Ta có: \(f\left(0\right)=a.0^2+b.0+c=c=1\)
\(f\left(1\right)=a.1^2+b.1+c=a+b+c=2\Rightarrow a+b+1=2\Rightarrow a+b=1\) (1)
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c=2\Rightarrow2\left(2a+b\right)+1=2\Rightarrow2\left(2a+b\right)=1\Rightarrow2a+b=\frac{1}{2}\) (2)
Lấy (2) trừ (1) ta được: \(a=\frac{-1}{2}\)
\(\Rightarrow b=1-\left(\frac{-1}{2}\right)=\frac{3}{2}\)
Vậy a = -1/2 , b = 3/2 , c = 1