There exist constants \(a,h\) and \(k\) such that
\(3x^2 + 12x + 4 = a(x - h)^2 + k\)
for all real numbers Enter the ordered triple \((a,h,k)\)
Note: I got (3,-2, -12) but it didnt work....
+36916 3 (x^2+4x) + 4 "complete the square for x "
3 (x+2)^2 - 12 + 4
3(x+2)^2 - 8
