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: 19

Answer

The equation $x={{y}^{2}}+3$
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Work Step by Step

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