Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 10 - Section 10.2 - The Ellipse - Exercise Set - Page 780: 72


See the explanation below.

Work Step by Step

a) To find the $y$ interpret , we put $x=0$ Thus, $\dfrac{y^2}{9}-\dfrac{0^2}{16}=1$ or, $y^2 =9$ or, $y = \pm 3$ Therefore, $y$ intercepts are: {$-3,3$} b)To find the $x$ interpret , we put $y=0$ Thus, $\dfrac{0^2}{9}-\dfrac{x^2}{16}=1$ or, $x^2 = -16$ or, $x = \pm \sqrt{-16}$ Therefore, there are no $x$ intercepts because of of no real numbers or a non-real solution.
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.