## Precalculus (6th Edition) Blitzer

$y$ is NOT a function of $x$
Take the square root of both sides to obtain: $\pm \sqrt{4x}=y \\\pm2\sqrt{x}=y$ Note that when you assign any positive 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 $x$.