Question 747686
The focus is below the vertex, so this vertex is a maximum.  You expect the coefficient on x^2 to be negative.  


A general equation for a parabola having vertex on the origin is {{{4py=x^2}}}, where p is the focal length, distance between the focus and the vertex.  In your exercise, {{{p=1/12}}}.  See your textbook and http://en.wikipedia.org/wiki/Parabola .


The equation for the described parabola is {{{highlight(y=(1/(4*p))*x^2)}}}, and as said, since the vertex is a maximum, we must have {{{(1/(4p))<0}}}, so we have:
{{{y=-(1/(4*(1/12)))x^2}}}
{{{y=-(12/4)x^2}}}
{{{highlight(y=-3x^2)}}}.