210 graphs of factorable rational functions

Graphs of Rational Functions
http://www.lahc.edu/math/precalculus/math_260a.html

Rational functions are functions of the form
R(x) = where P(x) and Q(x) are polynomials.
P(x)
Q(x)

A rational function is factorable if both P(x) and Q(x)
are factorable.
P(x)
Q(x)

are factorable.
Unless otherwise stated, the rational functions in this
section are assumed to be reduced factorable rational
functions.
P(x)
Q(x)

are factorable.
functions.
The principles of graphing rational functions are the
the same as for polynomials. We study the behaviors
and draw pieces of the graphs at important regions,
then complete the graphs by connecting them.
P(x)
Q(x)

are factorable.
functions.
The principles of graphing rational functions are the
the same as for polynomials. We study the behaviors
and draw pieces of the graphs at important regions,
then complete the graphs by connecting them.
However, the behaviors of rational functions are more
complicated due to the presence of the denominators.
P(x)
Q(x)

Vertical Asymptote

Vertical Asymptote
The function y = 1/x is not defined at x = 0.

Vertical Asymptote
The function y = 1/x is not defined at x = 0. So the
graph is not a continuous curve, it breaks at x = 0.

Vertical Asymptote
For small positive x's, y = 1/x is large.

Vertical Asymptote
The closer the x is to 0, the
smaller x is,

Vertical Asymptote
smaller x is, correspondingly
the larger y = 1/x is,

Vertical Asymptote
the larger y = 1/x is, hence
the higher the point (x, 1/x) is.

Vertical Asymptote
(1, 1)
x=0

Vertical Asymptote
(1, 1)
(1/2, 2)
x=0

Vertical Asymptote
(1, 1)
(1/2, 2)
(1/3, 3)
x=0

Vertical Asymptote
(1, 1)
(1/2, 2)
(1/3, 3)
The graph runs along x = 0 but
never touches x = 0 as shown.
x=0

Vertical Asymptote
(1, 1)
(1/2, 2)
(1/3, 3)
The boundary-line x = 0 is called a
vertical asymptote (VA).
x=0

Vertical Asymptote
negative x's, y = 1/x are negative
(1, 1)
(1/2, 2)
(1/3, 3)
x=0
vertical asymptote (VA). For "small"

Vertical Asymptote
negative x's, y = 1/x are negative so the corresponding
graph goes downward along the asymptote as shown.
(1, 1)
(1/2, 2)
(1/3, 3)
x=0

Vertical Asymptote
x=0
(1, 1)
(1/2, 2)
(1/3, 3)
negative x's, y = 1/x are negative so the corresponding
graph goes downward along the asymptote as shown.
(-1, -1)
(-1/2, -2)
(-1/3, -3)

Graph of y = 1/x
x=0
As x gets larger and larger, the corresponding y = 1/x
become smaller and smaller. This means the graph
gets closer and closer to the x-axis as it goes
further and further to the right
and to the left. To the right,
because y = 1/x is positive, the
graph stays above the
x-axis. To the left, y = 1/x is
negative so the graph stays
below the x-axis. As the graph
goes further to the left. It gets
(1, 1)
(2, 1/2) (3, 1/3)
(-1, -1)
(-2, -1/2)
(-3, -1/3)
closer and closer to the x-axis.
Hence the x-axis is a horizontal asymptote (HA).

Likewise, y = 1/x2 has x = 0 as a vertical asymptote.

However, since 1/x2 is always positive, the graph
goes upward along both sides of the asymptote.

Graph of y = 1/x2

We list the following facts about
vertical asymptotes of a reduced
rational function.
Graph of y = 1/x2

rational function.
Graph of y = 1/x2
I. The vertical asymptotes occur
at where the denominator is 0
(i.e. the roots of Q(x)).

rational function.
Graph of y = 1/x2
II. The graph runs along either
side of the vertical asymptotes.

rational function.
Graph of y = 1/x2
Whether the graph goes upward or downward
along the asymptote may be determined using the
sing-chart.

rational function.
Graph of y = 1/x2
Whether the graph goes upward or downward
along the asymptote may be determined using the
sing-chart. There are four different cases.

The four cases of graphs along a vertical asymptote:

+
e.g. y = 1/x

e.g. y = -1/x
+
e.g. y = 1/x
+

e.g. y = 1/x2e.g. y = -1/x
+
e.g. y = 1/x
+
+ +

e.g. y = 1/x2e.g. y = -1/x e.g. y = -1/x2
+
e.g. y = 1/x
+
+ +

e.g. y = 1/x2e.g. y = -1/x e.g. y = -1/x2
+
e.g. y = 1/x
+
+ +
Example A: Given the
following information of
roots, sign-chart and
vertical asymptotes,
draw the graph.

e.g. y = 1/x2e.g. y = -1/x e.g. y = -1/x2
+
e.g. y = 1/x
+
+ +
Example A: Given the
following information of
roots, sign-chart and
vertical asymptotes,
draw the graph.
++
root
VAVA

Horizontal Asymptotes

For x's where | x | is large (i.e. x is to the far right or
far left on the x-axis),

far left on the x-axis), the graph of a rational function
resembles the quotient of the leading terms of the
numerator and the denominator.

R(x) =
AxN + lower degree terms
BxK + lower degree terms
Specifically, if

R(x) =
Specifically, if
then for x's where | x | is large, the graph of R(x)
resembles (quotient of the leading terms).AxN
BxK

R(x) =
Specifically, if
BxK
The graph may or may not level off horizontally.

R(x) =
Specifically, if
BxK
The graph may or may not level off horizontally.
If it does, then we have a horizontal asymptote (HA).

We list all the possibilities of horizontal behavior below:

Theorem (Horizontal Behavior):

Given that R(x) =
the graph of R(x) as x goes to the far right (x  ∞) and
far left (x -∞) behaves similarly to AxN/BxK = AxN-K/B.
,

Given that R(x) =
I. If N > K,
,

Given that R(x) =
I. If N > K, then the graph of R(x) resembles the
polynomial AxN-K/B.
,

Given that R(x) =
polynomial AxN-K/B.
,
We write this as lim y = ±∞.x±∞

Given that R(x) =
polynomial AxN-K/B.
II. If N = K,
,

Given that R(x) =
polynomial AxN-K/B.
II. If N = K, then the graph of R(x) has y = A/B as a
horizontal asymptote (HA).
,

Given that R(x) =
polynomial AxN-K/B.
horizontal asymptote (HA). It is noted as lim y = A/B.x±∞
,

Given that R(x) =
polynomial AxN-K/B.
horizontal asymptote (HA). It is noted as lim y = A/B.
III. If N < K,
x±∞
,

Given that R(x) =
polynomial AxN-K/B.
III. If N < K, then the graph of R(x) has y = 0 as a
horizontal asymptote (HA) because N – K is negative.
x±∞
,

Given that R(x) =
polynomial AxN-K/B.
III. If N < K, then the graph of R(x) has y = 0 as a
horizontal asymptote (HA) because N – K is negative.
It is noted as lim y = 0.
x±∞
x±∞
,

Steps for graphing rational functions R(x) = P(x)
Q(x)

Steps for graphing rational functions R(x) =
I. (Roots) As for graphing polynomials, find the roots
of R(x) and their orders by solving R(x) = 0.
P(x)
Q(x)

P(x)
Q(x)
II. (VA) Find the vertical asymptotes (VA) of R(x) and
their orders by solving Q(x) = 0.

P(x)
Q(x)
Steps I and II give the sign-chart,

P(x)
Q(x)
Steps I and II give the sign-chart, the graphs around
the roots (using their orders)

P(x)
Q(x)
the roots (using their orders) and the graph along the
VA (upward or downward).

P(x)
Q(x)
VA (upward or downward). From these we construct
the middle portion of the graph (as in example A).

P(x)
Q(x)
Complete the graph with step III.

x±∞
P(x)
Q(x)
III. (HA) Use the above theorem to determine the
behavior of the graph to the far right and left, that is,
as .

x±∞
P(x)
Q(x)
III. (HA) Use the above theorem to determine the
behavior of the graph to the far right and left, that is,
as . (HA exists only if deg P < deg Q)

Example C:
Find the roots, VA and HA, if any, of R(x) =
Draw the sign-chart and sketch the graph.
x2 – 4x + 4
x2 – 1

Example C:
x2 – 4x + 4
x2 – 1
For it's root, set x2 – 4x + 4 = 0, i.e. x = 2 (ord = 2).

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.
x=2

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.
Both have order 1, so
the sign changes at
each of these values.
x=2

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.
the sign changes at
+–
x=2
++

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.
the sign changes at
++ –
x=2
+ +

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.
the sign changes at
As x±∞, R(x) resembles
x2/x2 = 1, i.e. it has y = 1
as the HA.
++ –
x=2
+ +

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.
the sign changes at
x2/x2 = 1, i.e. it has y = 1
as the HA.
++ –
x=2
++

Example C:
x2 – 4x + 4
x2 – 1
For VA, set Q(x) = 0, i.e. x2 – 1 = 0  x = ± 1.
the sign changes at
x2/x2 = 1, i.e. it has y = 1
as the HA. Note the
graph stays above the
HA to the far left,
and below to the far right.
++ –
x=2
++

Example D:
x2 – 2x – 3
x – 2

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
so x = -1, 3 are the roots of order 1.

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
For VA, set x – 2 = 0, i.e. x = 2.

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
For VA, set x – 2 = 0, i.e. x = 2.
Do the sign-chart.

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
For VA, set x – 2 = 0, i.e. x = 2.
x=3
Do the sign-chart.
x= –1

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
For VA, set x – 2 = 0, i.e. x = 2.
x=3
Do the sign-chart.
+–+–
x= –1

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
For VA, set x – 2 = 0, i.e. x = 2.
Do the sign-chart. Construct the
middle part of the graph.
x=3
+–+–
x= –1

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
For VA, set x – 2 = 0, i.e. x = 2.
As x ±∞, the graph of R(x)
resembles the graph of the
quotient of the leading terms
x2/x = x, or y = x.
x=3
+–+–
x= –1

Example D:
x2 – 2x – 3
x – 2
Set x2 – 2x – 3 = 0  (x – 3)(x + 1) = 0
For VA, set x – 2 = 0, i.e. x = 2.
x2/x = x, or y = x.
Hence there is no HA.
x=3
+–+–
x= –1

http://www.lahc.edu/math/precalculus/math_260a.html
From the last few sections, we see the importance of
the factored form of polynomial formulas.
This factoring problem is the same as the problem of
finding roots of polynomials.
In the next chapter, we give the possible (completely)
factored form of real polynomials.
This result is called the
“Fundamental Theorem of Algebra”.

210 graphs of factorable rational functions

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210 graphs of factorable rational functions