Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Appendix D - Graphing Techniques - Exercises - Page 447: 35

Answer

For $x^2 + y^2 = 12$, it has graphs that are symmetric to the $x$-axis, the $y$-axis and as well as the origin.

Work Step by Step

In $x^2 + y^2 = 12$, a) To test for symmetry w.r.t the $x$-axis replace $y$ with $-y$, we have $x^2 + (-y)^2 = 12$ $x^2 + y^2 = 12$ which the result is the same as the original equation, therefore, it is symmetric with respect to the $x$-axis b) To test for symmetry w.r.t the $y$-axis replace $x$ with $-x$, we have $(-x)^2 + y^2 = 12$ $x^2 + y^2 = 12$ which the result is the same as the original equation, therefore, it is symmetric with respect to the $y$-axis c) To test for symmetry w.r.t the origin replace $x$ with $-x$ and $y$ with $-y$, we have $(-x)^2 + (-y)^2 = 12$ $x^2 + y^2 = 12$ which the result is the same as the original equation, therefore, it is symmetric with respect to the origin
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.