## Precalculus (6th Edition) Blitzer

$y$ is NOT a function of $x$
Solve for $y$. Add $-x^2$ on both sides of the equation to obtain: $y^2=-x^2+25$ Take the square root of both sides to obtain: $y=\pm \sqrt{-x^2+25}$ Note that when you assign any real number value to $x$, the equation will give two values of $y$. Thus, the given equation does not represent $y$ as a function of $x$.