Answer
$f(x)=0.5(x^4 - 3 x^3 + x^2 + 3 x - 2).$
Work Step by Step
If $c$ is a zero of a function with multiplicity $b$ then $(x-c)^b$ is a “factor” of the function.
We can see that $-1$, $1$ and $2$ are zeros, and that the graph doesn't cross the x-axis at $1$ but it touches it, hence $1$'s multiplicity is even, e.g. $2$.
Therefore one possible solution is $f(x)=k(x+1)(x-1)^2(x-2)=k(x^4 - 3 x^3 + x^2 + 3 x - 2).$
We can also see that the graph contains $(0,-1)$, hence $f(0)=-1$.
Thus $k\cdot(0^3-3\cdot0^3+0^2+3\cdot0-2)=k\cdot(-2)=-1$, $k=0.5$, therefore $f(x)=0.5(x^4 - 3 x^3 + x^2 + 3 x - 2).$
