Nilai minimum fungsi y=x^2 – (m-3)x + (m+3) adalah 6.tentukan f(2)

Cari dulu nilai m nya :

Berarti :

x = -b/2a = (m-3)/2

F(x) = y = 6

x^2- (m-3)x + (m+3) = 6

((m-3)/2)^2 – (m-3).((m-3)/2 + m+3 = 6

((m^2-6m+9)/4) – ((m^2-6m+9)/2 + m + 3 = 6

((m^2- 6m + 9) – (2m^2- 12m + 18) + (4m+12)/4)) = 6

-m^2+ 10m + 3 = 24

-m^2+ 10m = 21

m^2- 10m + 21 = 0

(m-3) (m-7) = 0

m = 3 atau m = 7

Yg memenuhi m = 7

Jadi.F(x) = y = x^2 – 4x + 10

F(2) = 2^2 – 4(2) +10 = 6