Chapter 5: Exercises with Solutions

Section 5.2Vertex FormThe graph opens upward since a = 5 > 0, and the vertex is at (h, k), where h = − 2 5and k = 16 5 . Thus, the range is [k, ∞) = [ 165 , ∞) .71. First complete the square to transform the function into vertex form a(x−h) 2 +k:f(x) = −x 2 − 2x − 7= −1 ( x 2 + 2x + 7 )= −1 ( x 2 + 2x + 1 − 1 + 7 )= −1 (( x 2 + 2x + 1 ) − 1 + 7 )()= −1 (x + 1) 2 + 6= −1 (x + 1) 2 − 6The graph opens downward since a = −1 < 0, and the vertex is at (h, k), where h = −1and k = −6. Thus, the range is (−∞, k] = (−∞, −6].73. Substitute −6 for x in 9x 2 − 9x + 4 and simplify to get 382:f(−6) = 9(−6) 2 − 9(−6) + 4 = 38275. Substitute 11 for x in 4x 2 − 6x − 4 and simplify to get 414:f(11) = 4(11) 2 − 6(11) − 4 = 41477. Substitute x + 4 for x in −5x 2 + 4x + 2 and simplify:f(x + 4) = −5(x + 4) 2 + 4(x + 4) + 2= −5(x 2 + 8x + 16) + 4x + 16 + 2= −5x 2 − 36x − 6279. Substitute 4x − 1 for x in 4x 2 + 3x − 3 and simplify:f(4x − 1) = 4(4x − 1) 2 + 3(4x − 1) − 3= 4(16x 2 − 8x + 1) + 12x − 3 − 3= 64x 2 − 20x − 2Version: Fall 2007