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The module was a answer key at the last page.. This module is Module 9 of General Mathematics
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General Mathematics
Alternative Delivery Mode
Quarter 1 – Module 9 : Intercepts, Zeroes and Asymptotes of Rational Functions
First Edition, 2020
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Introductory Message
For the facilitator:
Welcome to the General Mathematics Alternative Delivery Mode (ADM) Module on
Determining the Intercepts, Zeroes and Asymptotes of Rational Functions!
This module was collaboratively designed, developed and reviewed by educators from
public institutions to assist you, the teacher or facilitator in helping the learners
meet the standards set by the K to 12 Curriculum while overcoming their personal,
social, and economic constraints in schooling.
This learning resource hopes to engage the learners into guided and independent
learning activities at their own pace and time. Furthermore, this also aims to help
learners acquire the needed 21st century skills while taking into consideration their
needs and circumstances.
In addition to the material in the main text, you will also see this box in the body of
the module:
As a facilitator you are expected to orient the learners on how to use this module.
You also need to keep track of the learners' progress while allowing them to manage
their own learning. Furthermore, you are expected to encourage and assist the
learners as they do the tasks included in the module.
For the learner:
Welcome to the General Mathematics Alternative Delivery Mode (ADM) Module on
Determining the Intercepts, Zeroes and Asymptotes of Rational Functions!
The hand is one of the most symbolized part of the human body. It is often used to
depict skill, action and purpose. Through our hands we may learn, create and
accomplish. Hence, the hand in this learning resource signifies that you as a learner
is capable and empowered to successfully achieve the relevant competencies and
skills at your own pace and time. Your academic success lies in your own hands!
This module was designed to provide you with fun and meaningful opportunities for
guided and independent learning at your own pace and time. You will be enabled to
process the contents of the learning resource while being an active learner.
Notes to the Teacher
This contains helpful tips or strategies that
will help you in guiding the learners.
At the end of this module you will also find:
The following are some reminders in using this module:
module. Use a separate sheet of paper in answering the exercises.
included in the module.
If you encounter any difficulty in answering the tasks in this module, do not
hesitate to consult your teacher or facilitator. Always bear in mind that you are
not alone.
We hope that through this material, you will experience meaningful learning and
gain deep understanding of the relevant competencies. You can do it!
References This is a list of all sources used in developing
this module.
I. Choose the letter of the best answer. Write the chosen letter on a separate
sheet of paper.
a. Range
b. Intercept
c. Domain
d. Zeroes
𝑥− 3
𝑥+ 3
a. All real numbers
b. All real numbers except – 3
c. All real numbers except 3
d. Cannot be determined
𝑥− 1
𝑥
a. x = - 1
b. x = 0
c. x = 1
d. All real numbers
graph of the function.
a. Asymptote
b. x – intercepts
c. y – intercepts
d. Range
a. Range
b. Intercept
c. Domain
d. Zeroes
1
𝑥
a. R =
b. R =
c. R = {𝑦|𝑦 ≠ 1 }
d. R =
a. A rational function is a quotient of functions.
b. Asymptotes are a common characteristic of rational functions.
c. An asymptote is a line that a graph approaches, but does not touch.
d. All of the above.
degree of the leading coefficient of the denominator of a rational
function, which of the following statements has to be true?
a. The graph has no asymptote
b. The graph of the function has slant asymptote
c. The graph of the function has a horizontal asymptote
d. None of the above
𝑥+ 5
3 𝑥
2
a. y = 3
b. y = 0
c. y = - 2
d. y = - 3
3 𝑥+ 1
𝑥− 5
a. x = 5
b. x = 3
c. x = 1
d. x = 0
𝑥
2
− 3 𝑥
𝑥+ 3
a. y = 3x
b. y = x – 6
c. y = x - 3
d. y = 3x + 6
the statement is ____________.
a. Always true
b. Sometimes true
c. Never true
d. Cannot be determined
3
3 +𝑥
a. does not exist
b. approaching at x = 3
c. approaching at y = - 3
d. approaching at y = 0
Lesson
Intercepts, Zeroes, and Asymptotes of
Rational Functions
In the previous lesson, you learned how to find domain and range of a rational
function. In this particular lesson, determining intercepts, zeroes and asymptotes of
rational functions will be done. Knowing fully the concept of the different properties
of rational function will be your guide to easily determine the behavior of a rational
function and it will prepare you for the next topic which is about graphing rational
function.
What’s In
Let’s recall first what you have learned from the previous lesson by answering the
following questions:
A. Which of the following is an example of rational function?
3 𝑥
2
𝑥− 1
𝑥
3
8
3
1
3 𝑥− 1
B. Find the domain and range of the functions.
𝑥
𝑥+ 3
3
𝑥− 4
𝑥+ 1
𝑥
2
− 1
Let us see if you got the correct answer in the activity, if your answer in question A
is number 1, you got it right you have a clear understanding of the concept of rational
function but if you are incorrect allow me to help you recall what a rational function
is, when two polynomial functions are expressed as a quotient and can be written in
the form 𝑓
𝑝(𝑥)
𝑞(𝑥)
and q(x) is a not the zero function it is called a rational function.
Numbers 2 and 3 are not examples of rational function, it is a rational equation and
rational inequality, respectively. Number 1 is written as the quotient of two
polynomial functions, so it is a rational function.
For activity B, let us review the meaning of domain and range of the function.
Domain is the set of first coordinates of a relation and it is the value of x that will
not make the denominator of the function equal to zero while Range is the set of
second coordinates. To determine the domain of rational function, simply equate the
denominator to zero and then solve for x, this value should be avoided so that the
function will not give an undefined or a meaningless function. Example find the
domain of F(x) =
x
x+ 3
, equating the denominator to zero, we have x + 3 = 0, so the
value of x = - 3, so the domain of the function are all real numbers except - 3 remember
we will avoid value/s that will make our denominator equal to zero, so if we will
substitute - 3 to our x in the denominator it will result to 0 and it will give us an
undefined function. In notation, D= (- ∞, −𝟑) ∪ (−𝟑, ∞)
To find the range of the function, change f(x) to y then, solve for x; remember range
are real values of y that will make a real value for the function. For example, find the
range of F(x) =
𝑥
𝑥+ 3
Changing F(x) to y, the new function is y =
𝑥
𝑥+ 3
By doing cross multiplication we have y(x+3) = x
Distributing y we now have xy + 3y = x
Simplifying the equation will give xy – x = 3y
Factoring the left side of the equation x(y – 1) = 3y
Dividing the equation by (y – 1)
𝑥(𝑦− 1 )
(𝑦− 1 )
3 𝑦
(𝑦− 1 )
Removing common factor, the value of x 𝑥 =
3 𝑦
𝑦− 1
Since we are looking for the value of y that will give a real value for the function so
we need to find value/s for y that will not make the denominator equal to 0.
Equating the denominator to zero y – 1 = 0
So, y = 1.
The range of the function F(x) =
𝑥
𝑥+ 3
is all real values of y except 1. In notation,
The following are the answers to Activity B
Activity
I – Connect Mo!
Connect the given statement/phrase in column A with the answer in column
B to complete the statement/phrase in column A. Write the letter of your
answer in a separate sheet of paper.
of a rational function …
of a function …
is also …
of a function … intersection of its
graph and an axis
, where g(x)
& h(x) are polynomials
What’s New
How was the activity? I believed that you connected it right. So, in this lesson, you
will know how to identify intercepts, zeroes and asymptotes of rational function.
What is It
INTERCEPTS AND ZEROES OF RATIONAL FUNCTIONS
The intercepts of the graph of a rational function are the points of intersection of its
graph and an axis.
The y-intercept of the graph of a rational function r(x) if it exists, occurs at r(0),
provided that r(x) is defined at x = 0. To find y-intercept simply evaluate the function
at x = 0.
The x-intercept of the graph of a rational function r(x), if it exists, occurs at the zeros
of the numerator that are not zeros of the denominators. To find x – intercept equate
the function to 0.
The zeroes of a function are the values of x which make the function zero. The
numbered zeroes are also x-intercepts of the graph of the function.
a. f(x) =
3 −𝑥
𝑥+ 1
b. f(x) =
3 𝑥
𝑥+ 3
c. f(x) =
𝑥
2
− 3 𝑥+ 2
𝑥
2
− 4
y-intercept
x-intercept
Figure 1. x and y intercepts using GeoGebra
zero of the
function
3 𝑥
𝑥+ 3
To find the x – intercept, simply equate the numerator to 0,
0 = 3x Equate the numerator to 0.
3x = 0 By Symmetric Property of Equality.
3 𝑥
3
=
0
3
Simplifying the fraction by multiplying
both sides by 1/3.
x = 0
So, the x – intercept is 0 or (0, 0).
To find the y – intercept, change the x value of the function to 0.
3 𝑥
𝑥+ 3
Substitute 0 to x values of the function.
3 ( 0 )
0 + 3
Simplifying the fraction.
0
3
= 0 The value of 𝑓(𝑥) or y – intercept.
So, the y – intercept is 0 or (0, 0).
𝑥
2
− 3 𝑥+ 2
𝑥
2
− 4
2
− 3 𝑥 + 2 = 0 Equate the numerator to 0.
(x – 2) (x – 1) = 0 By factoring.
x – 2 = 0 x – 1 = 0 Solve for x, by Zero product property.
x = 2 x = 1
So, the x – intercepts are x = 2 and x = 1. But by looking at the denominator
of the original function if we substitute 2 to the value of x,
x
2
2
The denominator will become 0, the function becomes meaningless.
So, we will only accept x – intercept at x = 1 or (1, 0).
To find the y – intercept:
𝑥
2
− 3 𝑥+ 2
𝑥
2
− 4
, change the x value of the function to 0.
( 0 )
2
− 3 ( 0 )+ 2
( 0 )
2
− 4
Simplify the fraction.
f(x) =
2
− 4
Reduce the fraction to lowest term.
1
2
The value of f(x) or y.
1
2
1
2
a. g(x) =
𝑥− 2
𝑥+ 6
b. H(x) =
𝑥− 3
𝑥
2
− 9
c. G(x) =
𝑥
2
+𝑥− 2
𝑥
2
− 4
To find the zeroes of a rational function, equate the function to 0 or solve for the x –
intercept of the function by equating the numerator to 0.
a. g(x) =
𝑥− 2
𝑥+ 6
x – 2 = 0 Equate the numerator to 0.
x = 2 Solve for x.
Thus, the zero of g(x) is 2.
b. 𝐻
𝑥− 3
𝑥
2
− 9
𝑥− 3
𝑥
2
− 9
Simplify by factoring the denominator.
𝑥− 3
(𝑥− 3 )(𝑥+ 3 )
Remove common factors.
1
𝑥+ 3
= 0 Equate the numerator to 0.
1 = 0 False statement.
So, there is no zero of the function. Which means that no point on the
graph touches the x – axis.