Mathematics B30
Module 2
Lesson 8
Mathematics B30
Rational, Inverse and Reciprocal Functions
179
Lesson 8
Mathematics B30
180
Lesson 8
Rational, Inverse and Reciprocal Functions
Introduction
The focus in Lesson 7 was the study of polynomial functions. In this lesson the study of
polynomial functions will be extended further to the study of rational functions, inverse
functions and reciprocal functions.
The same procedure is necessary for analyzing the three types of functions. Each function
has its own characteristics that will be used to sketch the graph of the function.
Once again, a graphing calculator can be used as a tool for analyzing the functions. As in
the other lessons, it is very important to understand the method before applying the
technology.
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Lesson 8
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Lesson 8
Objectives
After completing this lesson, you will be able to
•
define and illustrate rational functions.
•
sketch the graphs of rational functions with integral coefficients,
using the graphing calculator.
•
analyze the characteristics of the graphs of rational functions and
to identify the "zeros" of these graphs.
•
define, determine and sketch the graph of the inverse of a
function, where it exists.
•
define, determine and sketch the graph of the reciprocal of a function.
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Lesson 8
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Lesson 8
8.1 Rational Functions
Rational Function
A rational function is the quotient of two polynomials, p(x)
and q(x), in the form
p x 
where q x   0 .
f x  
qx 
In Lesson 7, it was shown how to analyze a polynomial function and then to graph the
function by determining the major characteristics or by using a graphing calculator.
In this lesson the same procedure will be applied to rational functions.
The first step in analyzing a rational function is to determine the zeros of the function or
the values that make the function equal to zero. These values are also referred to as the
x-intercepts.
The zeros of a rational function can be found by equating the rational function to
zero. In a rational expression, the expression is equal to zero when the numerator
is equal to zero.
Example 1
Find the zeros of the rational function, f  x  
x 3
.
x2
Solution:
f x  
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x 3
x2
Lesson 8
Equate the function to zero.
x 3
x2
0  x 3
x3
0
The numerator
will be equal to
zero.
The numerator of a rational function is a polynomial. Factor the polynomial first to find
the zeros of the function.
Example 2
2
1
Find the zeros of the rational function, f  x   x
.
2x
Solution:
2
1
f x   x
2x
Equate the function to zero.
Set the numerator to zero.
0
x2 1
2x
2
x 1  0
 x  1  x  1   0
x  1 x  1
The zeros of the function are x  1, 1 .
The denominator of a rational function also plays a major role in determining the shape of
the graph of the function.
In the same way that a rational number or expression cannot have a denominator that is
equal to zero, a rational function also cannot have the denominator equal to zero.
Any number that makes the value of the denominator equal to zero is non-permissible.
Another name for these values are excluded values.
The denominator of a rational function is in the form of a polynomial. The polynomial
must first be factored to determine its zeros. Any values that make any of the factors in
the denominator equal to zero are non-permissible.
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Lesson 8
Example 3
State the non-permissible values in the rational function f  x  
x 3
.
2 x 2  18 x
Solution:
Set the denominator equal to zero.
Set each factor equal to zero.
2 x 2  18 x  0
2 x( x  9)  0
2x  0
x 0
x 9  0
x 9
The non-permissible values for this rational function are x  0, 9 .
The domain of this rational function is all real numbers except x  0, 9 .
D  x  x  , x  0 , 9
Example 4
State the non-permissible values for the rational function, f  x  
4
.
x 2
Solution:
Set the denominator equal to zero.
x 2 0
x 2
The non-permissible value for this rational function is x  2 .
The domain of this rational function is all real numbers except x  2 .
D  x x  , x  2
A graph of the function f ( x ) 
near the excluded value, x = 2.
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shows what happens to f x  for values of x which are
x 2
187
Lesson 8
The following table shows the ordered pairs that satisfy this equation.
x
f(x)
1

4
3
0
1
2
4
1
1
2
2
2
8
undefined
1
2
3
8
4
4
5
2
4
3
The graph of this equation is:
f(x)
x
asymptote (x = 2)
What happens when x is equal to 2?
The vertical line, x = 2 , is called an asymptote. There is no point on this line that will
satisfy the function. Both of the curved lines will go increasingly closer to this line, but will
never touch it.
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Lesson 8
A graphing calculator can be used to graph this function.
Use the following key stroke pattern on the TI-83 graphing calculator.
CLEAR
•
Y=
4 ÷ ( X,T,  ,n – 2 )
ENTER
GRAPH
On the calculator screen the asymptote at x  2 appears as a solid vertical line.
Choose the TRACE function key and follow the curved lines to see what happens as
the value of x gets close to 2.
The two characteristics of a rational function that are helpful in determining the
graph of the function are the:
•
zeros of the function
•
asymptotes
The asymptotes in this case are vertical asymptotes and occur when the line with the
equation x = a is undefined.
Once the characteristics of a rational function have been determined, test points near the
asymptote(s) will give a more accurate picture of the graph.
Example 5
Analyze and sketch the graph of the rational function, f  x  
8
.
x 4
2
Solution:
•
There are no zeros of the function because there is no variable in the numerator.
Therefore the function does not cross the x-axis.
•
The asymptotes occur at x  2, 2 because the denominator is undefined at these
values.
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Lesson 8
A table of values is used to show the test points as the values of x approach the vertical
asymptotes.
x
4
3
2
f(x)
3
4
8
5
32
9
1
2
2
Undefined
1
2 1
0
32
8

7
3
2
1

Sketch the graph of the function, f  x  
1

8
3
1
1
2

32 Undefined
7
8
.
x 4
2
1
2
3
4
32
9
8
5
3
4
2
y
2
x
x  2
asymptotes
x2
Use the following key stroke pattern on the TI-83 Graphing
calculator:
CLEAR Y=
GRAPH
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÷
( X,T,  ,n ^ 2 – 4 ) ENTER
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Lesson 8
Exercise 8.1
1.
List the major characteristics of the rational functions.
a.
y
x
b.
y
x
c.
y
x
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Lesson 8
2.
3.
State the zeros and vertical asymptotes of the following rational functions.
a.
f x  
1
x
b.
f x  
x 3
x 1
c.
f x  
3
x 9
d.
f x  
4
x  x  30
e.
f x  
x
x 5
f.
f x  
1
x 3
g.
f x  
4x
x 2
h.
f x  
x2  6x 7
x 1
2
2
Sketch the graph of each function in Question #2.
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Lesson 8
8.2 Inverse of a Function
In this section you will learn the definition of the inverse of a function. You will also be
able to determine the equation of the inverse of any given function. The concept of the
inverse of a function will be used in developing logarithms.
Inverse of a Function
If f represents a function, then the inverse of f is represented by the symbol f 1 .
The graph of f 1 is the reflection of the graph of f about the line y  x .
Example 1
Sketch the inverse of the following graph y  f x  .
y
x
Solution:
y
Draw the line y  x .
y = f (x )
Sketch the graph y  f 1  x  .
y= x
x
–1
y = f ( x)
•
Note that f is a function, but f 1 is not a function in this case since the vertical line
test fails.
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Lesson 8
The  1 in the inverse function notation is not an exponent. f
1
or f  1 .
f x 
1
 x  does not mean
Example 2
Sketch the inverse of the following graph, y  f x  .
y
x
Solution:
y
Draw the line y  x .
y= f
–1
(x )
y= x
Sketch the graph y  f 1  x  .
y = f (x )
x
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Lesson 8
Activity 8.2
Hand this activity in with Assignment 4.
The graphs of four functions are given. Draw the line y  x and then sketch the
graph of the inverse of the function.
y
y
f
g
x
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Lesson 8
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Lesson 8
The equation of the inverse of a function can be found by studying the graph of the points
of the function.
y
(–2, 2)
(2, 3)
y= x
(3, 2)
x
(2, –2)
If a point is reflected about the line y  x then its coordinates are interchanged. This is
shown on the graph.
A point x, y  is on the graph of f if and only if  y, x  is on the graph of f 1 .
The inverse of a function is obtained by interchanging the components of the
ordered pairs of the function. That is, if x, y  is in the function f x  , then (y, x) is
in f
1
x  .
The equation of the inverse of a function can be found if the equation of the function is
given.
Given the equation y  2 x  1, a partial table of values is:
x
y
0
1
1
3
2
5
Interchange the x and y variables in the equation and then solve for y.
y  2x 1
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becomes

x  2y 1
2 y  x 1
x 1
y
2
197
Lesson 8
A partial table of values of the new equation is shown. The ordered pairs are interchanged
when comparing this table with the table of values for the original equation.
y
x 1
2
x
y
1
0
3
1
5
2
Inverse of a Function
f x  and f 1  x  are inverses if for every
x, y   f x  , there is a  y, x   f 1 x  .
Example 3
Determine the equation of the inverse of the function f x   4 x  1 . Sketch both the
function and its inverse.
Solution:
y  4x 1
Interchange x with y.
x  4 y 1
Solve for y.
4 y  x 1
x 1
y
4
Equation of the inverse of
f x  4 x 1 .
 
Make a table of values or use a graphing calculator to graph both f x  and f
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1
x  .
Lesson 8
y
f(x) = 4 x – 1
f –1 (x ) = x + 1
4
x
Use the following key stroke pattern to graph the function,
f x   4 x  1 .
CLEAR
Y=
4
X,T,  ,n
–
1
ENTER
GRAPH
To graph the inverse, f 1  x  of this function, use the DRAW function
key. To access this, use the following key stroke pattern.
•
•
•
2nd
PRGM
Arrow down to 8:DrawInv ENTER
( 4 X,T,  ,n – 1 ) ENTER
Both graphs will appear on the screen.
Example 4
Find the equation of the inverse of the function, f  x   1 
1
. Sketch the graph of
x
both f and f 1 .
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Lesson 8
Solution:
1
x
1
x 1
y
1
1 x
y
1  1  x  y
1
y
1 x
1
f 1  x  
1 x
y 1
Interchange variables.
Solve for y.
Multiply by y.
y
f –1(x )
f ( x)
f (x )
x
f –1(x )
The key stroke pattern on the T1-83 Graphing Calculator is
CLEAR
Y=
1
–
( 1 ÷ X,T,  ,n )
ENTER
2nd PRGM  (to) 8:DrawInv
(1 – ( 1 ÷ X,T,  n ) ) ENTER
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GRAPH
ENTER
Lesson 8
Example 5
Find the inverse of the function f  x   2 x 2  5 and sketch the graph of both
f and f 1 .
Solution:
Interchange variables.
Solve for y.
y  2x2  5
x  2 y2  5
2 y2  x  5
x 5
y2 
2
x 5
y
2
y
f (x )
x
f –1 (x )
Note that f
1
 x  is not a function in this case.
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Lesson 8
Use the following key stroke pattern:
CLEAR
Y=
2nd PRGM
ENTER
2
X,T,  ,n
ENTER
^
2
–
5
( 2 X,T,  ,n ^ 2
ENTER
–
GRAPH
5)
Exercise 8.2
1.
a.
b.
c.
d.
If f   2, 3 ,  1,  1, 0, 4 , 1, 5 , 2,  7  then find f 1 .
Sketch the graph of the inverse and use the vertical line test to determine if
the inverse is a function. f  2, 5 ,  1, 5 ,  1, 4 , 2,  4 
If f  2, 5 , 2, 4 , 3, 2 , 3, 1 then find f 1 .
If f  x   2 x 2  1 ,
i)
ii)
iii)
iv)
v)
2.
For each of the following functions, write the equation of the inverse.
a.
b.
c.
d.
e.
f.
g.
3.
is 0,  1 in f ?
is  1, 0  is f 1 ?
is (1, 1) in f 1 ?
is (9, 2) in f 1 ?
is (199, 10) in f 1 ?
f x  
1
x  4 
2
f x   2 x  5
f x   2 x 2  3
1x
f x  
x
f x   4 x  3
f x   3 x 2  6
f x   x 2  4 x
(Hint: Use the method of completing the square.)
Sketch each function and its inverse from Question #2.
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Lesson 8
8.3 Reciprocal of a Function
Reciprocal of a Function
If f(x) represents a function, then
1
is the reciprocal function.
f x 
Any relation that is a function has a reciprocal function. The relation can be linear,
quadratic, or any other type of polynomial.
Example 1
Determine the reciprocal function of f x   2 x  3 . Make a table of values and then
1
graph the relation determined by y and the graph of the relation determined by .
y
Solution:
y  2x  3
x
2
1
0
1
3
2
2
3
4
y
7
5
3
1
0
1
3
5
1
y

1
3
1
Undefined
1
1
3
1
5
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7

1
5

203
Lesson 8
Draw the graphs of the two functions.
y
1
f (x)
x
1
f (x )
f ( x)
Ver t ical Asym pt ot e
•
•
The root or zero of the function f(x) represents a critical point for the reciprocal
1
function
.
f x 
1
The root of f x  becomes the value of the vertical asymptote for
.
f x 
Another type of asymptote is a horizontal asymptote. This occurs when the value of y is
undefined.
From the graph in Example 1, the reciprocal function, y 
1
, is undefined when y  0 .
2x  3
y
1
f (x )
1
f (x )
horizon t al asympt ot e
f ( x)
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Lesson 8
The horizontal asymptote can also be determined algebraically by solving the equation in
terms of x, and finding the y values that make the equation undefined.
1
2x  3
2 xy  3 y  1
y
2 xy  3 y  1
x
3y 1
2y

The equation will be undefined when 2 y  0 , or in other words when y  0 .
Example 2
Given the function, f  x  
x
, sketch the graph of the function and the reciprocal
x2
function. Determine any critical points and vertical or horizontal asymptotes for
each.
Solution:
Determine the characteristics of the function, y 
x
.
x2
Roots of the function ( x-intercepts):
•
Set the numerator to zero.
x 0
Vertical asymptote:
•
Set the denominator to zero.
Horizontal asymptote:
•
Solve the equation for x.
x2 0
x  2
x
x2
xy  2 y  x
y
xy  x  2 y
x  y  1   2 y
2y
x
y 1
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Lesson 8
•
Set the denominator to zero.
y 1  0
y 1
Use test points to determine where the graph is drawn around the asymptotes.
•
•
•
Substitute the value x  4 into the equation. The value for y is y  2 .
This value is above the above the horizontal asymptote and to the left of the vertical
asymptote.
This is where the graph is drawn.
Sketch the graph of the function.
y
H or izont al
Asym pt ot e
x
Ver t ical
Asym pt ot e
Determine the characteristics of the reciprocal function, y 
Roots of the function ( x-intercepts):
•
Set the numerator to zero.
x2 0
x  2
Vertical asymptote:
•
Set the denominator to zero.
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x2
.
x
x 0
206
Lesson 8
Horizontal asymptote:
•
Solve the equation for x.
x2
x
xy  x  2
y
xy  x  2
x y  1  2
x
•
Set the denominator to zero.
2
y 1
y 1  0
y 1
Use test points to determine where the graph is drawn around the asymptotes.
•
•
•
Substitute the value x  2 into the equation. The value for y is y  2 .
This value is above the above the horizontal asymptote and to the right of the
vertical asymptote.
This is where the graph is drawn.
Sketch the graph of the function.
y
H or izont al
Asym pt ot e
x
Ver t ical
Asym pt ot e
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Lesson 8
A graphing calculator can be used to determine the sketch of the graph
of a function or a reciprocal function.
Use the following key stroke pattern for the TI-83 Graphing Calculator to
x
analyze the graph of the function, y 
.
x2
CLEAR Y= X,T,  ,n ÷ ( X,T,  ,n + 2 ) ENTER GRAPH
Use the following key stroke pattern to sketch the graph of the function,
x2
y
.
x
CLEAR Y= ( X,T,  ,n + 2 ) ÷ X,T,  ,n ENTER GRAPH
These two functions can be graphed on the same axis using Y1 for the first
equation and Y2 for the reciprocal equation.
Example 3
Analyze and sketch the graph of the reciprocal function of f  x   x 3  2 x 2 .
Solution:
Determine the reciprocal function.
y
1
x  2x2
3
Roots of the function ( x-intercepts):
•
There are no roots since the numerator does not contain a variable.
Vertical asymptotes:
•
Set the denominator to zero.
x3  2x2  0
x 2 x  2   0
x 0 x 2
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Lesson 8
Horizontal asymptote:
•
Solve the equation for x.
1
2
x  2x
y x 3  2 x 2   1
y
3
2
x  2x 
•
3
1
y
It is not easy to solve for x in this case. Another way of finding horizontal
asymptotes is to substitute very large test points into the equation. For example, if
1
x = 500 then y 
which is a number very close to zero. And if x were
124 500 000
larger, then y would be still closer to zero. Therefore, y = 0 is the horizontal
asymptote.
Use test points to determine where the graph is drawn around the asymptotes.
•
•
•
Substitute values into the equation to determine ordered pairs.
Some important ordered pairs are:
1 
 1

1,  1
 3, 
 2,  
16 
 9

These ordered pairs indicate if the graph of the function is above or below the
horizontal asymptote.
Sketch the graph of the function.
y
H or izont al
Asym pt ot e
x
Ver t ical
Asym pt ot es
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Lesson 8
Use the following key stroke pattern:
CLEAR Y= 1 (÷) ( X,T,  ,n ^ 3 – 2 X,T,  ,n ^ 2 )
ENTER GRAPH
Exercise 8.3
1.
2.
State the reciprocal function.
a.
f x   3 x  1
b.
f  x   x 2  4 x  12
c.
f x  
d.
f x   x 3  3 x 2  3 x  1
e.
f x  
x 5
x 3
x
x4
State the vertical and horizontal asymptotes of the given functions.
a.
y
3
x 7
b.
y
4x
x 2
c.
y
1 x
x
d.
y
5x
x  3 x  18
e.
y
x
3x  4
f.
y
3  5x
x 1
Mathematics B30
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Lesson 8
Determine the characteristics of the functions, f x  and the reciprocal functions
1
in Question #1.
f x 
3.
The characteristics are the x-intercepts, vertical asymptotes and horizontal
asymptotes.
4.
Graph both the functions and the reciprocal functions in Question #1.
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Lesson 8
Self Evaluation
1.
Analyze the following functions. Determine the inverse and the reciprocal function
for each. Sketch the graphs of both the inverse and the reciprocal function.
a.
f x  
1
x
b.
f x  
x
x2
c.
f x   x 2  1
d.
f x  
e.
f x   x 2  5 x  4
Mathematics B30
x 3
x 1
212
Lesson 8
Summary – Lesson 8
•
Create a summary of this lesson to assist you come examination time.
•
Each summary is to be sent in with the assignment to be evaluated.
•
Items to include in a summary:
•
definitions
•
formulas
•
calculator “shortcuts”
Mathematics B30
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Lesson 8
Mathematics B30
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Lesson 8
Answers to Exercises
Exercise 8.1 1.
a.
No x-intercepts
Vertical asymptote: x  2
b.
No x-intercepts
Vertical asymptotes: x  3 , 3
c.
x-intercepts: x  1, 1
Vertical asymptote: x  0
2.
Mathematics B30
a.
Zeros: none
Vertical asymptote: x  0
b.
Zeros: x  3
Vertical asymptote: x  1
c.
Zeros: none
Vertical asymptote: x  3 , 3
d.
Zeros: none
Vertical asymptotes: x  5, 6
e.
Zeros: x  0
Vertical asymptote: x  5
f.
Zeros: none
Vertical asymptote: x  3
g.
Zeros: x  0
Vertical asymptote: x  2
h.
Zeros: x  1,  7
Vertical asymptote: x  1
215
Lesson 8
3.
a.
b.
y
y
x
c.
x
d.
y
y
x
x
e.
f.
y
y
x
x
g.
h.
y
y
x
x
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Lesson 8
Exercise 8.2 1.
a.
1
f
 3,  2 ,  1,  1, 4 , 0 , 5, 1,  7, 2 
y
b.
The inverse is not a function.
x
2.
1
 5, 2 , 4 , 2 , 2, 3 , 1, 3 
c.
f
d.
i) yes
a.
y  2x  4
x 5
y
2
x 3
y
2
1
y
1 x
3x
y
4
x 6
y
3
b.
c.
d.
e.
f.
g.
ii) yes
iii) yes
iv) no
v) yes
x  y2  4 y
x  y 2  4 y  4   4
x  4  y  2
2
 x4  y2
y  2  x  4
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Lesson 8
3.
a.
y
b.
y
f (x )
f (x )
x
x
f –1(x )
c.
f –1(x )
y
d.
y
f ( x)
f –1(x)
f –1(x)
x
x
f (x)
e.
y
f.
y
f(x)
f ( x)
f –1(x)
x
x
f –1(x)
Mathematics B30
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Lesson 8
y
g.
f(x)
f –1(x)
x
Exercise 8.3 1.
2.
Mathematics B30
a.
1
1

f x  3 x  1
b.
1
1
 2
f  x  x  4 x  12
c.
1
x 3

f x  x  5
d.
1
1
 3
2
f x  x  3 x  3 x  1
e.
1
x4

f x 
x
a.
Vertical asymptote: x  7
Horizontal asymptote: y  0
b.
Vertical asymptote: x  2
Horizontal asymptote: y  4
c.
Vertical asymptote: x  0
Horizontal asymptote: y  1
d.
Vertical asymptote: x  6 , 3
Horizontal asymptote: y  0
219
Lesson 8
e.
Vertical asymptote: x  1
1
3
Horizontal asymptote: y 
3.
1
3
f.
Vertical asymptote: x  1
Horizontal asymptote: y  5
a.
f x 
x-intercepts: x  
1
f x 
x-intercepts: none
1
3
Vertical asymptote: none
Horizontal asymptote: none
1
3
Horizontal asymptote: y  0
Vertical asymptote: x  
b.
f x 
x-intercept: x  2, 6
Vertical asymptote: none
Horizontal asymptote: none
1
f x 
x-intercept: none
Vertical asymptote: x  2, 6
Horizontal asymptote: y  0
c.
f x 
x-intercept: x  5
Vertical asymptote: x  3
Horizontal asymptote: y  1
1
f x 
x-intercept: x  3
Vertical asymptote: x  5
Horizontal asymptote: y  1
Mathematics B30
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Lesson 8
d.
f x 
x-intercept: x  1
Vertical asymptote: none
Horizontal asymptote: none
1
f x 
x-intercept: none
Vertical asymptote: x  1
Horizontal asymptote: y  0
e.
f x 
x-intercept: x  0
Vertical asymptote: x  4
Horizontal asymptote: y  1
1
f x 
x-intercept: x  4
Vertical asymptote: x  0
Horizontal asymptote: y  1
4.
a.
y
x
Mathematics B30
221
Lesson 8
b.
y
1
f ( x)
1
f ( x)
x
1
f (x )
f (x)
c.
y
x
Mathematics B30
222
Lesson 8
d.
y
f (x )
1
f (x )
x
1
f ( x)
e.
y
f (x )
1
f (x )
f ( x)
1
f ( x)
Mathematics B30
223
x
Lesson 8
Answers to Self Evaluation
1.
a.
f
1
y
x   1
1
f (x )
x
x-intercept: none
Vertical asymptote: x = 0
Horizontal asymptote: y = 0
f –1(x )
x
1
x
f x 
x-intercept: x = 0
f –1(x )
Vertical asymptote: none
Horizontal asymptote: none
b.
f
1
x    2 x
y
x 1
x-intercept: x = 0
Vertical asymptote: x = 1
Horizontal asymptote: y  2
1
f (x )
1
f (x )
x
–1
f (x )
1
x2

f x 
x
f –1(x )
x-intercept: x  2
Vertical asymptote: x = 0
Horizontal asymptote: y = 1
Mathematics B30
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Lesson 8
c.
f
1
x   
x 1
x-intercept: x =  1
Vertical asymptote: none
Horizontal asymptote: none
1
1
 2
f x  x  1
x-intercept: none
Vertical asymptote: x  1 , x = 1
Horizontal asymptote: y = 0
d.
f
1
x  
x3
x 1
y
f –1(x )
x-intercept: x  3
1
f (x )
Vertical asymptote: x = 1
Horizontal asymptote: y = 1
1
x 1

f x  x  3
x
x-intercept: x = 1
Vertical asymptote: x  3
Horizontal asymptote: y = 1
1
f (x )
Mathematics B30
225
f –1(x )
Lesson 8
e.
5
9
f 1  x     x 
2
4
x-intercept: x = 4
Vertical asymptote: none
Horizontal asymptote: none
1
1
 2
f x  x  5 x  4
x-intercept: none
Vertical asymptote: x = 4, x = 1
Horizontal asymptote: y = 0
Mathematics B30
226
Lesson 8