Tentukan fungsi kuadrat yang melalui titik (4,2) (3,-2) dan (-1,2)

Fungsi kuadrat :
f(x) = y = ax² + bx + c
f(4) = 2 = 16a + 4b + c …(1)
f(-2) = 3 = 4a – 2b + c …(2)
f(-1) = 2 = a – b + c …(3)

• eliminasi pers (1) dan (2)
16a + 4b + c = 2
4a – 2b + c = 3
____________ _
12a + 6b = -1 …(4)

• eliminasi pers (1) dan (3)
16a + 4b + c = 2
a – b + c = 2
____________ _
15a + 5b = 0
3a + b = 0 …(5) × 6

• eliminasi pers (4) dan (5)
12a + 6b = -1
18a + 6b = 0
_________ _
-6a = -1
a = 1/6

• substitusi a = 1/6 ke pers (4)
12a + 6b = -1

12(1/6) + 6b = -1
2 + 6b = -1
6b = -3
b = -½

• substitusi a=1/6 dan b = -½ ke pers (3)
a – b + c = 2
1/6 – (-½) + c = 2
1/6 + ½ + c = 2
2/3 + c = 2
c = 4/3

jadi, pers tersebut :
f(x) = 1/6 x² – ½x + 4/3