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The quadratic ax^2 + bx + c can be expressed in the form 2(x - 4)^2 + 8. When the quadratic 3ax^2 + 3bx + 3c is expressed in the form n(x - h)^2 + k, what is h?

 Jul 14, 2019
 #1
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2(x - 4)^2  + 8   =

 

2 (x^2 - 8x + 16) + 8  =

 

2x^2 - 16x + 32 + 8  =

 

2x^2 - 16x  + 40    so........a = 2, b = -16 and c  =  40

 

So....3ax^2 +3bx + c   =   3(2)x^2 + 3(-16)x  + 40  =  6x^2  - 48x + 40

 

Factor out the 6  and we have

 

6 ( x^2  - 8x   +  40/6 )      T

 

Take 1/2 of - 8  = - 4.....square this = 16 and add and subtract within the parentheses

 

6 ( x^2  - 8x  + 16  + 40/6 - 16)    factor the first three terms  and simplify the rest

 

6 [ (x - 4)^2 +  20/3  -48/3 ]

 

6 [ ( x - 4)^2  - 28/3 ]

 

6 ( x - 4)^2  - 56

 

So h = 4

 

 

cool cool cool

 Jul 14, 2019

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