there exists 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).

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I think the answer is wrong

The answer Geno gave is almost right; he just made a small error in the final step:

Geno's answer:

3x^{2} + 12x + 4

= 3(x^{2} + 4x) + 4

= 3(x^{2} + 4x + 4) + 4 - 12

= 3(x + 2)^{2} - 12 This last line should be 3(x + 2)^{2} - 8 (as 4 - 12 = -8)

so a = 3, h = -2, and k = -8