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 2.1 - Page 224: 15

Answer

$y$ is not a function of $x$.

Work Step by Step

$x^{2}+y^{2}=16$ $y^{2}=16-x^{2}$ $y=±\sqrt{16-x^{2}}$ Since one $x$ value will give you two $y$ values, $y$ is not a function of $x$.
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