College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Chapter 2 Review - Exercises - Page 268: 39

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 $x$.
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