Examining the graph, you can see that it crosses the $x$ axis at $-2$ and $8$. From that, It must be something like $(x-8)(x+2)\cdot a$.
Further inspection shows the $x=3$ asymptote. So your "$a$" must be something like $1/(x-3)$.
So, your graph should come from the equation $(x^2 -6x -16)/(x-3)$.
EDIT: Forgot about something. Your graph can be changed by a multiplier $n$, being $n\gt0$, in this way: $n\bigg(\frac{x^2-6x-16}{x-3}\bigg)$ So, your graph can be changed by any multiplier, being it $\frac 1 2$ , $2$ or even $1$ (in this last case keeping the original answer). The roots will be kept ($-2$ and $8$, in this case) and the asymptote will be kept as well. So, a first look may not determine exactly which equation gives that graph, but it might get a clue to what the $n$ can be if further inspected.
Edit (2): As seen on the graph, the point at $x=4$ gives out $y=-24$. So, if the grid on the graph is 1 unit of $x$ by 1 unit of $y$, your "$n$" stated above must be $1/12$, giving the point ($4$,$-2$). So, your final answer should be $y=\frac{1}{12}\bigg(\frac{x^2 -6x-16}{x-3}\bigg)$