Answer
asymptotes: $y=\pm\frac{3}{4}(x-2)-2$
foci: $(2,-7),(2,3)$
Work Step by Step
Step 1. Rewriting the equation as $9(x^2-4x+4)-16(y^2+4y+4)=36-64-116=-144$ or $\frac{(y+2)^2}{9}-\frac{(x-2)^2}{16}=1$, we have $a=3,b=4, c=\sqrt {a^2+b^2}=5$ centered at $(2,-2)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(2,-5),(2,1)$ and asymptotes as $y+2=\pm\frac{a}{b}(x-2)$ or $y=\pm\frac{3}{4}(x-2)-2$
Step 3. We can graph the equation as shown in the figure with foci at $(2,-7),(2,3)$
