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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).

 Apr 11, 2020
 #1
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 Apr 11, 2020
 #2
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I think the answer is wrong

Guest Apr 11, 2020
 #3
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The answer Geno gave is almost right; he just made a small error in the final step:

 

Geno's answer:

   3x2 + 12x + 4

=  3(x2 + 4x) + 4

=  3(x2 + 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

Alan  Apr 12, 2020

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