Intermediate Algebra for College Students (7th Edition)

Published by Pearson

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

Answer

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.

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