Answer
The quadratic function is $f\left( x \right)=-{{x}^{2}}+x+2$.
Work Step by Step
To obtain the quadratic function at first we will find the coefficients a, b, c.
Since, $f\left( x \right)=a{{x}^{2}}+bx+c$.
So, $f\left( -2 \right)=-4$ gives,
$\begin{align}
& f\left( -2 \right)=a{{\left( -2 \right)}^{2}}+b\left( -2 \right)+c \\
& -4=4a-2b+c \\
& 4a-2b+c=-4
\end{align}$ (I)
$f\left( 1 \right)=2$ gives,
$\begin{align}
& f\left( 1 \right)=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\
& 2=a+b+c \\
& a+b+c=2
\end{align}$ (II)
$f\left( 2 \right)=0$ gives,
$\begin{align}
& f\left( 2 \right)=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\
& 0=4a+2b+c \\
& 4a+2b+c=0
\end{align}$ (III)
Equations (I), (II) and (III) give,
$\begin{align}
& 4a-2b+c=-4 \\
& a+b+c=2 \\
& 4a+2b+c=0
\end{align}$
Use the Gauss elimination method to obtain the solution of the above system of equations.
The matrix corresponding to the system of equations is as follows:
$\left[ \begin{matrix}
4 & -2 & 1 & -4 \\
1 & 1 & 1 & 2 \\
4 & 2 & 1 & 0 \\
\end{matrix} \right]$
${{R}_{1}}\to \frac{1}{4}{{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & \frac{1}{4} & -1 \\
1 & 1 & 1 & 2 \\
4 & 2 & 1 & 0 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -4 \right){{R}_{1}}$ give,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & \frac{1}{4} & -1 \\
0 & \frac{3}{2} & \frac{3}{4} & 3 \\
0 & 4 & 0 & 4 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{2}{3}{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & \frac{1}{4} & -1 \\
0 & 1 & \frac{1}{2} & 2 \\
0 & 4 & 0 & 4 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}+\left( -4 \right){{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & -\frac{1}{2} & \frac{1}{4} & -1 \\
0 & 1 & \frac{1}{2} & 2 \\
0 & 0 & -2 & -4 \\
\end{matrix} \right]$
${{R}_{3}}\to \frac{-1}{2}{{R}_{3}}$
$\left[ \begin{matrix}
1 & -\frac{1}{2} & \frac{1}{4} & -1 \\
0 & 1 & \frac{1}{2} & 2 \\
0 & 0 & 1 & 2 \\
\end{matrix} \right]$
Since the matrix is in the echelon form, so the system of the linear equations corresponding to the echelon form of the matrix is given as below:
$a-\frac{1}{2}b+\frac{1}{4}c=-1$ (IV)
$b+\frac{1}{2}c=2$ (V)
$c=2$ (VI)
Apply back the substitution method:
Equation (VI) gives,
$c=2$
Put the value of c in the equation (V) as follows:
It can be further simplified as,
$b=1$
Put the values of b and c in the equation (IV) to get,
$a-\frac{1}{2}\left( 1 \right)+\frac{1}{4}\left( 2 \right)=-1$
It can be further simplified as,
$a=-1$
Now, substitute the values of a, b, and c in the function $f\left( x \right)=a{{x}^{2}}+bx+c$; this gives,
$\begin{align}
& f\left( x \right)=\left( -1 \right){{x}^{2}}+\left( 1 \right)x+\left( 2 \right) \\
& =-{{x}^{2}}+x+2
\end{align}$
Hence, the required quadratic function is $f\left( x \right)=-{{x}^{2}}+x+2$.
