Solutions Key Quadratic Functions

CHAPTER 5 Solutions Key Quadratic Functions ARE YOU READY? PAGE 11 1. E. C. A. B 5. (.) (.)(.) 10. 6. ( 5) ( 5 )( 5 ) 5 7. 11 11 8. 1 16 1 9. 7 6 6 11. 75 75 5 11 15 11 1. (x - )(x - 6) x - 6x - x + 1 x - 8x + 1 15. (x + )(x + 7) x + 7x + x + 1 x + 9x + 1 10. ( 1 - ) (1 - ) (8) 16 _ 1. 5 18 9 1. (x + 9)(x - 9) x - 9x + 9x - 81 x - 81 16. (x - )(5x + 1) 10 x + x - 15x - 10 x - 1x - 5-1 USING TRANSFORMATIONS TO GRAPH QUADRATIC FUNCTIONS, PAGES 15- CHECK IT OUT! 1. x g(x) - x + 6x - 8 (x, g(x)) 1 g(1) -(1) + 6(1) - 8 - (1, -) g() -() + 6() - 8 0 (, 0) g() -() + 6() - 8 1 (, 1) g() -() + 6() - 8 0 (, 0) 5 g(5) -(5) + 6(5) - 8 - (5, -) a. g is f translated 5 units down. 17. x + 10 - x - - 10 x - x -1 19. (x - 1) 11 x - 11 1.. x 5 x _ 105 6 x 17 1 18. x - (1 - x ) x - 1 + x x + 1 x x 1 0. (x + 5) - 5x 1 x + 10-5x 1 -x -9 x.. b. g is f ranslated units left and units down. a. g is a horizontal compression of f by a factor of 1. b. g is f reflected across the x-axis and vertically compressed by a factor of 1. 15 Holt McDougal Algebra

a. g(x) 1 (x- ) - b. g(x)-(x+ 5 ) + 1 _ 5. d n(v) d(v) 0.09v 1 0.05v 15 Vertical compression by a factor of 1 ; the braking 15 distance wiil be less with optimally inflated new tires than with tires having more wear. THINK AND DISCUSS 1. Possible answer: a indicates a reflection, vertical stretch or vertical compression. h indicates a horizontal translation (left or right). k indicates a vertical translation (up or down).. Possible answer: The function for which a is greater will have a narrower graph.. x -1 0 1 g(x) -6-0 0 -.. x - -1 0 1 h(x) 0-1 0 8 5. d is f translated units right. 6. g is f translated units right and units up. EXERCISES GUIDED PRACTICE 1. vertex. x - -1 0 1 f(x) -1-6 - -6-1 7. h is f translated 1 unit left and units down. 15 Holt McDougal Algebra

8. g is a vertical stretch of f by a factor of. 9. h is a horizontal stretch of f by a factor of 8. 10. p is a vertical compression of f by a factor of 0.5. PRACTICE AND PROBLEM SOLVING 17. x - -1 0 1 f(x) 0 0 18. x - -1 0 1 g(x) 9 1 0 1 11. h is f reflected across the x-axis and horizontally compressed by a factor of 1 5. 19. x - -1 0 1 h(x) -1 - -1 5 15 1. g is a vertical stretch of f by a factor of.. 1. d is f reflected across the x-axis and vertically compressed by a factor of. 1. g(x) (x + ) 15. h(x) - x - 6 _ 16. Vertical compression by a factor of 15 ; the safe 59 working load is less for an old rope than for a newer rope of the same radius. 0. g is f translated units down. 1. h is f translated 5 units left.. j is f translated 1 unit right. 155 Holt McDougal Algebra

. g is f translated units left and units down.. h is f translated units up and units left. 5. h is f translated units right and 9 units down. 6. g is a vertical compression of f by a factor of 7. 7. h is f reflected across the x-axis and vertically stretched by a factor of 0. 8. j is a horizontal stretch of f by a factor of. 9. g(x) - 1 (x - 1) 0. h(x).5(x + ) + 1 1. Vertical translaton; at any given speed, the gas mileage for an SUV is 18 mi/gal less than for a compact car. a. Translation 10 units right and 00 units up. b. No; the largest pen Keille can build with an 80 ft roll has an area of 00 f t, and the largest pen she can build with a 0 ft roll has an area of 100 ft. Therefore, a roll that is twice as long allows her to build a pen with times the area.. p is f reflected across the x-axis and translated units right.. g is f vertically stretched by a factor of 8 and translated units left. 5. h is f vertically stretched by a factor of and translated units down. 6. p is f vertically compressed by a factor of 1 and translated units up. 7. g is f horizontally compressed by a factor of 1 and translated 1 unit up. 8. h is f reflected across the x-axis and horizontally stretched by a factor of. 9. C 0. B 1. A a. B(r) 5 - π r b. B is A reflected across the x-axis and translated 5 units up. c. A: D: {r 0 r.5}; R: {A 0 A 6.5π}. B: D: {r 0 r.5}; R: {B (5-6.5π) B 5}. Possible answer: The radius of the circle cannot be less than 0 or greater than half the side length of the square.. Horizontal line; linear or constant function.. Very narrow parabola opening upward with its vertex at (-5, 5). 5a. Vertical compression by a factor of 0.8 and translation.5 units right and 59 units up. b. y -6.08(t - ) + 95 TEST PREP 6. B 7. J 8. C 9. G 50. 5 CHALLENGE AND EXTEND 51. y -(x - ) + Translation 6 units up and 6 units right. 5a. f: horizontal compression by a factor of 1 and translation units down. g: vertical stretch by a factor of and translation units down. b. c. The functions are the same. d. g(x) (x) SPIRAL REVIEW 5. Yes; the price is justified because the volume of the large container is more than times the volume of the small container: V small 8π; V large 88π. 5. f(x) x 55. f(x) x 156 Holt McDougal Algebra

56. y + 5x 1 y -5x + 1 y - 5 x + 7 57. x - 1 y + -1-1 y -x - 5 y x + 10 5- PROPERTIES OF QUADRATIC FUNCTIONS IN STANDARD FORM, PAGES -0 CHECK IT OUT! 1. x a. (a) downward (b) x -b a -(-) (-) - -1 (c) f(-1) -(-1) - (-1) -(1) + - + The vertex is (-1, ). (d) The y-intercept is 0. (e) b. (a) upward (b) x -b a _ - (1) - (c) g ( - ) ( - ) + ( - ) - 1 9-9 The vertex is ( - - 1-1, - 1 ). (d) The y-intercept is -1. (e) a. x -b a -(-6) (1) f() ( ) - 6() + 9-18 + -6 The minimum is -6. D: ; R: {y y -6}. 6 b. x -b a 0 (-) 0 g(0) -(0) - - The maximum is -. D: ; R: {y y -}.. s -b a -.5 (-0.05) -.5-0.05 9 m(9) -0.05(9) +.5(9) - 0-0.05(01) + 10.05-0 0.05 The maximum mileage is 0.0 mi/gal at a speed of 9 mi/h. THINK AND DISCUSS 1. Possible answer: No; quadratic functions open in 1 direction. If they open upward, they have a minimum value. If they open downward, they have a maximum value.. Possible answer: The value of x increases faster than the value of x decreases.. EXERCISES GUIDED PRACTICE 1. minimum. x. x 0. x -5 5a. downward b. x -b a -(-) (-1) -1 c. f(-1) -(-1) - (-1) - 8-1 + - 8-7 The vertex is (-1, -7). d. The y-intercept is -8. e. 157 Holt McDougal Algebra

6a. upward b. x -b a -(-) 7a. downward b. x -b (1) a - (-1) c. g ( ) ( ) - ( ) + c. h() -() + () - 1 9-9 - + 8-1 + The vertex is (, ). - 1 The vertex is (, - 1 ). d. The y-intercept is. e. 8. x -b a _ 0 (1) 0 f(0) -1 The minimum is -1. D: ; R: {y y -1}. 10. x -b a - (-16) 1 d. The y-intercept is -1. e. h(1) -16(1) + (1) + -16 + + 0 The maximum is 0. D: ; R: {y y 0}. 11. x -b a -0.5 (-0.005) 5 h(5) -0.005(5) + 0.5(5) -.15 + 6.5.15 The maximum height is.15 m. PRACTICE AND PROBLEM SOLVING 1. x 0 1. x 1 1. x -1 9. x -b a - (-1) g ( ) - ( ) + ( ) - 9 - + 9-1 The maximum is 1. D: ; R: {y y 1 }. 15a. upward b. x -b a _ -1 (1) - 1 c. f ( - ) 1 ( - 1 ) + ( - 1 1-1 - ) - - 9 The vertex is ( - 1, - 9 ). d. The y-intercept is -. e. 16a. downward b. x -b a -6 (-) 1 c. g(1) -(1) + 6(1) - + 6 The vertex is (1, ). d. The y-intercept is 0. e. 17a. upward b. x -b a -(-) (0.5) c. h() 0.5( ) - () - 0.5() - - -6 The vertex is (, -6). d. The y-intercept is -. e. 158 Holt McDougal Algebra

18a. downward b. x -b a -8 (-) c. f() -() + 8() + 5-8 + 16 + 5 1 The vertex is (, 1). d. The y-intercept is 5. e. 19a. upward b. x -b a _ - () - 1 c. g ( - 1 ) ( - 1 ) + ( - 1 1 - The vertex is ( - 1-8 - 5, - 5 ). d. The y-intercept is -8. e. 0a. upward b. x -b a - (1) -1 _ ) - 8 c. h(-1) (-1) - 1 + (-1) - -1 + 1 - The vertex is (-1, -). d. The y-intercept is -1. e. 1a. downward b. x -b a 0 (-1) 0 c. f(0) - - (0 ) - The vertex is (0, -). d. The y-intercept is -. e. a. upward b. x -b a - (0.5) - c. g(-) 0.5(-) + (-) - 5.5-9 - 5-9.5 The vertex is (-, -9.5). d. The y-intercept is -5. e. a. upward b. x -b a -1 ( 1 ) - _ c. h(-) 1 (-) - + 1 - + 1 The vertex is (-, 1). d. The y-intercept is. e.. x -b a -7 (-) 7 f ( ) 7 - ( 7 ) + 7 ( 7 ) - - ( 9 16) + 9 - - 9 8 + 98 8-8.15 The maximum is.15. D: ; R: {y y.15}. 5. x -b a -6 (-1) g() 6() - ( ) 18-9 9 The maximum is 9. D: ; R: {y y 9}. 159 Holt McDougal Algebra

6. x -b a -(-) (1) h() () - () + - 8 + -1 The minimum is -1. D: ; R: {y y -1}. 8. x -b a -(-6) (-1) - g(-) -(- ) -6(-) + 1-9 + 18 + 1 10 The maximum is 10. D: ; R: {y y 10}. 9. x -b a _ -8 (1) - h(-) (-) + 8(-) + 16 16 - + 16 0 The minimum is 0. D: ; R: {y y 0}. _ 0. d -b a -0.657 (-0.0018) 18.5 7. x -b a 0 ( - 1 ) 0 f(0) - 1 (0) - - The maximum is -. D: ; R: {y y -}. T -0.0018(18.5) + 0.657(18.5) + 50.95-59.95 + 119.9 + 50.95 111 The maximum temperature in 00 is approximately 111. 1. Maximum height is 6 ft. Possible answer: The axis of symmetry is halfway between any points with the same y-value. Because the points (1, 8) and (, 8) have the same y-value, the axis of symmetry is x. Because the vertex lies on the axis of symmetry, the vertex of the graph is (, 6). Therefore, the maximum value of the function is 6. a. C(x) x( - x) b. x 0 8 1 16 C(x) 0 96 18 96 0 c. D: {0 x 16}; R:{y 0 y 18} Neither the width nor the area can be negative. d. x 8 cm a. t -b a -000 (-000) 0.75 h(0.75) -000(0.75) + 000(0.75) -56.5 + 115 56.5 mm b. 9.75 to 1. Possible answer: The ratio for spittle bugs is more than 67 times as great as the ratio for humans. c. x 1.8 9.75 x 168.75 m. A(x) 10x - x x -b a -10 (-1) 5 A(5) 10(5) - (5) 50-5 5 yd 5. min -.09771 6. max 1.1785 7. min -1.5 8. max 5.715 9. The axis of symmetry is halfway between any points with the same y-value. Halfway between -7 and is -. Therefore, the axis of symmetry is x -. 0. Yes; possible answer: a function such as f(x) - x - 5 may open downward and have a vertex below the x-axis. A function such as f(x) x + may open upward and have a vertex above the x-axis. 1a. t -b a -50 (-16) 1.6 s b. h(1.565) -16(1.565) + 50(1.565) + 6-9.065 + 78.15 + 6 5 ft TEST PREP. C. G. B 5. G 160 Holt McDougal Algebra

6. Because a is negative, the graph will open downward and have a maximum value. To find the maximum value, find the x-value of the vertex: x -b a -(-8) (-1) -. Then evaluate the function for x -. f(-) (-) - 8(-) + 0. The maximum value is 0. CHALLENGE AND EXTEND 7. Possible answer: f(x) x + x + ; g(x) - x - x +. 8a. (11, 8); possible answer: the point (-5, 8) is 8 units left of the axis of symmetry. The graph of the quadratic function must also pass through a point that is 8 units right of the axis of symmetry and has the same y-value as (-5, 8). This point has coordinates (11, 8). b. No; possible answer: you would need to know the coordinates of at least one other point on the function s graph to determine whether it opens upward or downward. 9. Possible answer: The function has no x-term, so b 0. Therefore, the axis of symmetry is x 0, and the vertex is (0, c). 50. If the value of the function is the same for different x-values, the axis of symmetry is halfway between these x-values. In this case, the axis of symmetry is x _ -1 + 1. SPIRAL REVIEW 51. 0 180 700 600 60 5. 5 0 5 0 9 5 _ 9 5 5 5 _ 5 5 5. 8 8 8 6 16 6 5. 0 16 19 19 55. f(0) (0 - ) + 1 9 + 1 10 f ( ) 1 ( 1 - ) + 1 5 + 1 9 f(-) (- - ) + 1 5 + 1 6 57. f(0) -(0 + 5) -(5) -0 f ( 1 + 5 ) ) - ( 1 ) - ( 11 - f(-) -(- + 5) -() -1 59. y mx + b - (1) + b - + b b -7 y x - 7 61. y mx + b 5 -() + b 5-6 + b b 11 y - 5 - (x - ) or y -x + 11 56. g(0) ( 0-1 ) ( ) 1 1 g ( ) 1 ( 1-1 ) (0) 0 g(-) ( - - 1 ) ( 5 5 ) 58. g(0) (0) - (0) + 8 0-0 + 8 8 g ( ) 1 ( 1 ) - ( 1 ) + 8 1 8 - + 8 9 8 g(-) (-) - (-) + 8-8 + 8 + 8 8 60. m -7-5 -1 - (-) -6 y mx + b -7-6(-1) + b -7 6 + b b -1 y -6x - 1 6. m _ 1-6 - - 5 6 y mx + b 1 5 6 (-) + b 1-5 + b b 8 y - 6 5 6 y 5 6 x + 8 (x - ) or 161 Holt McDougal Algebra

5- SOLVING QUADRATIC EQUATIONS BY GRAPHING AND FACTORING, PAGES -0 CHECK IT OUT! 1. x -b a -(-) (-1) -1 x - - -1 0 1 f(x) 0 0 The zeros of the function are x - and 1. a. f(x) x - 5x - 6 x - 5x - 6 0 (x - 6)(x + 1) 0 x - 6 0 or x + 1 0 x 6 or x -1 b. g(x) x - 8x x - 8x 0 x(x - 8) 0 x 0 or x - 8 0 x 0 or x 8. h(t) -16t + vt + h a. x - x - -16t + 8t + 0 x - x + 0-16t(t - ) (x - )(x - ) 0-16t(t - ) 0 x - x 0-16t 0 or t - 0 x t 0 or t The ball is in the air for s. b. 5 x 9 5 x - 9 0 (5x + )(5x - ) 0 5x + 0 or 5x - 0 x - or x 5 5 THINK AND DISCUSS 5. Possible answer: x 5 or x -5 x - 5 0 or x + 5 0 (x - 5)(x + 5) 0 x - 5 0 f(x) x - 5 1. Possible answer: The function has 1 distinct real zero.. Possible answer: Both linear and quadratic functions have exactly one y-intercept. Linear functions have at most one x-intercept, and quadratic functions have at most two x-intercepts.. Either the parabola opens downward and the maximum is less than 0, or the parabola opens upward and the minimum is greater than zero.. EXERCISES GUIDED PRACTICE 1. roots. x -b a _ - (1) - x -5 - - 0 1 f(x) 0-5 -9-5 0 The zeros of the function are x -5 and 1.. x -b a -6 (-1) x 1 5 f(x) - 0 1 0 - The zeros of the function are x and. _. x -b a 0 (1) 0 x - -1 0 1 f(x) 0-1 0 The zeros of the function are x -1 and 1. 5. f(x) x - 7x + 6 x - 7x + 6 0 (x - 6)(x - 1) 0 x - 6 0 or x - 1 0 x 6 or x 1 7. h(x) x + x x + x 0 x(x + ) 0 x 0 or x + 0 x 0 or x - 9. g(x) x - 6x - 16 x - 6x - 16 0 (x - 8)(x + ) 0 x - 8 0 or x + 0 x 8 or x - 11. h(t) -16t + 6 t + -16t + 6t + 0 - ( 16t - 6t - ) 0 -(16t + 1)(t - ) 0 16t + 1 0 or t - 0 t - 1 or t 16 The arrow is in the air at for s. 1. x - 6x -9 x - 6x + 9 0 (x - )(x - ) 0 x - 0 x 1. x 9 x - 9 0 (x + 7)(x - 7) 0 x + 7 0 or x - 7 0 x -7 or x 7 6. g(x) x - 5x + x - 5x + 0 (x - 1)(x - ) 0 x - 1 0 or x - 0 x 1 or x 8. f(x) x + 9x + 0 x + 9x + 0 0 (x + )(x + 5) 0 x + 0 or x + 5 0 x - or x -5 10. h(x) x + 1x + x + 1x + 0 (x + 1)(x + ) 0 x + 1 0 or x + 0 x - 1 or x - 1. 5 x + 0 0x 5x - 0x + 0 0 5 (x - x + ) 0 5(x -)(x - ) 0 x - 0 x 15. x or x x - 0 or x - 0 (x - )(x - ) 0 x - 7x + 1 0 f(x) x - 7x + 1 16 Holt McDougal Algebra

16. x - or x - x + 0 or x + 0 (x + )(x + ) 0 x + 8x + 16 0 f(x) x + 8x + 16 PRACTICE AND PROBLEM SOLVING 18. x -b a - (-1) 17. x or x 0 x - 0 or x 0 (x - )x 0 x - x 0 f(x) x - x x 0 1 f(x) - 0 1 0 - The zeros of the function are x 1 and. 19. x -b a _ -1 (1) - 1 x - -1-1 0 f(x) 0-6 -6.5-6 0 The zeros of the function are x - and. 0. x -b a _ 0 (1) 0 x - -1 0 1 f(x) 0-8 -9-8 0 The zeros of the function are x - and. 1. f(x) x + 11x + x + 11x + 0 (x + )(x + 8) 0 x + 0 or x + 8 0 x - or x -8. h(x) - x + 9x - x + 9x 0 -x(x - 9) 0 x 0 or x - 9 0 x 0 or x 9 5. g(x) x + 7x - 8 x + 7x - 8 0 (x -1)(x + 8) 0 x - 1 0 or x + 8 0 x 1 or x -8 7. h(t) -16t + 56-16t + 56 0-16( t - 16) 0-16(t + )(t - ) 0 t + 0 or t - 0 t - or t It will take s. 9. x 81 x - 81 0 (x + 9)(x - 9) 0 x + 9 0 or x - 9 0 x - 9 or x 9. g(x) x + x - 10 x + x - 10 0 (x - )(x + 5) 0 x - 0 or x + 5 0 x or x - 5. f(x) x - 15x + 5 x - 15x + 5 0 (x - 6)(x - 9) 0 x - 6 0 or x - 9 0 x 6 or x 9 6. h(x) x - 1x + 18 x - 1x + 18 0 (x - 6x + 9) 0 (x - ) 0 x - 0 x 8. x + 8x -16 x + 8x + 16 0 (x + ) 0 x + 0 x - 0. 9 x + 1x + 0 (x + ) 0 x + 0 x - 1. 6 x - 9 0 9 ( x - 1 ) 0 9(x + 1)(x - 1) 0 x + 1 0 or x - 1 0 x - 1 or x 1. 9 x 8x - 9x - 8x + 0 (7x - ) 0 7x - 0 x 7 5. Possible answer: x 6 or x x - 6 0 or x - 0 (x - 6)(x - ) 0 x - 8x + 1 0 f(x) x - 8x + 1 7. f(x) 6x - x - x + 6x 0 -x(x - 6) 0 x 0 or x - 6 0 x 0 or x 6 9. h(x) x - 1x + 6 x - 1x + 6 0 (x - 6 ) 0 x - 6 0 x 6 1. g(x) x - x + 11 x - x + 11 0 (x - 11 ) 0 x - 11 0 x 11. x - 10x + 5 0 (x - 5 ) 0 x - 5 0 x 5. Possible answer: x 5 or x -1 x - 5 0 or x + 1 0 (x - 5)(x + 1) 0 x - x - 5 0 f(x) x - x - 5 6. Possible answer: x or x x - 0 or x - 0 (x - )(x - ) 0 x - 6x + 9 0 f(x) x - 6x + 9 8. g(x) x - 5 x - 5 0 (x + 5)(x - 5) 0 x + 5 0 or x - 5 0 x -5 or x 5 0. f(x) x - 1 x - 1 0 ( x - ) 0 (x + )(x - ) 0 x + 0 or x - 0 x - or x. h(x) 0 + x - x - x + x + 0 0 - ( x - x - 0 ) 0 -(x - 6)(x + 5) 0 x - 6 0 or x + 5 0 x 6 or x -5. f(x) x - 11x + 0. g(x) x - 8x - 0 x - 11x + 0 0 x - 8x - 0 0 (x - 5)(x - 6) 0 (x - 10)(x + ) 0 x - 5 0 or x - 6 0 x - 10 0 or x + 0 x 5 or x 6 x 10 or x - 5. h(x) x + 18x + 8 x + 18x + 8 0 (x + 9x + 1) 0 (x + )(x + 7) 0 x + 0 or x+ 7 0 x - or x -7 6a. h(t) 7-16 t -16t + 7 9-16t + 6 0-16( t - ) 0-16(t + )(t - ) 0 t + 0 or t - 0 t - or t The woman will fall for s. 16 Holt McDougal Algebra

b. Possible answer: No; the relationship between the building height and jump time is quadratic, not linear. Therefore, a jump that is half as high will not last half as long. 7a. h(t) -16t + 16t + 5 b. -16t + 16t + 5 0 - ( 16t - 16t - 5 ) 0 -(t + 1)(t - 5) 0 t + 1 0 or t - 5 0 t - 1 or t 5 The juggler has 1.5 s. 8. x - x + 1 0 (x - 1 ) 0 x - 1 0 x 1 9. x + 6x -5 x + 6x + 5 0 (x + 1)(x + 5) 0 x + 1 0 or x + 5 0 x -1 or x -5 50. 5 x + 0x -16 51. 9 x + 6x -1 5x + 0x + 16 0 9x + 6x + 1 0 (5x + ) 0 (x + 1) 0 5x + 0 x - x + 1 0 x - 1 5 5. 5 x 5 5 x - 5 0 5 ( x - 9 ) 0 5(x + )(x - ) 0 x + 0 or x - 0 x - or _ x 5a. x -b a - (1) -1 f(-1) (-1) + (-1) - 8 1 - - 8-9 The vertex is (-1, -9). b. The y-intercept is -8. c. x + x - 8 0 (x + )(x - ) 0 x + 0 or x - 0 x - or x d. 5. x - 6 x x - x - 6 0 (x - )(x + ) 0 x - 0 or x + 0 x or x - 55a. x -b a _ 0 (1) 0 g(0 ) (0 ) - 16-16 The vertex is (0, -16). b. The y-intercept is -16. c. x - 16 0 (x + )(x - ) 0 x + 0 or x - 0 x - or x d. 56a. x -b a -(-1) (1) h ( 1 ) ( 1 ) - 1 1-1 1-1 - 1-9 The vertex is ( 1, -1 1 ). b. The y-intercept is -1. c. x - x - 1 0 (x - )(x + ) 0 x - 0 or x + 0 x or x - d. 57a. x -b a - (-) 1 58a. x -b a -(-5) 5 (1) f(1) -(1) + (1) -(1) + g ( ) 5 ( ) 5-5 ( 5 ) - 6 The vertex is (1, ). 5-5 - 6 b. The y-intercept is 0. - 9 c. -x + x 0 -x(x - ) 0 The vertex is x 0 or x - 0 ( 1, - 1 1 ). x 0 or x b. The y-intercept is -6. d. c. x - 5x - 6 0 (x - 6)(x + 1) 0 x - 6 0 or x + 1 0 x 6 or x -1 59a. x -b a _ -1 () - 1 6 6 ) ( - 1 6) - 1 d. h ( -1 6-1 1-1 6 - - 9 1 The vertex is ( - 1 6, - 1) 1. b. The y-intercept is -. 16 Holt McDougal Algebra

c. x + x - 0 (x - 1)(x + ) 0 x - 1 0 or x + 0 x 1 or x -1 1 d. 60a. The two legs are (x - ) and (x - ). The equation wil be (x - ) + (x - ) x. b. (x - ) + (x - ) x (x - x + ) +(x - 8x + 16) x x - 1x + 0 x x - 1x + 0 0 (x - )(x - 10) 0 x - 0 or x - 10 0 x or x 10 c. Possible answer: The solutions represent possible lengths in centimeters of the hypotenuse.if x 10, the triangle would have side lengths of 10 cm, 8 cm, and 6 cm. If x, the triangle would have side lengths of cm, 0 cm, and - cm. Because length cannot be negative, only the solution x 10 is reasonable. 61. x(x + 16) 80 x + 16x - 80 0 (x - )(x + 0) 0 x - 0 or x + 0 0 x since x > 0 The dimensions are ft by 0 ft. 6. x(x + 1) 10 x + x - 10 0 (x - 1)(x + 15) 0 x - 1 0 or x + 15 0 x 1 since x > 0 x + 1 (1) + 1 15 The dimensions are 1 cm by 15 cm. 6. (x + )(x - ) 50 x - x - 6 50 x - x - 56 0 (x - 8)(x + 7) 0 x - 8 0 or x + 7 0 x 8 since x > 0 x + (8) + 10; x - (8) - 5 The dimensions are 10 m by 5 m. 6. No; Possible anwer: if a function can be factored as a binomial squared, the two factors are identical. When each factor is set to zero, each equation will have the same solution. Therefore, the function has only one distinct zero. 65. Possible answer: The Zero Product Property states that if a product equals zero, then at least one of the factors must equal zero. The zeros of a function can be found by writing the function rule as a product of factors and setting it equal to zero. You can then determine the value that makes each factor equal to zero. These values are the zeros of the function. 66a. h(t) -16t + 16t + 5-16t + 16t + 5 0 - ( 16t - 16t - 5 ) 0 -(t + 1)(t - 5) 0 t + 1 0 or t - 5 0 t - 1 or t 5 The ball will stay in air for 1.5 s. b. d(1.5) 85(1.5) 106.5 The horizontal distance travelled is 106.5 ft. TEST PREP 67. B 68. J 69. C 70. x + x - 1 0 (x + 7)(x - ) 0 x + 7 0 or x - 0 x -7 or x The positive root is x. CHALLENGE AND EXTEND 71. ( x - x ) x x - x x x - x 0 x(x - ) 0 x 0 or x - 0 x 0 or x 7. x - x + 1 8 0 8 x - 6x + 1 0 (x - 1)(x - 1) 0 x - 1 0 or x - 1 0 x 1 or x 1 7. x + x + 0.1 0 (x + 0.)(x + 0.7) 0 x + 0. 0 or x + 0.7 0 x -0. or x -0.7 7. x 1 x x - 1 x 0 x ( x - 1 ) 0 x 0 or x - 1 0 x 0 or x 1 165 Holt McDougal Algebra

75a. (a + b)(a - ab + b ) a - a b + a b + a b - a b + b a + b b. 8 x + 7 (x) + ( ) (x + ) ((x) - (x)() + ) (x + ) (x - 6x + 9) c. a - b (a - b)(a + ab + b ) d. x - 1 x - 1 (x - 1) ((x) + (x)(1) + 1 ) (x - 1) (x + x + 1) SPIRAL REVIEW 76. ( 1. 10 8) ( 6.1 1 0 -) (1. 6.1) 1 0 5 78. 80. 8.5 1 0 5 _.5 10 6 1. 10 - (.5 1.) 106.5 10 10 1 7.5 n 5 7.5n 60 n 8 8. 6.8.5 r 90.5r 61 r 16 77. (.7 1 0 10) (. 1 0 ) (.7.) 1 0 1 79. 8.6 1 0 1.1 10-6.8 10 _ (.1.8 ) 10-9 0.65 10-9 6.5 10-10 81. 1..8 w 8.8.8w 10.56 w. 8. h is a vertical compression of f by a factor of 0.5. 8. d is f translated units up. 85. g is f translated 1 unit left. 5- COMPLETING THE SQUARE, PAGES 1-8 CHECK IT OUT! 1a. x - 0 5 x 5 x ± 5 a. ( b ) ( b. ( b x ±5 x ± 5 ) ( ) x + x + (x + ) ) ( - ) (-) x - x + (x - ) c. ( b ) ( ) 9 x + x + 9 x + ( a. x - 9x x - 9x x - 9x + ( 9 ) ) + ( ) 9 b. x + 8x + 16 9 (x + ) 9 x + ± 9 x + ±7 x or -11 x - 9x + 81 + 81 ( x - 9 ) 89 x - 9 ± 89 x - 9 ± 89 x 9 ± 89 b. x - x 7 x - 8x 9 x - 8x + ( -8 ) 9 + ( -8 ) x - 8x + 16 9 + 16 (x - ) 5 x - ± 5 x - ±5 x 9 or -1 a. f(x) x + x + 15 (x + x) + 15 ( x + x + ( ) ) + 15 - ( (x + x + 1) + 15-1 (x + 1 ) + 1 The vertex is (-1, 1). b. g(x) 5 x - 50x + 18 5 ( x - 10x) + 18 ( 5x - 10x + ( 10 ) ) ) + 18-5 ( 10 5( x - 10x + 5) + 18-15 5(x - 5) + The vertex is (5, ). ) 166 Holt McDougal Algebra

THINK AND DISCUSS 1. Possible answer: Take the square root of each side by applying the Square Root Property; rewrite the equation as a difference of squares and factor.. Possible answer: Factor the x -term and the x-term so that the coefficient of the x -term is 1. Complete the square of the x -term and the x-term. Factor the perfect square. Multiply the original coefficient of the x -term by the number added to complete the square. Subtract this product from the constant term.. EXERCISES GUIDED PRACTICE 1. ( b ). (x - ) 16 x - ± 16 x - ± x 6 or -. x - x + 1 (x - 1 ) x - 1 ± x 1 ± _ 6. x - 1x + ( -1 ) x - 1x + 6 (x - 6 ) 8. x - 6x - x - 6x + ( - 6 ) - + ( -6. x - 10x + 5 16 (x - 5 ) 16 x - 5 ± 16 x - 5 ± x 9 or 1 5. x + 1x + ( 1 ) x + 1x + 9 (x + 7 ) 7. x - 9x + ( -9 ) x - 9x + 81 ( x - 9 ) ) x - 6x + 9 - + 9 (x - ) 5 x - ± 5 x ± 5 9. x + 8 6x x - 6x -8 x - 6x + ( -6 ) -8 + ( -6 ) x - 6x + 9-8 + 9 (x - ) 1 x - ± 1 x - ±1 x or 10. x - 0x 8 x - 10x x - 10x + ( _ -10 ) + ( _ -10 ) x - 10x + 5 + 5 (x - 5 ) 9 x - 5 ± 9 x 5 ± 9 11. x - x x + x x + x + ( ) + ( ) x + x + + (x + ) 8 x + ± 8 x - ± 7 1. 10x + x x + 10x + ( 10 ) + ( 10 ) x + 10x + 5 + 5 (x + 5 ) 67 x + 5 ± 67 x -5 ± 67 1. x + 8x - 15 0 x + 8x 15 (x + x) 15 ( x + x + ( ) ) 15 + ( ) (x + x + ) 15 + 8 (x + ) (x + ) x + ± x - ± _ 6 1. f(x) x + 6x - (x + 6x) - ( x + 6x + ( 6 ) ) - - ( 6 ) (x + 6x + 9) - - 9 (x + ) - 1 The vertex is (-, -1). 15. g(x) x - 10x + 11 ( x - 10x) + 11 x - 10x + ( _ -10 _ ) + 11 - ( -10 (x - 10x + 5) + 11-5 (x - 5 ) - 1 The vertex is (5, -1). ) 167 Holt McDougal Algebra

16. h(x) x - x + 5 ( x - x) + 5 ( x - 8x) + 5 _ ( x - 8x + ( -8 ) ) + 5 - ( -10 (x - 8x + 16) + 5-8 (x - ) + 5 The vertex is (, 5). 17. f(x) x + 8x - 10 (x + 8x) - 10 ( x + 8x + ( 8 ) ) - 10 - ( 8 ) (x + 8x + 16) - 10-16 (x + ) - 6 The vertex is (-, -6). 18. g(x) x - x + 16 ( x - x ) + 16 ( x - x + ( - ( x - x + 9 ( x - The vertex is ( ) + 55, 55 ). ) ) + 16 - ( - + 16-9 ) 19. h(x) x - 1x - ( x - 1x ) - ( x - x ) - ) ( x - x + ( - ) ) - - ( - (x - x + ) - - 1 (x - ) - 16 The vertex is (, -16). PRACTICE AND PROBLEM SOLVING 0. (x + ) 6 x + ± 6 x + ±6 x or -8. (x - ) 5 x - ± 5 x ± 5. x + 10x + ( 10 ) x + 10x + 5 (x + 5 ) ) ) 1. x - 6x + 9 100 (x - ) 100 x - ± 100 x - ±10 x 1 or -7 _. x - 18x + ( -18 ) x - 18x + 81 (x - 9 ) 5. x - 1 x - 1 ( x - 1 x + ( -0.5 ) x + 1 16 ) 6. x + x 7 x + x + ( ) 7 + ( ) x + x + 1 7 + 1 (x + 1 ) 8 x + 1 ± 8 x -1 ± 7. x - x -1 x - x + ( - ) -1 + ( - ) x - x + -1 + (x - ) x - ± x ± 8. x - 8x x - x 11 x - x + ( - ) 11 + ( - ) x - x + 11 + (x - ) 15 x - ± 15 x ± 15 9. 8x x + 1 x - 8x -1 x - 8x + ( -8 ) -1 + ( -8 ) x - 8x + 16-1 + 16 (x - ) x - ± x - ± x 6 or 0. x + x - 5 0 x + x 5 x + x + ( x + x + 9 ( x + ) 5 + ( ) 5 + 9 ) 9 ± 9 - ± x 9 x + 1. x + 6x 1 x + x 1 x + x + ( ) 1 + 1 x + x + 1 1 + 1 (x + 1 ) x + 1 ± x -1 ± 168 Holt McDougal Algebra

. f(x) x - x + 1 ( x - x) + 1 ( x - x + ( - ) ) + 1 - ( - (x - x + ) + 1 - (x - ) + 9 The vertex is (, 9).. g(x) x + 1x + 71 (x + 1x) + 71 ( x + 1x + ( 1 ) ) ) + 71 - ( 1 (x + 1x + 9) + 71-9 (x + 7 ) + The vertex is (-7, ).. h(x) 9 x + 18x - (9x + 18x) - ) 9 (x + x) - 9 ( x + x + ( ) ) - - 9 ( ) 9 (x + x + 1) - - 9 9(x + 1 ) - 1 The vertex is (-1, -1). 5. f(x) x + x - 7 (x + x)- 7 ( x + x + ( ) ) - 7 - ( ) (x + x + ) - 7 - (x + ) - 11 The vertex is (-, -11). 6. g(x) x - 16x + ( x - 16x ) + _ ( x - 16x + ( -16 _ ) ) + - ( -16 (x - 16x + 6) + - 6 (x - 8 ) - 6 The vertex is (8, -6). 7. h(x) x + 6x + 5 (x + 6x) + 5 (x + x) + 5 ( x + x + ( ) ) + 5 - ( ) ) ( x + x + 9 + 5-9 (x + 1.5) + 0.5 The vertex is (-1.5, 0.5). ) 1 8a. h(x) 9000 x - 7 15 x + 500 1 ( x - 00x) + 500 9000 1 9000( x - 00x + ( -00 ) ) + 500-1 9000( -00 ) 1 9000 (x - 00x +,10,000) + 500-90 1 9000 (x - 100 ) + 10 b. The vertex is (100, 10). The vertex represents the distance from the left tower (100 ft) at which the height of the main cable reaches its lowest point (10 ft above the roadway). c. The distance is 100 00 ft. 9a. h(t) -16t + 1-16t + 1 0-16( t - 151. ) 0-16(t + 1.)(t - 1.) 0 t + 1. 0 or t - 1. 0 t -1. or t 1. It takes about 1. s. b. h(t) - 16t + 161-16t + 161 0-16(t - 100.75) 0-16(t + 10.0)(t - 10.0) 0 t -10.0 or t 10.0 It takes 1. - 10.0. s longer. 0. h(t) -16t + t + 6 - (16t + t)+ 6 -( 16t - 1.5t ) + 6-16( t - 1.5t + ( -1.5 ) ) + 6 + 16 ( -1.5 ) -16(t - 1.5t + 0.565) + 6 + 9-16(t - 0.75 ) + 15 The maximum height of 15 ft occurs 0.75 s after the ball is shot. 1. x - 1 x x ±. 8 x - 00 0 x 5 x ± 5 x ±5. 5 x 0 x 0 x 0. -x + 6-1 -x -7 x 7 x ± 7 1 x ± 169 Holt McDougal Algebra

5. (x + 1 ) 7 x + 1 ± 7 x -1 ± 7 7. ( x + x + ) 5 ± 5 ± _ 5 - ± 5 x x + 9. 9 x + 18x + 9 5 9(x + x + 1) 5 9(x + 1) 5 (x + 1 ) 5 9 x + 1 ± 5 9 5 6. ( x + 1 ) - 9 ( x + 1 16 0 ) 9 x + 1 x + 1 16 ± 9 16 ± x 1 or -1 8. x + 1x + 9 6 (x + 7 ) 6 x + 7 ± 6 x + 7 ±8 x 1 or -15 x + 1 ± x -1 ± 5 50. A is incorrect; possible answer: in the third step, the number is added inside the parentheses. Because the expression in parentheses is multiplied by, the total number added to the function rule is 8. Therefore, the number subtracted from the rule should be 8 instead of. 51. x + 8x -15 x + 8x + ( 8 ) -15 + ( 8 ) x + 8x + 16-15 + 16 (x + ) 1 x + ± 1 x -5 or - 5. x + x -1 x + x + ( ) -1 + ( ) x + x + 11-1 + 11 (x + 11 ) 100 x + 11 ± 100 x -1 or - 1 5. x + x 1 ( x + x ) 1 ( x + x + ( ) ) 1 + ( ) ( x + x + 9) 1 + ( x + ) 7 ( x + ) 7 9 x + ± 7 9 - ± 7 x 5. x 5x + 1 x - 5x 1 x - 5 x 6 x - 5 x + ( -5 x - 5 x + 5 ( x - 5 ) 6 + ( -5 ) 16 6 + 5 16 ) 11 16 ± _ 11 16 x or - x - 5 55. x - 7x - 0 x - 7x x - 7x + ( -7 ) + ( -7 ) x - 7x + 9 + 9 ( x - ) 7 57 x - 7 ± 57 x 7 ± 57 56. x x + 11 x - x 11 x - x + ( - ) 11 + ( - ) x - x + 11 + (x - ) 15 x - ± 15 x ± 15 57. x + 6x + 0 x + 6x - x + 6x + ( 6 ) - + ( ) 6 x + 6x + 9 - + 9 (x + ) 5 x + ± 5 x - ± 5 170 Holt McDougal Algebra

58. 5 x + 10x - 7 0 5 x + 10x 7 x + x 7 5 x + x + ( ) 7 5 + ( x + x + 1 7 5 + 1 ) (x + 1 ) 1 5 x + 1 ± 1 5 x + 1 ± 15 5 x -1 ± 15 5 59. x - 8x x - 8x + ( -8 ) + ( -8 ) x - 8x + 16 + 16 (x - ) 0 x - ± 0 x ± 10 60a. h 5 m h(t) 5-5 t 5 t - 5 0 5 ( t - 1 ) 0 5(t + 1)(t - 1) 0 t + 1 0 or t - 1 0 t -1 or t 1 when h 5 m, t 1 s; h 10 m h(t) 10-5 t 5 t - 10 0 5 ( t - ) 0 5 (t + ) (t - ) 0 t + 0 or t- 0 t - or t when h 10 m, t 1.1 s; h 0 m h(t) 0-5 t 5 t - 0 0 5 ( t - ) 0 5(t + )(t - ) 0 t + 0 or t - 0 t - or t when h 0 m, t s; h 0 h(t) 0-5 t 5 t - 0 0 5 ( t - 6) 0 5 (t + 6) (t - 6) 0 t + 6 0 or t - 6 0 t - 6 or t 6 when h 0 m, t.5 s. b. The height will be h 0 m. c. h 5 m, s 18t 18(1) 18 km/h; h 10 m, s 18t 18(1.1) 5.8 km/h; h 0 m, s 18t 18() 6 km/h; h 0 m, s 18t 18(.5).1 km/h. d. The dive has to be times as high. 61a. 7.5 + 6-16t + t + 5.5-16t + t 8 t - t - 1 t - t + ( ) - 1 + ( ) (t - 1 ) 1 t - 1 ± 1 t 1 ± t 0. or t 1.7 The ball ascends at t 0. s and descends over the player s head at t 1.7 s. b. v 10 ft 71 ft/s 1.7 s 6. 55 0 10500 ft 10500 10.5 ft The side length is about 100 ft. 6. f(x) x + x - f(x) (x + 1 ) - f(x) g(x) 6a. h(t) -16t + 0t + b. h(t) (-16t + 0t) + -16 ( t - 5 t ) + -16( t - 5-16 ( t - 5 t + ( -5 t + 5 6) + + 5-16 ( t - 8) 5 + 10 1 c. The maximum height is 1 ft. d. h(t) -16t + t + (-16t + t) + 8 ) ) + + 16 ( -5-16( t - t) + -16( t - t + ( ) ) + + 16 ( ) -16(t - t + 1) + + 16-16(t - 1) + 0 0-1 9 The ball will go 9 ft higher. 65. ±7.16 66. ±1.688 67. ±.19 68. -.0,.10 69. ±1.58 70. -1.16, 0.66 8 ) 71. The square root of a negative number is not a real number. 171 Holt McDougal Algebra

7. Possible answer: Factoring is useful for solving quadratic equations with integer roots when the coefficient of the x term is not a large number. Completing the square is useful for solving quadratic equations that cannot be factored easily. It involves rewriting part of an equation so that it can be factored as a perfect square binomial. TEST PREP 7. B 7. F 75. A 76. H 77. x - x 10 x - 0.5x 5 x - 0.5x + ( -0.5 ) 5 + ( -0.5 ) x - 0.5x + 0.065 5 + 0.065 (x - 0.5 ) 5.065 x - 0.5 ± 5.065 x.5 or - 78. x - 6x 16 x - 6x + ( -6 ) 16 + ( -6 x - 6x + 9 16 + 9 (x - ) 5 x - ± 5 x 8 or - CHALLENGE AND EXTEND 79. ( b ) 1 b ±1 b ± 81. x + bx + 7 ( x + b b ( ) x + 9 ) 9 b 6 ± b ±18 ) ) to both sides simplify both sides create a square binomial on the left side square root both sides solve for both values of x add ( b 80. x - bx + 16 ( x - b x + ) b ( ) b 8 ± b ±16 8. ( b ) ac b 8. f(x) x - x 5 + 19 x - 5x + 19 0 x - 5x -19 x - - 5x + ( 5 ) -19 + ( ± ac b ± ac - 5 ) x - 5x + 0-19 + 0 5) 1 x - 5 ± 1 x 5 ± 1 (x - 8. f(x) x + 6x + x + 6 x + 0 x + 6 x - x + 6 x + (_ 6 ) - + (_ 6 ) x + 6 x + 7 - + 7 ) x + ± x - ± (x + 85a. Let l be the length, and w be the width of the field. l + w 1800 l 1800 - w l 900-1.5w A l w (900-1.5w) w -1.5w + 900w -1.5( w - 600w ) -1.5( w - 600w + ( -600 ) ) + 1.5 ( -600-1.5(w - 600w + 90000) + 15000-1.5(w - 00 ) + 15000 The largest area of the field is 15,000 f t. b. w 00 ft; l 900-1.5w 900-1.5(00) 50 ft. ) The dimensions of the largest field are 50 ft by 00 ft. c. Let x be the side length of the square field. 5x 1800 x 60 A x (60 ) 19600 The largest area of the square field is 19,600 f t. SPIRAL REVIEW 86. {x x > 7} 87. {x -6 x 1} 88. {x x n for n N} 89. {x -1 x 5} 90. B 5 895 6 675 168 6 185 615 95 91. The matrix is. 9. The entry is b 1. 9. The value is 168; it represents the amount in dollars that the Hernandez family budgeted for housing. 9. x ; (, 0) 95. x 0; (0, -1) 96. x 0; (0,.5) 17 Holt McDougal Algebra

5-5 COMPLEX NUMBERS AND ROOTS, PAGES 50-55 CHECK IT OUT! 1a. -1 (1)(-1) 1-1 -1-1 i c. - 1-6 - 1 (6)(-1) - 1 6-1 - 1 9 7-1 - 1 () 7-1 -i 7 b. x + 8 0 x -8 x ± -8 x ±i a. x - 6i -8 + (0y)i x -8 x - -6i (0y)i -6 0y - 10 y a. f(x) x + x + 1 x + x + 1 0 x + x -1 x + x + -1 + (x + ) -9 x + ± -9 x - ± i b. g(x) x - 8x + 18 x - 8x + 18 0 x - 8x -18 x - 8x + 16-18 + 16 (x - ) - b. i + + i - i x - ± - x ± i b. -6 (6)(-1) 6-1 (6) -1 1i a. x -6 x ± -6 x ±6i c. 9 x + 5 0 9 x -5 x - 5 9 x ± - 5 9 x ± 5 i b. -8 + (6y)i 5x - i 6-8 8 5x - 5 x (6y)i -i 6 6y - 6 y - 6 6 5a. 9 - i 9 + i c. 0-8i 0 + 8i 8i THINK AND DISCUSS 1. Possible answer: If a quadratic equation has nonreal roots, the roots are complex conjugates. Because + i is nonreal, the other solution is its complex conjugate, - i.. Possible answer: A real number equal to a; an imaginary number equal to bi; yes; both are complex, because real numbers and imaginary numbers are both sunsets of complex numbers.. EXERCISES GUIDED PRACTICE 1. imaginary. 5-100 5 (100)(-1) 5 100-1 5(10) -1 50i. - - - ()(-1) - -1-16 -1 -i 6. x -9 x ± -9 x ±i 8. x -16 x - x ± - x ±i 10. -x + 6i (-y)i - 1 -x -1 x 7 6i (-y)i 6 -y - 1 y 11. - + (y)i -1x - i + 8 - -1x + 8 1x 1 x 1 (y)i -i y -1. 1-16 1 (16)(-1) 1 6-1 1 () -1 i 5. -1 (1)(-1) 1-1 1i 7. x + 7 0 x -7 x -6 x ± -6 x ±6i 9. x + 11 0 x -11 x ± 11 x ±11i 17 Holt McDougal Algebra

1. f(x) x - 1x + 5 x - 1x + 5 0 x - 1x -5 x - 1x + 6-5 + 6 (x - 6 ) -9 x - 6 ± -9 x 6 ± i 1. g(x) x + 6x + x + 6x - x + 6x + 9 - + 9 (x + ) -5 x + ± -5 x - ± 5i 1. - 9i 9i 16. 8i - - + 8i - - 8i 15. 5 + 5i 5-5i 17. 6 + i 6 - i PRACTICE AND PROBLEM SOLVING 18. 8-19. - 1-90 8 ()(-1) 8-1 8() -1 16i 0. 6-1 6 (1)(-1) 6 1-1 6-1 6() -1 1i. x + 9 0 x -9 x ± -9 x ±7i. x + 7 0 x -7 x -9 x ± -9 x ±i 6. 9x + (y)i - 5-1i + 9x - 5 9x 9 x 1 (y)i -1i y -1-1 (90)(-1) - 1 90-1 - 1-1 9 10-1 () 10-1 -i 10 1. -50 (50)(-1) 50-1 5-1 5i. 5 x -80 x -16 x ± -16 x ±i 5. 1 x - x -6 x ± -6 x ±8i 7. 5(x - 1) + (y)i -15i - 0 5(x - 1) -0 x - 1 - x - (y)i -15i y -15 y -5 8. f(x) x + x + x + x + 0 x + x - x + x + 1 - + 1 (x + 1 ) - x + 1 ± - x -1 ± i 9. g(x) x - x + 1 x - x + 1 0 x - x -1 x - x - 1 x - x + 9 6-1 + 9 6 ( x - 8) - 7 x - 6 8 ± - 7 6 ± i x _ 7 8 0. f(x) x + x + 8 x + x + 8 0 x + x -8 x + x + -8 + (x + ) -. i -i. -.5i + 1 1 -.5i 1 +.5i x + ± - x - ± i 1. g(x) x - 6x + 10 x - 6x + 10 0 x - 6x -10 x - x - 10 x - x + 1-10. - - i - + i i 5. 10-1 -1 + i 10-1 - i 10 + 1 (x - 1 ) - 7 x - 1 ± -7 x 1 ± i 1 6. No; the participant will not win a prize. Possible answer: The solutions to the equation 16x - x + 18 0 are imaginary so the distance between the puck and the bell never reaches 0. 7. 1-1i 8. -5 7 i 9. - 5 - i 0. -1 + i 1. 9 + i. 17 i 17 Holt McDougal Algebra

. ci + 1 -d + 6 - ci 1 -d + 6 d 5 ci -ci ci 0 c 0 5. c + i d + di i di d c d c ± d c ± 7. 1 x -7 x -81 x ± -81 x ±9i 9. 1 x + 7 0 1 x -7 x -1 x ± -1 x ±1i 51. x + 16 0 x -16 x -8 x ± -8 x ±i 5. x + 10x + 9 0 x + 10x -9 x + 10x + 5-9 + 5 (x + 5 ) - x + 5 ± - x -5 ± i 5. x - 1x + 0 x - 1x - x - 1x + 6 - + 6 (x - 6 ) -8 x - 6 ± -8 x 6 ± i 55. x + x -5 x + x + 1-5 + 1 (x + 1 ) - x + 1 ± - x -1 ± i 57. -19 x - x -19 + 1 x - x + 1-5 (x - 1 ) ± -5 x - 1 1 ± i 5 x. c + ci + di c ci di c d 1 d 6. 8 x -8 x -1 x ± -1 x ±i 8. x + 1.5 0 x -1.5 x -6.5 x ± -6.5 x ±.5i 50. x -0 x ± -0 x ±i 0 5. x - x + 8 0 x - x -8 x - x + -8 + (x - ) - x - ± - x ± i 56. x + 18-6x x + 6x -18 x + 6x + 9-18 + 9 (x + ) -9 x + ± -9 x - ± i 58. Never true. 59. Always true. 60. Always true. 61. Sometimes true. Possible answer: i is a complex number that is imaginary; is a complex number that is not imaginary. 6. Always true. 6. Sometimes true. Possible answer: The quadratic equation x - has no real solutions; the quadratic equation x has real solutions. 6. Sometimes true. Possible answer: The quadratic equation x has real, complex roots,; the quadratic equation x - has complex roots, but these roots are not real. 65. Sometimes true. Possible answer: The quadratic equation x - has roots that form a conjugate pair; the quadratic equation x has roots that do not form a conjugate pair. 66. f(x) x - 10x + 6 x - 10x + 6 0 x - 10x -6 x - 10x + 5-6 + 5 (x - 5 ) -1 x - 5 ± -1 x 5 ± i 67. g(x) x + x + 17 x + x + 17 0 x + x -17 x + x + 1-17 + 1 (x + 1 ) -16 x + 1 ± -16 x -1 ± i 68. h(x) x - 10x + 50 x - 10x + 50 0 x - 10x -50 x - 10x + 5-50 + 5 (x - 5 ) -5 x - 5 ± -5 x 5 ± 5i 69. f(x) x + 16x + 7 x + 16x + 7 0 x + 16x -7 x + 16x + 6-7 + 6 (x + 8 ) -9 x + 8 ± -9 x -8 ± i 70. g(x) x - 10x + 7 x - 10x + 7 0 x - 10x -7 x - 10x + 5-7 + 5 (x - 5 ) -1 x - 5 ± -1 x 5 ± i 175 Holt McDougal Algebra

71. h(x) x - 16x + 68 x - 16x + 68 0 x - 16x 68 x - 16x + 6-68 + 6 (x - 8 ) - x - 8 ± - x 8 ± i 7. Possible answer: No; the graph of the function does not cross the x-axis. Therefore, it has nonreal, complex zeros. Algebra must be used to determine these zeros. 7. The complex conjugate of a real number a is the number a. 7. Possible answer: You can use the Square Root Property or complete the square to solve for complex roots with imaginary parts. 75a. 08-16t + 11t 08-16(t - 7t) -1 t - 7t -1 + 9 t - 7t + 9 ) - ( t - 7 - ± t- 7 ± _ i t - 7 7 ± i t b. Based on the solution to part a, there are no real values of t for which the height of the ball is 08 ft. Therefore, the ball does not hit the roof. c. h(t) -16t + 11t -16( t - 7t ) -16 ( t - 7t + 9-16 ( t - 7 ) + 16 ( 9 ) ) + 196 The maximum height that the ball can hit is 196 ft. TEST PREP 76. D 77. F 78. A; f(x) x - x + 17 x - x + 17 0 x - x -17 x - x + 1-16 (x - 1 ) -16 x - 1 ± -16 x 1 ± i 79. G; - i - 5 9 + ci - 11 -i ci - c 80. A; - 1 6 x 6 x -6 x ± -6 x ±6i 81. When a < 0, the solutions are imaginary and complex. When a > 0, the solutions are real and complex. CHALLENGE AND EXTEND 8. 5a + b 1 5a 1 - b a 1 - b 5-5b 7 + a -5b 7 + ( 1 - b 5 ) -5b 7 + 5 - _ 1b 5-1 5 b 9 5 b - a 1 - b 1 - (-) 5 5 a + bi - i 8. Possible answer: A quadratic equation can have a single distinct root. For example, the only distinct root of (x - 1 ) 0 is x 1. A quadratic equation cannot have a single imaginary root because imaginary roots occur only as conjugate pairs. Because a quadratic equation can have single distinct real number root and because real numbers are complex, a quadratic equation can have a single distinct complex root. 8a. If b 0 and c 0, the equation has real solutions. If b 0 and c > 0, the equation has nonreal complex solutions. b. If c 0, the equation has real solutions. c. If c > 0 and c ( b ), the equation has real solutions. If c > 0 and c > ( b ), the equation has nonreal complex solutions. d. The solutions have imaginary pairs if c > ( b ). SPIRAL REVIEW 85. T 1-5 7-1 - - row 1 (- -) + (1 0) + (- ) 16 + 0-1 (- 1) + (1 -) + (- -) - - + - (- -) + (1 1) + (- ) 8 + 1-5 row (0 -) + (- 0) + (1 ) 0 + 0 + (0 1) + (- -) + (1 -) 0 + 9-7 (0 -) + (- 1) + (1 ) 0 - + -1 row ( -) + (0 -) + ( ) -8 + 0 + - ( 1) + (- -) + ( -) + 6 - ( -) + (- 1) + ( ) - - + - 176 Holt McDougal Algebra

86. -0-15 TV -5 8 10 1 row 1 (- 10) + (1 0) + (- -5) -0 + 0 + 10-0 (- 1) + (1-1) + (- 5) - - 1-10 -15 row (0 10) + (- 0) + (1-5) 0 + 0-5 -5 (0 1) + (- -1) + (1 5) 0 + + 5 8 row ( 10) + (0 -) + ( -5) 0 + 0-10 10 ( 1) + (- -1) + ( 5) + + 10 1 87. ST is not defined. 88. S 11-5 - 10 row 1 (1 1) + (- -5) 1 + 10-11 (1-5) + (-5 0) -5 + 0-5 row (- 1) + (- 0) - + 0 - (- -5) + (0 0) 10 + 0 10 89a. upward _ b. x -b a -1 ( 5) 1 -.5 c. f(-.5) 1 5 (-.5) + (-.5) - 10 1.5 -.5-10 -11.5 The vertex is (-.5, -11.5). d. The y-intercept is -10. e. 90a. downward b. x -b a 0 (-1) 0 c. f(0) -(0) + The vertex is (0, ). d. The y-intercept is. e. 91a. upward b. x -b a - () -1 _ c. f(-1) (-1) + (-1) - - - -5 The vertex is (-1, -5). d. The y-intercept is -. e. 9a. downward b. x -b a - (- 1 ) c. f() - 1 () + () + 1 9 - + 9 + 1 11 The vertex is (, 5.5). d. The y-intercept is 1. e. 9. x + 5x 1 9. 6 x -x + x + 5x - 1 0 6 x + x - 0 (x - )(x + 7) 0 (x - 1)(x + ) 0 x - 0 or x + 7 0 x - 1 0 or x + 0 x or x -7 x 1 or x - 95. x + 9 15x x - 15x + 9 0 (x - )(x - ) 0 x - 0 or x - 0 x or x 177 Holt McDougal Algebra

96. x 1 x - 1 0 (x + 1)(x - 1) 0 x + 1 0 or x - 1 0 x - 1 or x 1 97. x + 11x - x + 11x + 0 (x + )(x + 8) 0 x + 0 or x + 8 0 x - or x -8 98. x -7x x + 7x 0 x(x + 7) 0 x 0 or x + 7 0 x 0 or x -7 5-6 THE QUADRATIC FORMULA, PAGES 56-6 CHECK IT OUT! 1a. x -b ± b - ac a - ± - (1)(-7) _ (1) - ± _ 9 + 8 - ± 7 1b. x -b ± b - ac a -(-8) ± (-8) - (1)(10) (1) 8 ± 6-0 8 ± 6 ± 6. x -b ± b - ac a -(-1) ± (-1) - ()(8) () 1 ± 1-96 6 1 ± -95 6 a. b - ac (-) - (1)() 16-16 0 b - ac 0; the equation has 1 distinct solution. b. b - ac (-) - (1)(8) 16 - -16 b - ac < 0; the equation has distinct nonreal complex solutions. c. b - ac (-) - (1)(-) 16 + 8 b - ac > 0; the equation has distinct real solutions.. 0-16t - t + 00 t -b ± b -ac a -(-) ± (-) - (-16)(00) _ (-16) ± + 5600 - ± 601 - _ 1 ± 601-16 t -5.065 or t.975 The time cannot be negative, so the water lands on the target about.975 seconds after it is released. x 91t 91(.975) 9 The water will have traveled a horizontal distance of about 9 ft. THINK AND DISCUSS 1. Possible answer: If the function has x-intercepts. there are real zeros. If there is 1 x-intercept, the function has 1 distinct real zero. If there are no x-intercepts, the function has nonreal complex zeros.. Possible: For c < 16, the equation has real solutions. For c 16, the equation has 1 distinct real solution. For c > 16, the equation has nonreal complex solutions.. EXERCISES GUIDED PRACTICE 1. Possible answer: The value of the discriminant indicates the number and types of roots.. x -b ± b - ac a -7 ± 7 - (1)(10) _ (1) -7 ± 9-0 _ -7 ± - or -5 178 Holt McDougal Algebra

. x -b ± b - ac a -(-) ± (-) - ()(-1) _ () ± 16 + 1 6 ± 7 6 _ ± 7. x -b ± b - ac a -(-5) ± (-5) - ()(0) () 5 ± 5 6 5 ± 5 6 5 or 0 5. x -b ± b - ac a -(-5) ± (-5) - (-1)(6) _ (-1) 5 ± 5 + - 5 ± 7 - -6 or 1 6. x -b ± b - ac a -(-5) ± (-5) - ()(-6) _ () 5 ± 5 + 96 8 5 ± 11 8 or - 7. x -b ± b - ac a -0 ± 0 - ()(-19) () 0 ± 15 ± _ 8 ± _ 8 8. x -b ± b - ac a -(-) ± (-) - ()() () ± - ± i 5 _ 1 ± i 5 9. x -b ± b - ac a -6 ± 6 - (1)(1) _ (1) -6 ± 6-8 -6 ± i _ - ± i 10. x -b ± b - ac a - ± - ()() () - ± 16-6 6 - ± i _ 5 6 - ± i 5 11. x -b ± b - ac a - ± - (1)(10) _ (1) - ± 16-0 _ - ± i 6 - ± i 6 1. x -b ± b - ac a -7 ± 7 - (-5)(-) (-5) -7 ± 9-60 -10-7 ± i 11-10 1. x -b ± b - ac a -7 ± 7 - (10)() _ (10) -7 ± 9-160 _ 0-7 ± i 111 0 179 Holt McDougal Algebra

1. x - x + 1 0 b - ac (-) - ()(1) 16-16 0 b - ac 0; the equation has 1 distinct real solution. 15. x + x - 10 0 b - ac - (1)(-10) + 0 b - ac > 0; the equation has distinct real solutions. 16. - x + x - 0 b - ac - (-1)(-) - 16-1 b - ac < 0; the equation has distinct nonreal complex solutions. 17. Let x be the length of the shorter leg, then (x + 6) is the length of the longer leg. x + (x + 6 ) 5 x + x + 1x + 6 65 x + 1x - 589 0 x -b ± b - ac a -1 ± 1 - ()(-589) () -1 ± 856 - ± 11 x 1 or x -0 The length of the shorter leg is 1 in., and the length of the longer leg is 1 + 6 0 in. PRACTICE AND PROBLEM SOLVING 18. x -b ± b - ac a -(-10) ± (-10) - ()() _ () 10 ± 100-6 6 10 ± 8 6 or 1 19. x -b ± b - ac a -6 ± 6 - (1)(0) (1) -6 ± _ 6-0 _ -6 ± 6 0 or -6 0. x - x - 0 x -b ± b - ac a -(-) ± (-) - (1)(-) _ (1) ± 9 + 16 ± 5 or -1 1. x -b ± b - ac a -(-) ± (-) - (-1)(9) _ (-1) ± + 6 - ± 10 - -1 ± 10. x -b ± b - ac a -(-7) ± (-7) - ()(-8) _ () 7 ± 9 + 6 7 ± 11. x -b ± b - ac a -0 ± 0 - (7)(-) _ (7) 0 ± 8 1 ± _ 1 1 ± _ 1 7. x -b ± b - ac a -1 ± 1 - (1)(1) (1) -1 ± 1 - -1 ± i 5. x -b ± b - ac a -(-1) ± (-1) - (-1)(-1) (-1) _ 1 ± 1 - - -1 ± i 180 Holt McDougal Algebra

6. x -b ± b - ac a -0 ± 0 - ()(8) () 0 ± -6 ±8i ±i 7. x -b ± b - ac a -7 ± 7 - ()(-1) () -7 ± 9 + 10 _ -7 ± _ 17 8. x -b ± b - ac a -(-1) ± (-1) - (1)(-5) _ (1) 1 ± 1 + 0 1 ± 1 9. x -b ± b - ac a - ± - (-)(-) (-) - ± 16-8 -6 - ± i _ -6 ± i 0. b - ac 1. b - ac (-) - ()(5) (-) - ()(-8) - 0-6 9 + 6 7 b - ac < 0; b - ac > 0; the equation has the equation has distinct nonreal distinct real solutions. complex solutions.. b - ac (-16) - ()() 56-56 0 b - ac 0; the equation has 1 distinct real solution.. b - ac (-8) - ()(8) 6-96 - b - ac < 0; the equation has distinct nonreal complex solutions.. b - ac (-8) - ()(9) 78-78 0 b - ac 0; the equation has 1 distinct real solution. 5. b - ac (-8.5) - (.)(1.) 7.5-16.6 55.61 b - ac > 0; the equation has distinct real solutions. 6. -16t + (60-11) 0-16t + 9 0 t -b ± b - ac a -0 ± 0 - (-16)(9) (-16) _ 0 ± 16 - _ ±56 - -1.75 or 1.75 The walker will fall for 1.75 s. 7a. 5 t + 0t 585 5 t + 0t - 585 0 5 (t + t - 117) 0 t -b ± b - ac a - ± - (1)(-117) (1) - ± _ 8 - ± 9 or -1 It will take her 9 s to reach the bottom. b. 5 t + 0t 9.5 5 t + 0t - 9.5 0 5 (t + t - 58.5) 0 t -b ± b - ac a - ± - (1)(-58.5) _ (1) - ± _ 50 - ± 5 10 t 6 or t -10 It will take about 6 s to reach the bottom. 8. x -b ± b - ac a -(-) ± (-) - ()(-) _ () ± 16 + 6 ± 10 6 ± 10 181 Holt McDougal Algebra

9. x -b ± b - ac a -(-) ± _ (-) - ()(-1) () _ ± + 8 ± _ 1 ±. x -b ± b - ac a -(-5) ± (-5) - ()(-) _ () 5 ± 5 + 8 6 5 ± 7 6 0. x -b ± b - ac a -6 ± 6 - ()(5) () -6 ± 6-0 -6 ± i _ - ± i. x -b ± b - ac a -(-1) ± (-1) - (1)() (1) 1 ± 1-88 1 ± i 87 1. x -b ± b - ac a - ± (-) - ()(-1) _ () - ± 9 + 8 - ± 17 a. -0.17t + 187t + 61000 0 t -b ± b - ac a -187 ± 187 - (-0.17)(61000) _ (-0.17) -187 ± 769-0. t -61.7 or t 16 The flight took about 16 s. b. axis of symmetry: t -b a -187 (-0.17) 550 h(550) -0.17(550) + 187(550) + 61,000-515 + 10850 + 61,000 115 The highest altitude is about 11,000 m. 18 Holt McDougal Algebra

c. -0.17t + 187t + 61000 5000-0.17t + 187t + 56000 0 t -b ± b - ac a -187 ± 187 - (-0.17)(56000) _ (-0.17) -187 ± 1869-0. t 18 or x 951.6 The ship entered the thermosphere first at t 19 s then the mesosphere at t 95 s. -0.17t + 187t + 61000 50000-0.17t + 187t + 11000 0 t -b ± b - ac a -187 ± 187 - (-0.17)(11000) _ (-0.17) -187 ± 9-0. t 1156 After the mesosphere, the shop entered the stratosphere at t 1156 s. -0.17t + 187t + 61000 10000-0.17t + 187t + 51000 0 t -b ± b - ac a -187 ± 187 - (-0.17)(51000) _ (-0.17) -187 ± 6969-0. t 16.17 The ship entered the troposphere at t 16 s. 5. x - x - 10 0 (x - 5)(x + ) 0 x - 5 0 or x + 0 x 5 or x - 7. x + x - 15 0 (x - )(x + 5) 0 x - 0 or x + 5 0 x 1.5 or x -.5 8. x + x - 0 x -b ± b - ac a -() ± () - (1)(-) _ (1) - ± + 8 - ± -1 ± 6. x - 16 0 (x - )(x + ) 0 x - 0 or x + 0 x or x - 9. x - x - 1 0 (x - 7)(x + ) 0 x - 7 0 or x + 0 x 7 or x - 50. x - x - 1 0 x -b ± b - ac a -(-) ± (-) - ()(-1) _ () ± 16 + 16 8 ± 8 _ 1 ± 5. x 7 x ± 7 51. 6 x 150 x 5 x ± 5 x ±5 5. x - 16x + 6 0 (x - 8)(x - 8) 0 x -8 0 x 8 5. The problem does not have a meaningful solution because the roots of the quadratic equation are imaginary. 55. (5 - w)(0 - w) 66 500-50w - 0w + w 66 w - 90w + 0 (w - 5w + 117) 0 (w - )(w - 9) 0 w - 0 or w - 9 0 w or w 19.5 The width of the frame is in. 56. b - ac 0 8 - (1)(c) 0 6 - c 0 -c -6 c 16 58. b - ac 0 (c) - (1)(9) 0 c 196 c 9 c ± 9 c ±7 57. x + 1x - c 0 b - ac 0 1 - (1)(-c) 0 1 + c 0 c -1 c -6 59. Possible answer: The Quadratic Formula; this method would be easier because the absolute value of a is great and the value of c is not an integer. Therefore, it would be difficult to solve the equation by factoring or by completing the square. 18 Holt McDougal Algebra

60a. -16t + 19t + 5-16t + 19t + 1 0 t -b ± b - ac a -19 ± 19 - (-16)(1) _ (-16) -19 ± 5-19 ± 5 _ 17 t -0.05 or t 1. The ball is in the air for about 1. s. b. No, the ball will reach third base before the runner. Possible answer: When the ball reaches third base, the runner will have been running for about. s. The runner will have traveled only about 65 ft in this amount of time. TEST PREP 61. B 6. H; b - ac (-8) - ()(-1) 6 + 11 176 6. C; x -b ± b - ac a -(-6) ± (-6) - (1)(10) (1) 6 ± 6-0 6 ± i ± i CHALLENGE AND EXTENDED 6. F 65. Let x be the length of one leg, then the length of the other leg is 0-17 -x - x. x + ( - x) 1 7 x + 59-6x + x 89 x - 6x + 0 0 (x - x + 10) 0 (x - 8)(x - 15) 0 x - 8 0 or x - 15 0 x 8 or x 15 The length of the legs are 8 cm and 15 cm. 66a. Let x be the length of the rectangle, then the width of the rectangle will be 88 - x - x. d x + ( - x) x + 196-88x + x x - 88x + 196 ( x - x) + 196 (x - x + 8) + 196 - (8) (x - ) + 968 Therefore, the least possible value of the length of the diagonal is 968 1.1 cm. b. The dimensions of the rectangle are cm by cm. 67. Possible answer: x 68. Possibele anwer: x 1 69. Possible answer: x - 70a. x 1 + x -b + b - ac + -b - b - ac a a _ -b + b - ac - b - b - ac a _ -b b a - a x 1 x -b + b - ac -b - b - ac a a b - ac ) ( -b - b - ac) ( -b + (a) (a) (-b) -( b - ac) a -( b b - ac ) a _ ac c a a b. x - x - 15 0 71. The solutions are nonreal and complex. Regardless -1 ± i of the values of a, b and c, the roots are. SPIRAL REVIEW _ 7..1 cm 0.05 cm/day 60 days h(x) 0.05x 7. x - y -7 x - 6y 1 R 1 R R R - R 1 R 1 9 R R 1 R 1 + 6 R solution (-5, -1) 7. x + y 1 x + y 1 R R - R 1 R 1 R 1 - R solution (0, -16) - -7 1-6 -1 1-6 1 - -7 1-6 1 0 9-9 1-6 1 0 1-1 1 0-5 0 1-1 1 1 1 1 1 1 1 0 1-16 1 0 0 0 1-16 18 Holt McDougal Algebra

75. x + 5y -1 x - 7y 9 R 1 R R 1 1 R R R - R 1 R 1 19 R R 1 R 1 + 7 R solution (1, -1) 5-1 -7 9-7 9 5-1 1-7 9 76. x - 5x 1 x - 5x + 5 1 + 5 ( x - 5 ) 9 x - 5 ± 9 x 5 ± 9 77. x 16x - x - 16x - x - 8x - x - 8x + 16 - + 16 (x - ) 1 x - ± 1 x ± 1 5-1 1-7 9 0 19-19 1-7 9 0 1-1 1 0 0 0 1-1 78. x 5 x - 1-5x + x -1 x - 0.6. x - 0.6 + 0.09. + 0.09 (x - 0. ).9 x - 0. ±.9 x 0. ±.9 x ± 9 10 READY TO GO ON? PAGE 65 1. g is f translated units left and units down.. g is f reflected across the x-axis, vertically stretched by a factor of, and translated 1 unit right.. g is f vertically compressed by a factor of 1 and translated 1 unit up.. g(x) 9(x + ) 5. g(x) - x + 6a. upward b. x -b a -(-) (1) c. f() ( ) - () + - 8 + -1 The vertex is (, -1). d. The y-intercept is. e. 8a. upward b. x -b a -(-6) (1) c. h() - 6() 9-18 -9 The vertex is (, -9). d. The y-intercept is 0. e. 7a. downward b. x -b a - (-1) 1 c. g(1) -(1) + (1) - 1-1 + - 1 0 The vertex is (1, 0). d. The y-intercept is -1. e. 185 Holt McDougal Algebra

9. x -b a _ -0.5 (-0.0075) _ -0.5-0.015. h(.) -0.0075(.) + 0.5(.) + 5-8. + 16.665 + 5 1. The maximum height of the ball is about 1 ft. 10. x - 100 0 (x - 10)(x + 10) 0 x - 10 0 or x + 10 0 x 10 or x -10 11. x + 5x - 0 (x - )(x + 8) 0 x - 0 or x + 8 0 x or x -8 1. x + 8x 0 x(x + ) 0 x 0 or x + 0 x 0 or x - 1. x - 6x 0 x - 6x + 9 0 + 9 (x - ) 9 x - ± 9 x - or 10 1. x + 18x 15 x + 18x + 81 15 + 81 (x + 9 ) 96 x + 9 ± 96 x -9 ± 6 15. x + 1x 8 x + 1x + 9 8 + 9 (x + 7 ) 57 x + 7 ± 57 x -7 ± 57 16. f(x) x + x + 18 (x + x) + 18 (x + x + 1) + 18-1 (x + 1 ) - 6 The vertex is (-1, -6). 17. g(x) x - 1x + 9 ( x - 1x ) + 9 (x - 1x + 6) + 9-6 (x - 6 ) + The vertex is (6, ). 18. h(x) 5 x - 0x + 9 ( 5x - 0x ) + 9 5 ( x - x) + 9 5 (x - x + ) + 9-5() 5(x - ) - 11 The vertex is (, -11). 19. x -8 x -16 x ± -16 x ±i 0. x - 0x -15 x - 0x + 100-15 + 100 (x - 10 ) -5 x - 10-5 x 10 ± 5i 1. x - 8x + 0 0 x -b ± b - ac a -(-8) ± (-8) - (1)(0) (1) 8 ± -56 8 ± i 1 ± i 1. f(x) x + 1x + 6 + x + 1x + 8 x -b ± b - ac a -1 ± 1 - (1)(8) (1) -1 ± -8-1 ± i -6 ± i. x -b ± b - ac a -7 ± 7 - (1)(15) _ (1) -7 ± 9-60 - 7 ± i _ 11. x -b ± b - ac a -(-5) ± (-5) - ()() () 5 ± 5-5 ± 1 1.5 or 1 5. 00 t + 18t 0 t + 18t - 00 t - 18 ± (18) - ()(- 00) _ () t 8 s 186 Holt McDougal Algebra

5-7 SOLVING QUADRATIC INEQUALITIES, PAGES 66-7 CHECK IT OUT! 1a. a. -1 < x < b. x 0 or x.5 b. a. x - 6x + 10 x - 6x + 8 0 (x - )(x - ) 0 x - 0 or x - 0 x or x x-value test: 0-6(0) + 10-6() + 10 5-6(5) + 10 Therefore x or x. b. -x + x + 7 -x + x + 5 0-1(x + 1)(x - 5) 0 x + 1 0 or x - 5 0 x -1 or x 5 x-value test: -(-) + (-) + 7 < -(0) + (0) + 7 < -() + () + 7 < Therefore x < -1 or x >.5.. -5x + 150x - 5000 7500-5x + 150x - 5000 7500-5x + 150x - 1500 0-5(x - 50 + 500) 0 x -b ± b - ac a -(-50) ± (-50) - (1)(500) (1) 50 ± _ 500 5 ± 5 5 x 6. or x 1.8 x-value test: -5(0) + 150(0) - 5000 > 7500-5(0) + 150(0) - 5000 > 7500-5(0) + 150(0) - 5000 > 7500 Therefore there should be fewer than 1 peope or more than 6 people. THINK AND DISCUSS 1. Possible answer: Graphing a quadratic inequality is essentially the same as graphing a linear inequality: plot points on a boundary, connect them with a solid or dashed curve, and shade above or below accordingly. For a quadratic inequality, the boundary is a parabola, and for a linear inequality, the boundary is a line.. Possible answer: The intersection points are included when the inequality symbol is or. The intersection points are not included when the inequality symbol is > or <.. EXERCISES GUIDED PRACTICE 1. Possible answer: y < x + x +... 5. 0 x 5 6. x < 0 or x > 1 7. 1.5 x 8. x + 10x + 1 1 x + 10x - 11 0 (x - 1)(x + 11) 0 x - 1 0 or x + 11 0 x 1 or x -11 x-value test: (-15) + 10(15) + 1 1 (0) + 10(0) + 1 1 ( ) + 10() + 1 1 Therefore x -11 or x 1. 187 Holt McDougal Algebra