There exist constants a,h,k and such that 3x^2+72x+4=a(x-h)^2+k for all real numbers x Enter the ordered triple (a,h,k)
3x^2 + 72x + 4
3 ( x^2 + 24x + 4/3) complete the square on x inside the parentheses
3 ( x^2 + 24x + 144 + 4/3 - 144)
3 [ ( x + 12)^2 + 4/3 - 432/3 ]
3 [ (x + 12)^2 - 428/3 ]
3(x + 12)^2 - 428
3 ( x - -12)^2 - 428
(a , h , k) = (3 , -12 , -428)
