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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  Enter the ordered triple \((a,h,k)\)

 

Note: I got (3,-2, -12) but it didnt work....
 

 Apr 11, 2020
edited by Guest  Apr 11, 2020
 #1
avatar+23623 
+1

3 (x^2+4x)  + 4            "complete the square for x "

3 (x+2)^2  - 12 + 4

3(x+2)^2 - 8

 Apr 11, 2020
 #2
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im confused... i just got 3x^2 + 12x + 4... is the ordered triple (3,12,4)?

Guest Apr 11, 2020
 #3
avatar+23623 
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a (x-h)^2 + k

3 (x+2)^2 -8

 

a = 3    h = -2       k = -8

ElectricPavlov  Apr 11, 2020
edited by ElectricPavlov  Apr 11, 2020
 #4
avatar+398 
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Thx!

AnimalMaster  Apr 12, 2020

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