suatu fungsi dirumuskan f(x)=ax+b jika f(-2)=14 dan f(3)= -1,nilai f(7) adalah

Kelas 8 q

f(x) = ax + b

f(-2) = -2a + b = 14
f(3) = 3a + b = -1

3a + b = -1
-2a + b = 14
-------------------- -
5a = -15
a = -3

-2a + b = 14
-2 . 3 + b = 14
b = 14 + 6
b = 20

f(x) = -3x + 20
f(7) = -3 . 7 + 20
f(7) = -21 + 20
f(7) = -1

f(x) = ax+b

f(-2) = -2a+b = 14
f(3) = 3a+b = -1

gunakan metode eliminasi :

3a + b = -1
-2a + b = 14
——————— -
5a = -15
a = -3

kemudian, metode substitusi :

⇔-2a + b = 14
⇔-2(-3) + b = 14
⇔6 + b = 14
⇔b = 14 - 6
⇔b = 8

nilai f(7) :

f(x) = ax+b

⇔f(7) = -3(7) + 8
⇔f(7) = -21 + 8
⇔f(7) = -13