Rational Inequalities
There is one step added to the process of solving rational inequalities because a rational function can also change signs at its vertical asymptotes or at a break in the graph. For instance, look at the graph of the function
If we want to solve the inequality
Testing the intervals between each critical point to see if the values in that interval satisfy the function gives us:
Interval  Test Point  Positive/Negative?  Part of Solution set? 


4    no 
(3, 0)  2  +  yes 
(0, 3)  2    no 

4  +  yes 
Thus, the solutions to
Guided Practice
Questions
1)
2)
Solutions
1) To identify the graph of the inequality

For
f(x)≥2x+5x−1: 
To find the critical points, identify the value(s) which make the denominator = 0:
x≠1 
That gives us a vertical asymptote of
x=1 
The horizontal asymptote becomes apparent as x becomes truly huge and the "+5" and "1" no longer matter. At that point, we have
f(x)=2xx→f(x)=2 So the horizontal asymptote isy=2
 Now that you know the shape of the graph, simply shade the area above the lines, since the original function was f(x) is greaterthan function, and leave the lines solid since it was a greaterthan or equal to.
 The final graph should look like:
2) To graph

To identify domain limitations, find value(s) which make the denominator = 0: In this case, where
x2 , the only variable in the denominator, is added to 0, any value for x will be positive. So the domain is all real numbers.  With no limitations on the domain, there are no vertical asymptotes.

The horizontal asymptote:
y=0 becomes apparent as x becomes huge and the "+2" and "+1" no longer have an effect, giving:f(x)=xx2→f(x)=1x→f(x)=0 So the horizontal asymptote is 0
 Now that you know the shape of the graph, simply shade the area below the lines, since the original function was f(x) is lessthan function, and leave the lines solid since it was or equal to.
 The final graph should look like: