Answer
No.
Work Step by Step
Solve for $y$:
$y^2=x\\y=\pm \sqrt x.$
This relation is s not a function, because a distinct $x$-value corresponds to two different $y$-values.
For example, $(4, 2)$ and $(4, –2).$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 2 - Functions and Their Graphs - 2.1 Functions - 2.1 Assess Your Understanding - Page 57: 37
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