f(×)=ax+b jika f(2)=8 f(4)=12 tentukan f(10)

f(x) = ax+b
f(2) = 2a + b = 8 ( persamaan 1 )
f(4)= 4a + b = 12 ( persamaan 2)

persamaan 1 dan 2 di eliminasi
2a + b = 8
4 a +b = 12 -

-2 a= -4
a = -4/-2
a = 2

2a+b=8
2x2 +b = 8
4+b= 8
b = 8-4
b = 4

f(10) = ax + b
= 2 (10) + 4
= 20 + 4
= 24

f(x) = ax + b

f(2) = 8
8 = 2a + b ... (1)

f(4) = 12
12 = 4a + b ... (2)

Eliminasi b
8 = 2a + b
12 = 4a + b
------------------ (-)
-4 = -2a
a = 2

Substitusi a = 2 ke (1)
8 = 2(2) + b
b = 8 - 4
b = 4

f(x) = 2x + 4
f(10) = 2(10) + 4
f(10) = 24