I am guessing that you are trying to fit a straight line to the data?
One common method is least squares, that is, in this case, find $a_0,a_1$ to minimize $\sum_{i=1}^9 (a_0+a_1 x_i - y_i)^2$, where $(x_i, y_i)$ are your data above. If we let
$$D=\begin{bmatrix} 1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots\end{bmatrix}, b=
\begin{bmatrix} y_1 \\ y_2 \\ \vdots\end{bmatrix}, a = \begin{bmatrix} a_0 \\ a_1\end{bmatrix},$$
then we can write the problem as $\min_{a \in \mathbb{R}^2} || D a -b ||_2^2.$ The minimizing solution satisfies
$$D^T(Da-b) = 0,$$
so we have $a = (D^T D)^{-1}D^T y$. In your case, this (or rather Octave) gives $a=\binom{1.26406}{0.35936}$. So the line $x \mapsto 1.26406+0.35936 x$ is the best-fit line in a least squares sense.