Transcript Rational Functions

Rational Functions
An introduction
L. Waihman
A function is CONTINUOUS if you can draw the graph
without lifting your pencil.
A POINT OF DISCONTINUITY occurs when
there is a break in the graph.
Note the break in the graph
when x=3. Why?
Look at the equation of the graph.
Where is this equation undefined?
x2  9
x 3
We can factor the numerator and reduce
the fraction to determine that the graph
will be a line; however, the undefined point
remains, so there is a point of discontinuity here.
 x  3 x  3
x3
 x3
There are three basic kinds of discontinuity:
point, jump, and infinite.
The greatest integer function is an example
of a jump discontinuity.
Tangent, Cotangent, secant and cosecant
functions are all examples of infinite
discontinuities.
The previous function was an example of a
point of discontinuity.
A rational function is the quotient
of at least two polynomials.
The graphs of rational functions frequently
display infinite and point discontinuities.
Rational functions have vertical asymptotes
and may have horizontal asymptotes as well.
Let’s look at the parent function: f
1
x

 
x
If x = 0, then the entire function is undefined.
Thus, there is a vertical asymptote at x=0.
Looking at the graph, you can see that the value of the
function
, as the values of x
0 from the positive
side; and the value of the function
- , as the values
of x
0 from the negative side. These are the limits of
the function and are written as:
lim f ( x )  
x 0
lim f ( x )  
x 0
Domain
• The domain is then limited to:
 ,0   0,  
• To find the domain of a rational function,
set the denominator equal to zero.
• The denominator will always be all real
numbers except those values found by
solving this equation.
Determine the domain of these
rational functions:
x 2
1. f  x  
x 2
2. f  x  
D :  , 2    2,  
2  x 2  9
D:
x2
x3  1
3. f  x   2
x 4
D:
 ,0    0,  
 , 2    2,2    2,  
Recall that a vertical asymptote occurs when
there is a value for which the function is
undefined. This means, if there are no
common factors, anywhere the denominator
equals zero.
Since a rational function is a quotient of two,
polynomials, there will always be at least
one value for which the entire function is
undefined.
Remember that asymptotes are lines.
When you label a vertical asymptote, you
must write the equation of the vertical
line.
Just make x equal everything it couldn’t in
the domain.
State the vertical asymptotes:
x 2
1. f  x  
x 2
V.A. : x  2
x3  1
2. f  x   2
x 4
V.A. : x  2
Let’s say x is any positive number. As that value increases,
the value of the entire function decreases; but, it will
never become zero or negative.
So this part of the graph will never
cross the x-axis. We express this
using limit notation as:
lim f  x   0
x 
What if x is a negative number? As that
value decreases, the value of the entire
function increases; but, it will never
become zero or positive.
So this part of the graph will never
cross the x-axis, either and: lim f x  0
x 
 
Thus, the line y  0 is a horizontal asymptote.
As x
∞, f(x)
0, and as x
∞, f(x)
0.
Given: f  x  is a polynomial of degree n , g  x  is a
polynomial of degree m , and f  x  , 3 possible
g x
conditions determine
a horizontal asymptote:
•If n<m, then y  0 is a horizontal asymptote.
•If n>m, then there is NO horizontal asymptote.
•If n=m, then y  c is a horizontal asymptote, where c
is the quotient of the leading coefficients.
Horizontal Asymptotes
BOBO BOTN EATS D/C
Bigger On Bottom y=0
Bigger On Top – None
Exponents Are The Same;
Divide the Coefficients
Find the horizontal asymptote:
2x  1
1. f  x  
x2
x3  1
2. f  x  
x2
x2
3. f  x   2
x  x  20
Exponents are the same;
divide the coefficients
H.A. : y  2
Bigger on Top; None
H.A. : none
Bigger on Bottom; y=0
H.A. : y  0
Suppose that you were asked to graph: f  x   3x  5
x 1
1st, determine where the graph is undefined.
(Set the denominator  to zero and solve for the variable.)
x  1 0
x  1
There is a vertical asymptote here.
Draw a dotted line at: x  1
2nd , find the x-intercept by setting the numerator = to 0
and solving for the variable.
3x  5  0
3x  5
5
x
3
So, the graph crosses the x-axis at
5 
 3 ,0 


3rd , find the y-intercept by letting x=0 and solving for y.
3(0)  5
y
01
5
y
1
So, the graph crosses the y-axis at  0, 5 
4th , find the horizontal asymptote.
(Note: the exponents are the same so divide
the coefficients - EATS D/C)
3/1 = 3
The horizontal asymptote is:
y 3
Now, put all the information together and sketch
the graph:
Graph:
f  x   4x  3
x 5
1st, find the vertical asymptote.
2nd
, find the x-intercept .
3rd , find the y-intercept.
x 5
 3 
 4 ,0


 3 
 0, 5 


4th , find the horizontal asymptote. y  4
5th , sketch the graph.
1st,
x2  x  2
Graph: f ( x )  2
x  x 6
 x  1 x  2 
factor the entire equation: x  3 x  2



Then find the vertical asymptotes: x  3
x  2
2nd , find the x-intercepts:
3rd
, find the y-intercept:
 1, 0 
 2, 0 
 1
 0, 3 


4th , find the horizontal asymptote: y  1
5th , sketch the graph.
2x 2  3x
Graph:f ( x ) 
x 1
Notice that in this function, the degree of the numerator
is larger than the denominator. Thus n>m and there is no
horizontal asymptote. However, if n is one more than m,
the rational function will have a slant asymptote.
To find the slant asymptote, divide the numerator by the
2x  5
denominator:
x  1 2x 2  3x
2x 2  2x
 5x
5 x  5
5
The result is 2 x  5  x51 . Notice that as the values of x
increase, the fractional part decreases (goes to 0), so the
function approaches the line 2 x  5. Thus the line y  2 x  5
is a slant asymptote.
Graph:
2x 2  3x
x 1

x 2x  3
x 1
1st, find the vertical asymptote. x  1
2nd
, find the


,
0
0,
0
x-intercepts:   and  2 
3


3rd , find the y-intercept:  0, 0 
4th , find the horizontal asymptote.
none
5th , find the slant asymptote: y  2 x  5
6th , sketch the graph.