Hi y+8+3x^2=12x y = -3x^2 + 12x - 8 P(0,-8) and (1,1) on this parabola opening downward -3 < 0 Completing the Square y = -3(x-2)^2 + 12 - 8 y = -3(x-2)^2 + 4 V(2,4) line of symmetry x = 2