## College Algebra (10th Edition)

The given 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 $y$. Note that the given equation gives two different y-values for most values of $x$. Example: When $x=0$: $y^2=4-x^2 \\y^2=4-0^2 \\y^2=4 \\y = \pm \sqrt{4} \\y=\pm 2$ Therefore, the given equation does not define $y$ as a function of $x$.