Answer
$f(x)=-\frac{1}{2}(x^3-2x^2-x+2)$
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, hence
$f(x)=k(x+1)(x-1)(x-2)\\
f(x)=k(x^3-2x^2-x+2)$
We can also see that the graph contains $(0,-1)$, hence $f(0)=-1$.
Thus
$f(0)=k(0^3-2(0^2)-0+2)\\
-1=k(0-0-0+2)\\
-1=k(2)\\
-\frac{1}{2}=k$
Therefore, $f(x)=-\frac{1}{2}(x^3-2x^2-x+2).$
