+400 Find all real values of p such that
2(x+8)(x-4p)
has a minimum value of -32 over all real values of x.
+129895 2(x + 8)(x - 4p)
2 (x^2 + (8 -4p)x - 32p)
2x^2 + (16 - 8p)x - 64p (1)
The x coordinate of the vertex is [ - (16 - 8p) / (2*2) ] = [8p - 16) /4 = 2p - 4
So sub this into this equation
2(2p-4)^2 + (16 - 8p)(2p -4) - 64(2p -4) = -32
Simplifying
-8(p^2 + 12p - 28) = -32
p^2 + 12p - 28 = 4
p^2 + 12p - 32 = 0
p^2 + 12p = 32
p^2 + 12p + 36 = 32 + 36
(p + 6)^2 = 68 take both roots
p + 6 = sqrt (68) p+ 6 = -sqrt (68)
p = 2sqrt (17) - 6 p = -2sqrt(17) - 6
