## Algebra 2 (1st Edition)

$f(x)=-2(x+1)(x-2)(x-3)$
If the function has x-intercepts $a,b,c,$ then it has the form $f(x)=k(x-a)(x-b)(x-c)$. Here the x-intercepts are $-1,2,3$, hence $f(x)=k(x+1)(x-2)(x-3)$. The function also goes through $(0,-12)$, hence: $-12=k(1)(-2)(-3)=6k\\k=-2$ Hence $f(x)=-2(x+1)(x-2)(x-3)$