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: 48


Center: $(-4,-5)$ $;$ Radius: $6$ Domain: $[-10,2]$ $;$ Range: $[-11,1]$ The graph is:

Work Step by Step

$(x+4)^{2}+(y+5)^{2}=36$ 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 $(h,k)=(-4,-5)$ and that $r^{2}=36$ The center of the circle is $(-4,-5)$ The radius of the circle is: $r^{2}=36$ $\sqrt{r^{2}}=\sqrt{36}$ $r=6$ From the equation's graph (shown below), it can be seen that its domain is $[-10,2]$ and its range is $[-11,1]$
Small 1508190212
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.