+0

# probably last one for tonight

0
246
1

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
+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   Jul 30, 2019