Answer
$y=-10x^2+40x-20$
Work Step by Step
The matrices can be formed as:
$A=\begin{bmatrix}
1& 1 & 1 \\
2^2 & 2& 1 \\
3^2 & 3 & 1\\
\end{bmatrix}=\begin{bmatrix}
1& 1 &1 \\
4 & 2&1 \\
9 & 3 & 1\end{bmatrix}$
$x=\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}$
$b=\begin{bmatrix}
10\\
20\\
10 \\
\end{bmatrix}$
Apply matrices to the least square solution:
$x_0=(A^TA)^{-1}A^Tb$
$=(\begin{bmatrix}
1& 4& 9\\
1 & 2&3 \\
1 & 1 & 1\end{bmatrix}\begin{bmatrix}
1& 1 & 1\\
4& 2& 1 \\
9 & 3 & 1\\
\end{bmatrix} )^{-1}\begin{bmatrix}
1& 4&9 \\
1 & 2&3\\
1 & 1 & 1\end{bmatrix} \begin{bmatrix}
10\\
20\\
10 \\
\end{bmatrix}$
$=\begin{bmatrix}
98 & 36 & 14\\
36& 14 & 6 \\
14 & 6 & 3
\end{bmatrix}^{-1}\begin{bmatrix}
1 & 4 & 9\\
1& 2&3 \\
1 & 1&1 \\\end{bmatrix}\begin{bmatrix}
10\\
20\\
10 \\
\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}
3 & -12 & 10\\
-12 & 49 & -42 \\
10 & -42 & 38
\end{bmatrix}\begin{bmatrix}
1 & 4 & 9\\
1& 2&3 \\
1 & 1&1 \\\end{bmatrix}\begin{bmatrix}
10\\
20\\
10 \\
\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}
1 & -2 &1 \\
-5 & 8 & -3 \\
6& -6 & 2
\end{bmatrix}\begin{bmatrix}
10\\
20\\
10 \\
\end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}
-20 \\
80 \\
-40\end{bmatrix}$
$=\begin{bmatrix}
-10 \\
40 \\
-20
\end{bmatrix}$
The equation of the least squares parabola is $y=-10x^2+40x-20$
