# Rational Functions.ppt - PowerPoint Presentation by wuyunyi

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```									  RATIONAL
FUNCTIONS
A rational function is a function of the form:

px 
Rx                  where p and q
qx           are polynomials
px 
Rx  
What would the domain of a rational

qx 
function be?

We’d need to make sure the
denominator  0
2
Rx  
5x
3 x
x   : x  3
Find the domain.

x 3
H x  
x  2x  2
x   : x  2, x  2
If you can’t see it in your
x 1         head, set the denominator = 0
F x   2                and factor to find “illegal”
x  5x  4      values.
x  4x  1  0    x   : x  4, x  1
The graph of f  x   2 looks like this:
1
x

If you choose x values close to 0, the graph gets
close to the asymptote, but never touches it.

Since x  0, the graph approaches 0 but never crosses or
touches 0. A vertical line drawn at x = 0 is called a vertical
asymptote. It is a sketching aid to figure out the graph of
a rational function. There will be a vertical asymptote at x
values that make the denominator = 0
Let’s consider the graph f  x  
1
x
We recognize this function as the reciprocal function
from our “library” of functions.

Can you see the vertical asymptote?
Let’s see why the graph looks
like it does near 0 by putting in
some numbers close to 0.
1 1            The closer to 0 you get
f          10 for x (from positive
 10  1
10      direction), the larger the
function value will be     Try some negatives
 1      1                                 1    1
f           100      1              f           100
 10  100 
1
 100    1          f                       1
100
 10    1
100
10
Does the function f  x  
1 have an x intercept?    1
0
x                         x
There is NOT a value that you can plug in for x that
would make the function = 0. The graph approaches
but never crosses the horizontal line y = 0. This is
called a horizontal asymptote.

A graph will NEVER cross a
vertical asymptote because the
x value is “illegal” (would make
the denominator 0)

A graph may cross a horizontal
asymptote near the middle of
the graph but will approach it
when you move to the far right
or left
Graph Q x   3 
1 1
 3         vertical translation,
x  x
moved up 3
This is just the reciprocal function transformed. We can
trade the terms places to make it easier to see this.

The vertical asymptote
Q x   3 
1             remains the same because in
x             either function, x ≠ 0

The horizontal asymptote
f x  
1                will move up 3 like the graph
x                does.
Finding Asymptotes
VERTICAL ASYMPTOTES

There will be a vertical asymptote at any
“illegal” x value, so anywhere that would make
the denominator = 0
So there are vertical
x  2x  5
2
Rx   2                    asymptotes at x = 4
 x 1
x x 4 3x 4 0     and x = -1.

Let’s set the bottom = 0
and factor and solve to
find where the vertical
asymptote(s) should be.
HORIZONTAL ASYMPTOTES
We compare the degrees of the polynomial in the
numerator and the polynomial in the denominator to tell
us about horizontal asymptotes.
1<2
degree of top = 1
If the degree of the numerator is
less than the degree of the
2x  5
1
 
R x  2                               the x axis is a
denominator, (remember degree
x  3x  4 is the highest power on any is
horizontal asymptote. This x
term) the x axis = 0. horizontal
along the line y is a
asymptote.
degree of bottom = 2
HORIZONTAL ASYMPTOTES
The leading coefficient
is the number in front of   If the degree of the numerator is
the highest powered x       equal to the degree of the
term.                       denominator, then there is a
degree of top = 2   horizontal asymptote at:
y = leading coefficient of top
2x  4x  5
2
R x   2                      leading coefficient of bottom
1 x  3x  4

degree of bottom = 2

horizontal asymptote at:
2
y  2
1
OBLIQUE ASYMPTOTES
If the degree of the numerator is
greater than the degree of the
denominator, then there is not a
degree of top = 3      horizontal asymptote, but an
oblique one. The equation is
x  2 x  3x  5
3       2           found by doing long division and
Rx                        the quotient is the equation of
x  3x  4
2
the oblique asymptote ignoring
the remainder.

degree of bottom = 2
x  5  a remainder
x 2  3 x  4 x 3  2 x 2  3x  5   Oblique asymptote
at y = x + 5
SUMMARY OF HOW TO FIND ASYMPTOTES

Vertical Asymptotes are the values that are NOT in the
domain. To find them, set the denominator = 0 and solve.

To determine horizontal or oblique asymptotes, compare
the degrees of the numerator and denominator.
1. If the degree of the top < the bottom, horizontal
asymptote along the x axis (y = 0)
2. If the degree of the top = bottom, horizontal asymptote
at y = leading coefficient of top over leading coefficient
of bottom
3. If the degree of the top > the bottom, oblique
asymptote found by long division.
Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.

Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au

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