## 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: 16

#### Answer

The equation $x=4-3y-{{y}^{2}}$.

#### Work Step by Step

$x=4-3y-{{y}^{2}}$ (equation -1) \begin{align} & y=-\frac{b}{2a} \\ & =-\frac{\left( -3 \right)}{2\cdot \left( -1 \right)} \\ & =-\frac{3}{2} \end{align} Now for the x value put $y=-\frac{3}{2}$ in equation (1). \begin{align} & x=4-3\cdot \left( -\frac{3}{2} \right)-{{\left( -\frac{3}{2} \right)}^{2}} \\ & =4+\frac{9}{2}-\frac{9}{4} \\ & =\frac{4\cdot 4+9\cdot 2-9}{4} \\ & =\frac{16+18-9}{4} \end{align} And, $x=\frac{25}{4}$ The vertex is$\left( \frac{25}{4},-\frac{3}{2} \right)$. Now choose some value of x on both sides of the vertex and compute the corresponding y value. $\begin{matrix} x & y \\ 6 & -1 \\ 6 & -2 \\ 4 & 0 \\ 0 & 1 \\ \end{matrix}$ Now plot all the values of x and y on the rectangular coordinate system and join each point with one another to form a parabola as shown below

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