Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 11 - Section 11.3 - Hyperbolas - 11.3 Exercises - Page 805: 8

Answer

Hyperbola matches with the graph labeled as I

Work Step by Step

$9x^2-25y^2=225$ $\frac{x^2}{25}-\frac{y^2}{9}=1$ Hyperbola with horizontal transverse axis: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ So: $a^2=25$ $a=5$ Vertices: $(±a,0)=(±5,0)$ Hyperbola matches with the graph labeled as I
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