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)
3x2 + 12x + 4
= 3(x2 + 4x) + 4
= 3(x2 + 4x + 4) + 4 - 12
= 3(x + 2)2 - 12
a = 3 h = -2 k = -12
Sorry, but (3,-2,-12) isn't right.
