Ta có:\(f\left(x\right)=ax+b\)
\(\Rightarrow f\left(0\right)=a.0+b\)
\(\Rightarrow f\left(0\right)=b\)
Mà\(f\left(0\right)=-5\)
\(\Rightarrow b=-5\)
\(\Rightarrow f\left(x\right)=ax-5\)
\(\Rightarrow f\left(1\right)=a-5\)
Mà\(f\left(1\right)=-2\)
\(\Rightarrow a-5=-2\)
\(\Leftrightarrow a=3\)
\(\Rightarrow f\left(x\right)=3x-5\)
Vậy\(f\left(x\right)=3x-5\)
Linz