jika f adalah fungsi kuadrat yang grafiknya melalui titik (1,0),(4,0),dan (0,-4), maka nilai f (7) adalah...

titik (1,0) -> f(x) = ax^2 + bx + c
f(1) = a(1)^2 + b(1) + c
0 = a + b + c ...pers.1

titik (4,0) -> f(4) = a(4)^2 + b(4) + c
0 = 16a + 4b + c ....pers.2

titik (0,-4) -> f(0) = a(0)^2 + b(0) + c
-4 = c

substitusi pers.1
a + b -4 = 0
a + b = 4

subst.pers.2
16a + 4b + -4 = 0
16a + 4b = 4

eliminasi
(a + b = 4) × 4 = 4a + 4b = 16
(16a + 4b = 4) × 1 = 16a + 4b = 4

dikurang
4a + 4b = 16
16a + 4b = 4
------------------- -
-12a = 12
a = 12/(-12) = -1

nyari b
-1 + b = 4
b = 4+1 = 5

substitusi
f(x) = ax^2 + bx + c
f(7) = -1(7)^2 + 5(7) -4
= -49 + 35 -4
= -18

*Moga membantu :D