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

Published by Pearson

# Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set - Page 854: 27

#### Answer

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

#### Work Step by Step

$x=-\frac{1}{2}{{y}^{2}}$ (equation - 1) \begin{align} & y=-\frac{b}{2a} \\ & =-\frac{\left( 0 \right)}{2\cdot \left( \frac{1}{2} \right)} \\ & =0 \end{align} Now for the x value put $y=0$ in equation (1). \begin{align} & x=-\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. $\begin{matrix} x & y \\ 0 & 0 \\ -2 & 2 \\ -8 & 4 \\ -2 & -2 \\ -8 & -4 \\ \end{matrix}$

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