## Precalculus (6th Edition) Blitzer

a. $y=\pm3$ b. See explanations
a. The y-intercepts can be found by letting $x=0$. We have $y^2=9$ and $y=\pm3$ b. If we try to find the x-intercept by letting $y=0$, we get $x^2=-16$, which has no real solution. Thus there is no x-intercept.