persamaan fungsi kuadrat yg grafiknya melalui titik titik (1,3), (4,0), dan (2,-2) adalah...

y = ax^2 +bx + c. melalui
(1,3) => a(1)^2 + b(1)+ c = 3
a + b + c = 3.....(1)
(4,0) => a(4)^2 + b(4) + c = 0
16a +4b + c = 0.....(2)
(2,-2)=> a(2)^2 + b(2)+c = -2
4a + 2b + c = -2......(3)
ambil (1) dan (2), eliminasi c, maka
a + b + c = 3
16a + 4b + c = 0
--------------------- -
-15a - 3b = 3 => 5a + b = -1....(4)
ambil (1) dan (3), eliminasj c, maka
a + b + c = 3
4a +2b + c = -2
--------------------- -
-3a - b = 5
5a + b = -1....(4)
---------------------+
2a = 4
a = 4/2 = 2
5a + b = -1...(4)
5.2 + b = -1
10 + b = -1
b = -1 -10
b = -11
a + b + c = 3....(1)
2 + (-11) + c = 3
2 -11 + c = 3
-9 + c = 3
c = 3 +9 = 12
y = 2x^2 -11x +12 atau
f (x) = 2x^2 -11x +12