Answer
$f(x)=\frac{-1}{3}(x+2)x(x-2)=\frac{-1}{3}(x^3 - 4 x)$.
Work Step by Step
If $a$ is a zero of a function with multiplicity $b$ then $(x-a)^b$ is a “multiplier” of the function.
Hence our function looks like this: $f(x)=k(x+2)x(x-2)=k(x^3 - 4 x)$.
Since $(-4,16)$ is on the graph we plug in $x=-4$ to $f(x)$ to get $f(-4)=k((-4)^3 - 4\cdot (-4))=16\\k(-64+16)=16\\-48k=16\\k=\frac{-1}{3}$.
Therefore our function is: $f(x)=\frac{-1}{3}(x+2)x(x-2)=\frac{-1}{3}(x^3 - 4 x)$.
