Question 1039113
{{{y=a(x-h)^2+k}}} standard form with vertex  (h,k).


Two of your given points have the same y coordinate for different x coordinates, telling you where the axis of symmetry is:


{{{(-2+4)/2=2/2=1}}} the middle x value of those two points, giving {{{y=a(x-1)^2+k}}}.


You are given a y-intercept, which may give more information for the equation.
{{{y=a(0-1)^2+k}}}
{{{-4=a(0-1)^2+k}}}
{{{-4=a+k}}}



What you have found currently is this:
{{{system(y=a(x-1)^2+k,a+k=-4)}}}


Either of the other points should give more information.
Use  (4,-20),
{{{-20=a(4-1)^2+k}}}
{{{-20=9a+k}}}


Summarize again everything found:
{{{system(y=a(x-1)^2+k,a+k=-4,9a+k=-20)}}}
The two equations here with only variables a and k will permit finding values for a and k.
-
{{{a+k-(9a+k)=-4-(-20)}}}
{{{-8a=-4+20}}}
{{{8a=4-20=-16}}}
{{{highlight(a=-2)}}}
-
{{{a+k=-4}}}
{{{k=-4-a}}}
{{{k=-4-(-2)}}}
{{{k=-4+2}}}
{{{highlight(k=-2)}}}



EQUATION:  {{{highlight(y=-2(x-1)^2-2)}}}



{{{graph(300,300,-2,8,-8,2,-2(x-1)^2-2)}}}