suatu fungsi di tentukan dengan rumus f(x) = ax + b.jika f(3) = 11, dan f(2) = 8, tentukanlah nilai f(-2) tolong di jawab dengan jalannya ya

f(x)=ax+b
f(3)=3a+b=11
f(2)=2a+b=8
--------------- -
a=3
2a+b=8
2×3+b=8
6+b=8
b=2
jadi rumus fungsi f(x)=3x+2
maka f(-2)= 3×(-2)+2
=-6+2
= -4

MAPEL = MATEMATIKA
KELAS = 8
BAB = RELASI & FUNGSI

diketahui =
rumus = f(x) = ax + b
f(3) = 11
f(2) = 8

ditanya =
nilai f(-2) ???


f(3) = ax + b = 11
f(3) = 3a + b = 11

f(2) = ax + b = 8
f(2) = 2a + b = 8

3a + b = 11
2a + b = 8
_________-
a = 3

3a + b = 11
3(3) + b = 11
9 + b = 11
b = 11 - 9
b = 2

rumus fungsi
f(x) = ax + b
f(x) = 3x + 2

f(-2) = 3x + 2
f(-2) = 3(-2) + 2
f(-2) = -6 + 2
f(-2) = -4

semoga membantu