3.3 graphs of factorable polynomials and rational functions

Graphs of Factorable Polynomials
Function–Dance (origin unknown)

Following are some of the basic shapes of graphs
that we encounter often. The dotted tangent line is
for reference. Practice drawing them a few times.

We start with the graphs of the polynomials y = ±xN.

The graphs y = xeven
y = x2

y = x2y = x4
(1, 1)(-1, 1)

y = x2y = x4y = x6
(1, 1)(-1, 1)

y = x2y = x4y = x6
y = -x2
y = -x4
y = -x6
(1, 1)(-1, 1)
(-1,-1) (1,-1)
The graphs y = –xeven

y = x2y = x4y = x6
y = -x2
y = -x4
y = -x6
(1, 1)(-1, 1)
(-1,-1) (1,-1)
Plot these functions and zoom in on the
region around x = –1 to x = 1.
Note that the graphs in between the points
(1, 1) and (–1,1) drop lower as the power
increases. However the graphs switch
positions as they pass to the right of (1, 1)
or to the left of (–1,1). ( Why?)

y = x2y = x4y = x6
y = xE y = –xE
(1, 1)(-1, 1)
Graphs of even
ordered roots
y = ±xEven.
y = -x2
y = -x4
y = -x6
(-1,-1) (1,-1)

y = x3
The graphs y = xodd

y = x3
y = x5
(1, 1)
(-1, -1)
The graphs y = xodd

y = x3
y = x5
y = x7
(1, 1)
(-1, -1)
The graphs y = xodd

The graphs y = xodd
y = x3
y = x5
y = x7 y = -x3
y = -x5
y = -x7
(1, 1)
(-1, -1)
(-1, 1)
(1,-1)
The graphs y = –xodd

The graphs y = xodd
y = x3
y = x5
y = x7 y = -x3
y = -x5
y = -x7
y = xD y = –xD
(1, 1)
(-1, -1)
(-1, 1)
(1,-1)
The graphs y = –xodd
Graphs of odd
ordered roots
y = ±xodD

Facts about the graphs of polynomials:

• The graphs of polynomials are unbroken curves.

• Polynomial curves are smooth (no corners).

• Let P(x) = anxn + lower degree terms.

For large |x|, the leading term anxn dominates the
lower degree terms.

lower degree terms. For x's such that | x | are large,
the "lower degree terms" are negligible compared to
anxn.

anxn.
Hence, for x where |x| is "large", the graph of P(x)
resembles the graph y = anxn.

anxn.
resembles the graph y = anxn.This means there're four behaviors of
polynomial-graphs to the far left or far right
(as | x | becomes large).

anxn.
resembles the graph y = anxn.This means there're four behaviors of
polynomial-graphs to the far left or far right
(as | x | becomes large). These behaviors are based
on the sign the leading term anxn, and whether n is
even or odd.

I. The "Arms" of Polynomial Graphs

y = +xeven + lower degree terms:

y = +xeven + lower degree terms: y = –xeven + lower degree terms:

y = +xodd + lower degree terms:

y = +xodd + lower degree terms: y = –xodd + lower degree terms:

For factorable polynomials, we use the sign-charts to
sketch the central portion of the graphs.

Recall that given a polynomial P(x), it's sign-chart is
constructed in the following manner:

Construction of the sign-chart of polynomial P(x):

I. Find the roots of P(x) and their order respectively.

II. Draw the real line, mark off the answers from I.

III. Sample a point for its sign, use the the orders of
the roots to extend and fill in the signs.

III. Sample a point for its sign, use the the orders of
the roots to extend and fill in the signs.
(Reminder:
Across odd-ordered root, sign changes
Across even-ordered root, sign stays the same.)

Example A: Make the sign-chart of f(x) = x2 – 3x – 4
and graph y = f(x).

and graph y = f(x).
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
so x = 4 and x= –1 are
two roots of odd order.

and graph y = f(x).
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
The sign chart and the
graph of y = f(x) are
shown here.
y=(x – 4)(x+1)
x
y

and graph y = f(x).
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
shown here.
y=(x – 4)(x+1)
x
y
Note the sign-chart reflects the properties of the graph.

and graph y = f(x).
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
shown here.
y=(x – 4)(x+1)
x
y
I. The graph touches or crosses the x-axis at the roots.

and graph y = f(x).
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
shown here.
y=(x – 4)(x+1)
x
y
II. The graph is above the x-axis where the sign is "+".

and graph y = f(x).
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
shown here.
y=(x – 4)(x+1)
x
y
II. The graph is above the x-axis where the sign is "+".
III. The graph is below the x-axis where the sign is "–".

II. The “Mid-Portions” of Polynomial Graphs

Graphs of an odd ordered root (x – r)D at x = r.

+
+ +
order = 1 order = 1
order = 3, 5, 7..
r r
r
Graphs of an odd ordered root (x – r)D at x = r.
+
order = 3, 5, 7..
r
y = (x – r)1 y = –(x – r)1
y = (x – r)3 or 5.. y = –(x – r)3 or 5..

order = 2, 4, 6 ..
++
If the we know the roots of a factorable polynomial,
then we may construct the central portion of the
graph (the body) in the following manner using its
sign chart.
I. Draw the graph about each root using the
information about the order of each root.
II. Connect all the pieces together to form the graph.
x=r
r
Graphs of an even ordered root at (x – r)E at x= r.
order = 2, 4, 6 ..
y = (x – r)2 or 4.. y = –(x – r)2 or 4..

For example, given two
roots with their orders and
the sign-chart of a
polynomial,

the sign-chart of a
polynomial,
+
order=2 order=3

the sign-chart of a
polynomial, the graphs
around each root are as
shown.
+
order=2 order=3

the sign-chart of a
shown. Connect them to get
the whole graph.
+
order=2 order=3

the sign-chart of a
the whole graph.
Example B: Given P(x) = -x(x + 2)2(x – 3)2, identify
the roots and their orders. Make the sign-chart.
Sketch the graph about each root. Connect them to
complete the graph.
+
order=2 order=3

the sign-chart of a
the whole graph.
complete the graph.
The roots are x = 0 of order 1,
+
order=2 order=3

the sign-chart of a
the whole graph.
complete the graph.
The roots are x = 0 of order 1, x = -2 of order 2,
+
order=2 order=3

the sign-chart of a
the whole graph.
complete the graph.
The roots are x = 0 of order 1, x = -2 of order 2,
and x = 3 of order 2.
+
order=2 order=3

The sign-chart of P(x) = -x(x + 2)2(x – 3)2 is
++
x = 3
order 2
x = 0
order 1
x = -2
order 2

++
x = 3
order 2
x = 0
order 1
x = -2
order 2
By the sign-chart and the order of each root, we draw
the graph about each root.

++
x = 3
order 2
x = 0
order 1
x = -2
order 2
the graph about each root. (Note for x = 0 of order 1,
the graph approximates a line going through the point.)

++
x = 3
order 2
x = 0
order 1
x = -2
order 2
Connect all the pieces to get the graph of P(x).

x = -2
order 2
++
x = 0
order 1
x = 3
order 2
Connect all the pieces to get the graph of P(x).

Note the graph resembles y = -x5, it's leading term,
when viewed at a distance.

Note the graph resembles y = -x5, it's leading term,
when viewed at a distance.
-2
++
0 3

Graphs of Rational Functions
Recalled that rational functions are functions of the
form R(x) = where P(x) and Q(x) are
polynomials.
P(x)
Q(x)

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

polynomials.
P(x)
Q(x)
In this section we study the graphs of reduced
factorable rational functions.

polynomials.
P(x)
Q(x)
The main principle of graphing these functions is the
the same as polynomials.

polynomials.
P(x)
Q(x)
the same as polynomials. We analyze the behaviors
and draw pieces the graphs at important locations,
then complete the graphs by connecting them.

polynomials.
P(x)
Q(x)
the same as polynomials. We analyze the behaviors
and draw pieces the graphs at important locations,
then complete the graphs by connecting them.
However the behaviors of rational functions are more
complicated due to the presence of the denominators.

x=0
The graph of y = 1/x has an
asymptotes at x = 0 as
shown here.
We also call vertical asymptotes
“poles”. Therefore y = 1/x has a
pole of order 1 at x = 0.
+ + +– – –
Graph of y = 1/x2
Graph of y = 1/x
x=0
The graph of y = 1/x2 has a pole
of order 2 at x = 0 and its graph
is shown here. Similar to the
orders of roots, the even or odd
orders of the poles determine
the behaviors of the graphs.
They are shown below.
+ + + + + +
Graph of y = ±1/xN for N = 1, 2, 3..

e.g. y = –1/x1 or 3,..
+
e.g. y = 1/x1 or 3,..
+
Graphs of
odd ordered poles

e.g. y = –1/x1 or 3,..
+
e.g. y = 1/x1 or 3,..
+
+ +
Graphs of
even ordered polesodd ordered poles
Graphs of
e.g. y = 1/x2 or 4,.. e.g. y = –1/x2 or 4,..

e.g. y = –1/x1 or 3,..
+
e.g. y = 1/x1 or 3,..
+
+ +
Graphs of
Graphs of
e.g. y = 1/x2 or 4,.. e.g. y = –1/x2 or 4,..
We are examining the
“Mid–Portion” of the
graphs of rational functions.
We will look at the
“Arms” of these graphs later.

e.g. y = –1/x1 or 3,..
+
e.g. y = 1/x1 or 3,..
+
+ +
Example A: Given the
following information of
roots, sign-chart and
vertical asymptotes,
draw the graph.
++
root
VAVA
Graphs of
Graphs of
e.g. y = 1/x2 or 4,.. e.g. y = –1/x2 or 4,..

e.g. y = –1/x1 or 3,..
+
e.g. y = 1/x1 or 3,..
+
+ +
draw the graph.
++
root
VAVAGraph from right to left:
Graphs of
Graphs of
e.g. y = 1/x2 or 4,.. e.g. y = –1/x2 or 4,..

draw the graph.
++
root
VAVAGraph from right to left:
e.g. y = –1/x1 or 3,..
+
e.g. y = 1/x1 or 3,..
+
+ +
Graphs of
Graphs of
e.g. y = 1/x2 or 4,.. e.g. y = –1/x2 or 4,..

Horizontal Asymptotes

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

far left of 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 similar 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) resemble 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 a rational function R(x) = P(x)
Q(x)

Steps for graphing a rational function R(x) =
I. (Roots) Find the roots of P(x) and their orders
by solving P(x) = 0.
P(x)
Q(x)

P(x)
Q(x)
II. (Poles or VA) Find the vertical asymptotes (VA)
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 shape of the
graph around the roots, and the shape of the graph
along the poles.

P(x)
Q(x)
along the poles.
Steps I and II give the "mid-section“ of the graph.

P(x)
Q(x)
along the poles.
III. (HA) Use the last theorem to determine the
behavior of the graph to the right and left as x±∞.

P(x)
Q(x)
along the poles.
The horizontal asymptote exists only if the limit exists.

P(x)
Q(x)
along the poles.
The horizontal asymptote exists only if the limit exists.
Step III gives the “arms” of the graph.

Example B.
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 B.
x2 – 4x + 4
x2 – 1
For its roots, set x2 – 4x + 4 = 0, i.e.. x = 2 (ord = 2).

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

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

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

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

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

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

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

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

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

Example C.
x2 – 2x – 3
x – 2

Example C.
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 C.
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 C.
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 C.
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.

Example C.
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.
+–+–

Example C.
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
+–+–

Example C.
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
+–+–

Example C.
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
+–+–

Example C.
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.
x=3
+–+–

Example C.
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.
x=3
+–+–
Note the y– int (0, 3/2).
(0, 3/2)

3.3 graphs of factorable polynomials and rational functions

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