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Find all values of \(p\) such that

                       \(2(x+4)(x-2p)\)

has a minimum value of \(-18\) over all real values of x. (In other words, we cannot have x be nonreal.)

 Jul 30, 2019
 #1
avatar+104937 
+1

This one is a little tricky, Logic........

 

2(x + 4) (x - 2p)      expand and we have

 

2 ( x^2 + 4x - 2px - 8p)  =

 

2 ( x^2 + (4 -2p)x  - 8p)  =

 

2x^2  +  2(4-2p)x - 2*8p  =

 

2x^2 + (8 - 4p)x  - 16p    ⇒   (1)

 

The x coordinate of the vertex  =  - (8- 4p) / [2*2]  = [ 4p - 8] / 4   =  ( p- 2)   ⇒      (2)

 

Since we want the minimum value to =  -18....we need to   put (2) into (1) and set it = -18

 

So we have

 

2 (p - 2)^2  + (8 - 4p)(p - 2) - 16p  = -18       simplify

 

2 (p^2 - 4p + 4) + ( 8p - 4p^2 - 16 + 8p) - 16p  = -18

 

2p^2 - 8p + 8  + 16p - 4p^2 - 16 - 16p  = -18

 

-2p^2  -8p - 8  = -18

 

-2p^2 - 8p + 10  = 0      divide through by -2

 

p^2  + 4p - 5  =  0      factor

 

(p + 5) (p - 1)  = 0        set each factor to 0 and solve for p and we get that

 

p + 5  = 0         p - 1  = 0

p= -5                 p = 1

 

So.....our possible functions are

 

2x^2 + (8 - 4(-5))x  - 16(-5)  =  2x^2 + 28x + 80        

 

and

 

2x^2 + (8- 4(1))x - 16(1)  =  2x^2  + 4x  - 16

 

See the two graphs here : https://www.desmos.com/calculator/in40roikco

 

Notice that both have a minimum value =  -18

 

 

 

cool cool cool

 Jul 30, 2019

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