## Precalculus (6th Edition) Blitzer

See graph. $(0,-1)$ and $(3,2)$. See explanations.
Step 1. See graph. Step 2. We can find the intersections as $(0,-1)$ and $(3,2)$. Step 3. With the original equations, plug-in the coordinates; we have $(0-3)^2+(-1+1)^2=9, (-1)=(0)-1$ and $(3-3)^2+(2+1)^2=9, (2)=(3)-1$