## College Algebra 7th Edition

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.