Precalculus (6th Edition)

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

Chapter 10 - Analytic Geometry - Chapter 10 Test Prep - Review Exercises - Page 1001: 33

Answer

$$\eqalign{ & {\text{domain: }}\left[ { - 3,3} \right] \cr & {\text{range: }}\left[ { - 2,2} \right] \cr & {\text{Vertices: }}\left( { \pm 3,0} \right) \cr} $$

Work Step by Step

$$\eqalign{ & 4{x^2} + 9{y^2} = 36 \cr & {\text{Divide both sides by 36}} \cr & \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1 \cr & {\text{The equation of the ellipse is in the form }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,\,\left( {a > b} \right) \cr & a = 3,\,\,b = 2 \cr & \cr & {\text{The ellipse with center at the origin}} \cr & {\text{Vertices }}\left( { \pm a,0} \right) \cr & {\text{Vertices }}\left( { \pm 3,0} \right) \cr & \cr & {\text{The domain of the ellipse is }}\left[ { - a,a} \right] \cr & {\text{domain }}\left[ { - 3,3} \right] \cr & \cr & {\text{The range of the ellipse is }}\left[ { - b,b} \right] \cr & {\text{range }}\left[ { - 2,2} \right] \cr & \cr & {\text{Therefore,}} \cr & {\text{domain: }}\left[ { - 3,3} \right] \cr & {\text{range: }}\left[ { - 2,2} \right] \cr & {\text{Vertices: }}\left( { \pm 3,0} \right) \cr & \cr & {\text{Graph}} \cr} $$
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