Ta có :
\(f\left(x\right)=ax^2+bx+c\)
+) \(f\left(0\right)=2014\)
\(\Leftrightarrow a.0^2+b.0+c=2014\)
\(\Leftrightarrow c=2014\)
+) \(f\left(1\right)=2015\)
\(\Leftrightarrow a.1^2+b.1+c=2015\)
\(\Leftrightarrow a+b+2014=2015\)
\(\Leftrightarrow a+b=1\left(1\right)\)
+) \(f\left(-1\right)=2017\)
\(\Leftrightarrow a.\left(-1\right)^2+b\left(-1\right)+c=2017\)
\(\Leftrightarrow a-b+2014=2017\)
\(\Leftrightarrow a-b=3\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow\left(a+b\right)+\left(a-b\right)=1+3\)
\(\Leftrightarrow2a=4\)
\(\Leftrightarrow a=2\)
\(\Leftrightarrow b=0\)
Vậy \(f\left(x\right)=2x^2+2014\)
\(\Leftrightarrow f\left(2\right)=2.2^2+2014=2022\)
