+0  
 
0
510
4
avatar

Find all values of p such that2(x+4)(x-2p) has a minimum value of -18.

Please explain thoroughly and very well.

 Oct 31, 2014

Best Answer 

 #1
avatar+118609 
+10

$$\\y=2(x+4)(x-2p) \qquad p\in R\\
$This is a concave up parabola$\\
$Roots are $ x=-4 \;\;and\;\;x=2p\\
$Axis of symmetry is $ x=(2p+-4)/2 = p-2\\
Min:\\
y=2(p-2+4)(p-2-2p)\\
y=2(p+2)(-2-p)\\
y=-2(p+2)^2\\
$You want this value to be $-18\\
-18=-2(p+2)^2\\
9=(p+2)^2\\
p+2=\pm3\\
p=\pm3-2\\
p=1 \quad or \quad p=-5$$

.
 Oct 31, 2014
 #1
avatar+118609 
+10
Best Answer

$$\\y=2(x+4)(x-2p) \qquad p\in R\\
$This is a concave up parabola$\\
$Roots are $ x=-4 \;\;and\;\;x=2p\\
$Axis of symmetry is $ x=(2p+-4)/2 = p-2\\
Min:\\
y=2(p-2+4)(p-2-2p)\\
y=2(p+2)(-2-p)\\
y=-2(p+2)^2\\
$You want this value to be $-18\\
-18=-2(p+2)^2\\
9=(p+2)^2\\
p+2=\pm3\\
p=\pm3-2\\
p=1 \quad or \quad p=-5$$

Melody Oct 31, 2014
 #2
avatar+128475 
0

Nice one, Melody....!!!

 

 Oct 31, 2014
 #3
avatar+118609 
0

Thanks Chris :)

 Nov 1, 2014
 #4
avatar
0

Can you write this in a form that I can write as well?

Not in the complicated way/

 Nov 1, 2014

4 Online Users

avatar
avatar
avatar