Answer
The quadratic equation is $ y={{x}^{2}}-3$
Work Step by Step
Let the quadratic function be $ y=a{{x}^{2}}+bx+c $. Since it passes through the points $\left( -1,-2 \right),\left( 2,1 \right),\left( -2,1 \right)$, we put these points one by one in the quadratic function.
Since it passes through the point $\left( -1,-2 \right)$, the equation becomes:
$\begin{align}
& -2=a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\
& a-b+c=-2
\end{align}$ (I)
Next it passes through the point $\left( 2,1 \right)$, so the equation becomes:
$\begin{align}
& 1=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c \\
& 4a+2b+c=1
\end{align}$ (II)
It also passes through the point $\left( -2,1 \right)$, so the equation becomes:
$\begin{align}
& 1=a{{\left( -2 \right)}^{2}}+b\left( -2 \right)+c \\
& 4a-2b+c=1
\end{align}$ (III)
Then consider Equations (I), (II), and (III), and represent them as a system of linear equations:
$\begin{align}
& a-b+c=-2 \\
& 4a+2b+c=1 \\
& 4a-2b+c=1
\end{align}$
And the augmented matrix is
$\left[ \begin{matrix}
1 & -1 & 1 & -2 \\
4 & 2 & 1 & 1 \\
4 & -2 & 1 & 1 \\
\end{matrix} \right]$
Use the Gaussian elimination method to find the values of a, b, and c.
To find the row echelon form of the matrix, perform the elementary row operations.
${{R}_{2}}\to {{R}_{2}}+\left( -4 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -4 \right){{R}_{1}}$ Which gives
$\left[ \begin{matrix}
1 & -1 & 1 & -2 \\
0 & 6 & -3 & 9 \\
0 & 2 & -3 & 9 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{1}{6}{{R}_{2}}$ Which gives
$\left[ \begin{matrix}
1 & -1 & 1 & -2 \\
0 & 1 & -\frac{1}{2} & \frac{3}{2} \\
0 & 2 & -3 & 9 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+\left( -2 \right){{R}_{2}}$ That gives
$\left[ \begin{matrix}
1 & -1 & 1 & -2 \\
0 & 1 & -\frac{1}{2} & \frac{3}{2} \\
0 & 0 & -2 & 6 \\
\end{matrix} \right]$
${{R}_{3}}\to -\frac{1}{2}{{R}_{3}}$ Which gives
$\left[ \begin{matrix}
1 & -1 & 1 & -2 \\
0 & 1 & -\frac{1}{2} & \frac{3}{2} \\
0 & 0 & 1 & -3 \\
\end{matrix} \right]$
The above matrix is in row echelon form; represent the system of equations from the last matrix:
$ a-b+c=-2$ (IV)
$ b-\frac{1}{2}c=\frac{3}{2}$ (V)
$ c=-3$ (VI)
Then use the back substitution method and obtain the values of a, b, c:
$ c=-3,b=0,\text{ and }a=1$.
Therefore, the quadratic function is
$ y=1{{x}^{2}}+0x-3$
Thus, the quadratic function is $ y={{x}^{2}}-3$.
