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There exist 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 $(a,h,k).$

 May 5, 2020
 #1
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Please try the LaTex agian :(

 May 5, 2020
 #2
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Yes, I agree that the LaTeX has to be improved. Please take some time to do so.

Guest May 5, 2020
 #3
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Completing the square, we get (x + 6)^2 - 32, so (a,h,k) = (1,-6,-32).

 May 9, 2020

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