Question 752890
The model you tried to fit did not work because that is a form of equation for a circle.  


When you use the definition of a parabola to derive an equation, using a focal distance p from the vertex, you get a general equation {{{4py=x^2}}}.  This is a parabola with vertex on the origin, and vertex at a minimum. This derived equation can equivalently be written {{{y=(1/(4p))x^2}}}.  In your case, seeing the vertex is (0,0) and the focus is (0,-1/12), you should be able to know the value for p for the model.  If you SEE this value for p, then good!  Put in the values into {{{y=((1/(4p))x^2)}}} and simplify.








This should now be obvious.  Vertex is (0,0) and focus (0,-1/12).  p=1/12, and because the parabola opens DOWNWARD, the coefficient on x^2 must be LESS THAN ZERO, so you show a negative sign (instead of implying a positive sign).
{{{y=-1*(1/(4*(1/12)))x^2}}}