The graph of the equation y = Ax^2 + Bx + C , where are A, B, C integers, is shown below. Find A + B + C.
The vertex is (-1, -1)
So we have the form
4p ( y + 1) = (x + 1)^2
And the point (1,3) is on the graph so
4p ( 3 + 1) = ( 1 + 1)^2
16p = 4
p = 4/16 = 1/4
The equation is
y + 1 = ( x + 1)^2
y = x^2 + 2x + 1 - 1
y = x^2 + 2x
A = 1 B = 2 C = 0
A + B + C = 3
