Answer
Vertices: $V(0,±3)$
Foci: $F(0,±5)$
Asymptotes: $y=±\frac{3}{4}x$
Work Step by Step
$\frac{y^2}{9}-\frac{x^2}{16}=1$
$\frac{y^2}{3^2}-\frac{x^2}{4^2}=1$
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$a=3$
$b=4$
$c^2=a^2+b^2=3^2+4^2=9+16=25$
$c=5$
Vertices: $V(0,±a)=V(0,±3)$
Foci: $F(0,±c)=F(0,±5)$
Asymptotes: $y=±\frac{a}{b}x=±\frac{3}{4}x$
