## College Algebra (10th Edition)

Published by Pearson

# Chapter 3 - Section 3.1 - Functions - 3.1 Assess Your Understanding - Page 211: 42

#### Answer

The equation does not define $y$ as a function of $x$.

#### Work Step by Step

An equation defines $y$ as a function of $x$ if for every value of $x$, the equation gives only one corresponding value of $y$. Solve for $y$ in the given equation to obtain: $x^2-4y^2=1 \\-4y^2=1-x^2 \\y^2=\dfrac{1-x^2}{-4} \\\sqrt{y^2} = \pm \sqrt{\dfrac{1-x^2}{-4}} \\y= \pm \sqrt{\dfrac{1-x^2}{-4}}$ The equation gives two different values of $y$ for most values of $x$. Thus, the equation does not define $y$ as a function of $x$.

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