Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,-11). Express your answer in the form "ax^2+bx+c".
+36916 Vertex form of parabola
y = a ( x- 2)^2 + 4 sub in point to calulate a
-11 = a(-1)^2 + 4
a = -15
y = -15 (x-2)^2 + 4
= -15x^2 + 60x-56
