Definition
A rational function f has the
form

where g (x) and h (x) are
polynomial functions.
The domain of f is the set of
all real numbers except the values of x that make the denominator
h (x) zero.
In what follows, we assume
that g (x) and h (x) have no common factors.
Vertical
Asymptotes
Let

The domain of f is the set of
all real numbers except 3, since 3 makes the denominator zero and
the division by zero is not allowed in mathematics. However we
can try to find out how does the graph of f behave close to
3.
let us evaluate function f at
values of x close to 3 such that x < 3. The values are shown
in the table below:
| x |
1 |
2 |
2.5 |
2.8 |
2.9 |
2.99 |
2.999 |
2.99999 |
| f
(x) |
-1 |
-2 |
-4 |
-10 |
-20 |
-200 |
-2000 |
-2*105 |
Let us now evaluate f at values of x close
to 3 such that x > 3.
| x |
5 |
4 |
3.5 |
3.2 |
3.1 |
3.01 |
3.001 |
3.00001 |
| f
(x) |
1 |
2 |
4 |
10 |
20 |
200 |
2000 |
2*105 |
The graph of f is shown below.

Notes
1 - As x approaches 3 from the left or by
values smaller than 3, f (x) decreases without bound.
2 - As x approaches 3 from the right or by
values larger than 3, f (x) increases without bound.
We say that the line x = 3, broken line, is
the vertical asymptote for the graph of f.
In general, the line x = a is a vertical
asymptote for the graph of f if f (x) either increases or
decreases without bound as x approaches a from the right or from
the left. This is symbolically written as:
Horizontal
Asymptotes
Let

1 - Let x increase and find
values of f (x).
| x |
1 |
10 |
103 |
106 |
| f
(x) |
3 |
2.1 |
2.001 |
2.000001 |
2 - Let x decrease and find
values of f (x).
| x |
-1 |
-10 |
-103 |
-106 |
| f
(x) |
1 |
1.9 |
1.999 |
1.999999 |
As | x | increases, the numerator is
dominated by the term 2x and the numerator has only one term x.
Therefore f(x) takes values close to 2x / x = 2. See graphical
behavior below.

In general, the line y = b is a horizontal
asymptote for the graph of f if f (x) approaches a constant b as
x increases or decreases without bound.
How to find the horizontal
asymptote?
Let f be a rational function defined as
follows

Theorem
m is the degree of the
polynomial in the numerator and n is the degree of the polynomial
in the numerator.
case 1: For m < n , the
horizontal asymptote is the line y = 0.
case 2: For m = n , the
horizontal asymptote is the line y = am /
bn
case 3: For m > n , there
is no horizontal asymptote.
Example 1: Let f be a
rational function defined by

a - Find the domain of
f.
b - Find the x and y
intercepts of the graph of f.
c - Find the vertical and
horizontal asymptotes for the graph of f if there are
any.
d - Use your answers to parts
a, b and c above to sketch the graph of function f.
Answer to Example
1
a - The domain of f is the set
of all real numbers except x = 1, since this value of x makes the
denominator zero.
b - The x intercept is found
by solving f (x) = 0 or x+1 = 0. The x intercept is at the point
(-1 , 0).
The y intercept is at the
point (0 , f(0)) = (0 , -1).
c - The vertical asymptote is
given by the zero of the denominator x = 1.
The degree of the numerator is
1 and the degree of the denominator is 1. They are equal and
according to the theorem above, the horizontal asymptote is the
line y = 1 / 1 = 1
e - Although parts a, b and c
give some important information about the graph of f, we still
need to construct a sign table for function f in order to be able
to sketch with ease.
The sign of f (x) changes at
the zeros of the numerator and denominator. To find the sign
table, we proceed as in solving rational inequalities. The zeros
of the numerator and denominator which are -1 and 1 divides the
real number line into 3 intervals:
(- infinity , -1) , (-1 , 1) ,
(1 , + infinity).
We select a test value within
each interval and find the sign of f (x).
In (- infinity , -1) , select
-2 and find f (-2) = ( -2 + 1) / (-2 - 1) = 1 / 3 >
0.
In (-1 , 1) , select 0 and
find f(0) = -1 < 0.
In (1 , + infinity) , select 2
and find f (2) = ( 2 + 1) / (2 - 1) = 3 > 0.
Let us put all the information about f in a
table.
| x |
- inf
|
-1 |
|
1 |
+ inf
|
| f (x) |
+ |
0
x-intercepts
|
- |
V.A |
+ |
In the table above V.A means vertical
asymptote.
To sketch the graph of f, we start by sketching
the x and y intercepts and the vertical and horizontal asymptotes
in broken lines. See sketch below.

We now start sketching the graph of f starting
from the left.
In the interval (-inf , -1) f (x) is positive
hence the graph is above the x axis. Starting from left, we
sketch f taking into account the fact that y = 1 is a horizontal
asymptote: The graph of f is close to this line on the left. See
sketch below.

Between -1 and 1 f (x) is negative, hence the
graph of f is below the x axis. (0 , -1) is a y intercept and x
=1 is a vertical asymptote: as x approaches 1 from left f (x)
deceases without bound because f (x) < 0 in( -1 , 1). See
sketch below.

For x > 1, f (x) > 0 hence the graph is
above the x axis. As x approaches 1 from the right, the graph of
f increases without bound ( f(x) >0 ). Also as x increases,
the graph of f approaches y = 1 the horizontal asymptote. See
sketch below.

We now put all "pieces" of the graph of f
together to obtain the graph of f.

Matched Problem: Let f be a
rational function defined by
f (x) = (-x + 2) / (x + 4)
a - Find the domain of
f.
b - Find the x and y
intercepts of the graph of f.
c - Find the vertical and
horizontal asymptotes for the graph of f if there are
any.
d - Use your answers to parts
a, b and c above to sketch the graph of function f.