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There exist constants a, h, and k such that 3x^2 + 16x + 4 = a(x - h)^2 + k for all real numbers x. Enter the ordered triple (a,h,k)

 Jun 25, 2021
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3x^2   +  16x  +  4

 

3  ( x^2  +  (16/3)x  +  4/3)              complete  the  square inside  the parentheses

 

3 ( x ^2  +  (16/3)x  + 64/9  + 4/3 - 64/9)

 

3 [  ( x +  8/3)^2   +  12/9 - 64/9)  ]

 

3   [  ( x + 8/3)^2  -   52/9  ]  

 

3 ( x +  8/3)^2  -  52/3

 

3 ( x -  -8/3)^2  - 52/3

 

(a , h , k)  =    (3 ,- 8/3 , -52/3 )

 

 

cool cool cool

 Jun 25, 2021
edited by CPhill  Jun 25, 2021
edited by CPhill  Jun 25, 2021
edited by CPhill  Jun 25, 2021

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