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: 22

Answer

The graph is shown below.

Work Step by Step

$xy=-1$ Divide x on both sides of the equation $xy=-1$, $\begin{align} & \frac{xy}{x}=\frac{-1}{x} \\ & y=-\frac{1}{x} \end{align}$ Now substitute $x=-4$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=\frac{-1}{-4} \\ & =\frac{1}{4} \\ & =0.25 \end{align}$ Now substitute $x=-3$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=\frac{-1}{-3} \\ & =\frac{1}{3} \\ & =0.33 \end{align}$ Now substitute $x=-2$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=\frac{-1}{-2} \\ & =\frac{1}{2} \\ & =0.5 \end{align}$ Now substitute $x=-1$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=\frac{-1}{-1} \\ & =\frac{1}{1} \\ & =1 \end{align}$ Now substitute $x=1$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=-\frac{1}{1} \\ & =-1 \end{align}$ Now substitute $x=2$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=-\frac{1}{2} \\ & =-0.5 \end{align}$ Now substitute $x=3$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=-\frac{1}{3} \\ & =-0.33 \end{align}$ Now substitute $x=4$ into the equation $y=-\frac{1}{x}$, $\begin{align} & y=-\frac{1}{4} \\ & =-0.25 \end{align}$ Thus, the required table is, $\begin{matrix} x & -4 & -3 & -2 & -1 & 1 & 2 & 3 & 4 \\ y=-\frac{1}{x} & 0.25 & 0.33 & 0.5 & 1 & -1 & -0.5 & -0.33 & -0.25 \\ \end{matrix}$ Now plot the points and draw two curves.
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