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

Answer

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 $[-11,1]$
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.