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 - 13.3 Conic Sections: Hyperbolas - 13.3 Exercise Set - Page 871: 20

Answer

The graph is shown below.

Work Step by Step

$xy=-9$ Divide x on both sides: $xy=-9$, $\begin{align} & \frac{xy}{x}=\frac{-9}{x} \\ & y=-\frac{9}{x} \end{align}$ Now substitute $x=-4$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=\frac{-9}{-4} \\ & =\frac{9}{4} \\ & =2.25 \end{align}$ Now substitute $x=-3$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=\frac{-9}{-3} \\ & =\frac{9}{3} \\ & =3 \end{align}$ Now substitute $x=-2$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=\frac{-9}{-2} \\ & =\frac{9}{2} \\ & =4.5 \end{align}$ Now substitute $x=-1$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=\frac{-9}{-1} \\ & =\frac{9}{1} \\ & =9 \end{align}$ Now substitute $x=1$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=-\frac{9}{1} \\ & =-9 \end{align}$ Now substitute $x=2$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=-\frac{9}{2} \\ & =-4.5 \end{align}$ Now substitute $x=3$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=-\frac{9}{3} \\ & =-3 \end{align}$ Now substitute $x=4$ into the equation $y=-\frac{9}{x}$, $\begin{align} & y=-\frac{9}{4} \\ & =-2.25 \end{align}$ Thus, the required table is, $\begin{matrix} x & -4 & -3 & -2 & -1 & 1 & 2 & 3 & 4 \\ y=-\frac{9}{x} & 2.25 & 3 & 4.5 & 9 & -9 & -4.5 & -3 & -2.25 \\ \end{matrix}$ Now plot the points and draw two curves.
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