Question 824633
Compare with standard form, {{{y=a(x-h)+k}}}.
You have h=0, so y=ax^2+k and also k=0 because vertex is at the origin.
You have {{{y=ax^2}}} and want to just find a.


Notice the two given points are for the same y value.  With vertex at origin, the vertical line of symmetry is at x=0.  You can use either given point to find a.


{{{-5=a(2)^2}}}
Solve for a.