Answer
asymptotes: $y=\pm(x+3)+2$
foci: $(-3,2\pm\sqrt {10})$
Work Step by Step
Step 1. Rewrite the equation as $\frac{(y-2)^2}{5}-\frac{(x+3)^2}{5}=1$, we have $a=b=\sqrt 5, c=\sqrt {a^2+b^2}=\sqrt {10}$ centered at $(-3,2)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(-3,2\pm\sqrt 5)$ and asymptotes as $y-2=\pm\frac{a}{b}(x+3)$ or $y=\pm(x+3)+2$
Step 3. We can graph the equation as shown in the figure with foci at $(-3,2\pm\sqrt {10})$
