## Precalculus (10th Edition)

The equation does not define $y$ as a function of $x$.
Solve for $y$: $y ^2 = 4-x^2$ $y=± \sqrt{4-x^2}$ This relation is s not a function, because a distinct x - value corresponds to two different y-values. For example, (0, 2) and (0, –2)