Hi, the vertex form of a Parabola opening up(a>0) or down(a<0), where(h,k) is the vertex f(x)= 3x^2 + 6x +4 = 3(x+1)^2 - 3 + 4 |Completing the Square = 3(x+1)^2 +1 V(-1,1), a= 3 >0 opening Up, f(x) = 1 is a minimum