Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set - Page 854: 28

Answer

The graph for the equation $y=-\frac{1}{2}{{x}^{2}}$.

Work Step by Step

$y=-\frac{1}{2}{{x}^{2}}$ (equation - 1) $\begin{align} & x=-\frac{b}{2a} \\ & =-\frac{\left( 0 \right)}{2\cdot \left( -\frac{1}{2} \right)} \\ & =0 \end{align}$ Now for the y value put $x=0$ in equation (1), $\begin{align} & y=-\frac{1}{2}{{\left( 0 \right)}^{2}} \\ & =0 \end{align}$ The vertex is $\left( 0,0 \right)$. Now choose some value of x on both sides of the vertex and compute the corresponding y value. $\begin{matrix} x & y \\ 0 & 0 \\ -2 & -2 \\ 4 & -8 \\ 2 & -2 \\ -4 & -8 \\ \end{matrix}$
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