SOLUTION: the function f(x) = x^2 + 6x +20 + k(x^2 -3x -12) where k is real find the value of k and the value of p given that the minimum value of f(x) is p and that f(-2) =p the answers a

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Question 764499: the function f(x) = x^2 + 6x +20 + k(x^2 -3x -12) where k is real
find the value of k and the value of p given that the minimum value of f(x) is p and that f(-2) =p
the answers are 2/7 anf 80/7
i am happy if you just explain to me what to do, even briefly.
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
the function f(x) = x^2 + 6x +20 + k(x^2 -3x -12) where k is real
find the value of k and the value of p given that the minimum value of f(x) is p and that f(-2) =p
the answers are 2/7 and 80/7
i am happy if you just explain to me what to do, even briefly.

f(x) = x² + 6x + 20 + k(x² - 3x - 12)

f(x) = x² + 6x + 20 + kx² - 3kx - 12k

f(x) = x² + kx² + 6x - 3kx + 20 - 12k

f(x) = (1 + k)x² + (6 - 3k)x + 20 - 12k

We are told that f(-2) = p, so

f(-2) = (1 + k)(-2)² + (6 - 3k)(-2) + 20 - 12k = p

              (1 + k)(4) - 12  + 6k + 20 - 12k = p

                               4 + 4k + 8 - 6k = p

                                       12 - 2k = p

The minimum value of f(x) = ax² + bx + c is 

f(x) = (1 + k)x² + (6 - 3k)x + (20 - 12k)

 = 

 

                       

                       

                       


We are told this minimum value is equal to p, and we have above that

12 - 2k = p, so our equation is





Multiply through by the LCD of 4(1 + k)

(6 - 3k)² - 2(6 - 3k)² = (-8 + 10k)4(1 + k)

            -(6 - 3k)² = 4(-8 + 10k)(1 + k)

     -(36 - 36k + 9k²) = 4(-8 + 2k + 10k²)

       -36 + 36k - 9k² = -32 + 8k + 40k²

       -49k² + 28k - 4 = 0

        49k² - 28k + 4 = 0

             (7k - 2)² = 0

                7k - 2 = 0
 
                    7k = 2

                     k =  

Then since 12 - 2k = p 

                     p = 12 - 2k 

                     p = 12 - 2

                     p = 

                     p =  

Edwin
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