If you have some data $y$ and want to fit it to a curve of the form $1/(ax^2+bx+c)$, then what we can do is fit the data $1/y$ to the curve $ax^2+bx+c$. To do this in the least squares sense, we construct the following system of equations:
$$ \begin{pmatrix} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ \vdots & \vdots & \vdots \\ x_n^2 & x_n & 1\end{pmatrix} \begin{pmatrix} a \\ b \\ c\end{pmatrix} =
\begin{pmatrix}
1/y_1 \\ 1/y_2 \\ \vdots \\ 1/y_n
\end{pmatrix}
$$
The matix on the left-hand side is known as a Vandermonde matrix. Denote this matrix $A$
Since in general $n \neq 3$, to solve this we simply solve $A^TA \mathbf{p} = A^T\mathbf{y}$, where $\mathbf{p} = (a,b,c)^T$ is the parameter vector.
Thus, $$\mathbf{p} = (A^TA)^{-1}A^T\mathbf{y},$$ which may be computed in many ways (note that $A^TA$ is symmetric, so the LU factorization should be easy to compute).
EDIT: Regarding largeness of $\mathbf{y}$ (which should not matter for the data in your picture).
Note that we are solving $A^TA\mathbf{p} = A^T\mathbf{y}$. Look at $A^T\mathbf{y}$:
$$A^T \mathbf{y} = \begin{pmatrix}
\sum_{i=1}^n \frac{x_i^2}{y_1} \\
\sum_{i=1}^n \frac{x_i}{y_1} \\
\sum_{i=1}^n \frac{1}{y_1}
\end{pmatrix}$$
If $y_i$ is so small that you are concerned about numerical stability, solve the following problem instead:
$$10^m \hat{a} x^2 + 10^m \hat{b} x + 10^m \hat{c} = 1/y$$
for some $m > 0$. Then, you condition your Vandermonde matrix in such a way that $A^T\mathbf{y}$ becomes:
$$A^T \mathbf{y} = \begin{pmatrix}
\sum_{i=1}^n \frac{10^m x_i^2}{y_1} \\
\sum_{i=1}^n \frac{10^m x_i}{y_1} \\
\sum_{i=1}^n \frac{10^m}{y_1}
\end{pmatrix}$$
and the smallness is gone. Compute $\hat{a},\hat{b},\hat{c}$ in the traditional sense, then $a = 10^m\hat{a}$, and so forth.
Best Answer
You can use
where A,B,C,... are your points and "2" is the degree of your function. Be aware, that the parabola doesn't go through all your given points.