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: 17

Answer

$x={{y}^{2}}-4y+2$
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Work Step by Step

$x={{y}^{2}}-4y+2$ (equation - 1) $\begin{align} & y=-\frac{b}{2a} \\ & =-\frac{\left( -4 \right)}{2\cdot \left( 1 \right)} \\ & =2 \end{align}$ Now for the x value put $y=2$ in equation (1). $\begin{align} & x={{\left( 2 \right)}^{2}}-4\cdot 2+2 \\ & =4-8+2 \\ & =-2 \end{align}$ The vertex is$\left( -2,2 \right)$. Now choose some value of x on both sides of the vertex and compute the corresponding y value. $\begin{matrix} x & y \\ 7 & -1 \\ 2 & 0 \\ -1 & 1 \\ -2 & 2 \\ -1 & 3 \\ \end{matrix}$
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