Unit 6: Polynomial and Rational Functions
Objectives:
To review factor & remainder theory.
6.1 To define and illustrate polynomial and rational functions.
6.2 To sketch the graphs of polynomial and rational functions with integral coefficients, using
calculators or computers.
6.3 To analyze the characteristics of polynomial and rational functions and to identify the
‘zeros’ of these graphs.
6.4 To define, determine, and sketch the inverse of a function where it exists.
6.5 To define, determine, and sketch the reciprocal of a function.
Polynomial Functions
Notes:
A polynomial function is a relation in which the domain, or, x-values are related to the
range, or y values, by a polynomial with positive integral exponents of x.
Polynomial functions are usually written;
f(x) = axn + bxn-1 + cxn-2 +…+ z
Where a, b, c … are real numbers and n is a positive integer. Notice that the polynomial is
usually written with the exponents in descending order.
n = the degree of the polynomial
a = the leading coefficient
z = the constant term.
For Example:
f(x) = x5 - 3x4 - 5x3 + 15x2 + 4x - 12 is
• a polynomial of degree 5 with
• leading coefficient 1 and
5– 15
10
15
20
5
10
• constant
term -12.
2– 4
4
2
TI-82 Calculator: We can see the graph of a polynomial function by using the [Y=] key. To
graph the example above, press [Y=] and then enter the left hand side of the equation. Use
the exponent key [^] to enter the exponents.
The display will show: Y1 = x^5 - 3x^4 - 5x^3 + 15x^2 - 12
Press [GRAPH] to see the graph.
y
20
15
10
5
– 4
– 2
– 5
– 10
– 15
-133-
2
4
x
Long Division of Polynomials
In earlier classes we learned to use long division to find the quotient and remainder for
questions like; (x3 - 3x2 + 3x - 6) ÷ (x - 2)
the quotient
xx2  1
32
xx23
xx 36
the question

()xx32 2
xx2 3

() xx2  2
subtract x 2 times (x - 2)
bring down the next term
subtract -x times (x - 2)
bring down the next term
x 6
subtract 1 times (x - 2)

()x 2
the remainder
4
4
The solution is x2 - x + 1 remainder -4 or x 2  x  1 
x2
For Example Find the quotient (x4 - 16) ÷ (x - 2)
Solution:Write the question in long division form.
Fill in missing powers of x with the coefficient 0.
x3  2 x 2  4 x  2
x  2 x 4  0 x 3  0 x 2  0 x  16
( x 4  2 x 3 )
2 x3  0 x 2
(2 x 3  4 x 2 )
4 x2  0 x
(4 x 2  8 x)
8 x  16
(8 x  16)
0
Fill in 0x3, 0x2, and 0x for the missing terms.
-Remember that subtracting a negative
produces a positive value.
The solution is (x4 - 16) ÷ (x - 2) = x3 + 2x2 + 4x + 2
-134-
Synthetic Division is a shorter version of long division in the case where the divisor is of
the form x - a.
For Example: (x4 + 13x3 - 5x2 + 33x + 9) ÷ (x + 3)
In the division below, the first row represents;
-the constant from the divisor with the opposite sign 3
-the coefficients from the polynomial in descending powers of x. (6 13
-5
33
9)
The second row is obtained by multiplying the preceding factor in the third row by the
constant.
The third row starts with the first coefficient from the first row. The remaining terms are
obtained by adding the first two rows.
constant from divisor with opposite sign
13 53
39

18 15 30 9
65  10
30
3 × (-3)
10 × (-3)
(-5) × (-3)
6 × (-3)
coefficients from dividend
36
remainder
The quotient and remainder are; 6x3 - 5x2 + 10x + 3 remainder 0. The coefficients and
remainder come from the last row.
For Example: Find the quotient and remainder for; (x3 - 8) ÷ (x - 2)
Solution: Set up the problem as in the example above. Remember to use the coefficient 0 for the
missing powers of x.
2 1 0 0 8
First row: constant  2  and coefficients
2 4 8
1 2 4 0
3
(x - 8) ÷ (x - 2) = 1x2 + 2x + 4 R = 0
Third row is the solution & remainder.
For Example: Find the quotient and remainder for; (2x3 + 7x2 - 5) ÷ (x + 3).
Solution:Set up the problem as in the examples above. Remember to use the coefficient 0 for the
missing power of x.
3 2 7 0 5
First row: constant  3  and coefficients
6 3 9
Third row is the solution & remainder.
2
1 3 4
(2x3 + 7x2 - 5) ÷ (x + 3) = 2x2 + x - 3 remainder 4 or 2 x 2 x  3  x 4 3
-135-
Practice Questions 6-1: Division of Polynomials
1. For each of the following polynomial functions;
i) arrange the terms in descending orders of x.
ii) list the leading coefficient
iii) list the degree of the equation
iv) list the constant term
v) sketch the graph as it appears on the TI-82 calculator.
a. f(x) = 3x2 - 15 - 13x + x3
c. f(x) = 2x2 - 1 + 2x5 + x9 - 5x4 -3x3
b. f(x) = 3 - x3 - 6x2 + x4 + 2x
c. f(x) = 5x3 - 3x4 - 3 + x
2. Use long division to find the following quotients;
a. (2x3 + 5x2 - x + 1) ÷ (2x + 3)
c. (6x3 - 5x2 - 10x) ÷ (2x - 3)
b. (15x3 + x2 + 1) ÷ (3x + 2)
d. (5x4 + 9x3 - 7x2 + 11x - 2) ÷ (5x - 1)
3. Use synthetic division to find the following quotients;
a. (x3 - 2x2 + 2x - 5) ÷ (x - 1)
c. (6x4 + 15x3 + 28x + 6) ÷ (x + 3)
b. (x3 - 2x2 + 2x - 5) ÷ (x + 1)
d. (3x3 + 7x2 - 4x + 3) ÷ (x + 3)
Factor and Remainder Theories
Remainder Theorem: For any polynomial function f(x), the value of f(x) at x = c (f(c)) is equal
to the remainder of f(x) ÷ (x - c).
For Example: If f(x) = 3x3 + 12x2 - 7x - 6, find the value of f(-5) by synthetic division.
Solution: Divide (3x3 + 12x2 - 7x - 6) ÷ (x + 5) by synthetic division.
5 3
12 7
6
15 15
3 3 8
40
 46
Note that the sign changes twice from f ( 5) to
( x  5) to 5
The remainder is -46
Therefore, f(-5) = -46.
To check the answer calculate:
f(-5) = 3(-5)3 + 12(-5)2 - 7(-5) - 6,
= 3(-125) + 12(25) -(-35) - 6
= -375 + 300 + 35 - 6
= -46 3
Notice that if we combine the factor theorem with synthetic division, we do not have to
calculate any of the exponents. The only operations required are multiplication, and addition
of integers.
-136-
Factor Theorem is a useful corollary of the remainder theorem:
For any function f(x), (x - a) is a factor iff (if and only if) f(a) = 0
For Example: Is (x + 2) a factor of f(x) =x4 - 4x3 - x2 + 16x - 12 ?
Solution: Find f(-2) (notice the changed sign).
f(-2) = (-2)4 - 4(-2)3 - (-2)2 + 16(-2) - 12
= 16 - 4(-8) - 4 - 32 - 12
= 16 + 32 - 4 - 32 -12
=0
f(-2) = 0
 , (x + 2) is a factor of f(x) = x4 - 4x3 - x2 + 16x - 12
We can use the factor theorem to find all of the factors for a polynomial of the form
f(x) = xn + axn-1 + bxn-2 +…+ c (The leading coefficient is 1), by checking all of the possible
factors of the constant term.
In the example above the constant term is -12. The factors are ±1,±2, ±3, ±4, ±6, and ±12
f(-1) = (-1)4 - 4(-1)3 - (-1)2 + 16(-1) - 12
= 1 - 4(-1) - 1 - 16 - 12
= 1 + 4 - 1 - 16 - 12
= -24
(x + 1) is not a factor
f(1) = (1)4 - 4(1)3 - (1)2 + 16(1) - 12
= 1 - 4(1) + 1 + 16 - 12
= 1 - 4 + 1 + 16 - 12
=0
(x - 1) is a factor
f(2) = (2)4 - 4(2)3 - (2)2 + 16(2) - 12
= 16 - 4(8) - 4 + 32 - 12
= 16 - 32 - 4 + 32 - 12
=0
f(-3) = (-3)4 - 4(-3)3 - (-3)2 + 16(-3) - 12
= 81 - 4(-27) - 9 - 48 - 12
= 81 +108 - 9 - 48 - 12
= 120
f(3) = (3)4 - 4(3)3 - (3)2 + 16(3) - 12
= 81 - 4(27) - 9 + 48 - 12
= 81 - 108 - 9 + 48 - 12
=0
(x - 2) is a factor
(x + 3) is not a factor.
(x - 3) is a factor
The factors are (x - 1), (x + 2), (x - 2), and (x - 3). Since the function is of degree 4, we can
stop after finding 4 factors.
-137-
Practice Questions 6-2: Factor and Remainder Theories
1. Use the remainder theorem and synthetic division to find the value of f(c) for each
polynomial;
a. f(x) = x3 + 2x2 - 3x + 5, c = 3
c. f(x) = 3 - 5x + 2x2 - 3x3, c = -5
e. f(x) = x3 - 2x + 5, c = 1/2
b. f(x) = -3x4 + 2x3 - 3x2 - x + 1, c = -2
d. f(x) = 2x4 - x2 + 2x - 9, c = -1
f. f(x) = x5 - 3x3 + 2, c = 2
2. Use the factor theorem to determine if the first polynomial is a factor of the second.
a. x + 1, 3x4 + x3 - x2 - 2x - 3
c. x + 2, x10 + 2x9 + x + 2
e. x + 2 , 2x4 - x3 - 3x2 + 2x + 2
b. x - 2, x3 - x + 6
d. x + 2i, x4 + 2x3 + x2 + 8x - 12
e. x - 2 , 2x4 - x3 - 3x2 + 2x - 2
3. Use the factor and remainder theories to factor the following polynomials completely.
a. f(x) = x3 + 4x2 + x - 6
c. f(x) = x3 - 31x - 30
b. f(x) = x4 - 10x2 + 9
d. f(x) = x3 -5x2 - 18x + 72
Polynomial and Rational Functions
As we learned earlier, A polynomial function is a relation in which the domain, or, x values are related to the range, or y values, by a polynomial with positive integral exponents
of x.
Polynomial functions are usually written;
f(x) = axn + bxn-1 + cxn-2 +…+ z
Where a, b, c … are real numbers and n is a positive integer. Notice that the polynomial
is usually written with the exponents in descending order.
n = the degree of the polynomial
a = the leading coefficient
z = the constant term.
A rational function is usually defined as;
p( x)
q( x)
Where p(x) and q(x) are polynomial functions and q(x) ≠ 0
f ( x) 
For Example:
3x 2  5 x  2
f ( x) 
x2  4
is a polynomial function made from the functions p ( x )  3x 2  5x  2 and q( x )  x 2  4
Any function can be used to generate ordered pairs by substitution for x and calculating.
-138-
3x 2  5 x  2
find three ordered pairs;
x2  4
Solution: Choose three values for x to substitute into the function and solve each for f(x).
For Example: Given, f ( x ) 
3(0) 2  5(0)  2
(0) 2  4
002

04
2 1


4 2
3(1) 2  5(1)  2
(1) 2  4
35 2

1 4
4 4


3 3
f (0) 
f (1) 
The solutions would be; f (0) 
3(1) 2  5(1)  2
(1) 2  4
35 2

1 4
6
1


3
2
f ( 1) 
1
4
1
1
, f (1)  , and f ( 1)   , or  0,  ,
2
3
2
 2
1
 4

1,  and,  1,  
 3

2
With rational functions, the possibility exists that a particular value of x may have no
corresponding value of f(x). When we try to evaluate the function above when x = -2 we find;
f ( 2) 
3(2)2  5(2)  2 3(4)  10  2 20


 undefined
(2)2  4
44
0
For larger polynomial and rational functions we can use the remainder theorem and
synthetic division to find ordered pairs;
For Example: Find three ordered pairs for the function f(x) = 3x4 - 2x3 + x2 - 6.
Solution: Use the remainder theorem to find f(1), f(2), and f(3).
1 3 2 1 0 6
3
3 1 2
1 2 2
2 3 2 1 0
6 8 18
3 4 9 18
2
4
6
36
30
3 3 2 1 0
9 21 66
3 7 22 66
6
198
192
From the remainders; f(1) = -4, f(2) = 30, and f(3) = 192.
The ordered pairs would be; (1, -4), (2, 30), and (3, 192)
We can evaluate rational functions by evaluating the numerator and denominator
separately and then calculating the quotient.
For Example: Evaluate f ( x ) 
Solution: Calculate
2 x 2  3x  1
at x = -3
2 x2 1
2(-3)2 - 3(-3) +1 = 28 and 2(-3)2 - 1 = 17.
28
17
TI-82 Calculator: The calculator provides several methods for generating ordered pairs for a
function;
Then f (3) 
-139-
–0 10
5
– 4
2
Unit 6: Polynomial and Rational Functions
If the function is entered into Y1 (or any other Y variable) using the [Y=] key, we can
evaluate the function at various values of x by:
1. Use the [STO] key to store a value of x.
Press [2nd][VARS] (Y-Vars) and choose 1:Function then 1:Y1
Press [ENTER] to find the value of Y1 at that value of x.
We can then store different values of x to carry out further evaluations.
2. Use the CALC menu ([2nd][TRACE]) to evaluate the functions.
Choose 1:value and and then enter the value of x.
To evaluate Y2 etc. press the down arrow. The number of the function is displayed at
the upper right corner of the screen.
3. Press [2nd][GRAPH] to see a table of values for all of the functions entered in the Y=
menu.
The TblSet menu ([2nd][WINDOW]) can be used to modify the contents of the table.
If Indpnt: is set to Ask, then you can choose the values of x to be displayed in the
table.
Although methods 2 and 3 are easier to use, method 1 can be used in conjunction with
the [MATH] menu to display values as fractions.
Graphs of Polynomial and Rational Functions
Consider the graph of the function f(x) = x5 - 3x4 - 5x3 + 15x2 + 4x - 12.
y
10
Maxima (singular - maximum)
5
– 4
– 2
2
4
x
– 5
x - intercepts (roots)
– 10
Minima (singular - minimum)
y - intercepts
This graph has several features which can be used to identify it. Using the [TRACE] function
of a graphing calculator, we can locate each important point;
-140-
Unit 6: Polynomial and Rational Functions
x - intercepts: (-2, 0), (-1, 0), (1, 0), (2, 0), and (3, 0)
y - intercept: (0, -12)
Maxima: (-1.6, 10.3) and (1.5, 3.3)
Minima: (-0.1, -12.3) (2.6, -6.4)
2– 6
2
4
2– 4
4
2
These points are useful when sketching graphs;
For Example: Sketch the graph of f(x) = -x3 + x2 + 4x - 4
Solution: Draw the graph using a graphing calculator or by making a table of values.
Estimate the location of the intercepts, maxima, and minima.
Use those points to sketch the graph.
y
x - intercepts: (-2, 0), (1, 0), and (2, 0)
2
– 4 – 2
– 2
2
x
4
y - intercept: (0, -4)
Minimum: (-0.9, -6.1)
– 4
Maximum: (1.5, 0.9)
– 6
4– 4
2
2
2– 4
4
2
If we use a graphing calculator such as the TI-82, we can use the CALC menu to
determine the zeros, maxima and minima.
The graphs of rational functions have another kind of special point of interest.
2x
Consider the graph of f ( x )  2
x 4
Notice that this graph has two places where the y
y
values change from negative to positive as x values
4
increase. In this graph those points have been
indicated by vertical lines. They represent places
2
where the value of f(x) is undefined.
– 4 – 2
– 2
– 4
2
4
x
Asymptotes at x = -2 and x = 2
x - intercept (0, 0) and
y - intercept (0, 0)
No maxima or minima.
TI-82 Calculator: The graphing calculator shows an almost vertical line at the asymptotes.
Remember that these lines are not part of the graph. They are a result of the type of display
used in the calculator.
-141-
4– 4
2
2
2– 4
4
2
2– 4
4
2
2– 4
4
2
Unit 6: Polynomial
and Rational Functions
Practice Questions 6 - 3: Graphing Polynomial and Rational Functions
1. Examine the following graphs. List the location of the intercepts, any maxima or minima,
and the equation of any vertical and horizontal asymptotes.
y
a.
4– 4
2
2
2– 8
4
6
8
2
4
6
– 4
4
4
2
2
– 2
2– 2
4
4
2– 4
2x
44
2
– 4
2
– 2
– 4
– 4
y
4
2
2
2
4
6
4– 4
2
2
28– 6
4
2x
4
– 4
– 2
2
– 2
– 2
– 4
– 4
y
e.
4
2
2
2
4
x
– 6
x
4
y
f.
4
– 2
x
4
y
d.
4
– 8– 6– 4– 2
– 4
– 2
– 2
c.
4– 4
2
2
2– 4
4
2
y
b.
– 4
– 2
– 2
– 2
– 4
– 4
2
4 x
2. Sketch the graph of the following functions using a graphing calculator or a table of values.
Estimate the location of any of the characteristics of the graphs. (See question 1.) Note the
degree of each function.
-142-
Unit 6: Polynomial and Rational Functions
a. f(x) = 3x - 2
b. f(x) = -x2 + 6x - 5
2
c. f(x) = x - 4x + 3
d. f(x) = x3 + 3x2 - x - 1
e. f(x) = -x3 -4x2
f. f(x) = -x4 + 2x2 + 1
2
g.f(x) = x4 - x3 - 5x2 + 2
h. f ( x ) 
3x  7
1
x3
i. f ( x ) 
j. f ( x ) 
x3
3x  7
x3
k. f ( x )  2
x  2x 1
3. Compare the graphs from question 2 to the graphs from question 1. List the degree and/or
type of equation (polynomial or rational) and the sign of the leading variable for each graph
in question 1.
Characteristics of Polynomial Functions
As you may recall, the real roots of a function correspond to the x - intercepts. You may
also have noticed that, while a quadratic (degree 2) has either 2 real roots or 2 complex roots,
equations of degree greater than two can have both real and complex roots. The number and
nature of the roots can be determined using the following theories;
Fundamental Theorem of Algebra: A polynomial of degree n, where n is a positive number,
has exactly n roots. These roots may be multiple (the same number more than once), real
numbers, or complex numbers.
For Example: The function f(x) = -2x5 + 3x2 - 6x + 3 has exactly 5 roots.
Descartes’ Rule of Signs: The number of real roots of a polynomial can be determined using
the following rules; For the polynomial function f(x), with real coefficients.
1. The number of positive real roots is either the same as the number of its variations in
signs, or less than that by a multiple of 2.
2. The number of negative real roots is either the same as the number of variations in signs
of f(-x), or less than that by a multiple of 2.
For Example: The function f(x) = -2x5 + 3x2 - 6x + 3 has
3 variations in sign; -2x5  + 3x2, + 3x2 - 6x, and - 6x  + 3
 there are 3 or 1 positive real roots.
The function f(-x) = -2(-x)5 + 3(-x)2 - 6(-x) + 3 = 2x5 + 3x2 + 6x + 3 has
0 variations in sign
 there are no negative real roots.
 this equation has either 3 positive real roots and two complex roots or 1 positive real root
and 4 complex roots.
Rational Root Theorem: For the polynomial function f ( x )  ax n  bx n1  cx n2   z , where
a, b, c…z are integers, then any real rational root can be written in the form kh , where h is a
factor of z, and k is a factor of a.
-143-
Unit 6: Polynomial and Rational Functions
For Example: Given f ( x )  4 x 4  11x3  19 x 2  44 x  12 , List all of the possible rational roots.
Determine which of those roots are real by testing or using synthetic division.
Solution: All of the roots can be written as fractions where the denominator is a factor of 4 and
the numerator is a factor of 12.
List the factors of 4: {±1, ±2, ±4}
List the factors of 12: {±1, ±2, ±3, ±4, ±6, ±12}
 the possible roots are:
1 2 3 4 6 12 1 2 3 4 6 12 1 2 3 4 6 12
 , , , , , , , , , , , , , , , , ,
1 1 1 1 1
1
2 2 2 2 2
2
4 4 4 4 4
4


Simplified, with duplicates removed:
1 1 3
 ,  ,  , 1, 2, 3, 4, 6, 12
4 2 4


To determine which of these are roots, we can use the factor theorem and synthetic
division to test. (Starting with the easy ones: Integers
1 4 11 19 44 12
1 4 11 19 44
12
2 4 11 19 44 12
4
4 7 26
7 26 18
1 is not a root
18
30
4 15 4
4 15 4 48
-1 is not a root
48
 36
4
8 6 50 12
3 25 6
0
2 is a root
Once we know that 2 is a factor, we can rewrite our polynomial as;
(x - 2)(4x3 - 3x2 - 25x - 6) using the quotient from the division. We can now use the reduced
polynomial, 4x3 - 3x2 - 25x - 6, to find the rest of the factors
Try 2 again.
2 4 3 25
6
8 10 30
4 5 15  36
2 is not a double root

Try -2 again.
2 4 3 25 6

8 22
4 11 3
-2 is a root
6
0
The reduced polynomial is
4x2 – 11x – 3
Which can be factored to;
(4x + 1)(x – 3)
1
 is a root and 3 is a root.
4
1
The roots are  ,  2, 2, and 3 .
4
If some of the roots had been complex, we would have found fewer than 4 rational roots.
-144-
Unit 6: Polynomial and Rational Functions
Graphs of polynomial Functions by Degree:
Degree 1: Linear
Leading coefficient positive
Leading coefficient negative.
y
y
x
x
Degree 2: Quadratic
Leading coefficient positive
Leading coefficient negative
y
y
x
x
Degree 3: Cubic
Leading coefficient positive
y
Leading coefficient negative
y
y
y
x
x
x
x
Degree 4: Quartic
Leading coefficient positive
y
Leading coefficient negative
y
y
y
x
x
x
x
Once we know the general shapes of the graphs and the roots we can make an accurate sketch.
-145-
Unit 6: Polynomial and Rational Functions
For Example: Sketch the graph of f(x) = x3 - 4x2 - 3x + 12.
Solution: This is a Cubic equation with three possible roots.
By Descartes’ rule there are 2 or 0 real positive roots and 1 real negative root.
The possible roots are {±1, ±2, ±3, ±4, ±6, and ± 12}
Try 1
1 1 4 3 12
Try -1
1 1 4 3
3 1 4 3
12
Try 4
12
Try -3
3 1 4 3
12
12
3 21 54
1 7 18  42
-3 is not a root
Notice that we can not use
factor theorem to find
irrational roots.
3 3 18
1 1 6  6
3 is not a root
The reduced polynomial;
x2  3  0
4 0 12
1 0 3
0
410
is 5a root
5–
12
2 4 14
1 2 7  2
2 is not a root
Try 3
2 12 18
1 6 9  6
-2 is not a root
4 1 4 3
2 1 4 3
12
1 5 2
1 5 2 10
-1 is not a root
1 3 6
1 3 6 6
1 is not a root.
Try -2
2 1 4 3
Try 2
x 3
5– 10
10
5
The roots are {4,  3 }.
The y - intercept can be found when x = 0
f(x) = 03 - 4(0)2 - 3(0) + 12 = 12
Plot the information on the graph and sketch
a smooth cubic curve passing through them.
y
10
5
– 10
– 5
5
– 5
-146-
10 x
Unit 6: Polynomial and Rational Functions
Practice Questions 6-4: Characteristics of Polynomial Functions
1. For Each of the following functions;
i. Determine the total number of roots.
ii. Use Descartes rule to determine the number of real roots.
iii. Find as many of the roots as possible using factor theorem and the quadratic formula.
iv. Locate the y - intercept (x = 0)
v. Sketch the graph
a. f(x) = x3 - 9x2 + 27x - 27
c. f(x) = x4 + x3 - 7x2 - 13x - 6
e. f(x) = 2x - 3
b. f(x) = -x4 + 5x3 - 5x2 - 5x + 6
d. f(x) = x3 + x2 - x - 1
f. f(x) = -2x2 + 15x -25
Characteristics of Rational Functions
As we may have noticed from Practice Questions 6-3, rational functions can be
recognized from their graphs because they have discontinuities called asymptotes. Rational
functions do show all of the other characteristics of zeros(roots), intercepts and maxima or
minima.
To find the roots for a rational function, we need to solve the numerator of the rational
function;
x3
For Example: If f ( x )  2
, find the roots of the equation.
2 x  3x  1
Solution: The root is found by solving the numerator. x - 3 = 0  x = 3
Remember, the root is also the x-intercept.
To find the y - intercept, set x = 0 and solve for f(0) = (y).
For Example: If f ( x ) 
x3
, find the y - intercept.
2 x  3x  1
2
Solution: Set x = 0, and solve.
x 3
(0)  3
3
f ( x)  2
 2

 3
2 x  3x  1 (0)  3(0)  1 1
A vertical asymptote occurs when the denominator is equal to 0. To find the vertical
asymptotes, find the roots for the polynomial in the denominator.
For Example: If f ( x ) 
x3
, find the vertical asymptotes.
2 x  3x  1
2
Solution: The asymptotes are found by solving the denominator;
2x2 - 3x + 1 = 0
The denominator may be any sort of polynomial.
(2x - 1)(x - 1) = 0
Quadratics may be solved by factoring or using the
or the quadratic formula.
x  12 , and x = 1
The vertical asymptotes are x  12 , and x = 1.
-147-
Unit 6: Polynomial and Rational Functions
To sketch the graph of a rational function we must find the roots (if they exist), the y intercept, and the vertical asymptotes. Near the asymptotes we can use test points to
determine which direction the curve is moving.
x  3
2
For Example: Sketch the graph of f(x) = 2 x  3 x  1 .
Solution: For this function we have already determined that;
The root or x - intercept is x = 3
The y - intercept is y = -3.
Asymptotes occur at x  12 , and x = 1.
Since there are 2 asymptotes, we should use at least 4 test points. One slightly below and
one slightly above each.
Calculate: x = 0.2, x = 0.6, x = 0.9, and x = 0.1
5
1
2
3
4
5
Using the equation we find; f(0.2) = -5.83, f(0.6) = 30, f(0.9) = 26.25, and f(1.1) = -15.8
Sketch the graph using the points and asymptotes that we have found.
Notice that in this graph we have had to exaggerate the curve between the asymptotes.
The smallest value for f(x) that we found was f(0.9) = 26.25. Notice also, that the right hand
curve must cross the x-axis at x = 3.
y
35
30
25
20
15
10
5
– 5
– 4
– 3
– 2
– 1
– 5
1
2
3
4
TI-82 Calculator: To graph a rational function, we must be careful to make proper use of
brackets to separate the numerator and denominator.
In the example above, the proper formula to enter into Y1 is;
Y1=(x-3)/(2x2-3x+1)
Notice the use of brackets.
Without the brackets, the expression becomes;
Y1=x2-3/2x+1
A parabola.
-148-
5
x
Unit 6: Polynomial and Rational Functions
For Example: Sketch the graph of f ( x ) 
3x
.
x 9
2
Solution: Find the root(s) by solving the numerator:
3x = 0  x = 0
Find the y-intercept by setting x = 0 and solving:
3(0)
0
f (0)  2

0
(0)  9 9
Solve the denominator to find the asymptotes:
x2 - 9 = 0
x2 = 9
x = ±3  x = -3, x = 3
5 points near each asymptote to find the direction of the graph;
Use5–test
1– 5
2
3
4
5
1
2
3
4
3(4)
12 12
3(2)
6
6
f ( 4) 


 1.7 f ( 2) 


 1.2
2
2
(4)  9 16  9
7
(2)  9 4  9 5
3(2)
6
6
3(4)
12
12
f (2) 


 1.2
f (4) 

  1.7
2
2
(2)  9 4  9 5
(4)  9 16  9 7
Sketch the graph;
y
5
– 5– 4– 3– 2– 1
1 2 3 4 5 x
– 5
-149-
Unit 6: Polynomial and Rational Functions
Practice Questions 6-5: Characteristics of Rational Functions
1. 1. For Each of the following functions;
i. Find the roots (if they exist).
ii. Determine the y-intercept if it exists.
iii. Determine the location of any vertical asymptotes.
iv. Sketch the graph
a. f ( x ) 
2
x
4
2x  5
x3
d. f ( x ) 
x2
3 x  2
f. f ( x )  2
x  16
b. f ( x ) 
x
2x  5
3
e. f ( x )  2
x  6x  5
4 x2  9
g. f ( x )  2
x 4
c. f ( x ) 
The Inverse of a Function
For any function, f(x), the inverse of the function, f-1(x) , can be found by interchanging x
and y in the equation for the function.
For Example: If f ( x )  3x 2  9 find the inverse, f-1(x).
Solution: y = 3x2 - 9
Rewrite the function using y= notation.
x = 3y2 - 9
Interchange x and y.
2
3y = x + 9
Solve the inverse equation for y
x9
y2 
3
x9
x9
3
3x  27
Rationalize the denominator
y



3
3
3
3
3 x  27
Rewrite using f-1(x) notation.
3
The graph of the inverse of a function is the reflection of the original curve in the
f (x1)  
y = x line.
-150-
Unit 6: Polynomial and Rational Functions
For Example: Sketch the Inverse of the following
function
y
10
5
10
5
– 10
5
– 15
5
10
f(x)
f – 10
(x)
– 10
– 5
10 x
5
– 5
– 10
Solution: Draw in the line y = x, and sketch ythe reflection of f(x).
f(x)
10
y=x
5
– 10
– 5
5
10 x
f
– 5
–1
(x)
– 10
Notice that the graph of the inverse of a function may not be a function.
TI-82 Calculator: To graph the inverse of a function on the calculator, it is necessary to find the
equation of the inverse and enter that into the Y= register.
In the case of the function in the previous examples, the inverse is not a function;
f ( x )  3x 2  9 and f ( x )  
3 x  273
Each value of x produces 2 values for y because each square root produces both a
positive and a negative value. The calculator only uses the positive or principal square root.
To see the entire graph, it is necessary to enter a formula for both values;
Y1=
(3x+27)/3 and Y2=-
(3x+27)/3
One can also use the Y-VARS menu ([2nd][VARS]) to enter Y2=-Y1
-151-
Unit 6: Polynomial and Rational Functions
Reciprocal Functions
The reciprocal of a function; If f ( x ) is a function then the reciprocal of the function
1
1
would be
or  f ( x )  .
f ( x)
For Example: For the function f(x) = 4x2 - 9 the reciprocal would be;
In the case of a rational function if f ( x ) 
For Example: For the function f ( x ) 
1
f ( x)

1
.
4x  9
2
p( x)
1
q( x)
, then
.

f ( x) p( x)
q( x)
2x
1
3x 2  5 x  2
the
reciprocal
would
be;

3x 2  5 x  2
f ( x)
2x
To draw the graph of the reciprocal of a function we can use the characteristics of the
function to assist;
If we have a table of values for the original function we can create a table of values for
the reciprocal by taking the reciprocal of each y value.
Vertical Asymptotes of a reciprocal function occur at the roots of the original function.
Horizontal asymptotes occur where y-values become undefined for any value of x. To
find a horizontal asymptote, solve the equation for x (if possible)and set the denominator
equal to zero.
3
the horizontal asymptote can be found by;
2x 1
3
3 y
Solve y 
for x  x 
2x 1
2y
Set the denominator = 0; 2y = 0  y = 0
A horizontal asymptote occurs at y = 0
For Example: For the function m f ( x ) 
For Example: Given the function f(x) = x2 - 4, graph the function and its reciprocal, determine
the critical values and/or asymptotes for each.
Solution: Begin by graphing the function,
Roots; solve x2 - 4 = 0 x2 = 4  x = ± 2
y - intercept: set x = 0  y = (0)2 - 4
y = -4
No asymptotes.
Sketch the graph;
-152-
Unit 6: Polynomial and
Rational Functions
y
10
5
– 10
– 5
10 x
5
– 5
– 10
1
Find the equation of the reciprocal; 
f ( x)

1
x 4
2
Roots; solve the numerator - no roots
1
1

(0)  4 4
4
2
Vertical asymptotes: solve the denominator x - 4 = 0
 x = ±2
1
1
Horizontal asymptotes;
solve for x
 x2  4 
 x2   4
– 10
5
y
y
5–
f(x)
1 5
y - intercept; set x = 0
 y 
1

2
y  4 y2
1 4 y
1f(x)
 4y
 x
 x
y
y
y
Set the denominator equal to zero; y = 0 is the horizontal asymptote.
x2 
Sketch the graph of the reciprocal on the
same axes as the function;
y
5 x
– 5
1
f(x)
– 5
f(x)
– 10
-153-
Unit 6: Polynomial and Rational Functions
Practice Questions 6-6: Inverse and Reciprocal Functions
1. Sketch the inverse
of each of the following functions.
y
a.
y
b.
x
x
y
y
c.
d.
x
x
2. For each of the following functions, write the equation of the inverse (f-1(x));
a. f(x) = 3x - 2
b. f(x) = x2 - 9
1 x
d. f ( x ) 
x
f. f(x) = -x - 6
c. f(x) = (x - 2)2
e. f(x) = -3x2 - 5
3. Sketch the graph of each function and inverse from question 2.
4. For each of the functions listed in question 2 write the equation of the reciprocal.
5. For each of the functions in question 2, list the important characteristics of the graph of the
function, sketch the reciprocal and list its important characteristics. (Important characteristics
are; roots, y - intercept, maxima or minima, horizontal and vertical asymptotes,
-154-
Unit 6: Polynomial and Rational Functions
Unit 6 Solutions
Practice Questions 6-1: page 128
y
v.
1. a. i f(x) = x3 +3x2 -13x - 15
30
ii. leading coefficient = 3
20
iii. degree = 3
iv. constant = -15
10
10
2
4
6
8
– 10
2
4
6
8
1– 5
2
3
4
5
1
2
3
4
2 4 6 8 10 x
– 10– 8– 6– 4– 2
– 10
– 20
– 30
v.
b. i f(x) = x - x - 6x + 2x + 3
ii. leading coefficient = 1
iii. degree = 4
iv. constant = 2
4
3
2
y
10
8
6
4
5– 5
1
2
3
4
1
2
3
4
1– 5
2
3
4
5
1
2
3
4
2
1 2 3 4 5 x
– 5– 4– 3– 2– 1
– 2
– 4
– 6
– 8
– 10
v.
c. i f(x) = x9 + 2x5 - 5x4 - 3x3 + 2x2 - 1
ii. leading coefficient = 1
iii. degree = 9
iv. constant = -1
y
5
4
3
2
5– 5
1
2
3
4
1
2
3
4
1– 5
2
3
4
5
1
2
3
4
1
1 2 3 4 5 x
– 5– 4– 3– 2– 1
– 1
– 2
– 3
– 4
– 5
v.
d. i f(x) = -3x + 5x + x - 3
ii. leading coefficient = -3
iii. degree = 4
iv. constant = -3
4
3
y
5
4
3
2
1
– 5– 4– 3– 2– 1
– 1
– 2
– 3
– 4
– 5
-155-
1 2 3 4 5 x
Unit 6: Polynomial and Rational Functions
2. a. x + x - 2 R = 7 b. 5x2 - 3x + 2 R = -3 c. 3x2 + 2x - 2 R = -6 d. x3 + 2x2 - x + 2
3. a. x2 - x + 1 R = -4 b. x2 - 3x + 5 c. 6x3 - 3x2 + 9x + 1 R = 3 d. 3x2 - 2x + 2 R = -3
2
Practice Questions 6-2: Page 130
1. a. f(3) = 41 b. f(-2) = 73 c. f(-5) = 453 d. f(-1) = -10 e. f(1/2) = 33/8 f. f(2) = 10
2. a. f(-1) = 0 yes b. f(2) = 12 no c. f(-2) = 0 yes d. f(2i) = 0 yes e. f  2   4 no
f. f
2
 0 yes
3. a. (x - 1)(x + 2)(x + 3) b. (x + 1)(x - 1)(x + 3)(x - 3)
d. (x - 3)(x + 4)(x - 6)
c. (x + 1)(x + 5)(x - 6)
Practice Questions 6-3: Page 134
10
5
– 10
5
5– 10
10
5
1. a. Roots {0.6}; y -intercept ( -2.5); Max {(-3.5, -1.5)}; Min {(-1, -3.5)}
b. Roots {0, 3}; y -intercept (0); Max {(1.8, 0.5)}; Asymptotes x = -1, x = 4
c. Roots {-5.5, -2, 0, 2, 4.5}; y -intercept (0); Max {(-4.5, 2.5), (1, 0.5)}; Min {(-1, -0.4),
(3.5, -1)}
d. Roots {-5, -2, 1, 4}; y -intercept ( 2); Max {(-0.5, 2.5)}; Min {(-4, -4), (3, -4)}
e. Roots {-3}; y -intercept ( -3); Asymptotes x = -1
f. Roots {-2}; Asymptotes x = -4, x = 0
2. a. f(x) = 3x - 2
y
10
x - intercept
5
2
3
y - intercept {-2}
10
5
– 10
5
5– 10
10
5
– 10

– 5
5
10 x
– 5
– 10
b. f(x) = -x + 6x – 5
2
y
10
x - intercepts { 1, 5}
5
– 10
– 5
y - intercept {-5}
5
10 x
Maximum (3, 4)
– 5
– 10
-156-
Unit 6: Polynomial and Rational Functions
c. f(x) = x2 - 4x + 3
y
x - intercepts { 1, 3}
10
y - intercept {3}
5
Minimum (2, -1)
10
5
5
10
– 10
– 5
5
10 x
– 5
– 10
d. f(x) = x3 + 3x2 - x - 1
y
x - intercepts (-3.2, -0.46, 0.68)
10
y - intercept {-1}
5
Maximum (-2, 5)
10
5
5
10
– 10
– 5
5
10 x
Minimum (0, -1)
– 5
– 10
e. f(x) = -x3 - 4x2
y
10
x - intercepts (-4, 0)
5
10
5
5
10
– 10
y - intercept ( 0)
– 5
5
10 x
Minimum (-2.7, -9.4)
Maximum (0, 0)
– 5
– 10
f. f(x) = -x4 + 2x2 + 1
y
x - intercepts (-1.6, 1.6)
10
y - intercept (1)
5
Maxima (-1, 2), (1, 2)
– 10
– 5
5
10 x
Minimum (0, 1)
– 5
– 10
-157-
Unit 6: Polynomial and Rational Functions
g. f(x) = x - x - 5x + 2
4
3
2
y
x - intercepts (-1.6, -0.7, 0.62, 2.7)
10
y - intercept (2)
5
Minima (-1.2, -1.4), (2, -10)
10
5
– 10
5
5– 10
10
5
– 10
– 5
10 x
5
Maximum (0, 2)
– 5
– 10
2
3x7
h. f ( x) 
No x intercept
y
10
y - intercept (-2/7)
5
10
5
– 10
5
5– 10
10
5
– 10
Vertical Asymptote x = 7/3
– 5
5
10 x
– 5
– 10
i. f ( x ) 
1
x3
No x - intercept
y
10
y - intercept (- 1/3)
5
4
2
2
4
– 10
Vertical asymptote x = -3
– 5
5
10 x
– 5
– 10
j. f ( x ) 
x3
3x  7
x - intercept (-3)
y
y - intercept (-1)
4
Maximum (0.14, -0.96)
2
Vertical asymptotes x = -1, x = 3
– 4
– 2
2
4
x
– 2
– 4
-158-
k. f ( x ) 
x3
x  2x 1
Unit 6: Polynomial and Rational Functions
2
y
x - intercept (-3)
10
y - intercept (3)
5
Vertical asymptote x = 1
– 10
– 5
5
10 x
– 5
– 10
3. a. Polynomial, degree = 3, leading variable positive
b. Rational, numerator degree = 2, denominator degree = 2, leading variable positive
c. Polynomial, degree = 5, leading
5– 10
10
5 variable positive
5– 10
10
5 variable positive
d. Polynomial, degree = 4, leading
e. Rational, numerator degree = 0, denominator degree = 1, numerator negative
f. Rational, numerator degree = 1, denominator degree = 2, numerator positive
Details for rational functions given here are greater than expected from students.
Practice Questions 6-4: Page 139
y
1. a.
i. Degree = 3 - three roots
ii. 3 or 1 positive real roots, 0 negative
real roots
5– 10
10
5
iii. Roots {3}
5– 10
10
5
iv. y - intercept-27
10
v.
5
– 10
– 5
5
10 x
5
10 x
– 5
– 10
y
b.
i. Degree = 4 - four roots
ii. 3 or 1 positive real roots, 1 negative
real root
iii. Roots {-1, 1, 2, 3}
iv. y - intercept (6)
10
v.
5
– 10
– 5
– 5
– 10
-159-
Unit 6: Polynomial and Rational Functions
y
c.
i. Degree = 4 - four roots
ii. 1 positive real root: 3, or 1 negative
real roots
5– 10
10
5
iii. Roots {-2, -1, 3}
5– 10
10
5
iv. y - intercept (-6)
v.
5 x
– 5
– 10
– 20
– 30
– 40
y
d.
i. Degree = 3 - three roots
ii. 1 positive real root: 2, or 0 negative
real roots
5– 10
10
5
iii. Roots {-1, 1}
5– 10
10
5
iv. y - intercept (-1)
10
v.
5
– 10
– 5
5
10 x
5
10 x
5
10 x
– 5
– 10
y
e.
i. Degree = 1 - 1 roots
ii. 1 positive real root: 0 negative real
roots
5– 10
10
5
iii. Roots {1.5}
5– 10
10
5
iv. y - intercept (-3)
10
v.
5
– 10
– 5
– 5
– 10
y
f.
i. Degree = 2 - 2 roots
ii. 2 positive real root: 0 negative real
roots
iii. Roots {2.5, 5}
iv. y - intercept (-25)
10
v.
5
– 10
– 5
– 5
– 10
-160-
Unit 6: Polynomial and Rational Functions
Practice Questions 6-5: Page 141
y
1. a. i. No roots
10
iv.
ii. no y - intercept
iii. Vertical asymptote x = 0
5
10
5
– 10
5
5– 10
10
5
– 10
– 5
5
10 x
5
10 x
5
10 x
5
10 x
– 5
– 10
y
b. i. No roots
ii. y - intercept {-4/5}
iii. Vertical asymptote x = 5/2
10
iv.
5
10
5
– 10
5
5– 10
10
5
– 10
– 5
– 5
– 10
y
c. i. Root (0)
ii. y - intercept {0}
iii. Vertical asymptote x = 5/2
Horizontal asymptote y = 1/2
10
iv.
5
10
5
– 10
5
5– 10
10
5
– 10
– 5
– 5
– 10
y
d. i. Root (3)
ii. y - intercept {-3/2}
iii. Vertical asymptote x = -2
Horizontal asymptote y = 1
10
iv.
5
– 10
– 5
– 5
– 10
-161-
Unit 6: Polynomial and Rational Functions
y
e. i. No root
ii. y - intercept {-3/5}
iii. Vertical asymptote x = 1 and x = 5
10
iv.
5
10
5
– 10
5
5– 10
10
5
– 10
– 5
5
10 x
5
10 x
5
10 x
– 5
– 10
y
f. i. Root ( /3)
ii. y - intercept {-1/8}
iii. Vertical asymptote x = -4 and x = 4
2
10
iv.
5
10
5
– 10
5
5– 10
10
5
– 10
– 5
– 5
– 10
y
g. i. Root ( /2, /2)
ii. y - intercept {-9/4}
iii. Vertical asymptote x = -2 and x = 2
Horizontal asymptote y = 4
-3
3
10
iv.
5
– 10
– 5
– 5
– 10
Practice Questions 6-6:
Page 145
y
1.
y
a.
b.
x
x
-162-
Unit 6: Polynomial and Rational Functionsy
y
c.
d.
x
5– 5
5– 5
x
10
5
– 10
5
5– 10
10
5
2. a. f-1(x) = 3y - 2
3 x  15
3y
e. f-1(x) =
3.
a.
x9
b. f-1(x) =
x 2
c. f-1(x) =
d. f-1(x) =
1
x 1
f. f-1(x) = -x - 6
y
b.
5
10
5
10
5
– 10
5
5– 10
10
5
10
5
– 10
5
5– 10
10
5
5 x
– 5
– 10
– 5
5
10 x
5
10 x
5
10 x
– 5
– 5
y
10
c.
– 10
d.
y
10
5
5
10
5
– 10
5
5
10
15
10
5
– 10
5
5– 10
10
5
– 10
– 5
y
10 x
5
– 10
– 5
– 5
– 5
– 10
– 10
e. 10
f.
y
10
5
5
5
10
15
x
– 10
– 5
– 5
– 5
– 10
– 10
-163-
5– 5
5– 5
Unit 6: Polynomial
and Rational Functions
1
1
x
1
1
1
1
1
4. a.  f ( x )  
b.  f ( x )   2
c.  f ( x )  
d.  f ( x )  
2
3x  2
x 9
1 x
( x  2)
1
1
1
1
e.  f ( x )  
f.  f ( x )  
2
3 x  5
x  6
y
5. a. f(x):
5
Root (2/3)
y-intercept (-2)
(f
(x ))
1
:
5– 5
5– 5
No root
y-intercept (-0.5)
Vertical asymptote x = (2/3)
Horizontal asymptote y = 0
5 x
– 5
– 5
b. f(x):
Roots {-3, 3}
y-intercept (-9)
Minimum (0, -9)
(f
(x ))
1
(f
5– 10
10
5
:
5– 10
10
5
No root
y-intercept (-1/9)
Vertical asymptote x = -3, x = 3
Horizontal asymptote y = 0
c. f(x):
Root {2}
y-intercept (4)
Minimum (2, 0)
(x ))
1
y
5
5 x
– 5
– 5
y
10
5
5– 5
5– 5
:
No root
y-intercept (1/4)
Vertical asymptote x = 2
Horizontal asymptote y = 0
– 10
– 5
5
10 x
– 5
– 10
d. f(x):
y
5
Root {1}
No y-intercept
Vertical asymptote x = 0
Horizontal asymptote y = -1
(f
(x ))
1
:
No root
y-intercept (0)
Vertical asymptote x = 1
Horizontal asymptote y = -1
5 x
– 5
– 5
-164-
e. f(x):
No root
y-intercept -5
Maximum (0, -5)
(f
(x ))
Unit 6: Polynomial and Rational Functions
y
1
:
No root
y-intercept (-1/5)
Horizontal asymptote y = 0
0.25
0.2
0.15
0.1
10
5
– 10
5
– 15
5
10
0.05
– 4
– 2
– 0.05
2
4
x
– 0.1
– 0.15
– 0.2
– 0.25
y
f. f(x):
Root {-6}
y-intercept -6
(f
(x ))
10
5
1
:
No root
y-intercept (-1/6)
Vertical asymptote x = -6
Horizontal asymptote y = 0
– 15
– 10
x
– 5
– 5
– 10
-165-
Unit 6: Polynomial and Rational Functions
Unit 6 Review
1.
A polynomial function is a relation in which the domain, or, x-values are related to the
range, or y values, by a polynomial with positive integral exponents of x.
Polynomial functions are usually written;
f ( x )  ax n  bx n1  cx n2   z
Where a, b, c … are real numbers and n is a positive integer. Notice that the polynomial is
usually written with the exponents in descending order.
n = the degree of the polynomial
a = the leading coefficient
z = the constant term.
p( x)
q( x)
Where p(x) and q(x) are polynomial functions and q ( x )  0
2. A rational function is usually defined as; f ( x ) 
3. Synthetic Division is a compact version of long division of polynomials that may be used to
find quotients and remainders of the sort f(x) ÷ (x - a). Where a is an integer.
4. Remainder Theorem: For any polynomial function f(x), the value of f(x) at x = c (f(c)) is
equal to the remainder of f(x) ÷ (x - c).
5. Factor Theorem is a useful corollary of the remainder theorem:
For any function f(x), (x - a) is a factor iff (if and only if) f(a) = 0
6. Factor theorem and remainder theorem can be used to find values for f(a) where a is a real
number and to determine the roots for a polynomial function.
10
5
– 10
5
2– 4
4
2
7. The graphs of polynomial and rational functions have characteristic features which may be
used to describe them;
A polynomial Function
A rational Function
y
10
Maxima (singular - maximum)
Asymptote
5
– 4
– 2
2
4
x
– 5
x - intercepts (roots)
– 10
Minima (singular - minimum)
Asymptote
y - intercepts
-166-
Unit 6: Polynomial and Rational Functions
8. Several rules may be used to assist with analyzing functions to determine their
characteristics.
i. Fundamental Theorem of Algebra: A polynomial of degree n, where n is a positive
number, has exactly n roots. These roots may be multiple (the same number more than
once), real numbers, or complex numbers.
ii. Descartes’ Rule of Signs: The number of real roots of a polynomial can be determined
using the following rules; For the polynomial function f(x), with real coefficients.
1. The number of positive real roots is either the same as the number of its variations
in signs, or less than that by a multiple of 2.
2. The number of negative real roots is either the same as the number of variations in
signs of f(-x), or less than that by a multiple of 2.
iii. Rational Root Theorem: For the polynomial function f ( x )  ax n  bx n1  cx n2   z
where a, b, c…z are integers, then any real rational root can be written in the form kh ,
where h is a factor of z, and k is a factor of a.
9. Rational Functions have additional techniques for analyzing;
i. To find the roots for a rational function, we need to solve the numerator of the rational
function;
ii. A vertical asymptote occurs when the denominator is equal to 0. To find the vertical
asymptotes, find the roots for the polynomial in the denominator.
iii. Horizontal asymptotes occur where y-values become undefined for any value of x. To
find a horizontal asymptote, solve the equation for x (if possible)and set the denominator
equal to zero.
10. For any function, f(x), the inverse of the function, f-1(x) , can be found by interchanging x and
y in the equation for the function.
The graph of the inverse of a function is the reflection of the original curve in the
y = x line.
11. The reciprocal of a function; If f ( x ) is a function then the reciprocal of the function would
1
1
be
or  f ( x )  .
f ( x)
-167-
Unit 6: Polynomial and Rational Functions
Unit 6: Review Questions
1. Factor completely using synthetic division, factor and remainder theorems;
a. f(x) = x3 + x2 - 10x + 8
b. f(x) = x4 - 8x3 + 18x2 - 27
c. f(x) = x3 + x2 - 16x + 20
d. f(x) = x3 + 2x2 - 3x - 10
2. Sketch the graph of each of the polynomials in question 1. Identify all of the major
characteristics.
3. Identify the following as; rational function, polynomial function, or neither;
2x
a. f ( x )  x 2  9
b. f ( x )  2
3x  5tx  2
3x
c. f ( x )  3x 4  2 x3  7 x  9
d. f ( x ) 
2x  3
2
e. f ( x )  2  3
f. f ( x )  x12
x
4. Sketch each of the following. List the location of all intercepts and vertical asymptotes.
x
1
a. f ( x )  2
b. f ( x ) 
x 9
x6
1
x 3
c. f ( x )  2
d. f ( x )  2
x 9
x  2 x  15
4– 4
2
2
2– 4
4
2
5. For each of the following;
2– 4
4
2
2– 4
4
2
i. State the type (polynomial/rational)
and/or degree.
ii. List the sign of the leading coefficient or numerator.
iii. List the real roots and intercept.
iv. List the location of any maxima or minima
v. List the equation of any asymptotes.
y
a.
4
4
2
4– 4
2
2
2– 4
4
2
2
4– 4
2
2
2– 4
4
2
– 4 – 2
– 2
2
4
x
– 4 – 2
– 2
– 4
4
x
2
4
x
y
c.
4
4
2
2
– 4 – 2
– 2
2
– 4
y
c.
y
b.
2
4
x
– 4 – 2
– 2
– 4
– 4
-168-
Unit 6: Polynomial and Rational Functions
6. Sketch the inverse of each of the functions in question 5. Which of them are not functions?
7. Find the equation of the inverse of each of the following functions. Sketch the graph of the
inverse.
a. f(x) = 3x + 1
b. f(x) = -x + 2
3
d. f ( x ) 
x5
g. f(x) = 2x2 - 3
c. f(x) = 2(x - 3)2
e. f(x) = x3 - 8
8. Determine the reciprocal for each function in question 7. Sketch the graph of each reciprocal.
25
5
10
15
20
– 5
5– 10
10
5
9. Use Descartes rule to determine the number of positive and negative real roots of
f(x) = -x6 + 3x2 - 3x + 4.
Unit 6 Review Solutions
1. a. f(x) = (x + 4)(x - 1)(x - 2) b. f(x) = (x - 3)3(x + 1)2 c. f(x) = (x - 2)2(x + 5)
d. f(x) = (x - 2)(x + 2 + i)(x + 2 - i)
y
2.
a.
25
Roots: {-4, 1, 2}
20
y - intercept {8}
15
5– 30
5
10
15
20
25
2– 4
4
2
Maximum (-2.2, 24.2) From calculator.
10
Minimum (1.5, -1.5)
5
– 10
– 5
10 x
5
– 5
y
Roots: {-1, 3}
5
b.
– 4
– 2
– 5
– 10
2
4
x
y - intercept {-27}
Minimum (0, -27) From calculator.
– 15
– 20
– 25
– 30
-169-
Unit 6: Polynomial and Rational Functions
Roots {-5, 2}
y
c.
70
y-intercept {20}
60
50
4– 14
2
2
4
6
8
10
12
2– 4
4
2
Minimum (2, 0)
40
30
20
10
– 10– 8– 6– 4– 2
– 10
d.
2 4 6 8 10 x
y
Root {2}
4
2
– 4
– 2
– 2
y-intercept (-10)
2
4
x
Maximum (-2, -4)
– 4
– 6
Minimum (0.5, -11)
– 8
4– 4
2
2
2– 4
4
2
– 10
– 12
– 14
3. a. Neither
b. Rational
c. Polynomial
d. Neither
e. Rational
4. a. Vertical asymptotes x = -3, x = 3
y
4
Root {0}
y - intercept {0}
f. Rational
2
4– 4
2
2
2
4
6
8
– 4 – 2
– 2
2
x
4
– 4
b. Vertical asymptote x = 6
y
4
Horizontal asymptote y = 0
2
No root
2
– 2
y-intercept {-1/6}
– 4
-170-
4
6
8
x
Unit 6: Polynomial and Rational Functions
c. Vertical asymptote x = -3, x = 3
y
4
Horizontal asymptote y = 0
2
4– 4
2
2
2– 4
4
6
8
10
2
No root
– 4 – 2
– 2
y-intercept { /9}
-1
2
x
4
– 4
d. Vertical asymptotes x = -3, x = 5
y
4
Root {3}
2
y - intercept { /5}
1
– 4– 2
– 2
2
4
8 10 x
6
– 4
5. a. i. Polynomial: degree 6
ii. Leading Coefficient Positive
iii. Roots {-4, -1.5, 1, 2.7}
y - intercept {2}
iv. Maxima (0, 2), (4, 3)
Minima (-3, -2), (2, -1), (5, 2)
v. No Asymptotes
c. i. Rational:
ii. Numertor Positive
iii. Roots {-0.3}
2– 4
4
2
No y - intercept
2– 4
4
2
iv. No Maxima
No Minima
v. Vertical Asymptotes x = 0
Horizontal Asymptote y = 2
4– 4
2
2
2– 4
4
2
6
b. i. Polynomial: degree 3
ii. Leading Coefficient Positive
iii. Roots {-4}
y - intercept very large
iv. No Maxima
No Minima
v. No Asymptotes
d. i. Polynomial: degree 5
ii. Leading Coefficient Negative
iii. Roots {-4, -1}
y - intercept {-2}
iv. Maxima (-2, 1), (2, -1)
Minima (-4, 0), (0, -2)
v. No Asymptotes
y
y
a. not a function
b. function
4
4
2
2
4– 4
2
2
2– 4
4
2
– 4 – 2
– 2
2
4
x
4– 4
2
2
2– 4
4
2
– 4 – 2
– 2
y
x
2
4
x
y
d. Not a function
4
4
2
2
– 4 – 2
– 2
4
– 4
– 4
c. function
2
2
4
x
– 4 – 2
– 2
– 4
– 4
-171-
4– 4
2
2
2– 4
4
2
4– 4
2
2
2– 4
4
2
7. a. f (x1) 
Unit 6: Polynomial and Rational Functions
b. f(x1)   x  2
x 1

3 3
y
y
4
4
2
2
– 4 – 2
– 2
4– 4
2
2
2– 4
4
2
2
4
– 4 – 2
– 2
x
– 10
2
4
6
8
10
2
4
6
8
– 10
2
4
6
8
2
4
x
– 4
– 4
2x
3
2
c. f (x1) 
d. f ( x1) 
3  5x
x
y
y
4
2 4 6 8 10 x
– 10
– 8– 6– 4– 2
– 2
2
– 4
– 4 – 2
– 2
10
5
– 10
5
5– 10
10
5
2
4
x
– 6
4– 4
2
2
2– 4
4
2
– 8
– 4
– 10
e. f (x1)  3 x  8
f. f (x1) 
y
2x  6
2
10
y
4
5
2
– 10
– 5
10 x
5
– 5
4– 4
2
2
2– 4
4
2
– 10
8. a.  f ( x )  
1
– 4 – 2
– 2
4– 4
2
2
2– 4
4
2
4
x
2
4
x
– 4
1
3x  1
b.  f ( x )  
1
1
x  2
y
y
4
4
2
2
– 4 – 2
– 2
2
2
4
x
– 4 – 2
– 2
– 4
– 4
-172-
4– 4
2
2
2– 4
4
2
4– 4
2
2
2– 4
4
2
Unit 6: Polynomial and Rational Functions
c.  f ( x )  
1
1
2( x  3) 2
d.  f ( x )  
1
x 5
3
y
y
4
4
2
2
– 4 – 2
– 2
4– 4
2
2
2– 4
4
2
2
4
– 4 – 2
– 2
x
4– 4
2
2
2– 4
4
2
1
1
x 8
f.  f ( x )  
1
3
x
2
4
x
1
2x  3
2
y
y
4
4
2
2
– 4 – 2
– 2
4
– 4
– 4
e.  f ( x )  
2
2
4
x
– 4 – 2
– 2
– 4
– 4
9. 3 or 1 Positive real roots, 1 negative real root.
-173-