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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 $(a,h,k).$

 Aug 2, 2020
 #1
avatar+26620 
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So if we arrange the left side properly, we can get form on R   and see what a h k are

 

3x^2 + 12x + 4    divide by 3

3  (x^2 + 4x)   + 4        'complete the square'

3 ( x^2 +4x + 4)   + 4

 

3(x+2)^2   + 12 + 4

3 (x+2)^2  + 16

 Aug 2, 2020

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