Answer
The equation of the graph matches with the graph labeled as IV.
Work Step by Step
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{y^2}{1}-\frac{x^2}{9}=1$
$a^2=1$
$a=1$
Vertices: $(0,±a)=(0,±1)$
The hyperbola with vertices in $(0,±1)$ is the graph labeled as IV.
