f(A)=Ax+B jika f(6)=13 dan f(-2)=-11.maka f(7)=

f(6)= A(6)+B = 6A+B = 13 ... (i)
f(-2)= A(-2)+B = -2A+B = -11 ...(ii)
dari (i) dan (ii) diperoleh
6A+B = 13
-2A+B = -11
__________-
8A = 24
A = 24/8 = 3
substitusi A=3 ke persamaan (i) atau (ii)
misalnya kita substitusikan ke persamaan (i), maka diperoleh
6(3) + B = 13
18 + B = 13
B = 13 - 18 = -5

maka f(A) = 3x-5
nilai dari f(7) = 3(7) - 5 = 21 - 5 = 16

Mapel : Matematika

Kelas : 10

Bab : Relasi dan Fungsi
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f(x) = ax + b

f(6) = 13

f(-2) = -11

→ Ditanya f(7) =...?

Penyelesaian :

f(6) = 13

ax + b = 13

6a + b = 13..... (1)

__________

f(-2) = -11

ax + b = -11

-2a + b = -11........ (2)

Gunakan Metode eliminasi

6a + b = 13
-2a + b = -11
....................... -
8a = 24

a = 24/8

a = 3

→ tentukan nilai b
6a + b = 13
6(3) + b = 13
18 + b = 13
b = 13 - 18
b = -5

→ ax + b = 3x - 5

Maka, nilai f(7)
= 3x - 5

= 3(7) - 5

= 21 - 5

= 16 ✔