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 - Mid-Chapter Review - Mixed Review - Page 873: 9

Answer

Parabola.

Work Step by Step

$x={{\left( y+3 \right)}^{2}}+2$ It is written as: $\begin{align} & x={{\left( y+3 \right)}^{2}}+2 \\ & {{\left( y+3 \right)}^{2}}=x-2 \\ & {{\left( y+3 \right)}^{2}}=4\left( \frac{1}{4} \right)\left( x-2 \right) \end{align}$ The above equation is a standard form of a parabola ${{\left( y-k \right)}^{2}}=4a(x-h)$. Therefore, it is a parabola equation. Graph: The vertices of the parabola are $\left( h,k \right)=\left( 2,-3 \right)$, and the focus of the parabola is $\left( a,0 \right)=\left( \frac{1}{4},0 \right)$. Therefore, the graph of the parabola $x={{\left( y+3 \right)}^{2}}+2$ is shown below:
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