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There exist constants  [3x^2 + 12x + 4 = a(x - h)^2 + k\ and  such that
for all real numbers x Enter the ordered triple (a,h,k)

 Mar 6, 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

 Mar 6, 2020
 #2
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I get it, but the answer i put in was (3, -2, -12) wrong

Then, I found the same question with a different answer, but it was wrong too! What is going on?

 Mar 13, 2020
edited by Guest  Mar 13, 2020

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