Answer
asymptotes: $y=\pm\frac{1}{2}(x-1)-2$
foci: $(1,-2\pm2\sqrt {5})$
Work Step by Step
Step 1. From the given equation $\frac{(y+2)^2}{4}-\frac{(x-1)^2}{16}=1$, we have $a=2, b=4, c=\sqrt {a^2+b^2}=2\sqrt {5}$ centered at $(1,-2)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(1,-4),(1,0)$ and asymptotes as $y+2=\pm\frac{a}{b}(x-1)$ or $y=\pm\frac{1}{2}(x-1)-2$
Step 3. We can graph the equation as shown in the figure with foci at $(1,-2\pm2\sqrt {5})$
