+1206 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.)
+128474 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
