I thought I had answered this before. Perhaps it was on a different board. If y is a quadratic function of x then it can be written \(\displaystyle y= a(x- b)^2+ c\). When x= b, y= a(b- b)^2+ c= c so (b, c) is a point on the graph. Since a square is never negative, y is always c plus something so higher than c (if a is positive) or alway c minus something so lower than c (if a is negative). Here the highest point (the "vertex" of the parabola) is (-4, 4) so b= -4 and c= 4. We can write the equation as \(\displaystyle y= a(x+ 4)^2+ 4\). We also told that (0, -12) is on the parabola so \(\displaystyle -12= a(0+ 4)^2+ 4\). -12= 16a+ 4. Solve that for a.