Question

Jika fungsi y=ax^2+4x+3a mempunyai nilai maksimum-11, maka a^2-a adalah

Answer 1

Y' = 2 ax + 4 = 0
2a x = - 4
x = -2/a sub ke y = ax^2 + 4x + 3a  = -11

a(4/a^2 ) + 4(-2/a) + 3a = -11
4/a - 8/a + 3a = -11
-4/a + 3a = -11
3a^2 +11a - 4 = 0
(3a  -1)(a + 4) = 0
a = 1/3 atau a = - 4

nilai a^2 -a =
a= 1/3 --> a^2 - a = 1/9 - 1/3 =  - 2/9
a = -4 --> a^2 - a = 16 + 4 = 20