We can model a set of data points to a quadratic function using the least squares method.
Suppose we have a set of points
$\{(x_1,y_i)@i=1,2,...,n \}$
and we want to find a quadratic expression
$y=a+bx+cx^2$
to model this data.
The least squares method seeks to minimise the sum of the square errors
$E=\sum^nn_{i=1} e^2_i =\sum^n_{i=1} (y_i-(a+bx_i+cx^2_i))^2$
.
Differentiate with respect to
$a, \: b, \: c$
in turn, and put the results equal to zero.
$\frac{\partial E}{\partial a}=2 \sum^n_{i=1} (y_i-(a+bx_i+cx^2_i))=0$

$\frac{\partial E}{\partial b}=2 \sum^n_{i=1} x_i(y_i-(a+bx_i+cx^2_i))=0$

$\frac{\partial E}{\partial c}=2 \sum^n_{i=1} x^2_i(y_i-(a+bx_i+cx^2_i))=0$

From the first of these,
$\sum^n_{i=1} y_i=an+b \sum^n_{i=1}x_i+c \sum^n_{i=1} x^2_i$

From the second of these,
$\sum^n_{i=1} y_i x_i=a \sum^n_{i=1} x_i+b \sum^n_{i=1}x^2_i+c \sum^n_{i=1} x^3_i$

From the third of these,
$\sum^n_{i=1} y_i x^2_i=a \sum^n_{i=1} x^2_i+b \sum^n_{i=1}x^3_i+c \sum^n_{i=1} x^4_i$

We can write this system in matrix form as
$\begin{pmatrix}\sum^n_{i=1} y_i \\ \sum^n_{i=1} y_i x_i \\ \sum^n_{i=1} y_i x^2_i \end{pmatrix} = \left( \begin{array}{ccc} n & \sum^n_{i=1}x_i & \sum^n_{i=1} x^2 \\ \sum^n_{i=1} x_i & \sum^n_{i=1}x^2_i & \sum^n_{i=1} x^3_i \\ \sum^n_{i=1} x^2_i & \sum^n_{i=1}x^3_i & \sum^n_{i=1} x^4_i \end{array} \right) \begin{pmatrix}a\\b\\c\end{pmatrix}$
.
$\begin{pmatrix}a\\b\\c\end{pmatrix} ={\left( \begin{array}{ccc} n & \sum^n_{i=1}x_i & \sum^n_{i=1} x^2_i \\ \sum^n_{i=1} x_i & \sum^n_{i=1}x^2_i & \sum^n_{i=1} x^3_i \\ \sum^n_{i=1} x^2_i & \sum^n_{i=1}x^3_i & \sum^n_{i=1} x^4_i \end{array} \right)}^{-1} \begin{pmatrix}\sum^n_{i=1} y_i \\ \sum^n_{i=1} y_i x_i \\ \sum^n_{i=1} y_i x^2_i \end{pmatrix}$
.
For the data points
$(-5,2),(-4,7),(-3,9),(-2,12),(-1,13),(0,14),(1,14),(2,13),3,10),(4,8),(5,14)$
.
$\sum^n_{i=1}x_i=0, \: \sum^n_{i=1}y_i=106, \: \sum^n_{i=1}x_iy_i=20, \: \sum^n_{i=1}x^2_i=110, \: \sum^n_{i=1}x^3_i=0, \:\sum^n_{i=1}x^4_i=1958, \: \sum^n_{i=1}y_i x^2_i=688$
.
The matrix equation above becomes
$\begin{pmatrix}a\\b\\c\end{pmatrix} ={\left( \begin{array}{ccc} 11 & 0 & 110 \\ 0 & 110 & 0 \\ 110 & 0 & 1958 \end{array} \right)}^{-1} \begin{pmatrix}106 \\ 20 \\ 688 \end{pmatrix} =\begin{pmatrix}13.9\\0.18\\-0.43\end{pmatrix}$

(to 2 decimal places).
The model is
$y=14.9+0.18x-0.43x^2$
.