Answer
$ \displaystyle \frac{(x-2)^{2}}{4}-\frac{(y+3)^{2}}{9}=1$
Work Step by Step
The branches of the graph indicate that
the transverse axis is horizontal,
the center is not at the origin, so
the form of the equation is
$\displaystyle \frac{(x-h)^{2}}{a^{2}}- \displaystyle \frac{(y-k)^{2}}{b^{2}}=1$
From the graph,
center: $(2,-3)=(h,k)$
The vertices are at 2 units to the left/right of the center,
$a=2, a^{2}=4.$
$b=3, b^{2}=9.$
$ \displaystyle \frac{(x-2)^{2}}{4}-\frac{(y+3)^{2}}{9}=1$
