Precalculus Essentials 5th Edition Blitzer SOLUTIONS MANUAL Full download at: Precalculus Essentials 5th Edition Blitzer TEST BANK Full download at: Chapter 2 Polynomial and Rational Functions Section 2.1 Check Point Exercises 1. a. (5 − 2i) + (3 + 3i) = 5 − 2i + 3 + 3i c. −14 + −12 = −14 + i 12 2 2 = −14 + 2i 3 2 = −14 + 2i 3 2 2 = (5 + 3) + (−2 + 3)i = 8 + i b. (2 + 6i) − (12 − i) = 2 + 6i −12 + i 5. x2 − 2x + 2 = 0 a = 1, b = −2, c = 2 2 = −7 + i 3 = (2 −12) + (6 +1)i = −10 + 7i x = −b ± b − 4ac 2a 2. a. 7i(2 − 9i) = 7i(2) − 7i(9i) −(−2) ± x = (−2)2 − 4(1)(2) 2(1) = 14i − 63i2 = 14i − 63(−1) = 63 + 14i x = 2 ± 4 − 8 2 b. (5 + 4i)(6 − 7i) = 30 − 35i + 24i − 28i2 = 30 − 35i + 24i − 28(−1) = 30 + 28 − 35i + 24i = 58 −11i x = 2 ± −4 2 x = 2 ± 2i 2 x = 1 ± i The solution set is {1 + i, 1 − i}.
 3 i 5 3 1  15  7i  4  19  7i   i 5 5 22. 3 3 4 i   12. (4  8i)(3  i)  12  4i  24i  8i2 4  i 4  i 4  i  12  28i  8  4  28i 34i 16  i2 34i  13. (7 – 5i)(–2 – 3i)  14  21i  10i  15i 2 = –14 – 15 – 11i = –29 – 11i 17 12 3   i 17 17
2i 2i 1  i 2i  2i 2 2  2i 31. 5 16  3 81 = 5(4i) + 3(9i) 23.      1  i 1  i 1  i 1  i 1  1 2 = 20i + 27i = 47i 24. 5i  5i  2 i 32. 5 8  3 18 2  i 2  i 2  i  5i 8  3i 18  5i 4  2  3i 9  2 10i  5i 2 4  1  10i  19i 2  9i 2 2 25.  5 10i 5  1  2i 8i 8i 4 3i  33. 2  4  2  2i 2  4  8i  4i2 = 4 – 8i – 4 4  3i 4  3i 4  3i = –8i 32i  24i2 16  9 34. 5  9  (5  i 9 )2  5  3i 2  24 32i 25  25  30i 9i2 24 32    i 25 25 = 25 + 30i – 9 = 16 + 30i 2 2 6i 6i 3 2i 18i 12i 2 35. 3  7  3  i 7  26.    3  2i 3  2i 3  2i 9  4  9  6i 7  i2 7  12 18i   12  18 i 13 13 13  9  7  6i 7 27. 2 3i  2 3i  2 i  2  6i 7 2 2 2  i 2  i 2  i 36. 2   2  i 11  4  4i  3i 2 4  1  4  4i 11  i 2 11 7 4i  4  11  4i 11 5 7 4   i  7  4i 11
5 5 37. 8  32  8  i 32 28. 3 4i  3 4i  4 3i 24 24 8 i 16 2 4  3i 4  3i 4  3i  12 25i 12i 2 16  9  25i 25  i   4i 2 24 1 2    i 3 6 29. 64  25  i 64  i 25 38.  28  12 i 28  12 i 4 7  8i  5i  3i 32 32 32  12 2i 7   3  7 i 30. 81  144  i 81  i 144 = 9i – 12i 32 8 16 = –3i
39. 6  12  6 i 12 45. x2  6 x  10  0 48 48  6 i 4 3 48  6 2i 3 48 1 3    i 8 24 6  (6)2  4(1)(10) x 2(1) x  6  36 40 2 x  6  4 2 x  6  2i 2 40. 15  18  15 i 18  15 i 9 2 33 33 33  15 3i 2   5  2 i 33 11 11 46. x  3  i The solution set is 3  i, 3  i. x2  2 x  17  0 41. 8 3 5   i 8 (i 3  5 2  (2)2  4(1)(17) x 2(1)  2i 2 i 3  5 2  4 68  2 6  2i 10 x 2 2  64 42. 12 4 2   i 12 (i 4  2 ) x  2  2i  4i 2 3 2i  2 3  2i 6 x  2 8i 2 x  1  4i  4 3  2i 6 The solution set is {1  4i,1  4i}. 43. 3 5 4 12   3i 5 8i 3 47. 4 x2  8x  13  0  24i2 15  24 15 8 x 82  4(4)(13) 2(4)   
48. 2 x2  2 x  3  0 50. 3x2  4 x  6  0 (2) x (2)2  4(2)(3) 2(2) (4) x (4)2  4(3)(6) 2(3)  2  4 24 4  2  20 4  2 2i 5 4 2(1 i 5 ) 4  1 i 5 2 1 5  i 4 16 72 6 4  56 6 4 2i 14 6 2(2 i 14 ) 6 2 i 14 3 2 14     i 2 2 3 3 The solution set is 1 5 1 5   i,   i . 2 2 2 2 The solution set is  2 14 2 14  3  3 i, 3  3 i . 49. 3x2  8x  7  0 8 (8)2 4(3)(7) 51. 2  3i  1  i   3  i  3  i  2  2i  3i  3i2   32  i2  x 2(3)  8  64 84 6  8  20 6  8 2i 5  2  5i  3i2  9  i 2  7  5i  4i2  7  5i  4  11  5i 52. 8  9i 2  i   1  i  1  i  6  16  8i  18i  9i 2  12  i2 2(4 i 5 ) 6     16  10i  9i 2  1  i2 2 4 i 5 3 4 5   i 3 3  15  10i  8i  15  10i  8  23  10i
54. 4  i 2  1  2i 2  16  8i  i2   1  4i  4i2 59. f  x  x 2 19 2  x 3i 2  19  16  8i  i 2  1  4i  4i2  15  12i  3i2  15  12i  f 3i    2  3i 9i2  19 2  3i  18  12i  9 19 2  3i 55. 5 16  3 81 10   5 16 1  3 81 1 2  3i  5  4i  3  9i 10 2 3i    20i  27i 2  3i 2  3i  47i or 0  47i 20 30i 4  9i2 56. 5 8  3 18 30i 4  9  5 4 2 1  3 9 2 1  20  30i  5  2 2 i  3  3 2 i 13  10i 2  9i 2 20 30   i  10  9 i 2 13 13  19i 2 or 0  19i 2 60. f  x  x2  11 3  x 57. f  x   x2  2 x  2 2 2 4i  11 16i  11 f 1  i   1  i 2  2 1  i   2 f 4i  3  4i 3  4i  1  2i  i2  2  2i  2  1  i2  1  1  0  16 11 3  4i
  5 3  4i  5  3  4i 3  4i 3  4i 58. f  x   x2  2 x  5 f 1  2i   1  2i 2  2 1  2i   5  15 20i 9  16i2  1  4i  4i2  2  4i  5  4  4i 2  15 20i 9  16 15  20i  4  4  25  0  15  20 i 25 25 3 4    i 5 5
61. E    5i 3   12  28i  15i  35i2  12  13i   12  35  13i  47  13i The voltage of the circuit is (47 + 13i) volts. 78. false; Changes to make the statement true will vary. A sample change is: (3 + 7i)(3 – 7i) = 9 + 49 = 58 which is a real number. 79. false; Changes to make the statement true will vary. A sample change is: 7 3i  7 3i  5 3i  44 6i  22  3 i 62. E  IR  2  3i  3  5i  6  10i  9i  15i2  6  i  15 5  3i 80. true 5  3i 5  3i 34 17 17  6  i  15  21  i The voltage of the circuit is 21  i  volts. 63. Sum: 5  i 15   5  i 15 4 4 81. 2  i  3  i   6  2i  3i  i2 4  i  1 4  i  5  i 15  5  5  10 Product:  5  i 15  4  7 i 7  i 7  i  28  4i 49  i 2 5  i 15 5  i  25  5i 15  5i 15  15i2  28 4i 49  1  25  15  40 64. – 72. Answers will vary. 73. makes sense 74. does not make sense; Explanations will vary. Sample explanation: Imaginary numbers are not 82. 50 28 4   i 50 50 14 2   i 25 25 1 i  1 i undefined. 1  2i 1  2i 1  i 1  1  i 1  2i 75. does not make sense; Explanations will vary. Sample explanation: i  1 ; It is not a variable in this context. 76. makes sense 77. false; Changes to make the statement true will vary. A sample change is: All irrational numbers are complex numbers.  1  2i 1  2i 1  2i 1  1  i 1  2i  1  i 1  1  2i 1 
8 8 88.  x2  2 x  1  0 83. 2  i 2 1  x2  2 x  1  0 i i i 8  2 i b  b 2 4ac x 2a i 8i  i (2) x 2  8 (2)2  4(1)(1) 2(1) 8i 2 i   2 2  i 2  i 16i 8i 2 4  i2 2 2 2 2  1  2 16i 8 4  1 The solution set is {1 2}. 8 16i 5 8 16   i 5 5 84. domain: [0, 2) range: [0, 2] 89. The graph of g is the graph of f shifted 1 unit up and 3 units to the left. 85. 86. f ( x) = 1 at 1 2 and 3 . 2 87. 0  2( x  3)2  8 2( x  3)2  8 x  3)2  4 x  3   4 x  3  2 x  1, 5
Section 2.2 Check Point Exercises 3. f ( x ) = −x2 + 4 x +1 Step 1: The parabola opens down because a < 0. Step 2: find the vertex: 1. f ( x ) = − ( x −1)2 + 4 b 4 x =− =− = 2 a =−1 h=1 2 k =4 2a 2(−1)      f ( x ) = −  x − 1  + 4   f (2) = −22 + 4(2) +1 = 5 The vertex is (2, 5). Step 1: The parabola opens down because a < 0. Step 2: find the vertex: (1, 4) Step 3: find the x-intercepts: 0 = − ( x −1)2 + 4 Step 3: find the x-intercepts: 0 = −x2 + 4x +1 −b ± b 2 − 4ac x = 2a ( x −1)2 = 4 x −1 = ±2 x = 1 ± 2 −4 ± x = 42 − 4(−1)(1) 2(−1) x = 3 or x = −1 Step 4: find the y-intercept: x = −4 ± 20 −2 f (0) = − (0 −1)2 + 4 = 3 x = 2 ± 5 The x-intercepts are x ≈ −0.2 and x ≈ −4.2 . Step 5: The axis of symmetry is x = 1. Step 4: find the y-intercept: f (0) = −02 + 4(0) +1 = 1 Step 5: The axis of symmetry is x = 2. 2. f ( x ) = ( x − 2)2 +1 Step 1: The parabola opens up because a > 0. Step 2: find the vertex: (2, 1) Step 3: find the x-intercepts: 0 = ( x − 2)2 +1 4. f ( x) = 4 x2 −16 x +1000 a. a = 4. The parabola opens upward and has a minimum value. ( x − 2)2 = −1 −b 16 b. x = = = 2 x − 2 = −1 2a 8
x = 2 ± i The equation has no real roots, thus the parabola has no x-intercepts. f (2) = 4(2)2 −16(2) + 1000 = 984 The minimum point is 984 at x = 2 . Step 4: find the y-intercept: f (0) = (0 − 2)2 +1 = 5 Step 5: The axis of symmetry is x = 2. c. domain: (−∞, ∞) range: [984, ∞)
18. f ( x ) = ( x −1)2 − 2 vertex: (1, –2) x-intercepts: 0 = ( x −1)2 − 2 ( x −1)2 = 2 20. f ( x ) = ( x − 3)2 + 2 vertex: (3, 2) x-intercepts: 0 = ( x − 3)2 + 2 ( x − 3)2 = −2 x −1 = ± 2 x = 1 ± 2 y-intercept: f (0) = (0 −1)2 − 2 = –1 The axis of symmetry is x = 1. x − 3 = ±i 2 x = 3 ± i 2 No x-intercepts. y-intercept: f (0) = (0 − 3)2 + 2 = 11 The axis of symmetry is x = 3. 19. (−∞, ∞) range: [−2, ∞) f ( x ) = ( x −1)2 + 2 vertex: (1, 2) x-intercepts: 0 = ( x −1)2 + 2 ( x −1)2 = −2 21. (−∞, ∞) range: [2, ∞) y −1 = ( x − 3)2 y = ( x − 3)2 + 1 vertex: (3, 1) x-intercepts: x −1 = ± −2 0 = ( x − 3)2 + 1 x = 1 ± i 2 ( x − 3)2 = −1 No x-intercepts. y-intercept: f (0) = (0 −1)2 + 2 = 3 The axis of symmetry is x = 1. domain: (−∞, ∞) range: [2, ∞) x – 3 = ±i x = 3 ±i No x-intercepts. y-intercept: 10 y = (0 − 3)2 +1 = 10 The axis of symmetry is x = 3. domain: (−∞, ∞) range: [1, ∞)
22. y − 3 = ( x −1)2 y = ( x −1)2 + 3 vertex: (1, 3) x-intercepts: 0 = ( x −1)2 + 3 ( x −1)2 = −3 The axis of symmetry is x = –2. x −1 = ±i 3 x = 1 ± i 3 domain: (−∞, ∞) range: [−1, ∞) No x-intercepts y-intercept: y = (0 −1)2 + 3 = 4 24. f ( x) = 5 −  x − 1  4  2  The axis of symmetry is x = 1. 2 f ( x) = −  x − 1  + 5  2  4 vertex:  1 , 5  x-intercepts: 0 = −  x − 1 2 5  +  4 domain: (−∞, ∞)  x − 1 2 5 = range: [3, ∞)  2  4 23. f ( x) = 2 ( x + 2)2 −1 vertex: (–2, –1) x-intercepts: 0 = 2 ( x + 2)2 −1 2 ( x + 2)2 = 1 ( x + 2)2 = 1  x − 1 = ± 5 2 2 x = 1 ± 5 2 y-intercept: f (0) = −  0 − 1 2 5 + = 1 4 1 2 x + 2 = ± 1 2 x = −2 ± 1 = −2 ± 2 The axis of symmetry is x = . 2 2 2 y-intercept: f (0) = 2 (0 + 2)2 −1 = 7
25. 26. f ( x ) = 4 − ( x f ( x ) = − ( x + 4 vertex: (1, 4) x-intercepts: 0 = − ( x −1)2 + 4 ( x −1)2 = 4 x – 1 = ±2 x = –1 or x = 3 y-intercept: f ( x ) = − (0 −1)2 + 4 = 3 The axis of symmetry is x = 1. domain: (−∞, ∞) range: (−∞, 4] f ( x ) = 1 − ( x − 3)2 f ( x ) = − ( x − 3)2 +1 vertex: (3, 1) x-intercepts: 0 = − ( x − 3)2 + 1 ( x − 3)2 = 1 x – 3 = ± 1 x = 2 or x = 4 y-intercept: f (0) = − (0 − 3)2 +1 = −8 The axis of symmetry is x = 3. domain: (−∞, ∞) range: (−∞,1] 27. 28. f ( x ) = x2 − − 3 f ( x ) = (x2 − 2 x +1)− 3 −1 f ( x ) = ( x −1)2 − 4 vertex: (1, –4) x-intercepts: 0 = ( x −1)2 − 4 ( x −1)2 = 4 x – 1 = ±2 x = –1 or x = 3 y-intercept: –3 f (0) = 02 − 2 (0) − 3 = −3 The axis of symmetry is x = 1. domain: (−∞, ∞) range: [−4, ∞) f (x ) = x2 − 2 x − 15 f (x ) = (x2 − 2 x + 1) − 15 − 1 f (x ) = (x − 1)2 − 16 vertex: (1, –16) x-intercepts: 0 = ( x −1)2 −16 ( x −1)2 = 16 x – 1 = ± 4 x = –3 or x = 5 y-intercept: f (0) = 02 − 2 (0) −15 = –15 The axis of symmetry is x = 1. domain: (−∞, ∞) range: [−16, ∞)
29. f ( x ) = x2 + 3x −10 x-intercepts: f ( x ) =  x2 + 3x + 9  −10 − 9 0 = 2 7 2 81 x − −  4  4     4  8 2 f ( x ) =  x + 3 − 49  7 2 81  2  4 2  x − 4  = 8     vertex:  − 3 , − 49 7 2 81  2 4  x −  =     x-intercepts:  4  16 0 =  x + 3 2 − 49 4 x − 7 = ± 9 4 4 7 9 x + 3 2 = 49 x = ± 4 4  2  4 x 1 or x = 4 x + 3 = ± 7 2 2 2 y-intercept: 2 3 7 x =− 2 2 f (0) = 2 (0) − 7 (0) − 4 = –4 x = 2 or x = –5 y-intercept: f ( x ) = 02 + 3(0) −10 = −10 The axis of symmetry is x 3 . The axis of symmetry is x = 7 . 4 =− 2 domain: (−∞, ∞) range: − 81 , ∞  8  domain: (−∞, ∞) − 49 , ∞ a
31.  f ( x) = 2 x − x2 + 3  4  2 30.  f ( x ) = 2x2 − 7 x − 4 f ( x) = −x f ( x) = − (x + 2x + 3 2 − 2x +1)+ 3 +1 f ( x ) = 2  x2 − 7 x + 49  − 4 − 49 f ( x) = − ( x −1)2 + 4 vertex: (1, 4)  2 16  8   x-intercepts: f ( x ) = 2  x − 7 2 − 81 8 0 = − ( x −1)2 + 4 ( x −1)2 = 4 vertex:  7 , − 81  x – 1 = ±2 x = –1 or x = 3 y-intercept: f (0) = 2 (0) − (0)2 + 3 = 3 The axis of symmetry is x = 1.
( x + 3)2 = 6 x + 3 = ± 6 x = −3 ± 6 y-intercept: domain: (−∞, ∞) range: (−∞, 4] f (0) = (0)2 + 6(0) + 3 f (0) = 3 The axis of symmetry is x = −3 . 32. f ( x) = 5 − 4 x − x2 f ( x) = −x2 − 4 x + 5 f ( x) = − (x2 + 4 x + 4)+ 5 + 4 f ( x) = − ( x + 2)2 + 9 vertex: (–2, 9) x-intercepts: 0 = − ( x + 2)2 + 9 34. domain: (−∞, ∞) range: [−6, ∞) f ( x) = x2 + 4 x −1 f ( x) = ( x2 + 4x + 4) −1 − 4 2 ( x + 2)2 = 9 x + 2 = ± 3 x = –5, 1 y-intercept: f (0) = 5 − 4 (0) − (0)2 = 5 f ( x) = ( x + 2) − 5 vertex: (−2, −5) x-intercepts: 0 = ( x + 2)2 − 5 ( )2 x + 2 = 5 The axis of symmetry is x = –2. x + 2 = ± 5 x = −2 ± 5 y-intercept: domain: (−∞, ∞) range: (−∞,9] f (0) = (0)2 + 4(0) −1 f (0) = −1 The axis of symmetry is x = −2 . 33. f ( x) = x2 + 6 x + 3 f ( x) = ( x2 + 6 x + 9) + 3 − 9 f ( x) = ( x + 3)2 − 6 vertex: (−3, −6) x-intercepts: 0 = ( x + 3)2 − 6 domain: (−∞, ∞) range: [−5, ∞)
35. f ( x) = 2 x2 + 4 x − 3 f ( x) = 2( x2 + 2 x ) − 3 3 x − 1 3 2 13 = 3 f ( x) = 2( x2 + 2 x + 1) − 3 − 2 f ( x) = 2 ( x + 1)2 − 5 x − 1 3 2 13 = 9 vertex: (−1, −5) x − 1 = ± 13 x-intercepts: 0 = 2 ( x +1)2 − 5 3 9 x = 1 ± 13 3 3 2 ( x +1)2 = 5 ( x +1)2 = 5 2 x +1 = ± 5 2 y-intercept: f (0) = 3(0)2 − 2(0) − 4 f (0) = −4 The axis of symmetry is x = 1 . 3 x = −1 ± 10 2 y-intercept: f (0) = 2(0)2 + 4(0) − 3 f (0) = −3 The axis of symmetry is x = −1 . domain: (−∞, ∞) range: − 13 , ∞   3  36. domain: (−∞, ∞) range: [−5, ∞) f ( x) = 3x2 − 2 x − 4 f ( x) = 3 x2 − 2 x  − 4 37.  f ( x) = 2 x − x2 − 2 f ( x) = − x2 + 2 x − 2 f ( x) = − (x2 − 2 x + 1) − 2 + 1 f ( x) = − (x − 1)2 − 1 vertex: (1, –1) x-intercepts: 0 = − (x − 1)2 − 1  3   f ( x) = 3 x2 − 2 x + 1  − 4 − 1 (x − 1)2 = −1  3 9  3 x – 1 = ±i   f ( x) = 3 x − 1 2 − 13 3 x = 1 ±i No x-intercepts. y-intercept: vertex:  1 , − 13 
The axis of symmetry is x = 1. 40. f(x) = 2x2 – 8x – 3 a. a = 2. The parabola opens upward and has minimum value. b. x = −b = 8 = 2 2a 4 f(2) = 2(2)2 – 8(2) – 3 = 8 – 16 – 3 = –11 The minimum is –11 at x = 2 . domain: (−∞, ∞) range: (−∞, −1] 41. c. domain: (−∞, ∞) f ( x) = −4 x2 + 8x − 3 range: [−11, ∞) 38. f ( x) = 6 − 4 x + x2 a. a = -4. The parabola opens downward and has a maximum value. f ( x) = x2 − 4x + 6 b. x = −b = −8 = 1 2a −8 f ( x) = (x2 − 4 x + 4)+ 6 − 4 f ( x) = ( x − 2)2 + 2 f (1) = −4(1)2 + 8(1) − 3 = −4 + 8 − 3 = 1 The maximum is 1 at x = 1 . vertex: (2, 2) x-intercepts: 0 = ( x − 2)2 + 2 c. domain: (−∞, ∞) 42. f(x) = –2x2 – 12x + 3 range: (−∞,1] ( x − 2)2 = −2 a. a = –2. The parabola opens downward and has a maximum value. x − 2 = ±i 2 −b 12 x = 2 ± i 2 b. x = = = −3 2a −4 No x-intercepts y-intercept: f (0) = 6 − 4 (0) + (0)2 = 6 The axis of symmetry is x = 2. f(–3) = –2(–3)2 – 12(–3) + 3 = –18 + 36 + 3 = 21 The maximum is 21 at x = −3 . c. domain: (−∞, ∞) range: (−∞, 21] 43. f ( x) = 5x2 − 5x a. a = 5. The parabola opens upward and has a minimum value.
b. x = −b = 5 = 1 domain: (−∞, ∞) 2a f  1  = 5  1 10 2 − 5  1        range: [2, ∞)  2  5  2 5 5 39. f(x) = 3x2 – 12x – 1 = − 5 = − 10 = − 4 2 4 4 4 a. a = 3. The parabola opens upward and has a minimum value. The minimum is −5 4 at x = 1 . 2 b. x = −b = 12 = 2  −5  2a 6 c. domain: (−∞, ∞) range:  4 , ∞  f(2) = 3(2)2 – 12(2) – 1 = 12 – 24 – 1 = –13  The minimum is –13 at x = 2 . c. domain: (−∞, ∞) range: [−13, ∞)
x-coordinate of vertex = −b = −0.7 = 35 2a 2 (−0.01) 51. (h, k ) = (−10, −5) of vertex 2 f ( x ) = 2 ( x − h )2 + k y = −0.01x + 0.7 x + 6.1 2 = 2[x − (−10)]2 + (−5) y = −0.01(35) + 0.7(35) + 6.1 = 18.35 = 2 ( x +10)2 − 5 The maximum height of the shot is about 18.35 feet. This occurs 35 feet from its point of release.
2a 2 (−2) −4 The vertex is (8, 64). The maximum product is 64. This occurs when the two numbers are 8 and 16 − 8 = 8 . 62. Let x = one of the numbers Let 20 – x = the other number P ( x ) = x (20 − x ) = 20 x − x2 = −x2 + 20x x = − b = − 20 = − 20 = 10 When the width is x = 150 feet, the length is 600 − 2 (150) = 600 − 300 = 300 feet. The dimensions of the rectangular plot with maximum area are 150 feet by 300 feet. This gives an area of 150 ⋅ 300 = 45, 000 square feet. 2a 2 (−1) −2 The other number is 20 − x = 20 −10 = 10. The numbers which maximize the product are 10 and 10. The maximum product is 10 ⋅10 = 100.
y = 400 − 2 x 3 y = 400 − 2 x 3 3 2  400 2  y = 40 − x We need to maximize A = xy = x − x  . 3 3  A ( x ) = x (40 − x ) = −x2 + 40 x x = − b = − 40 = − 40 = 20. Rewrite A as a function of x. A ( x ) = x  400 − 2 x  = − 2 x2 + 400 x 2a 2 (−1) −2  3 3  3 3  When the length x is 20, the width y is y = 40 − x = 40 − 20 = 20. The dimensions of the rectangular region with maximum area are 20 yards by 20 yards. This gives an area of 20 ⋅ 20 = 400 square yards.
– 83. Answers will vary. 2a 2 (−2) −4 84. y = 2x2 – 82x + 720 When the height x is 3, the width is 12 − 2 x = 12 − 2 (3) = 12 − 6 = 6. A (3) = −2 (3)2 +12 (3) = −2 (9) + 36 = −18 + 36 = 18 The maximum cross-sectional area is 18 square inches. This occurs when the gutter is 3 inches deep and 6 inches wide. a. You can only see a little of the parabola.
b. a =2; b = –82 x = − b −82 = 20.5 87. y = 5x2 + 40x + 600 x = −b = −40 = −4 =− 2a 4 2a 10 y = 2(20.5)2 − 82(20.5) + 720 y = 5(–4)2 + 40(–4) + 600 = 840.5 −1681 + 720 = −120.5 = 80 – 160 + 600 = 520 vertex: (–4, 520) vertex: (20.5, –120.5) c. Ymax = 750 d. You can choose Xmin and Xmax so the x-value of the vertex is in the center of the graph. Choose Ymin to include the y-value of the vertex. 85. y = –0.25x2 + 40x x = −b = −40 = 80 88. y = 0.01x2 + 0.6x + 100 2a −0.5 −b −0.6 y = –0.25(80)2 + 40(80) = 1600 x = = = −30 2a 0.02 2 vertex: (80, 1600) y = 0.01(–30) + 0.6(–30) + 100 86. y = –4x2 + 20x + 160 x = −b = −20 = 2.5 = 9 – 18 + 100 = 91 The vertex is at (–30, 91). 89. a. The values of y increase then decrease. 2a −8 y = –4(2.5)2 + 20(2.5) + 160 = –2.5 + 50 +160 = 185 The vertex is at (2.5, 185). b. y = −0.48x2 + 6.17x + 9.57 c. x = −(6.17) ≈ 6 2(−0.48) y = −0.48(6)2 + 6.17(6) + 9.57 ≈ 29.3 According to the model in part (b), American Idol had the greatest number of viewers, 29.3 million, in Season 6.
2 ( x + 4)2 = −8 form f ( x ) = a ( x − h ) + k 2 ( x + 4)2 = −4 = a  x − (−3) 2 + (−1) Because the solutions to the equation are imaginary, we know that there are no x–intercepts. = a ( x + 3) −1 We use the point (−2, −3) on the graph to determine the value of a: f ( x ) = a ( x + 3)2 −1 −3 = a (−2 + 3)2 −1 −3 = a (1)2 −1 −3 = a −1 −2 = a Thus, the equation of the parabola is f ( x ) = −2 ( x + 3)2 −1 .
= ( x + 3)( x +1)( x −1)  π  π  440 − 2(110) = 220 π π The dimensions are 110 yards by 220 yards. π 105. Answers will vary. 110. f ( x) = x3 − 2 x − 5 f (2) = (2)3 − 2(2) − 5 = −1 f (3) = (3)3 − 2(3) − 5 = 16 The graph passes through (2, –1), which is below the x-axis, and (3, 16), which is above the x-axis. Since the graph of f is continuous, it must cross the x-axis somewhere between 2 and 3 to get from one of these 106. 3x + y 2 = 10 y 2 = 10 − 3x points to the other. 4 2 y =± 10 − 3x 111. f ( x) = x − 2 x +1 Since there are values of x that give more than one value for y (for example, if x = 0, then f (−x) = (−x)4 − 2(−x)2 +1 = x4 − 2 x2 +1 y =± 10 − 0 = ± 10 ), the equation does not define Since f (−x) = f ( x), the function is even. y as a function of x. Thus, the graph is symmetric with respect to the y-axis.
10. f ( x) = 2( x + 2)2 ( x − 3) The term is 2 ⋅ x2 ⋅ x, or 2 x3 . Since an > 0 and n is odd, the graph falls to the left and rises to the right. 2( x + 2)2 ( x − 3) = 0 x = −2 or x = 3 The x-intercepts are –2 and 3. f (0) = 2(0 + 2)2 (0 − 3) = −12 The y-intercept is –12. f (−x) = 2 ((−x) + 2)2 ((−x) − 3) = 2 (−x + 2)2 (−x − 3) No symmetry. Exercise Set 2.3 1. polynomial function; degree: 3 2. polynomial function; degree: 4 3. polynomial function; degree: 5 4. polynomial function; degree: 7 5. not a polynomial function 6. not a polynomial function 7. not a polynomial function 8. not a polynomial function 9. not a polynomial function 10. polynomial function; degree: 2 11. polynomial function Concept and Vocabulary Check 2.3 1. 5; −2 2. false 3. end; leading 4. falls; rises 5. rises; falls 6. rises; rises 12. Not a polynomial function because graph is not smooth. 13. Not a polynomial function because graph is not continuous. 14. polynomial function 15. (b) 16. (c) 17. (a) 18. (d) 7. falls; falls 8. true 9. true 10. x-intercept 11. turns around; crosses 12. 0; Intermediate Value 19. 20. 21. f ( x) = 5x3 + 7 x2 − x + 9 Since an > 0 and n is odd, the graph of f(x) falls to the left and rises to the right. f ( x) = 11x3 − 6 x2 + x + 3 Since an > 0 and n is odd, the graph of f(x) falls to the left and rises to the right. f ( x) = 5x4 + 7 x2 − x + 9 Since an > 0 and n is even, the graph of f(x) rises to 13. n −1 the left and to the