 1 x 0 3 1 3 Negative 2 − 5 2 + 5 x 1, x 2 51 5 Positive −42−2 0 468

Solution set: 3, 1 x −3 −2 −1 0 1

19. x2 8x 5 ≥ 0 x2 8x 5 0 Complete the square. x2 8x 16 5 16 x 42 21 x 4 ±21 x 4 ± 21

Critical numbers: x 4 ± 21 Test intervals: , 4 21, 4 21, 4 21, 4 21, Test: Is x2 8x 5 ≥ 0? Interval x-Value Value of x2 8x 5 Conclusion , 4 21 x 10 100 80 5 15 Positive 4 21, 4 21 x 0 0 0 5 5 Negative

4 21, x 2 4 16 5 15 Positive −4 − 21 −4 + 21 x Solution set: < 4 21 4 21, −10 −8 −6 −4 −2 0 2

20. 2x2 6x 15 ≤ 0 2 x2 6x 15 ≥ 0 6 ± 62 4215 6 ± 156 6 ± 239 3 39 x ± 22 4 4 2 2 3 39 3 39 Critical numbers: x , x 2 2 2 2 3 39 Test intervals: , ⇒ 2x2 6x 15 < 0 2 2 3 39 3 39 , ⇒ 2x2 6x 15 > 0 2 2 2 2 3 39 , ⇒ 2x2 6x 15 < 0 2 2 3 39 3 39 Solution interval: , , 3 − 39 3 + 39 2 2 2 2 22 22 x −2 −1 0 1 2 3 4 5 226 Chapter 2 Polynomial and Rational Functions

21. x3 3x2 x 3 > 0 22. x3 2x2 4x 8 ≤ 0

x2x 3 1x 3 > 0 x2x 2 4x 2 ≤ 0

x2 1x 3 > 0 x 2x2 4 ≤ 0

x 1x 1x 3 > 0 Critical numbers: x 2, x 2

3 2 Critical numbers: x ±1, x 3 Test intervals: , 2 ⇒ x 2x 4x 8 < 0 Test intervals: , 1, 1, 1, 1, 3, 3, 2, 2 ⇒ x3 2x2 4x 8 < 0 3 2 Test: Is x 1x 1x 3 > 0? 2, ⇒ x 2x 4x 8 > 0 Interval x-Value Value of Conclusion Solution interval: , 2 x 1x 1x 3 x , 1 x 2 135 15 Negative 01234 1, 1 x 0 113 3 Positive 1, 3 x 2 311 3 Negative 3, x 4 531 15 Positive

Solution set: 1, 1 3, x −1 0123 4

23. x3 2x2 9x 2 ≥ 20 x3 2x2 9x 18 ≥ 0 x2x 2 9x 2 ≥ 0 x 2x2 9 ≥ 0 x 2x 3x 3 ≥ 0 Critical numbers: x 2, x ±3 Test intervals: , 3, 3, 2, 2, 3, 3, Test: Is x 2x 3x 3 ≥ 0? Interval x-Value Value of x 2x 3x 3 Conclusion , 3 x 4 617 42 Negative 3, 2 x 0 233 18 Positive 2, 3 x 2.5 0.55.50.5 1.375 Negative

3, x 4 271 14 Positive x −4 −3 −2 −1012345 Solution set: 3, 2 3,

24. 2 x3 13x2 8x 46 ≥ 6 2 x3 13x2 8x 52 ≥ 0 x22x 13 42x 13 ≥ 0 2x 13x2 4 ≥ 0 13 Critical numbers: x 2 , x 2, x 2 13 Test intervals: , ⇒ 2x3 13x2 8x 52 < 0 −13 2 2 13, 2 ⇒ 2x3 13x2 8x 52 > 0 x 2 −8 −6 −4 −2 0 2 4 2, 2 ⇒ 2x3 13x2 8x 52 < 0 2, ⇒ 2x3 13x2 8x 52 > 0 13 Solution interval: 2 , 2 , 2, Section 2.7 Nonlinear Inequalities 227

25. 4 x2 4x 1 ≤ 0 26. x2 3x 8 > 0 2x 12 ≤ 0 The critical numbers are imaginary: 1 3 i23 Critical number: x ± 2 2 2 1 1 So the set of real numbers is the solution set. Test intervals: , , , 2 2 x − − − Test: Is 2x 12 ≤ 0? 3 2 1 0 123 Interval x-Value Value of 2x 12 Conclusion 1 , x 0 12 1 Positive 2 1 , x 1 12 1 Positive 2 1 Solution set: x 1 2 2 x −2 −1 0 1 2

27. 4 x3 6x2 < 0 28. 4 x3 12x2 > 0

2 x22x 3 < 0 4 x2x 3 > 0 3 Critical numbers: x 0, x 2 Critical numbers: x 0, x 3 3 3 ⇒ 2 Test intervals: , 0 , 0, 2 , 2, Test intervals: , 0 4x x 3 < 0 Test: Is 2x22x 3 < 0? 0, 3 ⇒ 4x2x 3 < 0 By testing an x-value in each test interval in the inequality, 3, ⇒ 4x2x 3 > 0 we see that the solution set is: , 0 0, 3 2 Solution interval: 3,

29. x3 4x ≥ 0 30. 2 x3 x4 ≤ 0 xx 2x 2 ≥ 0 x32 x ≤ 0 Critical numbers: x 0, x ±2 Critical numbers: x 0, x 2 Test intervals: , 2, 2, 0, 0, 2, 2, Test intervals: , 0 ⇒ x32 x < 0 Test: Is xx 2x 2 ≥ 0? 0, 2 ⇒ x32 x > 0 By testing an x-value in each test interval in the inequality, 2, ⇒ x32 x < 0 we see that the solution set is: 2, 0 2, Solution intervals: , 0 2,

31. x 12x 23 ≥ 0 32. x4x 3 ≤ 0 Critical numbers: x 1, x 2 Critical numbers: x 0, x 3

Test intervals: , 2, 2, 1, 1, ) Test intervals: , 0 ⇒ x4x 3 < 0 Test: Is x 12x 33 ≥ 0? 0, 3 ⇒ x4x 3 < 0

By testing an x-value in each test interval in the inequality, 3, ⇒ x4x 3 > 0 we see that the solution set is: 2, Solution intervals: , 0 0, 3 or , 3 228 Chapter 2 Polynomial and Rational Functions

33. y x2 2x 3 (a) y ≤ 0 when x ≤ 1 or x ≥ 3.

6 (b) y ≥ 3 when 0 ≤ x ≤ 2.

−57

−2

1 34. y x2 2x 1 (a) y ≤ 0 (b) y ≥ 7 2

1 2 1 12 x 2x 1 ≤ 0 x2 2x 1 ≥ 7 2 2 x2 4x 2 ≤ 0 x2 4x 12 ≥ 0 −10 14 4 ± 42 412 x 6x 2 ≥ 0 x 21 −4 y ≥ 7 when x ≤ 2, x ≥ 6. 4 ± 8 2 ± 2 2 y ≤ 0 when 2 2 ≤ x ≤ 2 2.

1 3 1 ≥ ≤ ≤ ≤ 35. y 8x 2x (a) y 0 when 2 x 0, 2 x < .

8 (b) y ≤ 6 when x ≤ 4.

−12 12

−8

36. y x3 x2 16x 16 (a) y ≤ 0 (b) y ≥ 36

3 2 3 2 48 x x 16x 16 ≤ 0 x x 16x 16 ≥ 36 x2x 1 16x 1 ≤ 0 x3 x2 16x 20 ≥ 0

2 −12 12 x 1x 16 ≤ 0 x 2x 5x 2 ≥ 0

y ≤ 0 when < x ≤ 4, 1 ≤ x ≤ 4. y ≥ 36 when x 2, 5 ≤ x < . −24

1 1 37. x > 0 38. 4 < 0 x x 1 x2 1 4x > 0 < 0 x x 1 Critical numbers: x 0, x ±1 Critical numbers: x 0, x 4 Test intervals: , 1 , 1, 0 , 0, 1 , 1, 1 4x Test intervals: , 0 ⇒ < 0 1 x2 x Test: Is > 0? x 1 1 4x 0, ⇒ > 0 4 x By testing an x-value in each test interval in the inequality, we see that the solution set is: , 1 0, 1 1 1 4x , ⇒ < 0 4 x x −22−1 0 1 1 Solution interval: , 0 , 4

1 4 x −1 0 1 Section 2.7 Nonlinear Inequalities 229

x 6 x 12 39. 2 < 0 40. 3 ≥ 0 x 1 x 2 x 6 2x 1 x 12 3x 2 < 0 ≥ 0 x 1 x 2 6 2x 4 x ≥ < 0 0 x 1 x 2 Critical numbers: x 2, x 3 Critical numbers: x 1, x 4 6 2x Test intervals: , 1, 1, 4, 4, Test intervals: , 2 ⇒ < 0 x 2 4 x Test: Is < 0? 6 2x x 1 2, 3 ⇒ > 0 x 2 x By testing an -value in each test interval in the inequality, 6 2x we see that the solution set is: , 1 4, 3, ⇒ < 0 x 2 x Solution interval: 2, 3 −2 −110 2345

x −2 −1 0123

3x 5 5 7x 41. > 4 42. < 4 x 5 1 2x 3x 5 5 7x 41 2x 4 > 0 < 0 x 5 1 2x 3x 5 4x 5 1 x > 0 < 0 x 5 1 2x 15 x 1 > 0 Critical numbers: x , x 1 x 5 2 x x Critical numbers: 5, 15 1 ⇒ 1 x Test intervals: , < 0 Test intervals: , 5, 5, 15, 15, 2 1 2x 1 1 x 15 x , 1 ⇒ > 0 Test: Is > 0? x 5 2 1 2x x By testing an x-value in each test interval in the inequality, ⇒ 1 1, < 0 we see that the solution set is: 5, 15 1 2x 1 5 Solution intervals: , 1, x 2 36912 15 18

− 1 2 x −2 −1 021 230 Chapter 2 Polynomial and Rational Functions

4 1 5 3 43. > 44. > x 5 2x 3 x 6 x 2 4 1 5x 2 3x 6 > 0 > 0 x 5 2x 3 x 6x 2 42x 3 x 5 2x 28 > 0 > 0 x 52x 3 x 6x 2 7x 7 Critical numbers: x 14, x 2, x 6 > 0 x 52x 3 2x 28 Test intervals: , 14 ⇒ < 0 3 x 6x 2 Critical numbers: x 1, x 5, x 2 2x 28 14, 2 ⇒ > 0 3 x 6x 2 Test intervals: , 5, 5, , 2 2x 28 2, 6 ⇒ < 0 3 x 6x 2 , 1, 1, 2 2x 28 6, ⇒ > 0 7x 1 x 6x 2 Test: Is > 0? x 52x 3 Solution intervals: 14, 2 6,

By testing an x-value in each test interval in the inequality, − − 3 14 2 6 we see that the solution set is: 5, 2 1, x −15−10 −5 0 5 10 −3 2 x −5 −4 −3 −2 −1 0

1 9 1 1 45. ≤ 46. ≥ x 3 4x 3 x x 3 1 9 1x 3 1x ≤ 0 ≥ 0 x 3 4x 3 xx 3 4x 3 9x 3 3 ≤ 0 ≥ 0 x 34x 3 xx 3 30 5x Critical numbers: x 3, x 0 ≤ 0 x x 3 4 3 3 Test intervals: , 3 ⇒ > 0 3 xx 3 Critical numbers: x 3, x , x 6 4 3 3, 0 ⇒ < 0 3 3 xx 3 Test intervals: , , , 3, 3, 6, 6, 4 4 3 0, ⇒ > 0 30 5x xx 3 Test: Is ≤ 0? x x 3 4 3 Solution intervals: , 3 0, By testing an x-value in each test interval in the inequality, x 3 we see that the solution set is: 4, 3 6, −41−3 −2 −1 0

−3 4 3 x −4 −2 0 2 468 Section 2.7 Nonlinear Inequalities 231

x2 2x x2 x 6 47. ≤ 0 48. ≥ 0 x2 9 x xx 2 x 3x 2 ≤ 0 ≥ 0 x 3x 3 x Critical numbers: x 0, x 2, x ±3 Critical numbers: x 3, x 0, x 2 Test intervals: x 3x 2 Test intervals: , 3 ⇒ < 0 , 3, 3, 2, 2, 0, 0, 3, 3, x xx 2 x 3x 2 Test: Is ≤ 0? 3, 0 ⇒ > 0 x 3x 3 x By testing an x-value in each test interval in the inequality, x 3x 2 0, 2 ⇒ < 0 we see that the solution set is: 3, 2 0, 3 x x x 3x 2 −3 −2 −1 0123 2, ⇒ > 0 x Solution intervals: 3, 0 2,

x −30−2 −1 123

5 2x 3x x 49. < 1 50. ≤ 3 x 1 x 1 x 1 x 4 5 2x 3xx 4 xx 1 3x 4x 1 1 < 0 ≤ 0 x 1 x 1 x 1x 4 5x 1 2xx 1 x 1x 1 x2 4x 12 < 0 ≤ 0 x 1x 1 x 1x 4 5x 5 2x2 2x x2 1 x 6x 2 < 0 ≤ 0 x 1x 1 x 1x 4 3x2 7x 6 Critical numbers: x 4, x 2, x 1, x 6 < 0 x x 1 1 x 6x 2 Test intervals: , 4 ⇒ < 0 3x 2x 3 x 1x 4 < 0 x 1 x 1 x 6x 2 4, 2 ⇒ > 0 2 x 1x 4 Critical numbers: x , x 3, x ±1 3 x 6x 2 2, 1 ⇒ < 0 2 2 x 1x 4 Test intervals: , 1, 1, , , 1, 1, 3, 3, 3 3 x 6x 2 1, 6 ⇒ > 0 3x 2x 3 x 1x 4 Test: Is < 0? x 1 x 1 x 6x 2 6, ⇒ < 0 By testing an x-value in each test interval in the inequality, x 1x 4 2 we see that the solution set is: , 1 3, 1 3, Solution intervals: , 4, 2, 1, 6, −2 3 x 1 x −1 0 1 2 3 4 −4 −2 024 6

3x (a) y ≤ 0 when 0 ≤ x < 2. 51. y x 2 (b) y ≥ 6 when 2 < x ≤ 4.

−612

−4 232 Chapter 2 Polynomial and Rational Functions

2x 2 52. y (a) y ≤ 0 (b) y ≥ 8 x 1 2x 2 2x 2 14 ≤ 0 ≥ 8 x 1 x 1 y ≤ 0 when 1 < x ≤ 2. 2x 2 8x 1 ≥ 0 −15 15 x 1

−6 6x 12 ≥ 0 x 1 6x 2 ≥ 0 x 1

y ≥ 8 when 2 ≤ x < 1.

2x2 5x 53. y 54. y x2 4 x2 4

6 (a) y ≥ 1 5x ≥ 1 x2 4 −66 5x x2 4 −2 ≥ 0 x2 4 ≥ ≤ ≥ (a)y 1 when x 2 or x 2. x 4x 1 ≥ 0 This can also be expressed as x ≥ 2. x2 4 (b)y ≤ 2 for all real numbers x. y ≥ 1 when 1 ≤ x ≤ 4.

This can also be expressed as < x < . (b) y ≤ 0 4 5x ≤ 0 x2 4 −66

y ≤ 0 when < x ≤ 0. −4

55. 4 x2 ≥ 0 56. x2 4 ≥ 0 2 x2 x ≥ 0 x 2x 2 ≥ 0 Critical numbers: x ±2 Critical numbers: x 2, x 2 Test intervals: , 2, 2, 2, 2, Test intervals: , 2 ⇒ x 2x 2 > 0 Test: Is 4 x2 ≥ 0? 2, 2 ⇒ x 2x 2 < 0 By testing an x-value in each test interval in the inequality, 2, ⇒ x 2x 2 > 0 we see that the domain set is: 2, 2 Domain: , 2 2,

57. x2 7x 12 ≥ 0 58. 144 9x2 ≥ 0 x 3x 4 ≥ 0 9 4 x4 x ≥ 0 Critical numbers: x 3, x 4 Critical numbers: x 4, x 4 Test intervals: , 3, 3, 4, 4, Test intervals: , 4 ⇒ 94 x4 x < 0 Test: Is x 3x 4 ≥ 0? 4, 4 ⇒ 94 x4 x > 0 By testing an x-value in each test interval in the inequality, 4, ⇒ 94 x4 x < 0 we see that the domain set is: , 3 4, Domain: 4, 4 Section 2.7 Nonlinear Inequalities 233

x x 59. ≥ 0 60. ≥ 0 x2 2x 35 x2 9 x x ≥ 0 ≥ 0 x 5x 7 x 3x 3 Critical numbers: x 0, x 5, x 7 Critical numbers: x 3, x 0, x 3 Test intervals: , 5, 5, 0, 0, 7, 7, x Test intervals: , 3 ⇒ < 0 x 3x 3 x Test: Is ≥ 0? x 5x 7 x 3, 0 ⇒ > 0 x 3x 3 By testing an x-value in each test interval in the inequality, we see that the domain set is: 5, 0 7, x 0, 3 ⇒ < 0 x 3x 3 x 3, ⇒ > 0 x 3x 3 Domain: 3, 0 3,

61. 0.4x2 5.26 < 10.2 62. 1.3x2 3.78 > 2.12 0.4x2 4.94 < 0 1.3x2 1.66 > 0 0.4x2 12.35 < 0 Critical numbers: ±1.13 Critical numbers: x ±3.51 Test intervals: , 1.13, 1.13, 1.13, 1.13, Test intervals: , 3.51, 3.51, 3.51, 3.51, Solution set: 1.13, 1.13 By testing an x-value in each test interval in the inequality, we see that the solution set is: 3.51, 3.51

63. 0.5x2 12.5x 1.6 > 0 64. 1.2x2 4.8x 3.1 < 5.3 ± 2 1.2x2 4.8x 2.2 0 12.5 12.5 4 0.5 1.6 < The zeros are x . 2 0.5 Critical numbers: 4.42, 0.42 Critical numbers:x 0.13 , x 25.13 Test intervals: , 4.42, 4.42, 0.42, 0.42, Test intervals: , 0.13 , 0.13, 25.13 , 25.13, Solution set: 4.42, 0.42 By testing an x-value in each test interval in the inequality, we see that the solution set is: 0.13, 25.13

1 2 65. > 3.4 66. > 5.8 2.3x 5.2 3.1x 3.7 1 2 5.83.1x 3.7 3.4 > 0 > 0 2.3x 5.2 3.1x 3.7 1 3.42.3x 5.2 23.46 17.98x > 0 > 0 2.3x 5.2 3.1x 3.7 7.82x 18.68 Critical numbers: x 1.19, x 1.30 > 0 2.3x 5.2 23.46 17.98x Test intervals: , 1.19 ⇒ < 0 Critical numbers: x 2.39, x 2.26 3.1x 3.7 Test intervals: , 2.26, 2.26, 2.39, 2.39, 23.46 17.98x 1.19, 1.30 ⇒ > 0 3.1x 3.7 By testing an x-value in each test interval in the inequality, we see that the solution set is: 2.26, 2.39 23.46 17.98x 1.30, ⇒ < 0 3.1x 3.7 Solution interval: 1.19, 1.30 234 Chapter 2 Polynomial and Rational Functions

2 2 2 2 67. s 16t v0t s0 16t 160t 68. s 16t v0t s0 16t 128t (a) 16t2 160t 0 (a) 16t2 128t 0 16tt 10 0 16tt 8 0 t 0, t 10 16t 0 ⇒ t 0 It will be back on the ground in 10 seconds. t 8 0 ⇒ t 8

(b) 16t2 160t > 384 It will be back on the ground in 8 seconds. 2 16t2 160t 384 > 0 (b) 16t 128t < 128 2 16t2 10t 24 > 0 16t 128t 128 < 0 t2 10t 24 < 0 Critical numbers: 4 2 2, 4 2 2 t 4t 6 < 0 Test intervals: , 4 22, 4 22, 4 22, 4 < t < 6 seconds 4 22,

Solution set: 0 seconds ≤ t < 4 22 seconds and 4 22 seconds < t ≤ 8 seconds

69. 2 L 2W 100 ⇒ W 50 L 70. 2L 2W 440 ⇒ W 220 L LW ≥ 500 LW ≥ 8000 L50 L ≥ 500 L220 L ≥ 8000 L2 50L 500 ≥ 0 L2 220L 8000 ≥ 0 By the Quadratic Formula we have: By the Quadratic Formula we have: Critical numbers: L 25 ± 55 Critical numbers: L 110 ± 1041 Test: Is L2 50L 500 ≥ 0? Test: Is L2 220L 8000 ≥ 0? Solution set: 25 55 ≤ L ≤ 25 55 Solution set: 110 1041 ≤ L ≤ 110 1041 13.8 meters ≤ L ≤ 36.2 meters 45.97 feet ≤ L ≤ 174.03 feet

71. R x75 0.0005x and C 30x 250,000 72. What is the price per unit? P R C When x 90,000: 75x 0.0005x2 30x 250,000 2,880,000 R $2,880,000 ⇒ $32 per unit 90,000 0.0005x2 45x 250,000 When x 100,000: P ≥ 750,000 3,000,000 0.0005x2 45x 250,000 ≥ 750,000 R $3,000,000 ⇒ $30 per unit 100,000 0.0005x2 45x 1,000,000 ≥ 0 Solution interval: $30.00 ≤ p ≤ $32.00 Critical numbers:x 40,000, x 50,000 (These were obtained by using the Quadratic Formula.) Test intervals: 0, 40,000, 40,000, 50,000, 50,000, By testing x- values in each test interval in the inequality, we see that the solution set is 40,000, 50,000 or 40,000 ≤ x ≤ 50,000. The price per unit is R p 75 0.0005x. x For x 40,000, p $55. For x 50,000, p $50. Therefore, for 40,000 ≤ x ≤ 50,000, $50.00 ≤ p ≤ $55.00. Section 2.7 Nonlinear Inequalities 235

73. C 0.0031t3 0.216t 2 5.54t 19.1, 0 ≤ t ≤ 23

(a) 80 (d)C will be between 85% and 100% t C when t is between 37 and 42. These 36 83.2 values correspond to the years 2017 to 2022.

023 37 85.4 0 38 87.8 (b)C will be greater than 75% when t C 39 90.5 t 31, which corresponds to 2011. 24 70.5 40 93.5 26 71.6 41 96.8 28 72.9 42 100.4 30 74.6 43 104.4

32 76.8 (e)85 ≤ C ≤ 100 when 36.82 ≤ t ≤ 41.89 or 37 ≤ t ≤ 42. 34 79.6 (f) The model is a third-degree polynomial and as t → , C → . (c)C 75 when t 30.41.

1 1 1 74. (a) 75. d 4681012 R R1 2 Load 2223.9 5593.9 10,312 16,378 23,792 2 R1 2R RR1 2 R1 R 2 R1 L

2R1 25,000 R 2 R 20,000 1 15,000 Since R ≥ 1, we have 10,000 2R 5,000 1 Maximum safe load ≥ 1 d 2 R1 4681012 2R Depth of the beam 1 ≥ 1 0 2 R1 (b) 2000 ≤ 168.5d 2 472.1 R 2 1 ≥ 2472.1 ≤ 168.5d 2 0. 2 R1 14.67 ≤ d2 Since R1 > 0, the only critical number is R1 2. The ≥ 3.83 ≤ d inequality is satisfied when R1 2 ohms. The minimum depth is 3.83 inches.

76. (a) N 0.03t 2 9.6t 172 (b) and (d) 220 ⇒ t 5 N 320 So the number of master’s degrees earned by women N = 320 280 exceeded 220,000 in 1995. 240 2 (c) N 0.03t 9.6t 172 200 N = 220 320 ⇒ t 16.2 (in thousands) 160 Master's degrees earned Master's degrees t So the number of master’s degrees earned by women 261014 18 ↔ will exceed 320,000 in 2006. Year (0 1990) 236 Chapter 2 Polynomial and Rational Functions

77. True 78. True x3 2x2 11x 12 x 3x 1x 4 The y-values are greater than zero for all values of x. The test intervals are , 3, 3, 1, 1, 4, and 4, .

79. x2 bx 4 0 80. x2 bx 4 0 To have at least one real solution,b2 16 ≥ 0. This To have at least one real solution, occurs when b ≤ 4 or b ≥ 4. This can be written as b2 414 ≥ 0 , 4 4, . b2 16 ≥ 0. This inequality is true for all real values of b. Thus, the interval for b such that the equation has at least one real solution is , .

81. 3x2 bx 10 0 82. 2x2 bx 5 0 To have at least one real solution, b2 4310 ≥ 0. To have at least one real solution, b2 120 ≥ 0 b2 425 ≥ 0 b 120b 120 ≥ 0 b2 40 ≥ 0. Critical numbers: b ±120 ±230 This occurs when b ≤ 210 or b ≥ 210. Thus, the interval for b such that the equation has at least Test intervals: one real solution is , 210 210, . , 230, 230, 230, 230, Test: Is b2 120 ≥ 0? Solution set: , 230 230,

83. (a) If a > 0 and c ≤ 0, then b can be any real number. If 84. (a) x a, x b a > 0 and c > 0, then for b2 4ac to be greater than or (b) − ++ b b ac equal to zero, is restricted to < 2 or −−+ b > 2 ac. + − + x (b) The center of the interval for b in Exercises 79–82 is 0. a b

85. 4x2 20x 25 2x 52 86. x 32 16 x 3 4x 3 4 x 7x 1

87. x2x 3) 4x 3 x2 4x 3 88. 2 x 4 54x 2xx3 27 x 2x 2x 3 2xx 3x2 3x 9

1 89. Area length width 90. A rea 2 base height 1 2x 1 x 2 b 3b 2 2 3 2 2x x 2b b Review Exercises for Chapter 2 237

Review Exercises for Chapter 2

1. (a)y 2x2 (b) y 2x2 Vertical stretch Vertical stretch and a reflection in the x-axis

y y

4 4 3 3 2 2 1 x x −4 −3 −2 −11234 −4 −3 −2 −11234 −1 −2 −3 −3 −4 −4

(c)y x2 2 (d) y x 22 Vertical shift two units upward Horizontal shift two units to the left (a) y y 4 4 3

1 1 x x −4 −3 −2 −11234 −4 −3 −2 −11234 −1 −1 −2 −2 −3 −3 −4 −4

2. (a) y x 2 4 (b) y 4 x 2 Vertical shift four units downward Reflection in the x-axis and a vertical shift y four units upward 3 y

2 5

x 3 −4 −3 −1 134 −1 2 −2 1 x −4 −3 −1 134 −1 − 5 −2 −3

(c) y x 3 2 1 2 (d) y 2 x 1 Horizontal shift three units to the right 1 Vertical shrink (each y-value is multiplied by 2 , y and a vertical shift one unit downward 5 y 4 3 4 2 3 1 2 x 1 −3 −2 −1 13452 −1 x −4 −3 −2 234 −2 −3 −2 −3 −4 238 Chapter 2 Polynomial and Rational Functions

3. gx x2 2x y 4. fx 6x x2 y 7 2 x2 6x 9 9 10 x 2x 1 1 6 8 2 5 x 32 9 x 1 1 4 6 3 Vertex: 1, 1 Vertex: 3, 9 4 2 Axis of symmetry: x 1 Axis of symmetry: x 3 x x −3 −2 −1 1 23456 − 2 −1 2 2 2 4 8 10 0 x 2x x x 2 0 6x x x 6 x −2 −2 x-intercepts: 0, 0, 2, 0 x-intercepts: 0, 0, 6, 0

5. f x x2 8x 10 6. hx 3 4x x2 x2 8x 16 16 10 x2 4x 3 x 42 6 y x2 4x 4 4 3 y Vertex: 4, 6 x 22 7 10 2 2 Axis of symmetry: x 4 x 2 7 8

x 6 0 x 42 6 −8 −4 2 Vertex: 2, 7 −2 4 x 42 6 Axis of symmetry: x 2 −4 2 x 4 ±6 0 3 4x x2 x −6 −2 2 4 6810 x 4 ± 6 0 x2 4x 3 x-intercepts: 4 ± 6, 0 4 ± 42 413 x 21 4 ± 28 2 ± 7 2 x-intercepts: 2 ± 7, 0

7. ft 2t2 4t 1 8. f x x2 8x 12 2t2 2t 1 1 1 x2 8x 16 16 12 2t 12 1 1 x 42 4 2t 12 3 y Vertex: 4, 4 y 6 8 Vertex: 1, 3 5 Axis of symmetry: x 4 6 4 2 Axis of symmetry: t 1 3 0 x 8x 12 4 2 0 2t 12 3 2 1 0 x 2 x 6 t x 2 − 2 t 1 3 −3 −2 −1 123456 x-intercepts: 2, 0, 6, 0 2 4810 −2 3 − t 1 ± 4 2 6 t 1 ± 2 6 t-intercepts: 1 ± , 0 2 Review Exercises for Chapter 2 239

9. hx 4x2 4x 13 10. f x x2 6x 1 4x2 x 13 x2 6x 9 9 1

1 1 2 4x2 x 13 x 3 8 4 4 Vertex: 3, 8 1 4x2 x 1 13 4 Axis of symmetry: x 3 2 1 2 0 x 6x 1 4x 12 y 2 6 ± 62 411 20 x 1 21 Vertex: , 12 2 15 6 ± 32 3 ± 22 y 1 10 2 Axis of symmetry: x 2 ± 5 x-intercepts: 3 2 2, 0 2 1 2 0 4x 12 x x −2 2 4810 2 −3 −2 −1 123 −2 1 2 x 3 −4 2 −6

No real zeros −8 x-intercepts: none

11. hx x2 5x 4 12. fx 4x2 4x 5 25 25 1 1 5 x2 5x 4 4x2 x 4 4 4 4 4 5 2 25 16 1 2 x y 4x 1 y 2 4 4 2 x 12 2 −8 −6 −4 −22 2 5 41 1 10 x −2 4x 4 2 4 2 8 −4 5 41 1 Vertex: , Vertex: , 4 2 4 2 2 5 1 x − − − − Axis of symmetry: x − Axis of symmetry: x 8 6 4 2 246 2 10 2 −2 0 x2 5x 4 0 4x2 4x 5 5 ± 41 4 ± 8i 1 By the Quadratic Formula, x . By the Quadratic Formula, x ± i. 2 8 2 5 ± 41 The equation has no real zeros. x-intercepts: , 0 2 x-intercepts: None

1 y 13. fx x2 5x 4 3 4 1 25 25 x2 5x 4 2 3 4 4 x −8 −6 −4 −22 1 5 2 41 x 3 2 4 −4

1 5 2 41 − x 6 3 2 12 0 x2 5x 4 5 41 5 ± 41 Vertex: , By the Quadratic Formula, x . 2 12 2 5 5 ± 41 Axis of symmetry: x x-intercepts: , 0 2 2 240 Chapter 2 Polynomial and Rational Functions

1 y 14. f x 6x2 24x 22 2 14 3x2 12x 11 12 10 3x2 4x 4 4 11 8 6 2 3x 2 34 11 4 2 2 3 x 2 1 x –6 –4 –2 46810 Vertex: 2, 1 Axis of symmetry: x 2 0 3x2 12x 11 12 ± 122 4311 12 ± 12 3 x 2 ± 23 6 3 3 x-intercepts: 2 ± , 0 3

15. Vertex: 4, 1 ⇒ fx ax 42 1 16. Vertex: 2, 2 ⇒ fx ax 2 2 2 Point: 2, 1 ⇒ 1 a2 42 1 Point: 0, 3 ⇒ 3 a0 22 2 2 4a 3 4a 2 1 1 4a 2 a 1 1 2 a Thus, f x 2 x 4 1. 4 1 2 f x 4 x 2 2

17. Vertex: 1, 4 ⇒ fx ax 12 4 18. Vertex: 2, 3 ⇒ fx ax 2 2 3 Point: 2, 3 ⇒ 3 a2 12 4 Point: 1, 6 ⇒ 6 a1 2 2 3 1 a 6 9a 3 Thus, fx x 12 4. 3 9a 1 3 a 1 2 f x 3 x 2 3

19. (a) (b) 2 x 2y 200 (c) A rea 100x x2 x y 100 x2 100x 2500 2500 y y 100 x x 502 2500 2 x Area xy x 50 2500 x100 x The maximum area occurs at the vertex when x 50 and y 100 50 50. The dimensions with the 2 100x x maximum area are x 50 meters and y 50 meters.

20. R 10p2 800p (a) R20 $12,000 (b) The maximum revenue occurs at the vertex of the parabola. R25 $13,750 b 800 $40 2a 210 R30 $15,000 R40 $16,000 The revenue is maximum when the price is $40 per unit. The maximum revenue is $16,000. Review Exercises for Chapter 2 241

21. C 70,000 120x 0.055x2 22. 26 0.107x2 5.68x 48.5 The minimum cost occurs at the vertex of the parabola. 0 0.107x2 5.68x 74.5 b 120 5.68 ± 5.682 40.10774.5 Vertex: 1091 units x 2a 20.055 20.107

Approximately 1091 units should be produced each day to x 23.7, 29.4 y yield a minimum cost. The age of the bride is 27 approximately 24 years 26 when the age of the groom 25 24 is 26 years. 23 Age of groom 22 x 20 21 22 23 24 25 Age of bride

23. y x3, fx x 43 24. y x3, f x 4x3 25. y x4, fx 2 x4

y y y

5 3 3 4 2 3 2 1 1 1 x x x − − − − − −2 1234 6 7 3 2 1 1 23 3 2 1 23 −1 −1

−3 −2 −2 −4 −3 −3

Transformation: Reflection in f x is a reflection in the x-axis Transformation: Reflection in the x-axis and a horizontal shift and a vertical stretch of the graph the x-axis and a vertical shift four units to the right of y x3. two units upward

4 4 5 5 5 1 5 26. y x , f x 2 x 2 27. y x , f x x 3 28. y x , f x 2x 3

y y y 5 6 8 4 5 3 6 4 2 4 3 1 2 x 1 −2 1 34567 x x −3 −2 −1 123456 −624−4 −2 6

−2 −3 −5 f x is a shift to the right two Transformation: Horizontal shift f x is a vertical shrink and a units and a vertical stretch of the three units to the right vertical shift three units upward graph of y x4. of the graph of y x5.

2 1 3 29. f x x 6x 9 30. f x 2x 2x The degree is even and the leading coefficient is negative. The degree is odd and the leading coefficient is positive. The graph falls to the left and falls to the right. The graph falls to the left and rises to the right.

3 4 2 5 2 31. g x 4 x 3x 2 32. h x x 7x 10x The degree is even and the leading coefficient is positive. The degree is odd and the leading coefficient is negative. The graph rises to the left and rises to the right. The graph rises to the left and falls to the right. 242 Chapter 2 Polynomial and Rational Functions

33. f x 2x2 11x 21 20 34. fx xx 3 2 3

2 2 −9 9 xx 0 2x 11x 21 0 3 −6 6 2x 3x 7 Zeros:x 0 of multiplicity 1 (odd multiplicity) 3 − Zeros:x , 7, all of 40 −5 2 x 3 of multiplicity 2 multiplicity 1 (odd multiplicity) (even multiplicity) Turning points: 1 Turning points: 2

35. f t t3 3t 3 36. f x x3 8x2 10 −10 10 0 t3 3t 0 x3 8x2 −5 4 0 tt2 3 0 x2x 8 Zeros:x 0 of multiplicity 2 Zeros:t 0, ±3 all of −3 −80 (even multiplicity) multiplicity 1 (odd multiplicity) Turning points: 2 x 8 of multiplicity 1 (odd multiplicity) Turning points: 2

37. f x 12x3 20x2 10 38. gx x4 x3 2x2 3 0 12x3 20x2 0 x4 x3 2x2 −4 5 − 2 2 0 4x23x 5 5 5 0 x x x 2 x2x 1x 2 Zeros:x 0 of multiplicity 2 −5 −3 (even multiplicity) Zeros:x 0 of multiplicity 2 (even multiplicity) 5 x 3 of multiplicity 1 x 1 of multiplicity 1 (odd multiplicity) (odd multiplicity) x 2 of multiplicity 1 (odd multiplicity) Turning points: 2 Turning points: 3

39. fx x3 x2 2 40. gx 2x3 4x2 (a) The degree is odd and the leading coefficient is (a) The degree is odd and the leading coefficient, 2, is negative. The graph rises to the left and falls to positive. The graph rises to the right and falls to the left. the right. (b) gx 2x3 4x2 (b) Zero: x 1 0 2x3 4x2 (c) x2x x 3 2 1 012 0 2 2 0 x2x 2 fx 34 10 0 2 2 6 The zeros are 0 and 2. (d) y (c) 4 x 3 2 1 01 3 gx 18 0206 2 1 (−1, 0) x (d) y −4 −3 −2 1234 4 3 − 3 2 −4 (−2, 0) (0, 0) x −4 −3 −1 1342 −1 −2 −3 −4 Review Exercises for Chapter 2 243

41. fx xx3 x2 5x 3 42. hx 3x2 x4 (a) The degree is even and the leading coefficient is (a) The degree is even and the leading coefficient,1 , is positive. The graph rises to the left and rises to negative. The graph falls to the left and falls to the right. the right. (b) gx 3x2 x4 (b) Zeros: x 0, 1, 3 0 3x2 x4 (c) x 4 3 2 1 01 2 3 0 x23 x2 fx 100 018 8 0 0 10 72 The zeros are 0,3, and 3. (c) (d) y x 2 1 012 (−3, 0) 3 (1, 0) x h x 4 202 4 −4 −2 −1 1342 (0, 0) (d) y 4 (0, 0) 3 −15 2 −18 (− 3, 0 ( ( 3, 0 ( −21 x −4 −3 −1 134 −1 −2 −3 −4

43. (a) fx 3x3 x2 3 44. (a) f x 0.25x3 3.65x 6.12

x 3 2 1 01 2 3 x 6 5 4 3 2 fx 87 25 1 352375 fx 25.98 6.88 4.72 10.32 11.42

(b) The zero is in the interval 1, 0. x 1 0 1 2 34 Zero: x 0.900 fx 9.52 6.12 2.72 0.82 1.92 7.52

(b) The only zero is in the interval 5, 4. It is x 4.479.

45. (a) fx x4 5x 1 46. (a) f x 7x4 3x3 8x2 2

x 3 2 1 01 23 x 3 2 1 0 12 fx 95 25 5 1 5 5 65 fx 416 58 2 24106

(b) There are two zeros, one in the interval 1, 0 and (b) There are zeros in the intervals 2, 1 and 1, 0. one in the interval 1, 2 They are x 1.211 and x 0.509. Zeros: x 0.200, x 1.772

4 47. 8x 5 48. 3 3 x 2 ) 24x2 x 8 3x 2 ) 4x 7 8 24 x2 16x 4x 3 29 1 5 x 8 3 1 5 x 10 4x 7 4 29 2 3x 2 3 33x 2 24x2 x 8 2 Thus, 8x 5 . 3x 2 3x 2 244 Chapter 2 Polynomial and Rational Functions

49. 5 x 2 50. 3x2 3 x2 3x 1 ) 5x3 13x2 x 2 x2 1 ) 3x4 0x3 0x2 0x 0 5 x3 15x2 5x 3x4 3x2 2 x2 6x 2 3x2 0 2 x2 6x 2 3x2 3 0 3 5x3 13x2 x 2 3x4 3 Thus, 5x 2. 3x2 3 x2 3x 1 x2 1 x2 1

51. x2 3x 2 52. 3x2 5x 8 x2 0x 2 ) x4 3x3 4x2 6x 3 2x2 0x 1 ) 6x4 10x3 13x2 5x 2 x4 0x3 2x2 6x4 0x3 3x2 3x3 2x2 6x 10x3 16x2 5x 3x3 0x2 6x 10x3 0x2 5x 2 x2 0x 3 16x2 0x 2 2 x2 0x 4 16x2 0x 8 1 10 x4 3x3 4x2 6x 3 1 6x4 10x3 13x2 5x 2 10 Thus, x2 3x 2 . 3x2 5x 8 x2 2 x2 2 2x2 1 2x2 1

53. 2 6 4 27 18 0 54. 5 0.1 0.3 0 0.5 12 16 22 8 0.5 4 20 0.1 0.8 4 19.5 6 8 11 4 8 0.1x3 0.3x2 0.5 19.5 Thus, 0.1x2 0.8x 4 x 5 x 5 6x4 4x3 27x2 18x 8 6x3 8x2 11x 4 . x 2 x 2

55. 4219 38 24 56. 33 20 29 12 8 44 24 9 33 12 3 11 4 0 2 11 6 0 3x3 20x2 29x 12 2x3 19x2 38x 24 3x2 11x 4 Thus, 2x2 11x 6. x 3 x 4

57. f x 20x 4 9x3 14x2 3x 3 (a) 120 9 14 3 0 (b) 4 20 9 14 3 0 20 11 3 0 15 18 3 0 20 11 3 0 0 20 24 4 0 0 3 Yes,x 1 is a zero of f. Yes,x 4 is a zero of f. (c) 0209 14 3 0 (d) 120 9 14 3 0 0 0 0 0 20 29 15 12 20 9 14 3 0 20 29 15 12 12

Yes,x 0 is a zero of f. No,x 1 is not a zero of f. Review Exercises for Chapter 2 245

58. f x 3x3 8x2 20x 16 (a) 438 20 16 (b) 43 8 20 16 12 16 16 12 80 240 3 4 4 0 3 20 60 224 Yes,x 4 is a zero of f. No,x 4 is not a zero of f. 2 (d) 13 8 20 16 (c) 3 3 8 20 16 2 4 16 3 11 9 3 6 24 0 3 11 9 25 2 No,x 1 is not a zero of f. Yes,x 3 is a zero of f.

59. f x x4 10x3 24x2 20x 44 (a) 31 10 24 20 44 (b) 11 10 24 20 44 3 21 135 465 1 9 33 53 1 7 45 155 421 1 9 33 53 9

Thus, f 3 421. f 1 9

60. gt 2t5 5t4 8t 20 (a) 42 5 0 0 8 20 8 52 208 832 3296 2 13 52 208 824 3276 Thus, g4 3276. (b) 2 2 5 0 0 8 20 22 52 4 10 42 102 8 20 2 5 22 52 4 10 42 102 0 Thus, g2 0.

61. f x x3 4x2 25x 28; Factor: x 4 62. f x 2x3 11x2 21x 90 (a) 414 25 28 (a) 62 11 21 90 4 32 28 12 6 90 1 8 7 0 2 1 15 0 Yes,x 4 is a factor of f x. Yes,x 6 is a factor of f x. (b) x2 8x 7 x 7x 1 (b) 2x2 x 15 2x 5x 3 The remaining factors of f are x 7 and x 1. The remaining factors are 2x 5 and x 3. (c) f x x3 4x2 25x 28 (c) f x 2x 5x 3x 6 5 x 7x 1x 4 (d) Zeros: x 2, 3, 6 (d) Zeros: 7, 1, 4 (e) 50

(e) 80 −75

−85 −100

−60 246 Chapter 2 Polynomial and Rational Functions

63. f x x 4 4x3 7x2 22x 24 64. f x x4 11x3 41x2 61x 30 Factors: x 2, x 3 (a) 2111 41 61 30 2 18 46 30 (a) 214 7 22 24 2 12 10 24 1 9 23 15 0 1 6 5 12 0 51 9 23 15 5 20 15 31 6 5 12 1 4 3 0 3 9 12 Yes,x 2 and x 5 are both factors of f x. 1 3 4 0 (b) x2 4x 3 x 1x 3 Both are factors since the remainders are zero. The remaining factors are x 1 and x 3. (b) x2 3x 4 x 1x 4 (c) f x x 1x 3x 2x 5 The remaining factors are x 1 and x 4. (d) Zeros: x 1, 2, 3, 5 (c) f x x 1x 4x 2x 3 (e) 4 (d) Zeros: 2, 1, 3, 4

−612 (e) 40

−8 −3 5

−10

65. 6 4 6 2i 66. 3 25 3 5i 67. i2 3i 1 3i

68. 5i i2 1 5i 69. 7 5i 4 2i 7 4 5i 2i 3 7i

2 2 2 2 2 2 2 2 2 70. i i i i 2 i 2i 2 2 2 2 2 2 2 2 2

71. 5i13 8i 65i 40i2 40 65i 72. 1 6i5 2i 5 2i 30i 12i2 5 28i 12 17 28i

73. 10 8i2 3i 20 30i 16i 24i2 74. i6 i3 2i i18 12i 3i 2i2 4 46i i20 9i 20i 9i2 9 20i

6 i 6 i 4 i 3 2i 3 2i 5 i 75. 76. 4 i 4 i 4 i 5 i 5 i 5 i 24 10i i2 15 3i 10i 2i2 16 1 25 i2 23 10i 17 7i 17 26 23 10 17 7i i 17 17 26 26 Review Exercises for Chapter 2 247

4 2 4 2 3i 2 1 i 1 5 1 4i 52 i 77. 78. 2 3i 1 i 2 3i 2 3i 1 i 1 i 2 i 1 4i 2 i1 4i 8 12i 2 2i 1 4i 10 5i 4 9 1 1 2 8i i 4i2 8 12 9 i 2 9i i 1 i 13 13 2 9i 2 9i 8 12 18 81i 2i 9i2 1 i i 13 13 4 81i2 21 1 9 83i 9 83i i 13 13 85 85 85

79. 3x2 1 0 80. 2 8x2 0 3 x2 1 8 x2 2 1 1 x2 x2 3 4 1 1 x ± x ± i 3 2 1 3 ± i ± i 3 3

81. x2 2x 10 0 82. 6x2 3x 27 0 x2 2x 1 10 1 b ± b2 4ac x 2a x 12 9 3 ± 32 4627 x 1 ±9 26 x 1 ± 3i 3 ± 639 12 3 ± 3i71 1 71 ± i 12 4 4

83. f x 3xx 22 84. f x x 4x 92 85. f x x2 9x 8 86. f x x3 6x Zeros: x 0, x 2 Zeros: x 9, 4 x 1x 8 xx2 6 Zeros: x 1, x 8 Zeros: x 0, ±6i

87. f x x 4x 6x 2ix 2i 88. f x x 8x 52x 3 ix 3 i Zeros: x 4, x 6, x 2i, x 2i Zeros: x 5, 8, 3 ± i

89. fx 4x3 8x2 3x 15 90. fx 3x4 4x3 5x2 8 ± ± ± ± ± 1 ± 3 ± 5 ± 15 ± ± ± ± ± 1 ± 2 ± 4 ± 8 Possible rational zeros: 1, 3, 5, 15, 2, 2, 2, 2 , Possible rational zeros: 1, 2, 4, 8, 3, 3, 3, 3 ± 1 ± 3 ± 5 ± 15 4, 4, 4, 4 248 Chapter 2 Polynomial and Rational Functions

91. f x x3 2x2 21x 18 92. f x 3x3 20x2 7x 30 Possible rational zeros: ±1, ±2, ±3, ±6, ±9, ±18 Possible rational zeros: ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30, ± 1, ± 2, ± 5, ± 10 112 21 18 3 3 3 3 1 3 18 1320 7 30 1 3 18 0 3 23 30 x3 2x2 21x 18 x 1x2 3x 18 3 23 30 0 S o, f x 3x3 20x2 7x 30 x 1x 6x 3 x 13x2 23x 30 The zeros of f x are x 1, x 6, and x 3. x 13x 5x 6 0 x 13x 5x 6. 5 Zeros: x 1, 3, 6

93. f x x3 10x2 17x 8 94. f x x3 9x2 24x 20 Possible rational zeros: ±1, ±2, ±4, ±8 Possible rational zeros: ±1, ±2, ±4, ±5, ±10, ±20 1110 17 8 51 9 24 20 1 9 8 5 20 20 1 9 8 0 1 4 4 0 x3 10x2 17x 8 x 1x2 9x 8 S o, f x x3 9x2 24x 20 x 1x 1x 8 x 5x2 4x 4 x 12x 8 x 5x 22. The zeros of f x are x 1 and x 8. Zeros: x 5, 2

95. f x x4 x3 11x2 x 12 Possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12 311 11 1 12 3 12 3 12 1 4 1 4 0 41 4 1 4 4 0 4 1 0 1 0 x4 x3 11x2 x 12 x 3x 4x2 1 The real zeros of f x are x 3, and x 4.

96. f x 25x 4 25x3 154x2 4x 24 ± ± ± ± ± ± ± ± ± 1 ± 2 ± 3 ± 4 ± 6 ± 8 ± 12 Possible rational zeros: 1, 2, 3, 4, 6, 8, 12, 24, 5, 5, 5, 5, 5, 5, 5 , ± 24 ± 1 ± 2 ± 3 ± 4 ± 6 ± 8 ± 12 ± 24 5 , 25, 25, 25, 25, 25, 25, 25, 25 325 25 154 4 24 75 150 12 24 25 50 4 8 0 22550 4 8 50 0 8 25 0 4 0 So, f x 25x 4 25x3 154x2 4x 24 x 3x 225x2 4 x 3x 25x 25x 2. ± 2 Zeros: x 3, 2, 5 Review Exercises for Chapter 2 249

2 97. f x 3 x 3 x 4 x 3i x 3i Since 3i is a zero, so is 3i. 3x 2x 4x2 3 Multiply by 3 to clear the fraction. 3x2 14x 8x2 3 3x4 14x3 17x2 42x 24

4 3 2 2 ± Note: f x a 3x 14x 17x 42x 24 , where a is any real nonzero number, has zeros 3, 4, and 3i.

98. Since 1 2i is a zero and the coefficients are real, 99. fx x3 4x2 x 4, Zero: i 1 2i must also be a zero. Since i is a zero, so is i. f x x 2x 3x 1 2ix 1 2i i 1 4 1 4 x2 x 6x 12 4 i 1 4i 4 x2 x 6x2 2x 5 1 4 i 4i 0 x 4 x3 3x2 17x 30 i 1 4 i 4i i 4i 1 4 0 fx x ix ix 4, Zeros: x ± i, 4

100. hx x3 2x2 16x 32 101. gx 2x 4 3x3 13x2 37x 15, Zero: 2 i Since 4i is a zero, so is 4i. Since 2 i is a zero, so is 2 i 4i 1 2 16 32 2 i 2 3 13 37 15 4i 16 8i 32 4 2i 5i 31 3i 15 1 2 4i 8i 0 2 1 2i 13 5i 6 3i 0

4i 1 2 4i 8i 2 i 2 1 2i 13 5i 6 3i 4i 8i 4 2i 10 5i 6 3i 1 2 0 2 5 3 0 hx x 4ix 4ix 2 gx x 2 ix 2 i2x2 5x 3 Zeros: x ± 4i, 2 x 2 ix 2 i2x 1x 3 ± 1 Zeros: x 2 i, 2, 3

102. fx 4x4 11x3 14x2 6x 103. fx x3 4x2 5x x4x3 11x2 14x 6 xx2 4x 5 One zero is x 0. Since 1 i is a zero, so is 1 i. xx 5x 1 1 i 4 11 14 6 Zeros: x 0, 5, 1 4 4i 11 3i 6 4 7 4i 3 3i 0

1 i 4 7 4i 3 3i 4 4i 3 3i 4 3 0 f x xx 1 ix 1 i4x 3 xx 1 ix 1 i4x 3 3 Zeros: 0, 4, 1 i, 1 i 250 Chapter 2 Polynomial and Rational Functions

104. gx x3 7x2 36 217 0 36 2 18 36 1 9 18 0

The zeros of x2 9x 18 x 3x 6 are x 3, 6. The zeros of gx are 2, 3, 6. gx x 2x 3x 6

105. gx x 4 4x3 3x2 40x 208, Zero: x 4 106. fx x4 8x3 8x2 72x 153 4 1 4 3 40 208 3 1 8 8 72 153 4 0 12 208 3 33 123 153 1 0 3 52 0 1 11 41 51 0 41 0 3 52 3 1 11 41 51 4 16 52 3 24 51 1 4 13 0 1 8 17 2 2 g x x 4 x 4x 13 By the Quadratic Formula, the zeros of x2 8x 17 are 2 By the Quadratic Formula the zeros of x 4x 13 are 8 ± 82 4117 8 ± 4 ± x 4 ± i. x 2 3i. The zeros of g x are x 4 of multiplicity 21 2 2, and x 2 ± 3i. The zeros of fx are 3, 3, 4 i, 4 i. gx x 42x 2 3ix 2 3i fx x 3x 3x 4 ix 4 i x 42x 2 3ix 2 3i

107. gx 5x3 3x2 6x 9 108. hx 2x5 4x3 2x2 5 gx has two variations in sign, so g has either two or no hx has three variations in sign, so h has either three or positive real zeros. one positive real zeros. 5 3 2 gx 5x3 3x2 6x 9 h x 2 x 4 x 2 x 5 gx has one variation in sign, so g has one negative 2x5 4x3 2x2 5 real zero. hx has two variations in sign, so h has either two or no negative real zeros.

109. fx 4x3 3x2 4x 3 110. gx 2x3 5x2 14x 8 (a) 143 4 3 (a) 825 14 8 4 1 5 16 88 592 4 1 5 2 2 11 74 600 Since the last row has all positive entries, Since the last row has all positive entries,x 8 is an x 1 is an upper bound. upper bound. (b) 42 5 14 8 (b) 1 4 3 4 3 4 8 52 152 5 1 1 4 2 13 38 144 4 4 5 17 4 Since the last row entries alternate in sign,x 4 is a Since the last row entries alternate in sign, lower bound. 1 x 4 is a lower bound. Review Exercises for Chapter 2 251

5x 3x2 8 111. f x 112. f x 113. f x x 12 1 3x x2 10x 24 Domain: all real numbers x 1 3x 0 8 except x 12 x 4x 6 3 x 1 Domain: all real numbers x 1 x except x 4 and x 6 3 Domain: all real numbers x 1 except x 3

x2 x 2 4 2x2 5x 3 114. f x 115. fx 116. fx x2 4 x 3 x2 2 Domain: all real numbers Vertical asymptote: x 3 Vertical asymptote: none Horizontal asymptote: y 0 Horizontal asymptote: y 2

2x 10 x3 4x2 x2x 4 117. hx 118. hx x2 2x 15 x2 3x 2 x 2x 1 2x 5 Vertical asymptotes: x 2, x 1 x 3x 5 Horizontal asymptotes: none 2 , x 5 x 3 Vertical asymptote: x 3 Horizontal asymptote: y 0

5 4 119. fx 120. fx x2 x (a) Domain: all real numbers x except x 0 (a) Domain: all real numbers x except x 0 (b) No intercepts (b) No intercepts (c) Vertical asymptote: x 0 (c) Vertical asymptote: x 0 Horizontal asymptote: y 0 Horizontal asymptote: y 0 (d) (d) x ±3 ±2 ±1 x 3 2 1 123 y 4 4 5 5 3 2 4 423 y 9 4 5

y y

1 4 x 3 − 112 2

−2 1 x −3 −3 −2 −1 1234

−2 −3 252 Chapter 2 Polynomial and Rational Functions

2 x x 2 x 3 121. gx 122. hx 1 x x 1 x 2 (a) Domain: all real numbers x except x 1 (a) Domain: all real numbers x except x 2 (b) x-intercept: 2, 0 (b) x-intercept: 3, 0 3 y-intercept: 0, 2 y-intercept: 0, 2 (c) Vertical asymptote: x 1 Horizontal asymptote: y 1 (c) Vertical asymptote: x 2 Horizontal asymptote: y 1 (d) x 1 023 (d) x 1 01345 1 5 y 2 2 4 2 4 3 1 2 y 3 2 202 3 y y 6 5 4 (0, 2) 4 (−2, 0) 2 3 x 3 (0, 2 ( −2

−4 x −2 −11 456 −6 (3, 0) −8 −2 −3

x2 2x 123. px 124. f x x2 1 x2 4 (a) Domain: all real numbers x (a) Domain: all real numbers x (b) Intercept: 0, 0 (b) Intercept: 0, 0 (c) Horizontal asymptote: y 1 (c) Horizontal asymptote: y 0

(d) (d) x ±3 ±2 ±1 0 x 2 1 012 1 2 2 1 9 4 1 y 2 5 0 5 2 y 10 5 2 0

y y

4 3

3 2

2 1 (0, 0) x 123 x −1 −3 −2 −123(0, 0) −2

−2 −3

x 125. fx x2 1 (a) Domain: all real numbers x (d) y x 2 1 012 (b) Intercept: 0, 0 2 2 1 1 2 y 5 2 0 2 5 (c) Horizontal asymptote: y 0 1 (0, 0) x 12 −1

−2 Review Exercises for Chapter 2 253

4 6x2 126. hx 127. f x x 12 x2 1 (a) Domain: all real numbers x except x 1 (a) Domain: all real numbers x (b)y- intercept: 0, 4 (b) Intercept: 0, 0 (c) Vertical asymptote: x 1 (c) Horizontal asymptote: y 6 Horizontal asymptote: y 0 (d) (d) x ±3 ±2 ±1 0 x 2 1 0234 27 24 4 4 y 5 5 3 0 y 9 1 4419 y y 4 7 2 6 (0, 0) 5 x −6 −4 −2 246 (0, 4) 3

1 x −8 −3 −2 −1 2345

2x2 6x2 11x 3 128. y 129. f x x2 4 3x2 x ± (a) Domain: all real numbers x except x 2 3x 12x 3 2x 3 1 , x (b) Intercept: 0, 0 x3x 1 x 3 1 (c) Vertical asymptotes: x 2, x 2 (a) Domain: all real numbers x except x 0 and x Horizontal asymptote: y 2 3 3 (d) (b)x- intercept: , 0 x ±5 ±4 ±3 ±1 0 2 y-intercept: none 50 8 18 2 y 21 3 5 3 0 (c) Vertical asymptote: x 0 y y Horizontal asymptote: 2 (d) 6 x 2 1 1 234 4 y 7 1 5 2 511 2 4 (0, 0) x −6 −4 46 y

2 x −8468−6 −4 −2 −2 3 (2, 0 ( −4 −6 −8 254 Chapter 2 Polynomial and Rational Functions

6x2 7x 2 2x3 2x 130. f x 131. fx 2x 4x2 1 x2 1 x2 1 (a) Domain: all real numbers x 2x 1 3x 2 3x 2 1 , x 2x 1 2x 1 2x 1 2 (b) Intercept: 0, 0 1 (a) Domain: all real numbers x except x ± (c) Slant asymptote: y 2x 2 (d) (b)y- intercept: 0, 2 x 2 1 012 2 16 16 x-intercept: , 0 y 5 1 01 5 3 y 1 (c) Vertical asymptote: x 2 3 3 2 Horizontal asymptote: y 2 1 (0, 0) x (d) − − − 2 3 2 1 123 x 3 2 1 0123 y 11 8 1 4 −2 5 3 502 3 5 −3

x −323−2 −1 (2, 0 ( 3

x2 1 132. f x x 1 (a) Domain: all real numbers x except x 1 (b) y-intercept: 0, 1 (c) Vertical asymptote: x 1 x2 1 2 Using long division, f x x 1 . x 1 x 1 Slant asymptote: y x 1

(d) 3 1 x 6 2 2 2 04 37 13 5 17 y 5 5 2 2 1 5

4 (0, 1)

x −66−4 −2 2 4 Review Exercises for Chapter 2 255

3x3 2x2 3x 2 3x3 4x2 12x 16 133. fx 134. fx 3x2 x 4 3x2 5x 2 3x 2x 1x 1 x 2x 23x 4 3x 4x 1 x 23x 1 3x 2x 1 x 23x 4 , x 2 3x 4 3x 1 1 23 1 x , x 1 (a) Domain: all real x except x 2 or x 3 3x 4 3 4 (a) Domain: all real numbers x except x 1, x (b)y- intercept: 0, 8 3 4 2 x-intercepts: , 0, 2, 0 (b)x- intercepts:1, 0 and , 0 3 3 1 1 y-intercept: 0, (c) Vertical asymptote: x 2 3 Using long division, 4 (c) Vertical asymptote: x 3x2 10x 8 5 3 f x x 3 . 3x 1 3x 1 1 Slant asymptote: y x 3 Slant asymptote: y x 3 (d) (d) x 3 2 0123 x 4 1 0124 y 44 12 1 14 y 96 21 1 16 13 5 2 02 5 13 4 8 2 0 11

y y

4 4 (4, 0 ( 3 2 3

2 2 x (3, 0 ( −646−4 −2 1 − 1 −2 (2, 0) (0, 2 ( (1, 0) x −2234−1 −6 −2 (0, −8)

C 0.5x 500 528p 135. C , 0 < x 136. C , 0 ≤ p < 100 x x 100 p 0.5 (a) 4000 Horizontal asymptote: C 0.5 1 As x increases, the average cost per unit approaches the horizontal asymptote, C 0.5 $0.50. 0 100 0

52825 (b) When p 25, C $176 million. 100 25 52850 When p 50, C $528 million. 100 50 52875 When p 75, C $1584 million. 100 75 (c) As p → 100, C → . No, it is not possible. 256 Chapter 2 Polynomial and Rational Functions

137. (a) (c) Because the horizontal margins total 4 inches, x must be 2 in. greater than 4 inches. The domain is x > 4.

y 200 2 in. 2 in. (d)

2 in. x

(b) The area of print is x 4y 4, which is 432 30 square inches. 0 The minimum area occurs when x 9.477 inches, so x 4y 4 30 22 9.477 7 30 y y 4 9.477 inches. x 4 9.477 4 30 The least amount of paper used is for a page size of about y 4 x 4 9.48 inches by 9.48 inches. 30 4x 4 y x 4 4x 14 y x 4 22x 7 y x 4 22x 7 2x2x 7 Total area xy x x 4 x 4

18.47x 2.96 138. y , 0 < x 139. 6 x2 5x < 4 0.23x 1 6 x2 5x 4 < 0 The limiting amount of CO2 uptake is determined by the horizontal asymptote, 3x 42x 1 < 0 4 1 18.47 Critical numbers: x , x y 80.3 mgdm2hr. 3 2 0.23 4 4 1 1 Test intervals: , 3, , 3, 2 , 2, 90 Test: Is 3x 42x 1 < 0? By testing an x-value in each test interval in the 4 1 inequality, we see that the solution set is: 3, 2

0 100 0

140. 2 x2 x ≥ 15 141. x3 16x ≥ 0 2 x2 x 15 ≥ 0 xx 4x 4 ≥ 0 2x 5x 3 ≥ 0 Critical numbers: x 0, x ±4 5 Critical numbers: x 2, x 3 Test intervals: , 4 , 4, 0 , 0, 4 , 4, Test intervals: , 3 ⇒ 2x 5x 3 > 0 Test: Is xx 4x 4 ≥ 0? 5 ⇒ 3, 2 2x 5 x 3 < 0 By testing an x- value in each test interval in the inequality, 5 ⇒ we see that the solution set is: 4, 0 4, . 2, 2x 5 x 3 > 0 5 Solution interval: , 3 2, Review Exercises for Chapter 2 257

2 3 142. 12x3 20x2 < 0 143. ≤ x 1 x 1 4 x23x 5 < 0 2x 1 3x 1 5 ≤ 0 Critical numbers: x 0, x 3 x 1x 1 ⇒ 3 2 Test intervals: , 0 12x 20x < 0 2x 2 3x 3 ≤ 0 5 ⇒ 3 2 x 1x 1) 0, 3 12x 20x < 0 5 ⇒ 3 2 x 5 3, 12x 20x > 0 ≤ 0 x 1x 1 5 Solution interval: , 0 0, 3 Critical numbers: x 5, x ±1 Test intervals: , 5, 5, 1, 1, 1, 1, x 5 Test: Is ≤ 0? x 1x 1 By testing an x-value in each test interval in the inequality, we see that the solution set is: 5, 1 1,

x 5 x2 7x 12 144. < 0 145. ≥ 0 3 x x Critical numbers: x 5, x 3 x 4x 3 ≥ 0 x x 5 Test intervals: , 3 ⇒ < 0 3 x Critical numbers: x 4, x 3, x 0 x 5 Test intervals: , 4, 4, 3, 3, 0, 0, 3, 5 ⇒ > 0 3 x x 4x 3 Test: Is ≥ 0? x 5 x 5, ⇒ < 0 3 x By testing an x-value in each test interval in the inequality, Solution intervals: , 3 5, we see that the solution set is: 4, 3 0,

1 1 146. > 147. 50001 r2 > 5500 x 2 x 1 r2 > 1.1 1 1 > 0 x 2 x 1 r > 1.0488

Critical numbers: x 2, x 0 r > 0.0488 1 1 Test intervals: , 0 ⇒ > 0 r > 4.9% x 2 x 1 1 0, 2 ⇒ < 0 x 2 x 1 1 2, ⇒ > 0 x 2 x Solution interval: , 0 2,

10001 3t 148. P 149. False. A fourth-degree polynomial 150. False. (See Exercise 123.) 5 t can have at most four zeros and The domain of 10001 3t complex zeros occur in conjugate 2000 ≤ 1 5 t pairs. f x x2 1 20005 t ≤ 10001 3t is the set of all real numbers x. 10,000 2000t ≤ 1000 3000t 1000t ≤ 9000 t ≥ 9 days 258 Chapter 2 Polynomial and Rational Functions

151. The maximum (or minimum) value of a quadratic 152. Answers will vary. Sample answer: function is located at its graph’s vertex. To find the Polynomials of degree n > 0 with real coefficients can vertex, either write the equation in standard form or be written as the product of linear and quadratic factors use the formula with real coefficients, where the quadratic factors have b b no real zeros. , f . 2a 2a Setting the factors equal to zero and solving for the If the leading coefficient is positive, the vertex is a variable can find the zeros of a polynomial function. minimum. If the leading coefficient is negative, the To solve an equation is to find all the values of the vertex is a maximum. variable for which the equation is true.

153. An asymptote of a graph is a line to which the graph becomes arbitrarily close as x increases or decreases without bound.

Problem Solving for Chapter 2

1. fx ax3 bx2 cx d ax2 ak bx ak2 bk c x k) ax3 bx2 cx d ax3 akx2 ak bx2 cx ak bx2 ak2 bkx ak2 bk cx d ak2 bk cx ak3 bk2 ck ak3 bk2 ck d Thus, f x ax3 bx2 cx d x kax2 ak bx ak2 bx c ak3 bk2 ck d and f k ak3 bk2 ck d. Since the remainder r ak3 bk2 ck d, f k r.

a2 2. (a) (d) 3x3 x2 90; a 3, b 1 ⇒ 9 y y3 y2 b3 3 2 12 9 3x 9x 9 90 3 2 ⇒ ⇒ 212 3x 3x 810 3x 9 x 3 a2 4 336 (e) 2x3 5x2 2500; a 2, b 5 ⇒ b3 125 480 4 4 4 2x3 5x2 2500 5 150 125 125 125 2x 3 2x 2 2x 6 252 80 ⇒ 4 ⇒ x 10 5 5 5 7 392 a2 49 (f) 7x3 6x2 1728; a 7, b 6 ⇒ 8 576 b3 216 49 49 49 9 810 7x3 6x2 1728 216 216 216 10 1100 7x 3 7x 2 7x 392 ⇒ 7 ⇒ x 6 6 6 6 (b) x3 x2 252 ⇒ x 6 a2 100 a2 1 (g) 10x3 3x2 297; a 10, b 3 ⇒ (c) x3 2x2 288; a 1, b 2 ⇒ b3 27 b3 8 100 100 100 1 1 1 3 2 x3 2x2 288 10x 3x 297 8 8 8 27 27 27 3 2 x 3 x 2 x 10x 10x ⇒ 10x ⇒ 36 ⇒ 3 ⇒ x 6 1100 10 x 3 2 2 2 3 3 3 Problem Solving for Chapter 2 259

3. V l w h x2x 3 x2x 3 20 x3 3x2 20 0 Possible rational zeros: ±1, ±2, ±4, ±5, ±10, ±20 x 213 0 20 x + 3 2 10 20 x 1 5 10 0 x 2x2 5x 10 0 Choosing the real positive value for x we have:x 2 and x 3 5. The dimensions of the mold are 2 inches 2 inches 5 inches. 5 ± 15i x 2 or x 2

f x rx 4. False. Since f x dxqx rx, we have qx . dx dx f x f 1 The statement should be corrected to read f 1 2 since qx . x 1 x 1

9 4 5. (a) y ax2 bx c 6. (a) Slope 5 3 2 0, 4: 4 a02 b0 c Slope of tangent line is less than 5. 4 c 4 1 (b) Slope 3 4, 0: 0 a42 b4 4 2 1 0 16a 4b 4 44a b 1 Slope of tangent line is greater than 3. 4.41 4 0 4a b 1 or b 1 4a (c) Slope 4.1 2.1 2 1, 0: 0 a12 b1 4 Slope of tangent line is less than 4.1. 4 a b f 2 h f 2 (d) Slope 4 a 1 4a 2 h 2 4 1 3a 2 h2 4 3 3a h 4h h2 a 1 h b 1 41 5 4 h, h 0 y x2 5x 4 (e) Slope 4 h, h 0 (b) Enter the data points 0, 4, 1, 0, 2, 2, 4, 0, 6, 10 and use the regression feature to obtain 4 1 3 2 y x 5x 4. 4 1 5 4 0.1 4.1 The results are the same as in (a)–(c). (f) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at 2, 4 is 4. 260 Chapter 2 Polynomial and Rational Functions

1 7. f x x kqx r 8. (a) z m z (a) Cubic, passes through 2, 5 , rises to the right 1 1 1 i One possibility: 1 i 1 i 1 i 1 i 1 1 f x x 2x2 5 i 2 2 2 x3 2x2 5 1 (b) zm (b) Cubic, passes through 3, 1, falls to the right z 1 1 3 i One possibility: 3 i 3 i 3 i 2 f x x 3 x 1 3 i 3 1 i x3 3x2 1 10 10 10 1 1 (c) z m z 2 8i 1 2 8i 2 8i 2 8i 2 8i 1 2 i 68 34 17

9. a bia bi a2 abi abi b2i2 a2 b2 Since a and b are real numbers,a2 b2 is also a real number.

ax b ax 10. fx 11. fx cx d x b2 d (a) b 0 ⇒ x b is a vertical asymptote. Vertical asymptote: x c a causes a vertical stretch if a > 1 and a vertical shrink if 0 < a< 1. For a > 1, the graph a Horizontal asymptote: y becomes wider as a increases. When a is negative c the graph is reflected about the x-axis. (i) a > 0, b < 0, c > 0, d < 0 (b)a 0. Varying the value of b varies the vertical Both the vertical asymptote and the horizontal asymptote of the graph of f. For b > 0, the graph asymptote are positive. Matches graph (d). is translated to the right. For b < 0, the graph is reflected in the x-axis and is translated to the left. (ii) a > 0, b > 0, c < 0, d < 0 Both the vertical asymptote and the horizontal asymptote are negative. Matches graph (b).

(iii) a < 0, b > 0, c > 0, d < 0 The vertical asymptote is positive and the horizontal asymptote is negative. Matches graph (a).

(iv) a > 0, b < 0, c > 0, d > 0 The vertical asymptote is negative and the horizontal asymptote is positive. Matches graph (c). Problem Solving for Chapter 2 261

12. (a) 50 (c) Age,x Near point, y Age,x Near point,y Quadratic Rational Model Model 16 3.0 16 3.0 3.66 3.05 32 4.7 0 70 0 32 4.7 2.32 4.63 44 9.8 44 9.8 11.83 7.58 50 19.7 50 19.7 19.97 11.11 60 39.4 60 39.4 38.54 50.00 y 0.0313x2 1.586x 21.02 The models are fairly good fits to the data. The quadratic 1 model seems to be a better fit for older ages and the (b) 0.007x 0.44 50 y rational model a better fit for younger ages. 1 (d) For x 25, the quadratic model yields y 0.9325 inches y 0.007x 0.44 and the rational model yields y 3.774 inches. 0 70 0 (e) The reciprocal model cannot be used to predict the near point for a person who is 70 years old because it results in a negative value y 20. The quadratic model yields y 63.37 inches. 262 Chapter 2 Polynomial and Rational Functions

Chapter 2 Practice Test

1. Sketch the graph of f x x2 6x 5 and identify the vertex and the intercepts.

2. Find the number of units x that produce a minimum cost C if C 0.01x2 90x 15,000.

3. Find the quadratic function that has a maximum at 1, 7 and passes through the point 2, 5.

4 4. Find two quadratic functions that have x-intercepts 2, 0 and 3, 0 .

5. Use the leading coefficient test to determine the right and left end behavior of the graph of the polynomial function fx 3x5 2x3 17.

6. Find all the real zeros of f x x5 5x3 4x.

7. Find a polynomial function with 0, 3, and 2 as zeros.

8. Sketch f x x3 12x.

9. Divide 3x4 7x 2 2x 10 by x 3 using long division.

10. Divide x3 11 by x2 2x 1.

11. Use synthetic division to divide 3x5 13x4 12x 1 by x 5.

12. Use synthetic division to find f6 given fx 7x3 40x2 12x 15.

13. Find the real zeros of fx x3 19x 30.

14. Find the real zeros of fx x4 x3 8x2 9x 9.

15. List all possible rational zeros of the function fx 6x3 5x2 4x 15.

3 20 2 10 16. Find the rational zeros of the polynomial f x x 3 x 9x 3 .

17. Write fx x4 x3 5x 10 as a product of linear factors.

18. Find a polynomial with real coefficients that has 2, 3 i, and 3 2i as zeros. Practice Test for Chapter 2 263

19. Use synthetic division to show that 3i is a zero of fx x3 4x2 9x 36.

x 1 20. Sketch the graph of fx and label all intercepts and asymptotes. 2x

8x2 9 21. Find all the asymptotes of fx . x2 1

4x2 2x 7 22. Find all the asymptotes of fx . x 1

23. Given z1 4 3i and z 2 2 i, find the following: (a) z1 z2

(b) z1z2 (c) z1 z2

24. Solve the inequality: x2 49 ≤ 0

x 3 25. Solve the inequality: ≥ 0 x 7