## Algebra 1

Published by Prentice Hall

# Chapter 9 - Quadratic Functions and Equations - 9-2 Quadratic Functions - Practice and Problem-Solving Exercises - Page 545: 32

#### Answer

Please see the graph.

#### Work Step by Step

The graph was made using graphing software. $y=\frac{1}{2}x^2+8x-20$ vertex: $x=-b/2a$ $x = -(8)/2*(.5)$ $x = -8/1$ $x = -8$ $y=\frac{1}{2}x^2+8x-20$ $y=\frac{1}{2}*8^2+8*-8-20$ $y=.5*64-64-20$ $y=32-64-20$ $y=-52$ $(-8,-52)$ Two points on the curve: $x=0$ $y=\frac{1}{2}*0^2+8*0-20$ $y= .5*0+0 - 20$ $y = -20$ $(0,-20)$ Since the distance from 0 to the axis of symmetry is 8 units, we also know $(8,-52)$ is on the curve.

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