Answer
Vertices: $(±3,0)$
Asymptotes:
$y=±\frac{4}{3}x$
Work Step by Step
Standard form when transverse axis is horizontal:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$a^2=9$
$a=3$
$b^2=16$
$b=4$
Vertices when transverse axis is horizontal: $(±a,0)=(±3,0)$
Asymptotes when transverse axis is horizontal:
$y=±\frac{b}{a}x$
$y=±\frac{4}{3}x$
