College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.8 - Page 319: 45

Answer

Center: $(-3,2)$ $;$ Radius: $2$ Domain: $[-5,-1]$ $;$ Range: $[0,4]$ The graph is:
1508188643

Work Step by Step

$(x+3)^{2}+(y-2)^{2}=4$ The standard form of the equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is its radius. From the equation given, it can be seen that $(h,k)=(-3,2)$ and that $r^{2}=4$ The center of the circle is $(-3,2)$ The radius of the circle is: $r^{2}=4$ $\sqrt{r^{2}}=\sqrt{4}$ $r=2$ From the equation's graph (shown below), it can be seen that its domain is $[-5,-1]$ and its range is $[0,4]$
Small 1508188643
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