## Algebra 2 (1st Edition)

$f(x)=8(x+11)(x-3)$.
If the x-intercepts of a graph are $a,b$, then the function has the form $f(x)=k(x-a)(x-b)$. The x-intercepts are -11, 3, hence the quadratic function becomes $f(x)=k(x+11)(x-3)$. The point (1,-192) is on the graph, hence if we plug in the values we get $-192=k(12)(-2)$. This yields k=8, hence $f(x)=8(x+11)(x-3)$.