\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\)
\(\Rightarrow c=2010\)
\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\)
\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\)(1)
\(f\left(-1\right)=2012\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=2012\)
\(\Rightarrow a-b+2010=2012\Rightarrow a-b=2\)(2)
Từ (1) và (2) suy ra \(\hept{\begin{cases}a=\frac{1+2}{2}=\frac{3}{2}\\b=\frac{1-2}{2}=\frac{-1}{2}\end{cases}}\)
\(\Rightarrow f\left(x\right)=\frac{3}{2}x^2-\frac{1}{2}x+2010\)
\(\Rightarrow f\left(-2\right)=\frac{3}{2}.4+\frac{1}{2}.2+2010=2017\)
- Ta có: \(f\left(x\right)=a.x^2+b.x+c\)
+ \(f\left(0\right)=a.0^2+b.0+c=c=2010\) (1)
+ \(f\left(1\right)=a.1^2+b.1+c=a+b+c=2011\) (2)
+ \(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c=2012\) (3)
- Thay \(c=2010\)vào đa thức (2), (3), ta có:
\(\hept{\begin{cases}a+b=2011-2010\\a-b=2012-2010\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=1\\a-b=2\end{cases}}\)
- Ta lại có: \(a-b=2\)
\(\Leftrightarrow a=b+2\)
- Thay \(a=b+2\)vào đa thức: \(a+b=1\), ta có:
\(b+2+b=1\)
\(\Leftrightarrow2b=-1\)
\(\Leftrightarrow b=-\frac{1}{2}=-0,5\)
- Thay \(b=-0,5\)vào đa thức: \(a+b=1\), ta có:
\(a-0,5=1\)
\(\Leftrightarrow a=1,5\)
Vậy hàm số \(y=f\left(x\right)\)có dạng: \(y=f\left(x\right)=1,5x^2-0,5x+2010\)
\(\Rightarrow f\left(-2\right)=1,5.\left(-2\right)^2-0,5.\left(-2\right)+2010=2017\)
Vậy \(f\left(-2\right)=2017\)
