Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.2 - The Hyperbola - Exercise Set - Page 982: 51

Answer

See graph; domain $(-\infty,-3]\cup[3,\infty)$, range $(-\infty,\infty)$
1585157310

Work Step by Step

Step 1. From the given equation $\frac{(x)^2}{9}-\frac{(y)^2}{16}=1$, we have $a=3, b=4, c=\sqrt {a^2+b^2}=5$ centered at $(0,0)$ with a horizontal transverse axis. Step 2. We can find the vertices as $(\pm3,0)$, foci as $(\pm5,0)$, and asymptotes as $y=\frac{b}{a}(x)$ or $y=\pm\frac{4}{3}(x)$ Step 3. We can graph the equation as shown in the figure with domain $(-\infty,-3]\cup[3,\infty)$ and range $(-\infty,\infty)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.