Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 670: 41

Answer

Refer to the graph below.
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Work Step by Step

RECALL: The graph of $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$ where $a\gt b$, is a vertical ellipse that has: its center at $(0, 0)$ major axis length = $2a$ minor axis length = $2b$ Before the given inequality can be graphed, there is a need to graph first the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$. The equation $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$ can be written as $\dfrac{x^2}{3^2}+\dfrac{y^2}{4^2}=1$ therefore is has $a=4$ and $b=3$. Thus, ots graph is a vertical ellipse whose center is at $(0, 0)$ and whose major axis is 8 units long and whose minor axis is 3 units long. Graph the ellipse. (Refer to the graph below.) The graph of the inequality $\dfrac{x^2}{9} + \dfrac{y^2}{16^2} \le1$ includes the ellipse itself and the region inside the ellipse. Thus, shade the region inside the ellipse. (Refer to the graph in the answer part above.)
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