Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.2 - Basics of Functions and Their Graphs - Exercise Set - Page 176: 15


$y$ is NOT a function of $x$

Work Step by Step

Solve for $y$. Add $-x^2$ on both sides of the equation to obtain: $y^2=-x^2+16$ Take the square root of both sides to obtain: $y=\pm \sqrt{-x^2+16}$ Note that when you assign any real number value to $x$, the equation will give two values of $y$. Thus, the given equation does not represent $y$ as a function of $x$.
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