diketahui rumus fungsi f(x)=ax+b.jika f(1)=4 dan f(3)=14,nilai f(-2) adalah

f(x)=ax+b
f(1)=a.1+b=4
=1a +b=4
f(3)=a.3+b=14
3a+b=14
3a+b=14
a+b=4. -
2a=10
a=10/2=5
subtitusi
ax+b=4
5.1+b=4
b=4-5=-1
f(-2)=ax+b=5.-2+(-1)=-11

Jawab:
f(x)=ax+b
f(1)=a.1+b=4
f(1)=1a+b=4......(1)

f(x)=ax+b
f(3)=a.3+b=14
f(3)=3a+b=14....(2

eliminasi nilai b
1a+b=4
3a+b=14
1a-3a=-2a
4-14=-10
-2a=-10
a=-10/-2
a=5
Subtitusi nilai a ke salah satu persamaan
f(1)=1a+b=4
f(1)=1.5+b=4
f(1)=5+b=4
b=4-5
b=-1
f(x)=ax+b
f(x)=5x+-1

f(-2)=5.-2+-1
= -10+-1
=-11