GM 04 Q1 FD rev - d14fikpiqfsi71.cloudfront.net
Transcript of GM 04 Q1 FD rev - d14fikpiqfsi71.cloudfront.net
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GENERAL MATHEMATICS | UNIT 4
Rational Functions
Table of Contents
Introduction ...................................................................................................................................... 3
Test Your Prerequisite Skills .......................................................................................................... 4
Objectives ........................................................................................................................................ 5
Lesson 1: Introduction to Rational Functions
- Warm Up! ............................................................................................................................. 5
- Learn about It! ..................................................................................................................... 6
- Letβs Practice! ....................................................................................................................... 7
- Check Your Understanding! ............................................................................................. 10
Lesson 2: Representing Rational Functions through Tables, Graphs, and Equations
- Warm Up! ........................................................................................................................... 12
- Learn about It! ................................................................................................................... 13
- Letβs Practice! ..................................................................................................................... 16
- Check Your Understanding! ............................................................................................. 22
Lesson 3: Domain and Range of a Rational Function
- Warm Up! ........................................................................................................................... 23
- Learn about It! ................................................................................................................... 24
- Letβs Practice! ..................................................................................................................... 27
- Check Your Understanding! ............................................................................................. 34
Lesson 4: Solving Problems Involving Rational Functions
- Warm Up! ........................................................................................................................... 35
- Learn about It! ................................................................................................................... 35
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- Letβs Practice! ..................................................................................................................... 37
- Check Your Understanding! ............................................................................................. 41
Challenge Yourself! ....................................................................................................................... 41
Performance Task ......................................................................................................................... 42
Wrap-up ......................................................................................................................................... 44
Key to Letβs Practice! ...................................................................................................................... 45
References ..................................................................................................................................... 47
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GRADE 11| GENERAL MATHEMATICS
UNIT 4
Rational Functions You have been studying functions in the previous lessons. One kind of functions that are widely used to model and solve many problems in the business world is the rational functions. Some examples of real-world scenarios that involve rational functions include average speed, concentrations of mixtures, average sales over time, and average costs over time. Many practical situations can be represented and analyzed using mathematical equations. For example, a person's Body Mass Index or BMI can give us an idea of the state of his or her health. This can be calculated by dividing the person's weight in kilograms by the square of his or her height in meters.
Another quantity that can be represented using rational functions is the expenses of a factory for a product. For example, a bag company needs to spend β±20 000 monthly to start their business operation. For every bag they produce, they need to
spend another β±400. How much do they spend to produce a bag? How much will they spend in all if they need to make 200 bags a day for a month? The average cost πΆ(π₯) can be calculated by finding the sum of the fixed monthly cost and the variable cost, then dividing the sum by the number of bags produced. In this unit, we will learn about rational functions and its application to real-life situations.
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Before you get started, answer the following items on a separate sheet of paper. This will help you assess your prior knowledge and practice some skills that you will need in studying the lessons in this unit. Show your complete solution.
1. Identify the following expressions are polynomial or not.
a. 3π₯& + 2 b. β*
&π₯& β π₯ + β5
c. π₯ + &-
d. 2- β 1 e. 4π₯& β ππ₯ β 10 f. 7π₯3 β βπ₯ + 7
2. Graph the following polynomial
functions. a. π(π₯) = 3π₯ + 5 b. π(π₯) = 4π₯ β 1 c. π(π₯) = π₯& + 2π₯ β 5 d. π(π₯) = 3π₯& + π₯ β 2 e. π(π₯) = π₯3 β 4π₯& + 7
3. Evaluate the given polynomial functions for the corresponding values of π₯.
a. π(π₯) = 2π₯& β 5π₯ + 7; π₯ = β1 b. π(π₯) = β3π₯3 + 5 β 6π₯ β 1;
π₯ = β4 c. π(π₯) = 3π₯7 β π₯& + 7π₯ β 5; π₯ =
6 d. π(π₯) = βπ₯7 + 2π₯3 + π₯& β π₯ +
2; π₯ = β3 e. π(π₯) = 4π₯3 + 3π₯& β 5π₯ + 1;
π₯ = 3
Test Your Prerequisite Skills
β’ Identifying whether a given expression is a polynomial or not β’ Graphing polynomial functions β’ Evaluating polynomial functions
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At the end of this unit, you should be able to
β’ represent real-life situations using rational functions; β’ represent a rational function through its table of values, graph, and equation; β’ determine the vertical and horizontal asymptotes of a rational function; β’ find the domain and range of a rational function; and β’ solve problems involving rational functions.
Letβs go out for a trip!
Materials Needed: activity sheet, pen, and paper Instructions: 1. This activity is to be done in pairs. 2. You are to complete the given table using the information as follows.
Your family vacation is on a place 150 kilometers away from your house. You will determine the time it will take for your trip at the given rates as follows.
Objectives
Lesson 1: Introduction to Rational Functions
Warm Up!
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Speed (π) Time (π) 30 kph 40 kph 50 kph 75 kph 80 kph
Given a speed π₯, how will you determine your travel time? What mathematical model can you form from the data?
3. The first group who answers the activity correctly wins the game.
In the Warm Up! activity, you were able to form a mathematical model to determine the travel time for a distance of 150 km with a speed of π₯ kph. This situation can be expressed
as π(π₯) = *:;-
. This function is an example of a rational function.
Note that whenever a variable appears in a polynomial, its exponent should always be a whole number.
Let us consider the function π(π₯) = 7-. This is a rational function since both the numerator
and denominator are polynomials, and the denominator is not equal to zero. Meanwhile,
Definition 1.1: A rational function is a function
of the form π(π₯) = <(-)=(-)
or π¦(π₯) = <(-)=(-)
where π(π₯)
and π(π₯) are polynomials, and π(π₯) is not equal to zero.
Learn about It!
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the function π(π₯) = -B&;
is not a rational function. Observe that the numerator is a
polynomial, but the denominator is zero. Therefore, π(π₯) = -B&;
is not a rational function.
How about the function π(π₯) = β-B:-
? The expression in the numerator βπ₯ + 5 can be
written as (π₯ + 5)CD. This shows that the exponent of the variable π₯ is not a whole number,
so we conclude that the expression is not a polynomial. Hence, π(π₯) is not a rational function.
Example 1: Is π(π₯) = 3
-a rational function?
Solution:
Step 1: Identify whether the numerator and the denominator are both polynomials. The numerator and the denominator are both polynomials. Step 2: Check if the denominator is not equal to zero. The denominator is not equal to zero.
Thus, π(π₯) = 3-is a rational function.
Try It Yourself! Is π(π₯) = -
;a rational function?
Letβs Practice!
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Example 2: Determine whether the function π(π₯) = -DB:-EF-B7
is rational or not.
Solution:
Step 1: Identify whether the numerator and the denominator are both polynomials. The numerator and the denominator are both polynomials. Step 2: Check if the denominator is not equal to zero. The denominator is not equal to zero.
Hence, π(π₯) = -DB:-EF-B7
is a rational function.
Try It Yourself!
Determine whether the function π(π₯) = -EB3-Dβ:-B&
is rational or not.
Example 3: Is the function π(π₯) = β&-B*-E
a rational function?
Solution: The expression in the numerator β2π₯ + 1 can be written as (2π₯ + 1)CD. This
shows that the exponent of the variable π₯ is not a whole number, so we conclude that the expression is not a polynomial.
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Try It Yourself!
Is the function π(π₯) = -B*β7-B*
rational?
Real-World Problems Example 4: Suppose a bag company operates on a
fixed monthly cost of β±20 000. In addition to this, the company spends β±400 as production cost for each bag it makes. Represent the situation using a function. Is the function rational or not?
Solution: The average cost πΆ(π₯) can be calculated by finding the sum of the fixed
monthly cost and the variable cost, then dividing the sum by the number of bags produced.
Let π₯ be the number of bags the company makes. The fixed monthly cost is
20000, and the variable cost is 400π₯. Using this information, the average cost is computed as follows.
πΆ(π₯) =20000 + 400π₯
π₯
The numerator and the denominator of the function πΆ(π₯) = &;;;;B7;;--
are
both polynomials. Thus, the function is rational.
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Try It Yourself!
The length of a road is 15 km more than 50 times its width π₯. The length of half of the road can then be represented
asπ(π₯) = :;-B*:&
. Is this function rational?
1. Complete the table below.
Is the expression on the numerator
a polynomial?
Is the expression on the
denominator a polynomial?
Is the function rational or not?
a. π(π₯) = :-
b. π(π₯) = -EB3-Dβ:-B&
c. π(π₯) = β-B*-
d. π(π₯) = -B:;
e. π(π₯) = -B*7-B&
Check Your Understanding!
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2. Determine whether the following functions are rational or not.
a. π(π₯) = ;-
b. π(π₯) = -DB*-Dβ-B7
c. π(π₯) = β-B7-
d. π(π₯) = -B*;;
e. π(π₯) = 7-B-HD
7-B&
3. The length of the road is 2 cm more than 5 times its width π₯. One-fourth of the
length can then be represented asπ(π₯) = :-B&7
. Is the given function rational or not?
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Solve and Graph! Materials Needed: chalk, graphing board, paper, and pen Instructions: 1. This activity will be done per group with four members. 2. Substitute the given values of π₯ for finding the corresponding values for π¦.
π(π₯) = 2π₯ + 1 Input (π) β3 β2 β1 1 2 3 Output (π) 3. Plot the points on a Cartesian Plane.
Warm Up!
Lesson 2: Representing Rational Functions through Tables, Graphs, and Equations
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In the Warm Up! activity, you were able to graph a linear function through solving for the values of π¦ with their corresponding π₯-values. Most of the functions you have been working on have graphs that are straight lines just like what you did in the activity. Some graphs have curved lines for quadratic or polynomial functions. These graphs represent certain equations and values when plugged into a given equation. Similarly, rational functions can also be represented using a table of values, a graph, or an equation.
Let us construct a table of values for π(π₯) = 3-, and sketch the graph of the function.
Solution:
Step 1: Assign some values of π₯ as inputs. Note that π₯ = 0 is not included because it will make the function undefined.
Input (π) β3 β2 β1 1 2 3
Output (π)
Step 2: To complete the table, evaluate the function π(π₯) = 3- for each given value of
π₯.
Learn about It!
Definition 2.1: A table of values is composed of values (π₯) and π(π₯) or π¦ that satisfy the given function.
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For π₯ = β3:
π(π₯) =3π₯
π(β3) =3β3
π(β3) = β1
For π₯ = β2:
π(π₯) =3π₯
π(β2) =3β2
π(β2) = β1.5 Evaluating at the other values of π₯ results in the table as follows:
Input (π) β3 β2 β1 1 2 3
Output (π) β1 β1.5 β3 3 1.5 1
The value of π(π₯) keeps decreasing for negative values of π₯ but suddenly becomes positive for positive values of π₯. Letβs observe more closely by making another table of values for the function.
Input (π) 0.001 0.001 0.1 1 2 3 10 100 1000
Output (π) 3000 300 30 3 1.5 1 0.3 0.03 0.003
Notice that as π₯ becomes closer to 0, the value of π(π₯) increases without any bound. This will produce us the vertical asymptote of the graph.
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In this case, the vertical asymptote of the graph is at π₯ = 0. Similarly, as π₯ becomes larger, the value of π(π₯) becomes smaller, eventually approaching 0. This yields the horizontal asymptote of the graph, which is π¦ = 0. A similar trend can be observed for the negative values of π₯. Can you find the asymptotes of the graph of π(π₯) based on the following table of values?
Input (π) β1000 β100 β10 β3 β2 β1 β0.1 β0.01 β0.001
Output (π) β0.003 β0.03 β0.3 β1 β1.5 β3 β30 β300 β3000 Step 3: Plot the ordered pairs on a Cartesian plane, then connect them with a
smooth curve to obtain the graph of the rational function.
Definition 2.2: An asymptote is a line that a curve approaches, but does not intersect.
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This graph shows us the vertical and horizontal asymptotes of the function. The curves of the graph approach the lines π₯ = 0and π¦ = 0 but do not intersect them.
Example 1: Given the rational function π(π₯) = π₯β3π₯+2, construct a table of values,
and sketch the graph of the function. Solution:
Step 1: Assign some values of π₯ as inputs. Notice that the chosen values of π₯ are consecutive integers, but β2 is not included because it will result in a denominator of zero.
Input (π) β7 β6 β5 β4 β3 β1 0 1 2 3
Output (π)
Step 2: To complete the table, evaluate the function π(π₯) = -F3-B&
for each given value
of π₯.
Input (π) β7 β6 β5 β4 β3 β1 0 1 2 3 Output (π) 2 2.25 2.67 3.5 6 β4 β1.5 β0.67 β0.25 0
Step 3: Plot the ordered pairs on a Cartesian plane, then connect them with a
smooth curve to obtain the graph of the rational function.
Letβs Practice!
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It is mentioned earlier that if π₯ = β2, the denominator will become 0. Thus,
the graph will never intersect the line π₯ = β2. This implies that the vertical asymptote of the graph is the line π₯ = β2.
Meanwhile, the horizontal asymptote of the function can be identified by
determining the value that π(π₯) approaches as π₯ increases or decreases without bound.
The table of values below contains several negative values of π₯ along with the
corresponding values of π(π₯). This function value approaches 1. Input (π) β10 β100 β1000 β10000 10 100 1000 10000
Output (π) 1.625 1.05 1.005 1.0005 0.58 0.95 0.995 0.9995
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Thus, the horizontal asymptote of the function is π¦ = 1.
Try It Yourself!
Make a table of values for π(π₯) = π₯+3π₯+1, then sketch its graph.
So far, we have determined the asymptotes of the given rational functions by substituting several values of π₯ and observing the trends in the resulting values of π(π₯). However, this method is slow and unreliable because it is easy to make mistakes when evaluating complicated rational functions. For this reason, we will be using some rules that will help us identify the asymptotes of a given rational function. Remember that these rules apply to all rational functions π(π₯) = <(-)
=(-) where π(π₯) and π(π₯) have no common factors. When
the numerator and the denominator have common factor(s), hole(s) at the zero(s) is(are) produced. Rules for Identifying the Asymptotes of a Rational Function Vertical asymptotes The graph of the rational π(π₯) has vertical asymptotes at the zeros of its denominator π(π₯). Horizontal asymptotes We can determine the horizontal asymptotes of π(π₯) by comparing the degrees of π(π₯) and π(π₯). Let us call these values π and π, respectively.
a. If π < π, then the horizontal asymptote of π(π₯) is the line π¦ = 0. b. If π = π, then the horizontal asymptote of π(π₯) is the line π¦ = PQ
RS, where the
expressions πU and πW are the leading coefficients of π(π₯) and π(π₯), respectively.
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c. If π > π,then the graph of π(π₯) has no horizontal asymptotes. Oblique Asymptotes If π = π + 1, divide the numerator by the denominator. The oblique asymptote is the quotient with the remainder ignored and set equal to π¦. Example 2: What is the vertical asymptote of π(π₯) = 2
π₯β1?
Solution:
Step 1: Reduce the rational function to lowest terms. The function is already in its lowest term. Step 2: Set the denominator to 0, and solve for π₯.
π₯ β 1 = 0
π₯ = 1
Thus, the vertical asymptote of π(π₯) = 2π₯β1 is π₯ = 1.
Try It Yourself!
What is the vertical asymptote of π(π₯) = π₯β74π₯ ?
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Example 3: What is the horizontal asymptote of π(π₯) = π₯+1π₯β2?
Solution: The function is already in its lowest term. Furthermore, the degree π of the
numerator π(π₯) and the degree π of the denominator π(π₯) are equal. Thus, the horizontal asymptote of π(π₯) is the line π¦ = PQ
RS or π¦ = 1.
. Try It Yourself!
What is the horizontal asymptote of π(π₯) = π₯β3π₯+5?
Real-World Problems Example 4: You and your friends are on a summer vacation trip
that is 120 kilometers away from your school. Write a function that describes the time it takes to make the trip as a function of your speed. Create a table of values for the average speed of 40 kph, 45 kph, 50 kph, 60 kph, and 80 kph.
Solution: Recall that the product of speed and time represents distance. If π· is the
distance, π is the speed, and π‘ is the time, then
π· = ππ‘ If both sides of the equation are divided by π, we obtain
π‘ =π·π
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Since the location of your summer trip is 120 kilometers away from your
school, we can write the function that describes the time it takes to make the trip as a function of your speed as
π(π) =120π
To construct the table of values for this function, evaluate the function
π(π) = *&;\
for each given value of the speed π.
Input (π) 40 45 50 60 80
Output (π) 3 2.67 2.4 2 1.5
Try It Yourself!
Sketch the graph of the function obtained in Example 4. Determine the vertical and the horizontal asymptote.
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1. Construct a table of values for the following rational functions, then sketch the graph.
a. π(π₯) = 6π₯
b. π(π₯) = π₯β43
c. π(π₯) = π₯+56
d. π(π₯) = 4π₯β7π₯+5
2. Complete the given table. Write NONE on the corresponding cell if the asymptote
does not exist.
Function Vertical Asymptote
Horizontal Asymptote
Oblique Asymptote
a. π(π₯) = π₯1
b. π(π₯) = π₯β105
c. π(π₯) = π₯+102
d. π(π₯) = π₯2β4π₯
e. π(π₯) = 2π₯β1π₯β2
3. The first location in your schoolβs field trip is 100 kilometers away from your school. Write a function that describes the time it takes to make the trip as a function of the speed of the bus. Create a table of values for the average speed of 50 kph, 55 kph, 60 kph, and 70 kph. Sketch the graph of the function, and determine the vertical and the horizontal asymptote.
Check Your Understanding!
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Locate the ends! Materials Needed: graphing papers, blue and orange color pens
Instructions:
1. This activity will be done by pairs. 2. Your teacher will hand you four different kinds of graph. 3. Locate the farthest left and right points. Mark them, and color the π₯ βaxis
blue. 4. Identify the lowest and highest points and mark the corresponding π¦ βaxis
orange. The graphs are shown as follows:
Warm Up!
Lesson 3: Domain and Range of a Rational Function
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The concept of domain and range is important as you learn how to graph rational functions. The activity in Warm Up! gave you a glimpse about the domain and range
Learn about It!
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given a graph. The blue marked π₯ βaxis indicates the domain while the orange marked π¦ βaxis is the range. These graphs, however, do not represent rational functions. So how then do we determine the domain and range of a rational function? In Lesson 2, you have learned how to represent functions through table of values,
graphs, and equations. Recall that the function π(π₯) = 3- can be represented by the
graph as follows. If we inspect the graph, we find that the function has a vertical asymptote at π₯ = 0, which implies that the graph does not contain a point for which π₯ = 0. Therefore, the domain of this function is the set of all real numbers π₯ such that π₯ β 0. In symbols, we write this as π·: {π₯|π₯ β 0}.
Definition 3.1: The domain of a function is the set of all values of π₯ that have corresponding values of π¦. It contains all values that go into the function.
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Similarly, the values of π¦ are all nonzero real numbers. Therefore, the range of the function is the set of all real numbers except π¦ = 0. In symbols, we write this as π : {π¦|π¦ β 0}. Remember that the domain and range of a function are obtained by finding and excluding the appropriate restrictions. In the case of rational functions, these restrictions can be found at the vertical and horizontal asymptotes because there are no points plotted along them. To find the domain of a rational function, perform the following steps:
1. Make sure that the fraction is in its simplest form. When the numerator and the denominator have common factor(s), hole(s) at the zero(s) is(are) produced.
2. Find the vertical asymptotes of the function by setting the denominator equal to zero and solving the resulting equation.
3. Exclude the asymptote/s obtained in the previous step from the set of real numbers. Reject the zero as well of the common factor that is canceled. The remaining elements of the set of real numbers comprise the domain of the function.
A similar procedure can be used to determine the range of a rational function:
1. Make sure that the fraction is in its simplest form. 2. Find the horizontal asymptotes of the given rational function by comparing the
degrees of the numerator and the denominator.
Definition 3.2: The range of a function is the set of all values of π¦ that can be obtained from the possible values of π₯. It contains all possible values of the function.
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3. Exclude the asymptote(s) obtained in the previous step from the set of real numbers. If there is a common factor canceled, say (π₯ β π), reject also the value for π(π). The remaining elements of the set of real numbers comprise the range of the function.
Example 1: Determine the domain and range of π(π₯) = π₯β3π₯+2.
Solution:
Step 1: Reduce the rational function to its simplest form. The function is already in its simplest form.
For the domain of the function:
Step 2.a: Find the vertical asymptotes of the function by setting the denominator equal
to zero, and solving the resulting equation.
π₯ + 2 = 0π₯ = β2
Step 2.b: Exclude the asymptote/s obtained in the previous step from the set of real
numbers. The remaining elements of the set of real numbers comprise the domain of the function.
Thus, the domain of this function is the set of all real numbers π₯ such that π₯ β β2. In symbols, π·: {π₯|π₯ β β2}.
Letβs Practice!
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For the range of the function: Step 3.a: Find the horizontal asymptotes of the given rational function by comparing
the degrees of the numerator and the denominator.
The numerator and the denominator of the function have the same degree. Recall that if π = π, then the horizontal asymptote of π(π₯) is the line π¦ = PQ
RS,
where the expressions πU and πW are the leading coefficients of π(π₯) and π(π₯), respectively. For the given rational function, this expression is equal to 1.
Step 3.b: Exclude the asymptote/s obtained in the previous step from the set of real
numbers. The remaining elements of the set of real numbers comprise the range of the function.
Thus, the range of this function is the set of all real numbers π¦ such that π¦ β 1. In symbols, π : {π¦|π¦ β 1}.
Try It Yourself!
Determine the domain and range of π(π₯) = 2π₯β1π₯β4 .
Example 2: Find the domain and range of π(π₯) = (π₯β1)(π₯+2)
(π₯β1)(π₯+3).
Solution:
Step 1: Reduce the rational function to its simplest form.
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The numerator and denominator have a common factor of π₯ β 1. Recall that when the numerator and the denominator have common factor(s), hole(s) at the zero(s) is(are) produced. Hence, the simplified form of the function is
π(π₯) = (π₯+2)(π₯+3).
For the domain of the function:
Step 2.a: Find the vertical asymptotes of the function by setting the denominator equal
to zero and solving the resulting equation.
π₯ + 3 = 0π₯ = β3
Step 2.b: Exclude the asymptote/s obtained in the previous step from the set of real
numbers. Reject the zero as well of the common factor that is canceled. Equating the common factor π₯ β 1 to zero yields to π₯ = 1. This is not included
in the domain of the function. The remaining elements of the set of real numbers comprise the domain of
the function.
Thus, the domain of this function is the set of all real numbers π₯ such that π₯ β β3, 1. In symbols, π·: {π₯|π₯ β β3, 1}.
For the range of the function: Step 3.a: Find the horizontal asymptotes of the given rational function by comparing
the degrees of the numerator and the denominator.
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The numerator and the denominator of the function have the same degree. Recall that if π = π, then the horizontal asymptote of π(π₯) is the line π¦ = PQ
RS,
where the expressions πU and πW are the leading coefficients of π(π₯) and π(π₯), respectively. For the given rational function, this expression is equal to 1.
Step 3.b: Exclude the asymptote/s obtained in the previous step from the set of real
numbers. If there is a common factor canceled, say (π₯ β π), reject also the value for π(π). The remaining elements of the set of real numbers comprise the range of the function.
The common factor canceled is π₯ β 1. Hence, we reject also the value for π(1).
π(π₯) =(π₯ + 2)(π₯ + 3)
π(1) =(1 + 2)(1 + 3)
π(1) =34
Thus, the range of this function is the set of all real numbers π¦ such that
π¦ β 1, 37. In symbols, π : eπ¦|π¦ β 1, 3
7f.
Try It Yourself!
Determine the domain and range of π(π₯) = (π₯β3)(π₯+4)(π₯+4)(π₯β1).
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Example 3: Determine the asymptote of the function π(π₯) = π₯2+4π₯β12π₯2β4 .
Solution:
Step 1: Reduce the rational function to its simplest form.
π₯& + 4π₯ β 12π₯& β 4 =
(π₯ β 2)(π₯ + 6)(π₯ + 2)(π₯ β 2)
π₯& + 4π₯ β 12π₯& β 4 =
(π₯ + 6)(π₯ + 2)
The simplified form of the function is π(π₯) = (π₯+6)
(π₯+2).
For the domain of the function:
Step 2.a: Find the vertical asymptotes of the function by setting the denominator equal to zero and solving the resulting equation.
π₯ + 2 = 0
π₯ = β2
Step 2.b: Exclude the asymptote/s obtained in the previous step from the set of real numbers. Reject the zero as well of the common factor that is canceled.
Equating the common factor π₯ β 2 to zero yields to π₯ = 2. This is not included
in the domain of the function. The remaining elements of the set of real numbers comprise the domain of
the function.
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Thus, the domain of this function is the set of all real numbers π₯ such that π₯ β β2, 2. In symbols, π·: {π₯|π₯ β β2, 2}.
For the range of the function: Step 3.a: Find the horizontal asymptotes of the given rational function by comparing
the degrees of the numerator and the denominator.
The numerator and the denominator of the function have the same degree. Recall that if π = π, then the horizontal asymptote of π(π₯) is the line π¦ = PQ
RS,
where the expressions πU and πW are the leading coefficients of π(π₯) and π(π₯), respectively. For the given rational function, this expression is equal to 1.
Step 3.b: Exclude the asymptote/s obtained in the previous step from the set of real
numbers. If there is a common factor canceled, say (π₯ β π), reject also the value for π(π). The remaining elements of the set of real numbers comprise the range of the function.
The common factor canceled is π₯ β 2. Hence, we reject also the value for π(2).
π(π₯) =(π₯ + 6)(π₯ + 2)
π(2) =(2+ 6)(2 + 2)
π(2) =84 = 2
Thus, the range of this function is the set of all real numbers π¦ such that π¦ β 1, 2. In symbols, π : {π¦|π¦ β 1, 2}.
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Try It Yourself!
Determine the domain and range of π(π₯) = π₯2+3π₯β18π₯2+π₯β12 .
Real-World Problems Example 4: Suppose a driver goes to a gas station to refuel.
He noticed that the station sells gasoline at β±40 per liter, and he has β±500 pesos in his wallet. Represent the number of liters of gasoline that the driver can purchase. What is the domain of the given function?
Solution: The number of liters of gasoline that the driver can purchase can be
represented by the function π(π₯) = -7;
, where π₯ is the amount of money that
the driver will spend. This gives us certain restrictions based on the amount that the driver has on hand. Therefore, we can only input values from 0 to 500 into the function. The domain of the function is π·:{π₯|0 β€ π₯ β€ 500}.
Try It Yourself!
Records from a factory producing electronic components show that on average, new employees can assemble π(π‘) components per day after π‘ days of training, where
π(π‘) = i:jjB:
; π‘ β₯ 0. What is the domain and range of the
function?
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1. Compute for the domain and range of the following rational functions.
Function Domain Range a. π(π₯) = 3
π₯
b. π(π₯) = π₯β62
c. π(π₯) = π₯+52
d. π(π₯) = π₯2+93
e. π(π₯) = π₯2+π₯β906
2. Solve for the vertical and horizontal asymptote of the following functions. Show
your complete solutions.
Function Vertical Asymptote
Horizontal Asymptote
a. π(π₯) = 6π₯
b. π(π₯) = π₯β43
c. π(π₯) = π₯+37
d. π(π₯) = π₯2+62
e. π(π₯) = π₯2+π₯β1005
3. After a drug is injected into a patientβs bloodstream, the concentration πΆ of the drug
in the bloodstream in π‘ minutes after the injection is given by πΆ(π‘) = &;jjDB&
; π‘ β₯ 0.
What is the domain and range of the given function?
Check Your Understanding!
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Boards Up! Materials Needed: drill boards, whiteboard marker, pen, and paper Instructions: 1. This game is to be in groups with 3 members each. 2. Your teacher will present 5 different rational functions. 3. For every rational function given, the first member of the group will evaluate 3
different values of π₯. These values are the first 3 multiples of 5. 4. The second member will determine the asymptotes of the rational functions. 5. The third member will identify the domain and range of the function. 6. The first group who completed the answers correctly gets the point.
The game in the Warm Up! activity allowed you to recall different concepts you have learned in this unit. Let us walk through these concepts again.
Recall that a rational function is of the form π(π₯) = <(-)=(-)
, where π(π₯) and π(π₯) are both
polynomials and π(π₯) β 0. A rational function may also be represented using a table of values, a graph, or an equation.
Lesson 4: Solving Problems Involving Rational Functions
Learn about It!
Warm Up!
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You have also learned about asymptotes. Recall that an asymptote is a line that a curve approaches, but does not intersect. There are rules that we follow to determine the asymptotes of the graph of a rational function. Vertical asymptotes The graph of the rational π(π₯) has vertical asymptotes at the zeros of its denominator π(π₯). Horizontal asymptotes We can determine the horizontal asymptotes of π(π₯) by comparing the degrees of π(π₯) and π(π₯). Letβs call these values π and π, respectively.
a. If π < π, then the horizontal asymptote of π(π₯) is the line π¦ = 0. b. If π = π, then the horizontal asymptote of π(π₯) is the line π¦ = PQ
RS, where the
expressions πU and πW are the leading coefficients of π(π₯) and π(π₯), respectively. c. If π > π,then the graph of π(π₯) has no horizontal asymptotes.
Oblique Asymptotes If π = π + 1, divide the numerator by the denominator. The oblique asymptote is the quotient with the remainder ignored and set equal to π¦. Moreover, you have learned how to identify the domain and range of a function. The domain of a function is the set of all values of π₯ that have corresponding values of π¦. It contains all the valid inputs of the function. On the other hand, the range of a function is the set of all values of π¦ that can be obtained from the possible values of π₯. It contains all the valid outputs of the function. Using these concepts, let us try to solve some word problems involving rational functions.
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Example 1: Suppose a company that manufactures pencils in Quezon City operates on a
fixed monthly cost of β±10 000. In addition to this, the company spends β±5 as production cost for each pencil. Represent the situation using rational functions.
Solution: The average cost πΆ(π₯) can be calculated by finding the sum of the fixed
monthly cost and the variable cost, then dividing the sum by the number π₯ of pencils produced. If the fixed monthly cost is β±10 000 and the variable cost is β±5, then the situation can be represented as
πΆ(π₯) = 10000+5π₯π₯ .
Try It Yourself! Suppose a company that manufactures bags operate on a fixed monthly cost of β±15 000. In addition to this, the company spends β±150 as production cost for each bag it makes. Represent the situation using rational functions.
Example 2: To join a tennis club, you have to pay a membership fee of β±2 000, plus β±100
per session. What is the average cost function and the average cost per session if you attend 10 sessions?
Solution: The average cost function can be expressed as the sum of the membership
fee and the variable cost, divided by the number of sessions. Let π₯ be the number of sessions. If the fixed monthly cost is β±2 000 and the variable cost is β±100, then the average cost function is
Letβs Practice!
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πΆ(π₯) = 2000+100π₯π₯ .
To determine the average cost if you attend 10 sessions, substitute π₯ = 10 into the average cost function.
πΆ(π₯) =2000 + 100π₯
π₯
πΆ(10) =2000 + 100(10)
10
πΆ(10) =2000 + 1000
10
πΆ(10) =300010
πΆ(10) = 300
Therefore, the average cost over 10 sessions is β±300.
Try It Yourself!
To join a golf club, you have to pay a membership fee of β±4 000, plus β±200 per session. What is the average cost function and the average cost per session if you attend 5 sessions?
Example 3: Suppose you are staying at a hotel that charges β±7 000 upon check-in, plus
β±1000 daily. What is the average cost function and the average daily cost if you stay there for 7 days?
Solution: The average cost function can be expressed as the sum of the check-in fee
and the variable cost, divided by the number of days. Let π₯ be the number of days. If the fixed monthly cost is β±7 000 and the variable cost is β±1000, then the average cost function is
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πΆ(π₯) = 7000+1000π₯π₯ .
To determine the average cost if you attend 7 days, substitute π₯ = 7 into the average cost function.
πΆ(π₯) =7000 + 1000π₯
π₯
πΆ(7) =7000 + 1000(7)
7
πΆ(7) =7000 + 7000
7
πΆ(7) =140007
πΆ(7) = 2000
Therefore, the average daily cost of a seven-day hotel stay is β±2000.
Try It Yourself! Suppose you are staying at a hotel that charges β±5 500 upon check-in, plus β±500 daily. What is the average cost function and the average daily cost if you stay there for 10 days?
More Real-World Problems Example 4: A car rental company charges a
fixed price of β±5000. For every kilometer traveled, an additional β±20 will be charged. What is the average cost per kilometer if you rent a car and drive it for 25 kilometers? Assume that the fuel, maintenance, and other relevant costs will not be charged to you.
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Solution: The average cost πΆ(π₯) can be calculated by adding the fixed rental cost and
the additional charge per kilometer, then dividing the sum by the number of kilometers traveled.
πΆ(π₯) =5000 + 20π₯
π₯
To determine the average cost per stay, substitute π₯ = 25 to the average cost function.
πΆ(π₯) =5000 + 20π₯
π₯
πΆ(25) =5000 + 20(25)
25
πΆ(25) =5000 + 500
25
πΆ(25) =550025
πΆ(25) = 220
Therefore, the average rental cost per kilometer for a 25-kilometer trip is β±220.
Try It Yourself! A van rental company charges a fixed β±4 000 fee, plus β±30 per kilometer traveled. What is the average cost per kilometer if you rent a car and drive it 10 kilometers? Assume that the fuel, maintenance, and other relevant costs will not be charged to you.
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Read and analyze the following problems. Show your solutions. 1. Suppose a company that manufactures teddy bear operates on a fixed monthly
cost of β±200 000. In addition to this, the company spends β±150 as production cost for each teddy bear it makes. Represent the situation using rational functions. What will be the average cost per teddy bear that will be produced?
2. Suppose you are staying at a hotel that charges β±3 000 upon check-in, plus β±200 daily. What is the average cost function and the average daily cost if you stay there for 12 days?
3. A van rental company charges a fixed β±3 000 fee, plus β±20 per kilometer traveled. What is the average cost per kilometer if you rent a car and drive it for 15 kilometers? Assume that the fuel, maintenance, and other relevant costs will not be charged to you.
4. Suppose you rent a bike and you need to pay an initial fee of β±100 and another β±15 per hour. How much will you spend for a total of 4 hours? What is the average cost per hour?
1. Suppose you are given a rational function π(π₯). Is the reciprocal of π(π₯) always a rational function? Support your answer with an example.
Challenge Yourself!
Check Your Understanding!
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2. Is it possible for a function to have more than one vertical asymptote? Describe how this can be done, or explain why it is not possible.
3. Given the rational function π(π₯) = -B:-DFl
, what should be the value of π so that the graph of the function has a hole at π₯ = β5?
4. Suppose you are asked to write a rational function π(π₯) that has a vertical
asymptote at π₯ = 3, a horizontal asymptote at π¦ = 1, and zero at π₯ = 5, what will be the function?
5. What is the difference between the functions f(x) = *;p
and f(x) = βpB:pif π₯ = 2?
You are a research analyst of a company that is about to venture a new business. You are to choose 3 different business ventures and explore the average cost function of the product they produce. Create a presentation of these functions which includes the following:
a. at least 5 sample number of products and represent these data through a table of values, graphs, and equations
b. domain and range of the functions c. asymptotes of the graph
Present the data to the research supervisor. He/she will assess the data presentation using the rubrics as follows.
Performance Task
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Performance Task Rubric
Criteria Below
Expectation (0β49%)
Needs Improvement
(50β74%)
Successful Performance
(75β99%)
Exemplary Performance
(99+%)
Punctuality and Neatness
The data is submitted late and has many erasures.
The data is submitted late and has few erasures.
The data is submitted on time but has few erasures.
The data is submitted on time and has no erasure.
Accuracy of Computations
There are significant errors in computations that lead to the wrong relativity of data.
There are few errors in the computation, but there is no clear reference in the analysis of the data.
All computations are correct, and the data are analyzed properly.
All computations are correct and with a complete solution. The data are analyzed with clear references.
Presentation
The students did not present the output.
The students presented the output but are not confident with their work.
The student presented the data and are confident with their work.
The students presented the data and are very confident with their work.
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Key Terms & Definitions
Key Terms Definitions
Rational Function
A rational function is a function of the form π(π₯) = <(-)=(-)
or
π¦(π₯) = <(-)=(-)
where π(π₯) and π(π₯) are polynomials, and π(π₯) is
not equal to zero.
Table of Values A table of values is composed of values (π₯) and π(π₯) or π¦ that satisfy the given function.
Asymptote An asymptote is a line that a curve approaches, but does not intersect.
Domain The domain of a function is all the set of all values of π₯ that have corresponding values of π¦. It contains all values that go into the function.
Range The range of a function is the set of all values of π¦ that can be obtained from the possible values of π₯. It contains all possible values of the function.
Wrap-up
can be represented through
Rational Functions
graph table of values equation
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Lesson 1
1. No. The denominator is zero. 2. Yes, it is a rational function. 3. No, the function is not rational. 4. Yes, it is a rational function.
Lesson 2
1. π(π₯) = π₯+3π₯+1
π β3 β2 0 1 2 3 4
π(π) 0 β1 3 2 1.67 1.5 1.4
2. π₯ = 0 3. π¦ = 1
Key to Letβs Practice!
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4. Graph:
Vertical Asymptote: π₯ = 0 Horizontal Asymptote: π¦ = 0
Lesson 3:
1. π·: {π₯|π₯ β 4} π : {π¦|π¦ β 2}
2. π·: {π₯|π₯ β β4, 1} π : {π¦|π¦ β 1, 1.4}
3. π·: {π₯|π₯ β β4, 3} π : {π¦|π¦ β 1, 1.29}
4. π·:{π₯|π₯ β₯ 0} π : {π¦|0 β€ π¦ < 75}
Lesson 4: 1. πΆ(π₯) = *:;;;B*:;-
-
2. πΆ(π₯) = 7;;;B&;;--
; β±1 000
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3. πΆ(π₯) = ::;;B:;;--
; β±1 050 4. πΆ(π₯) = 7;;;B3;-
-; β±430
Purplemath. βAdding and Subtracting Fractions.β Accessed April 21, 2018.
http://www.purplemath.com/modules/fraction4.htm Monterey Institute. βFinding Domain and Range.β Accessed April 21, 2018. http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_
RESOURCE/U17_L2_T3_text_final.html
References