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: 31

Answer

The graph for the equation $x=-2{{y}^{2}}-4y+1$.
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Work Step by Step

$x=-2{{y}^{2}}-4y+1$ (equation - 1) $\begin{align} & y=-\frac{b}{2a} \\ & =-\frac{\left( -4 \right)}{2\cdot \left( -2 \right)} \\ & =-1 \end{align}$ Now for the x value put $y=-1$ in equation (1), $\begin{align} & x=-2{{\left( -1 \right)}^{2}}-4\cdot \left( -1 \right)+1 \\ & =-2+4+1 \\ & =3 \end{align}$ The vertex is $\left( 3,-1 \right)$. Now choose some value of x on both sides of the vertex and compute the corresponding y value. $\begin{matrix} x & y \\ 3 & -1 \\ 1 & -2 \\ 1 & 0 \\ -5 & -3 \\ 5 & 1 \\ \end{matrix}$
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