## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$f(x)=3 (x+3)(x-1)(x-4)$
Let us consider that $a$ is a zero of a function with multiplicity $b$. Then this factor of the function can be expressed as: $(x-a)^b$. Therefore, we can write the equation of the function as: $f(x)=k (x+3)(x-1)(x-4)~~~~(1)$ Since, $(0,36)$ lies on the graph, we have: $k(3)(-1)(-4) =36\\12k =36 \\k=3$ Thus, equation (1) becomes: $f(x)=3 (x+3)(x-1)(x-4)$