There exist constants [3x^2 + 12x + 4 = a(x - h)^2 + k\ and such that 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
I get it, but the answer i put in was (3, -2, -12) wrong
Then, I found the same question with a different answer, but it was wrong too! What is going on?
