Answer
No.
Work Step by Step
We are given:
$x+y^{2}=14$
We solve for $y$:
$y^{2}=14-x$
$y=\pm\sqrt{14-x}$
Since $y$ can have two values for the same $x$ value (the positive and negative), $y$ does not describe a function of $x$.
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 2, Functions - Chapter 2 Review - Exercises - Page 268: 39
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