Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 10 - Analysis of Variance - Chapter 10 Test Prep - Review Exercises - Page 1000: 14

Answer

$${\text{hyperbola}}$$

Work Step by Step

$$\eqalign{ & 9{x^2} - 16{y^2} = 144 \cr & {\text{Divide both sides by 144}} \cr & \frac{{9{x^2}}}{{144}} - \frac{{16{y^2}}}{{144}} = \frac{{144}}{{144}} \cr & \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 \cr & {\text{The equation is written in the form }}\frac{{{x^2}}}{{{b^2}}} - \frac{{{y^2}}}{{{a^2}}} = 1\, \cr & {\text{Therefore,}} \cr & {\text{The graph of the equation is a hyperbola}} \cr} $$
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.