Answer
$f(x)=2(x+3)(x-1)(x-4)=2(x^3 - 2 x^2 - 11 x + 12)$.
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: $f(x)=k(x+3)(x-1)(x-4)=k(x^3 - 2 x^2 - 11 x + 12)$.
Since $(6,180)$ is on the graph we plug in $x=6$ to $f(x)$ to get $f(6)=k(6^3 - 2\cdot 6^2 - 11\cdot6 + 12)=180\\k(216-72-66+12)=180\\90k=180\\k=2$.
Therefore our function is: $f(x)=2(x+3)(x-1)(x-4)=2(x^3 - 2 x^2 - 11 x + 12)$.
