Answer
asymptotes: $y=\pm\frac{3}{4}x$
foci: $(0, \pm5)$
Work Step by Step
Step 1. Rewriting the equation as $\frac{y^2}{9}-\frac{x^2}{16}=1$, we have $a=3, b=4, c=\sqrt {a^2+b^2}=5$ centered at $(0,0)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(0, \pm3)$ and asymptotes as $y=\frac{a}{b}x=\pm\frac{3}{4}x$
Step 3. We can graph the equation as shown in the figure with foci at $(0, \pm5)$
