[tex]f(x)=\frac{2(x+7)(x-3)^2(x+1)}{3(x-4)^3(x+5)(x+1)}[/tex]
f(x) is not defined where the term of the denominator is zero.
This happens when x = 4, x = -5 or x = -1
The x-intercepts is the points where f(x) = 0
this happens when x = -7 and x = 3
Notice that x = -1 can't be a x-intercept because the function is not defined at this point.
