Consider the function $f(x)=2x^2(x+1)$. Basic analysis allows you to show that this function sends $[-1,0]$ onto $[-1/2,0]$, so that if $a_0 \in [-1,0]$, then $a_n$ also for all $n$.
We now seperate the values in $[-1,0)$ into $3$ cases : either $a_0<-1/2$, either $a_0=-1/2$, either $a_0>-1/2$.
The second case is easy as $-1/2$ is fixed under the function $f$. In the first case, we notice that if $a_0=-1$, the sequence gets stuck on $0$ and stays there. If $-1<a_0<-1/2$, then $a_1 >-1/2$ and we are in the third case. We now only need to show that if $a_0>-1/2$ then the sequence converges. This is simple, for you can show that for all $x \in (-1/2,0)$, $f(x)<x$. The sequence is thus decreasing and bounded, and so converges.
You can also show that it always converge to $-1/2$ except when $a_0=-1$. Give it a try!