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: 37

Answer

$$\eqalign{ & {\text{Vertices}}:\left( { - 3, - 4} \right){\text{ and }}\left( { - 3,0} \right) \cr & {\text{Asymptotes: }}y = \pm \frac{2}{3}\left( {x + 3} \right) - 2 \cr & {\text{domain: }}\left( { - \infty , + \infty } \right) \cr & {\text{range: }}\left( { - \infty , - 4} \right] \cup \left[ {0,\infty } \right) \cr} $$

Work Step by Step

$$\eqalign{ & \frac{{{{\left( {y + 2} \right)}^2}}}{4} - \frac{{{{\left( {x + 3} \right)}^2}}}{9} = 1 \cr & {\text{The equation is written in the form }}\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr & \frac{{{{\left( {y + 2} \right)}^2}}}{4} - \frac{{{{\left( {x + 3} \right)}^2}}}{9} = 1,{\text{ then }}a = 2,\,\,b = 3,\,\,\,h = - 3,\,\,\,k = - 2 \cr & \cr & {\text{Characteristics:}} \cr & {\text{Transverse axis: horizontal}} \cr & {\text{vertices: }}\left( {h,k \pm a} \right) \cr & {\text{vertices: }}\left( { - 3, - 2 \pm 2} \right) \cr & {\text{vertices: }}\left( { - 3, - 4} \right){\text{ and }}\left( { - 3,0} \right) \cr & \cr & {\text{asymptotes: }}y = \pm \frac{a}{b}\left( {x - h} \right) + k \cr & {\text{asymptotes: }}y = \pm \frac{2}{3}\left( {x + 3} \right) - 2 \cr & \cr & {\text{domain: }}\left( { - \infty , + \infty } \right) \cr & {\text{The range is: }}\left( { - \infty ,k - a} \right] \cup \left[ {k + a,\infty } \right) \cr & {\text{The range is: }}\left( { - \infty , - 4} \right] \cup \left[ {0,\infty } \right) \cr & \cr & {\text{Graph}} \cr} $$
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