There exist constants a, h, and k such that 3x^2+12x+4=a(x-h)^2+k for all real numbers x. Enter the ordered triple (a,h,k)
This is an exercise in completing the square: 3x^2 + 12x + 4 = 3(x + 1)^2 + 1, so (a,h,k) = (3,-1,1).
