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

off-topic
 Mar 10, 2020
 #1
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Completing the square, we get (x + 6)^2 - 32, so (a,h,k) = (1,-6,-32).

 Mar 10, 2020
 #2
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Its incorrect,

 Mar 10, 2020

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