Jika fungsi f(x) = x2- (p+2)x + 2(p+2) dan f(x) = x2 - 4px + 8p, Mempunyai titik minimum yang sama, tentukan nilai p

f(x) = x^2 - 4px + 8p
f'(x) = 0
2x - 4p = 0
2x = 4p
x = 2p

f(x) = x^2 - (p + 2)x + 2(p + 2)
f(2p) = (2p)^2 - (p + 2)(2p) + 2(p + 2)
f(2p) = 4p^2 - 2p^2 - 4p + 2p + 4
f(2p) = 2p^2 - 2p + 4

misal 2p = y
p = y/2

f(p) = 2(p/2)^2 - 2(p/2) + 4
= 2(p^2/4) - 2p + 4
= p^2 / 2 - 2p + 4
= 0,5p^2 - 2p + 4
f'(p) = p - 2 = 0
p = 2

f(x) = x^2 - (p + 2)x + 2(p + 2)
xp = -b/(2a) = -(-(p + 2))/2(1) = (p + 2)/2

f(x) = x^2 - 4px + 8
xp = -b/(2a) = -(-4p)/2(1) = 4p/2

Karena memiliki titik minimum yang sama maka otomatis sumbu simetri nya (xp) juga sama
(p + 2)/2 = 4p/2
p + 2 = 4p
2 = 3p
p = 2/3