Rational Functions and their graphs.....

RATIONAL FUNCTIONS AND
THEIR GRAPHS

Where p and q are polynomial functions and q is not the zero polynomial.
The domain consists of all real numbers except those for which the
denominator is 0.

Find the domain of the following rational functions:
All real numbers except -4 and 4.
All real numbers.

Finding Asymptotes
VERTICAL
ASYMPTOTES
( )
4
3
5
2
2
2
−
−
+
+
=
x
x
x
x
x
R
Let’s set the bottom = 0 and
factor and solve to find where
the vertical asymptote(s) should
be.
( )( ) 0
1
4 =
+
− x
x
So there are vertical
asymptotes at x = 4 and x = -1.

If the degree of the numerator is less than
the degree of the denominator, the x axis
is a horizontal asymptote. This is along
the line y = 0.
We compare the degrees of the polynomial in the numerator and the
polynomial in the denominator to tell us about horizontal asymptotes.
( )
4
3
5
2
2
+
−
+
=
x
x
x
x
R
degree of bottom = 2
HORIZONTAL ASYMPTOTES
1
1 < 2

degree of bottom = 2
HORIZONTAL ASYMPTOTES
degree of top = 2
The leading coefficient is the
number in front of the highest
powered x term.
1
2
=
( )
4
3
5
4
2
2
2
+
−
+
+
=
x
x
x
x
x
R
1
2
=
y

( )
4
3
5
3
2
2
2
3
+
−
+
−
+
=
x
x
x
x
x
x
R
If the degree of the numerator is greater
than the degree of the denominator, then
there is not a horizontal asymptote, but an
oblique one. The equation is found by
doing long division and the quotient is the
equation of the oblique asymptote ignoring
the remainder.
OBLIQUE ASYMPTOTES - Slanted
degree of top = 3
5
3
2 2
3
+
−
+ x
x
x
4
3
2
−
− x
x
remainder
a
5+
+
x
Oblique asymptote at y = x
+ 5

SUMMARY OF HOW TO FIND ASYMPTOTES
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.

STRATEGY FOR GRAPHING A RATIONAL FUNCTION

SKETCH THE GRAPH OF
10
5
3
2
)
(
+
−
=
x
x
x
f

10
5
3
2
)
(
+
−
=
x
x
x
f
The vertical asymptote is x = -2
The horizontal asymptote is y = 2/5

10
5
3
2
)
(
+
−
=
x
x
x
f
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10

SKETCH THE
GRAPH OF:
g(x) =
1
x −1
Vertical asymptotes at??
Horizontal asymptote at??
y = 0

f (x) =
2
x
x = 0
Horizontal asymptote at??
y = 0

h(x) =
−4
x
x = 0
y = 0

SKETCH THE
GRAPH OF:
y =
1
x + 3
−2
Vertical asymptotes at?? x = 1
y = 0
Hopefully you remember,
y = 1/x graph and it’s asymptotes:
Vertical asymptote: x = 0
Horizontal asymptote: y = 0

We have the function:

y =
1
x + 3
−2
But what if we simplified this and combined like terms:
y =
1
x + 3
−
2(x + 3)
x + 3
y =
1− 2x − 6
x + 3
y =
−2x − 5
x + 3
Now looking at this:
Vertical Asymptotes??
x = -3
Horizontal asymptotes??
y = -2

SKETCH THE
GRAPH OF:
h(x) =
x2
+ 3x
x
x = 0
h(x) =
x(x + 3)
x

FIND THE ASYMPTOTES OF EACH
FUNCTION:
y =
x2
+ 3x − 4
x

y =
x2
+ 3x −28
x3
−11x2
+28x

y =
x2
x
+
3x
x
−
4
x

y = x + 3−
4
x
Vertical Asymptote:
x = 0
y =
(x + 7)(x − 4)
x(x − 7)(x − 4)
Hole at x = 4
Vertical Asymptote:
x = 0 and x = 7
Horizontal Asymptote:
y = 0

WHAT MAKES A FUNCTION CONTINUOUS?
Continuous functions are predictable…
1) No breaks in the graph
A limit must exist at every x-value or the
graph will break.
2) No holes or jumps
The function cannot have undefined points
or vertical asymptotes.

Key Point:
Continuous functions can be
drawn with a single,
unbroken pencil stroke.

CONTINUITY OF POLYNOMIAL AND RATIONAL
FUNCTIONS

DISCONTINUITY
Discontinuity: a point at
which a function is not
continuous

DISCONTINUITY
Two Types of Discontinuities
1) Removable (hole in the graph)
2) Non-removable (break or vertical asymptote)
A discontinuity is called removable if a function can be made
continuous by defining (or redefining) a point.

DISCONTINUITY
2
2
( )
3 10
x
f x
x x
+
=
− −
Find the intervals on which these function are continuous.
2
( 2)( 5)
x
x x
+
=
+ −
1
( 5)
x
=
−
2 0
x + =
2
x = −
Vertical Asymptote:
5 0
x − =
5
x =
Removable
discontinuity
Non-removable
discontinuity

DISCONTINUITY
2
2
( )
3 10
x
f x
x x
+
=
− −
( , 2) ( 2, 5) (5, )
− −  −  

DISCONTINUITY
2
2 , 2
( )
4 1, 2
x x
f x
x x x
− 

= 
− + 

2
lim( 2 )
x
x
−
→
−
2
2
lim( 4 1)
x
x x
+
→
− +
(2)
f
4
= −
3
= −
4
= −
( , 2] (2, )
−  
Continuous on:

Determine the value(s) of x at which the function is
discontinuous. Describe the discontinuity as
removable or non-removable.
2
2
1
( )
5 6
x
f x
x x
−
=
− −
2
2
4 5
( )
25
x x
f x
x
− −
=
−
2
2
10 9
( )
81
x x
f x
x
+ +
=
−
2
2
4
( )
2 8
x
f x
x x
−
=
− −
(A) (B)
(C)

DISCONTINUITY
2
2
1
( )
5 6
x
f x
x x
−
=
− −
(A)
( 1)( 1)
( 6)( 1)
x x
x x
− +
=
− +
1
x = −
6
x = Non-removable discontinuity

DISCONTINUITY
(B)
9
x = −
9
x =
Removable discontinuity
( 9)( 1)
( 9)( 9)
x x
x x
+ +
=
+ −
2
2
10 9
( )
81
x x
f x
x
+ +
=
−

DISCONTINUITY
(C)
5
x =
5
x = − Non-removable discontinuity
( 5)( 1)
( 5)( 5)
x x
x x
− +
=
− +
2
2
4 5
( )
25
x x
f x
x
− −
=
−

DISCONTINUITY
(D)
2
x = −
4
x =
Removable discontinuity
( 2)( 2)
( 4)( 2)
x x
x x
− +
=
− +
2
2
4
( )
2 8
x
f x
x x
−
=
− −

CONCLUSION
Continuous functions have no breaks, no holes, and no jumps.
If you can evaluate any limit on the function using only the substitution
method, then the function is continuous.

Rational Functions and their graphs.....

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