See Problem 6.

Rewrite each equation in vertex form.

46. y = x 2 + 4 x + 1 y equals , x squared , plus 4 x plus 1
47. y = 2 x 2 − 8 x + 1 y equals , 2 x squared , minus 8 x plus 1
48. y = − x 2 − 2 x + 3 y equals negative , x squared , minus 2 x plus 3
49. y = x 2 + 4 x − 7 y equals , x squared , plus 4 x minus 7
50. y = 2 x 2 − 6 x − 1 y equals , 2 x squared , minus 6 x minus 1
51. y = − x 2 + 4 x − 1 y equals negative , x squared , plus 4 x minus 1

B Apply

52. Think About a Plan The area of the rectangle shown is 80 square inches. What is the value of x?

    

    • How can you write an equation to represent 80 in terms of x?
    • How can you find the value of x by completing the square?

Find the value of k that would make the left side of each equation a perfect square trinomial.

53. x 2 + k x + 25 = 0 x squared , plus k x plus 25 equals 0
54. x 2 − k x + 100 = 0 x squared , minus k x plus 100 equals 0
55. x 2 − k x + 121 = 0 x squared , minus k x plus 121 equals 0
56. x 2 + k x + 64 = 0 x squared , plus k x plus 64 equals 0
57. x 2 − k x + 81 = 0 x squared , minus k x plus 81 equals 0
58. 25 x 2 − k x + 1 = 0 25 x squared , minus k x plus 1 equals 0
59. x 2 + k x + 1 4 = 0 x squared , plus k x plus , 1 fourth , equals 0
60. 9 x 2 − k x + 4 = 0 9 x squared , minus k x plus 4 equals 0
61. 36 x 2 − k x + 49 = 0 36 x squared , minus k x plus 49 equals 0
62. Geometry The table shows some possible dimensions of rectangles with a perimeter of 100 units. Copy and complete the table.
    1. Plot the points (width, area). Find a model for the data set.
    2. What is another point in the data set? Use it to verify your model.
    3. What is a reasonable domain for this function? Explain.
    4. Find the maximum possible area. What dimensions yield this area?
    5. Find a function for area in terms of width without using the table. Do you get the same model as in part (a)? Explain.

Solve each quadratic equation by completing the square.

63. x 2 + 5 x − 3 = 0 x squared , plus 5 x minus 3 equals 0
64. x 2 + 3 x = 2 x squared , plus 3 x equals 2
65. x 2 − x = 5 x squared , minus x equals 5
66. x 2 + x − 1 = 0 x squared , plus x minus 1 equals 0
67. 3 x 2 − 4 x = 2 3 x squared , minus 4 x equals 2
68. 5 x 2 − x = 4 5 x squared , minus x equals 4
69. x 2 + 3 4 x = 1 2 x squared , plus , 3 fourths , x equals , 1 half
70. 2 x 2 − 1 2 x = 1 8 2 , x squared , minus , 1 half , x equals , 1 eighth
71. 3 x 2 + x = 2 3 3 x squared , plus x equals , 2 thirds
72. − x 2 + 2 x + 4 = 0 negative , x squared , plus 2 x plus 4 equals 0
73. − x 2 − 6 x = 2 negative , x squared , minus 6 x equals 2
74. − 0 . 25 x 2 − 0 . 6 x + 0 . 3 = 0 negative 0 . 25 , x squared , minus 0 . 6 x plus 0 . 3 equals 0
75. Football The quadratic function h = − 0 . 01 x 2 + 1 . 18 x + 2 h equals negative 0 . 01 , x squared , plus 1 . 18 x plus 2  models the height of a punted football. The horizontal distance in feet from the point of impact with the kicker's foot is x, and h is the height of the ball in feet.
    1. Write the function in vertex form. What is the maximum height of the punt?
    2. The nearest defensive player is 5 ft horizontally from the point of impact. How high must the player reach to block the punt?
    3. Suppose the ball was not blocked but continued on its path. How far down the field would the ball go before it hit the ground?

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