Precalculus (10th Edition)

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: 41



Work Step by Step

Solve for $y$: $2x^2+3y^2=1\\3y^2=1-2x^2\\y^2=\frac{1-2x^2}{3}\\y=\pm\frac{1-2x^2}{3}.$ This relation is s not a function, because a distinct $x$-value corresponds to two different $y$-values. For example, $(0, \frac{1}{3})$ and $(0, –\frac{1}{3}).$
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