Question 65556
for f(x)=1/x^2-2x-8 I need to find the interval/intervals where f(x)<0
f(x)=1/(x-4)(x+2)
You have vertical asymptotes at x=-2 and at x=4
Draw a number line.
Mark x=-2 and x=4 on the line
That breaks the line into 3 intervals;  
Determine if f(x)<0 in each of those intervals, as follows.
If x=-10 f(-10)=1/(-)(-) which is positive; so no solutions in (-inf,-2)
If x=0 f(0)=1/(-)(+) which is negative; so (-2,4) is part of the solution
If x=10 f(10)=1/(+)(+) which is positive; so no solutions in (4,inf)
SOLUTION: -2<x<4
Cheers,
Stan H.