Chapter 5: Exercises with Solutions

Chapter 5Quadratic Functions5.6 Solutions1. The graph opens downward since a = −1 > 0, and the vertex is at (h, k), whereh = − b(−2a = −3 2 and k = f(h) = f 3 )= 9 . Thus, the maximum value of the2 4function is 9 4 .3. The vertex is (h, k), where h = − b(−2a = −1 6 and k = f(h) = f 1 )= − 716 12 .5. The graph opens downward since a = −3 > 0, and the vertex is at (h, k), whereh = − b(−2a = −3 2 and k = f(h) = f 3 )= 11 . Thus, the maximum value of the2 4function is 11 4 .7. The vertex is (h, k), where h = − b(−2a = −1 2 and k = f(h) = f 1 )= 33 2 4 .9. The graph opens upward since a = 1 > 0, and the vertex is at (h, k), whereh = − b(−2a = −9 2 and k = f(h) = f 9 )= − 81 . Thus, the minumum value of the2 4function is − 814 .11. The graph opens downward since a = −3 < 0, and the vertex is at (h, k), whereh = − b2a = 4 ( ) 43 and k = f(h) = f = 10(3 3 . Thus, the range is (−∞, k] = −∞, 10 ].313. The graph opens upward since a = 4 > 0, and the vertex is at (h, k), whereh = − b(−2a = −9 8 and k = f(h) = f 9 )= − 209 . Thus, the range is [k, ∞) =[8 16− 209 )16 , ∞ .15. The axis of symmetry is x = h, where h = − b2a = −1 4 .17. If the y-axis is oriented with the positive direction upward, and the 0 mark isset at ground level, then the height (in feet) of the ball after t seconds is given by thefunctionh(t) = −16t 2 + 8t + 182Version: Fall 2007