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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)

 Jun 30, 2020
 #1
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 3x2 + 12x + 4

=  3(x2 + 4x) + 4

=  3(x2 + 4x + 4) + 4 - 12

=  3(x + 2)2 - 12

 

a = 3     h = -2     k = -12

 Jul 1, 2020
 #3
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Sorry, but (3,-2,-12) isn't right.

 Jul 1, 2020

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