Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-2 Quadratic Functions - Practice and Problem-Solving Exercises - Page 545: 26


The maximum height reached is $\frac{221}{16}$ft or 13.8 ft. The range is $0\leq h \leq \frac{221}{16}$

Work Step by Step

$h = -16t^{2} + 30t + 6$ The standard form for a quadratic equation is $y = ax^{2} + bx + c$ So a= -16, b= 30, and c= 6 Axis of symmetry: In this question the Axis of symmetry gives us the time to reach the maximum height. The formula for axis of symmetry is $x= \frac{-b}{2a}$ $x= \frac{-(30)}{2(-16)}$ $x= \frac{-30}{-32}$ $x= \frac{15}{16}$ Vertex: In this question the vertex gives us the maximum height of the ball. Plug in the x value of the axis of symmetry to find the y value of the vertex. $h = -16t^{2} + 30t + 6$ $h = -16(\frac{15}{16})^{2} + 30(\frac{15}{16}) + 6$ y= $\frac{221}{16}$ The vertex is ($\frac{15}{16}$, $\frac{221}{16}$) Since the ball starts at ground level of zero and goes up to $\frac{221}{16}$ the range is $0\leq h \leq \frac{221}{16}$
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