Answer
The equation matches with the graph labeled as IV.
Work Step by Step
It has the equation of a parabola with vertical axis: $x^2=4py$
$12y+x^2=0$
$x^2=-12y$
$4p=-12$
$p=-3$
$p\lt0$. Parabola opens downwards.
When $y=-1$
$x^2=-12(-1)=12$
$x=±2\sqrt 3\approx±3.46$
So, the parabola passes through the points $(±3.46,-1)$
The equation matches with the graph labeled as IV.