diketahui ordinat titik puncak fungsi kuadrat f(x)=ax2+bx+c adalah 6 . jika f(1) = f(5) = 0 , maka 4a+2b+6c =..

Fungsi kuadrat

f(x)= ax² + bx + c
f(1)= 0  --> a + b + c = 0
f(5)= 0 ---> 25a + 5b + c = 0
kurangkan  24a + 4b = 0
4(6a +b) = 0
6a + b = 0
b = - 6a
a + b + c = 0
a - 6a + c = 0
c = 5a

f(x)= ax² + bx + c --<  f(x) = ax²-6a x + 5a
ordinat titik puncak = 6
f '(x)= 0
2ax - 6a=0
2ax = 6a
x = 3
f(3)  = a(9) -6a(3) + 5a = 6
9a - 18 a + 5a = 6
-4a  = 6
a = - 3/2
b = -6a = 9
c = 5a = -15/2

4a + 2b + 6c = 4(-3/2) + 2(9) + 6(-15/2)
4a + 2b + 6c = -6 + 18 - 45
4a + 2b + 6c = - 33