One of the things you might have noticed is that an nth degree polynomial has at most n-1 "critical" points where the graph levels out (maybe but not necessarily where the graph "turns"). Since this graph has 4 such "critical points" (at x=-3, x= -1. something, x= 0. something, and x= 2) I would try a 5th degree polynomial.
Further, we can see from the graph that the polynomial is 0 at x= -3, x= -1, and x= 2. That means that the polynomial is of the form [tex]y= a(x+3)(x+1)(x- 2)Q(x)[tex] where a is a number and Q(x) is a quadratic polynomial.
To go further, you would have to specify exactly where the turning point ("-1.something", "0.something") and what the y values are there.