There exist constants a, h, and k such that 3x^2 + 16x + 4 = a(x - h)^2 + k for all real numbers x. Enter the ordered triple (a,h,k)
3x^2 + 16x + 4
3 ( x^2 + (16/3)x + 4/3) complete the square inside the parentheses
3 ( x ^2 + (16/3)x + 64/9 + 4/3 - 64/9)
3 [ ( x + 8/3)^2 + 12/9 - 64/9) ]
3 [ ( x + 8/3)^2 - 52/9 ]
3 ( x + 8/3)^2 - 52/3
3 ( x - -8/3)^2 - 52/3
(a , h , k) = (3 ,- 8/3 , -52/3 )
