## College Algebra (10th Edition)

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