Answer
$$\color{blue}{\bf{f(x)= -\dfrac{1}{2}x^3 -\dfrac{1}{2}x^2+x }}$$
Work Step by Step
We are given three zeros of a polynomial and the value of the function for a given ${x}$:
Our zeros are $\bf{ -2 }$, $\bf{ 1 }$, and $\bf{ 0 }$, so let's make them into factors of our polynomial.
If $\bf{ -2 }$ is a zero then:
$( -2 +n)=0$
$n= 2 $
so our factor is $\bf{(x +2 )}$ because when $x=\bf{ -2 }$, $(x +2 )=0$
If $\bf{ 1 }$ is a zero then:
$( 1 +n)=0$
$n= -1 $
so our factor is $\bf{(x -1 )}$ because when $x=\bf{ 1 }$, $(x -1 )=0$
If $\bf{ 0 }$ is a zero then:
$( 0 +n)=0$
$n= 0 $
so our factor is $\bf{(x )}$
Now we have three factors of our 3rd degree polynomial:
$\bf{(x +2 )(x -1 )(x )}$, which, multiplied by some unknown factor $a$, make up our function:
$f(x)=a(x +2 )(x -1 )(x )$
If $f( -1 )= -1 $ then
${(-1 +2 )(-1 -1 )(-1 )}$ times some number $a$, equals $ -1 $ or:
$f( -1 )= -1 =a (-1 +2 )(-1 -1 )(-1 ) $
$-1 =a (1 )(-2 )(-1 ) $
$-1 =a (2) $
$\bf{a= -\dfrac{1}{2} }$
Now that we have the value of $a$, we can find $f(x)$
$f(x)= -\dfrac{1}{2} (x +2 )(x -1 )(x ) $
$f(x)= -\dfrac{1}{2} (x^2+x-2 )(x ) $
$f(x)= -\dfrac{1}{2} (x^3+x^2-2x ) $
$$\color{blue}{\bf{f(x)= -\dfrac{1}{2}x^3 -\dfrac{1}{2}x^2+x }}$$
