SELF-INSTRUCTIONAL PACKETS (SIPacks)
MATHEMATICS Grade 9 – SSC
School
PAMPANGA HIGH SCHOOL
Teaching Dates/ Week
August 24 - 28 (Week 1)
Teacher
IVAN T. SALAS
Quarter
FIRST
I. OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations,
inequalities and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various
situations, formulate real-life problems involving quadratic equations, inequalities and
functions, and rational algebraic equations and solvethem using a variety of strategies.
C. Learning Competencies
The Learners
➢ Illustrate quadratic equations. (M9AL-Ia-1)
D. Objectives
At the end of the period, the learners will be able to:
➢ define, identify, and give examples of quadratic equations;
➢ write quadratic equations in standard form and identify the real numbers a, b and c.
II. CONTENT
Topic/Lessons: Illustrations of Quadratic Equations
Learning Resources:
References: Learner’s Material for Mathematics Grade 9 pp. 11-55
Teacher’s Guide for Mathematics pp. 14-18
Patterns and Practicalities
III.
PROCEDURE:
DAY 1
A. Reviewing previous lesson or presenting the new lesson
Hi! I hope you’re doing well. I bet you already miss being in school.
Don’t worry; you will soon see your friends and teachers in school when things finally get
better.
For the meantime, let us have our discussion through this lesson I have prepared for you.
To start with, recall your knowledge about linear equations.
In Activity 1 below are mathematical equations, can you identify which among these are
linear equations? If you’re ready, read the direction thoroughly and accomplish Activity 1.
Page 1 of 148
Activity 1: LINEAR OR NOT?
On the space provided, write LINEAR if the
given is a linear equation. Otherwise, write
NOT LINEAR.
1. x2 – 5x + 3 = 0
___________
2. 2s + 3t = -7
___________
3. 6x( x – 2) = 8
___________
4. 8k – 3 = 12
___________
2
5. 9r –25 = 0
___________
6. c = 12n – 5
___________
7. ½ x2 = 3x
___________
8. ¾ (h + 6) = 0
___________
9. (x + 9) (x – 2) = 0 ___________
10. 9 – 4x = 15
___________
2
11. ( x + 7) = 1
___________
12. 4m2 + 4m +1 = 0
___________
13. 10( x – y) = 3
___________
2
14. 7 + 3x = - 6x
___________
15. x2 – 15x = 7
___________
B. Establishing a purpose for the lesson
A linear equation is an algebraic
equation in which each term has
an exponent of one and the
graphing of the equation results in
a straight line.
An example of linear equation is
y = mx + b.
You may check your answers by
turning to Page 9.
Give yourself one (1) point for
every correct answer.
I hope you got them all correctly!
.
Definition
Let us have another short activity!
This activity is entitled “Guess My Anagram.”
What is an anagram? Read the definition.
Let me give you an example.
The word LISTEN can be rearranged into the word SILENT.
The word BINARY can be rearranged into the word BRAINY.
An anagram is a word or
phrase that's formed by
rearranging the letters of
another word or phrase,
typically using the original
letters exactly once.
Did you understand? Are you now ready?
If you are, read the direction carefully and accomplish Activity 2.
Activity 2: GUESS MY ANAGRAM.
Think of other words you can form out of the following
anagrams. Write your answers on the space provided.
1. ACID QUART
__________________
2. CON DES
__________________
3. ERG EDE
__________________
4. QUAT IONE
__________________
5. DARTS AND
__________________
6. MORF
__________________
You may check your answers
by turning to Page 9
Give yourself one (1) point for
every correct answer.
I hope you got them all
correctly.
You can proceed to the next
part of the discussion
C. Presenting examples/instances of the lesson
In Activity 1, you were able to identify which among the given equations are linear and which are not.
The equations below are the ones you identified as NOT LINEAR. Check them out!
Page 2 of 148
Not Linear Equations
How are these equations different from those
which are linear?
x2 – 5x + 3 = 0
6x( x – 2) = 8
What characteristics do these equations have?
What kind of equations are they?
I hear you. Very good! These are called quadratic
equations.
9r –25 = 0
2
½ x2 = 3x
(x + 9) (x – 2) = 0
In this Lesson, we will be learning more on these
kinds of equations.
( x + 7)2 = 1
4m2 + 4m +1 = 0
On the other hand, the words you formed out of
the six (6) anagrams in Activity 2 are related with
this topic. You’ll come across these words as we
go on with our discussion.
7 + 3x2 = - 6x
x2 – 15x = 7
Figure 1.
Are you ready to learn more about these equations?
Let us now find out what a quadratic equation is!
Definition
Quadratic Equationin one variable is a mathematical sentence of degree 2. It can
be written in the standard form ax2 + bx + c = 0, where a, b,&c are real numbers
and a ≠ 0.
In the quadratic equation ax2 + bx + c = 0, ax2is called the quadratic term, bx is
the linear term, while c is the constant term.
From the definition, any equation that can be written in the following standard form
ax2 + bx + c = 0 where a, b,&c are real numbers and a ≠ 0 is called a quadratic equation.
Let us have some examples. Refer to figure 1 “Not Linear Equations” above.
Examples
x2 – 5x + 3 = 0
9r2 –25 = 0
4m2 + 4m +1 =
0
Quadratic equations
in standard form
½ x2 = 3x
7 + 3x2 = - 6x
x2 – 15x = 7
Quadratic equations
NOT in standard form
6x( x – 2) = 8
(x + 9) (x – 2) =
0
( x + 7)2 = 1
Quadratic equations
not in simplified form
The equations x2 – 5x + 3 = 0, 9r2 –25 = 0, ½ x2 = 3x, 4m2 + 4m +1 = 0, 7 + 3x2 = - 6x, and x2 –
15x = 7 are obviously quadratic equations. Each of these equations contains a variable in the
second degree or is raised to the second power.
Meanwhile, 6x( x – 2) = 8, (x + 9) (x – 2) = 0 and ( x + 7)2 = 1 are also quadratic equations. It’s
just that they are not expressed in their simplified form.
D. Discussing new concepts and practicing new skills #1
DAY 2
From the given examples of quadratic equations above, x2 – 5x + 3 = 0, 9r2 –25 = 0, and 4m2 +
4m +1 = 0 are written in the standard form. Recall that from the definition, the standard form of
a quadratic equation is expressed as ax2 + bx + c = 0.These equations are in the standard form
since the quadratic terms, linear and constant terms are written accordingly and each equation
is equal to zero.
Page 3 of 148
Here is how to write quadratic equations in standard form.
Examples
Write in standard form and determine a, b, and c
1.
This quadratic equation is already in standard
form, thus, a = 1, b =-5, and c = 3
2.
This is also a quadratic equation in the standard
form. However, this equation does not contain a
linear term.
Hence, a = 9, b = 0, c = -25
Note that quadratic equations may or may not contain a linear or a constant term. But quadratic
equations can’t go without a quadratic term. If that is the case, quadratic equations may be
written in one of these forms: 𝒂𝒙𝟐 = 𝟎, 𝒂𝒙𝟐 + 𝒃𝒙 = 𝟎, and 𝒂𝒙𝟐 + 𝒄 = 𝟎. These forms are called
incomplete quadratic.
For quadratic equations that does not contain a linear term, b = 0. While quadratic equations
that does not contain a constant term, c = 0.
Now try these.Write in standard form and determine a, b, and c.
a. 4x2 + 4x +1 = 0
b. x2 + 12 = 0
Your answers must be 4x2 + 4x +1 = 0, a = 4, b = 4, c = 1 and x2 + 12 = 0, a = 1, b = 0, c =12
If you get the correct answers you may proceed to example 3. If not, review examples 1 and 2
and try again.
3.½ x2 = 3x
This quadratic not in standard form. To write it in standard form;
½ x2 – 3x = 3x – 3x
Add -3x to both sides so that the right hand side (RHS) equals 0
½ x2 – 3x = 0
This is now the standard form.
a = ½, b = -3, c = 0 Since the constant term is missing, c = 0.
Now, try to write 7 + 3x2 = - 6x. in standard form nad identify the real numbers a, b and c
You should get 3x2 + 6x + 7,and a = 3, b = 6, c = 7as the real numbers.
If you get the correct answer you may proceed with Example 4.
If not, I am sorry but you have to go back to example number3 and try all over again.
E. Discussing new concepts and practicing new skills #1
DAY 3
Let’s have more complex examples.
Let us write the standard forms of the equations 6x( x – 2) = 8, (x + 9) (x – 2) = 0 and
(x + 7)2 = 1 and identify the real numbers a, b and c.
4.
6x (x – 2) = 8
This is quadratic not in simplified form.
6x2 - 12x = 8
Apply the distributive property to simplify the left hand side (LHS)
of the quation.
6x2 - 12x - 8 = 8 - 8
Add -8 to both sides so that the RHS equals 0.
6x - 12x – 8 = 0
`
2
a = 6, b = -12, c = -8
This is now the standard form.
These are the real numbers a, b, c
Page 4 of 148
5.
(x + 9) (x – 2) = 0
x2 - 2x + 9x -18 = 0
x2 + 7x – 18 = 0
This is quadratic not in simplified form. To write it in standard form;
Apply the FOIL method to simplify the LHS of the equation.
Combine similar terms.
x2 + 7x – 18 = 0
This is now the standard form.
a = 1, b = 7, c = -18
Theseare the real numbers a, b, c.
Quick
Review:
FOIL
Method
Now, it’s your turn. Try to write (x – 4) (x + 7) = 0 and identify the real numbers a, b and c.
You should get x2 + 3x – 21 = 0, and a = 1, b = 3, c = -21 as the real numbers.
If you get the correct answer you may proceed to example 6.
If not, I am sorry but you have to go back to example number 5 and try again.
6.
(x + 7)2 = 1
This is quadratic not in simplified form.
x2 + 14x + 49 =1
Apply the special product squaring binomial
to simplify the LHS
x2 + 14x + 49 -1 = 1 – 1 Add -1 to both sides so that the RHS becomes 0.
x2 + 14x + 48 = 0
This is now the standard form.
a = 1, b = 14, c = 48 These are the real numbers a, b, and c.
Quick Review: Squaring Binomial
Now it’s your turn, try this. Write (x +5)2 = 4 in standard form and determine a, b, and c
Your answers must be x2 +10x + 21 = 0, and a = 1, b =10, and c = 21.
If you get it correctly you may proceed to the next part of the lesson. If not, review
example 6 and try again.
For more about the standard form of quadratic equations visit;
https://www.youtube.com/watch?v=UZTvYYoOrmI
F. Developing mastery
DAY 4
Now, let us apply what you have learned from our discussion by doing the next activities.
Read the direction carefully. Copy the balloon in your notebook and then color it.
Page 5 of 148
(𝒙 + 𝟑) − 𝟕 = 𝟎
𝟐(𝒙 + 𝟑)𝟐 = 𝟎
𝒙𝟑 − 𝟕 = 𝟎
(
𝒙 𝒙 + 𝟑) + 𝟓 = 𝟎
One of the much-awaited attractions in
Pampanga is the Hot Air Balloons Festival in
Angeles City, Clark. What makes this festival
famous and fascinating is because of its
colorful giant balloons. The organizers of
this event are now looking for a more
colorful balloon designs. They are asking for
your help to color the balloon they have
designed as shown belowfollowing this
instruction.
𝟑𝒙 − 𝟐 = 𝟎
ACTIVITY 3: COLOR ME.
Identify whether each of the following
equations found in the balloon is quadratic,
linear or neither. Color each equation using
the following legend: green for quadratic,
blue for linear and orange for neither.
Did you enjoy the activity? Were you able to identify which equations are linear,
quadratic or neither? Refer to page 9 for the key to correction and enrichment activity.
Here is another activity to check if you were able to follow how to write quadratic
equations in standard form.
Read the direction carefully and write your answers on your notebook following the
format below.
Activity 4: IT’S YOUR TURN!
Write each quadratic equation in standard form, ax2 + bx + c = 0 then identify the values of
a, b, &c.
Standard Form
a
b
c
2
1. x + 4x = 8
_________________
____
____
____
2.
(x + 3)2 = 0
_________________
____
____
____
3.
9 = -11x2 –25x
_________________
____
____
____
4.
(x+2)(x–2)= - 2x
_________________
____
____
____
5.
x (x + 2) = 0
_________________
____
____
____
You may check your answers by turning to Page 10 for the key to
correction.
Score
Description
16 - 20
Very Good
12 - 15
Good
How many correct answers did you get?
8 - 11
Fair
Rate the result using the table.
4-7
Turn to page 8
sec. J and work
on the enrichment
activities
Give yourself one (1) point for every correct answer.
0-3
Page 6 of 148
G. Finding practical application of the concepts and skill in daily living.
Quadratic equations are actually used in everyday life. Their vast applications cover almost
every man’s activities as when calculating areas, determining a product's profit or formulating
the speed or height of an object and many more. The pictures shown above are just some of its
many practical applictations.
Can you think of another practical application of quadratic equations?
H. Making generalizations and abstractions about the lesson
Ok, let us have a summary of our lesson.
What is a quadratic equation?
Quadratic Equation in one variable is a mathematical sentence of degree 2. It can be
written in the standard form ax2 + bx + c = 0, where a, b, & c are real numbers and a ≠ 0.
In a quadratic equation, ax2 is called the quadratic term, bx is the linear term, while c is
the constant term.
The incomplete forms of quadratic equations are the following:
𝒂𝒙𝟐 = 𝟎, 𝒂𝒙𝟐 + 𝒃𝒙 = 𝟎, and 𝒂𝒙𝟐 + 𝒄 = 𝟎.
I.
Evaluating learning
DAY 5
Direction: Use yellow papers to answer the evaluation. Provide necessary solutions for
your answers. Use the format shown below to label your paper and it is to be
submitted to your Math teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 1 Parent’s Signature: __________________
EVALUATION 2.1
Please read the direction carefully, answer each item with care and with all honesty.
I. Which of the following equations are quadratic equations? Write Y if it is and N if not.
1. 3x – 2 = 0.
2
____
2. x + 3x = 0.
____
3. 3x – 2 = 0.
____
4. x (x + 3 ) – 5 = 0
____
II. Multiple Choice: Write the letter and words of the correct answer on the space provided
before the number
____ 5. Which of these equations illustrates quadratic equation?
A. (x + 3) + 8 = 0
C. x + y = 0
2
B 2(x +3) = 0
D. x3 = 2x
Page 7 of 148
____ 6. It is a polynomial equation of degree 2?
A. linear equation
C. quadratic inequality
B. quadratic equation
D. linear inequality
____ 7. Which of the following is the standard form of quadratic equation?
A. ax + b = 0
C. ax2 + bx + c = 0
B. ax + b = 0
D. ax2 + bx + c = 0
____ 8. Which of the following real number in the quadratic equation form ax2 + bx + c = 0
cannot be zero?
A. a
C. c
B. b
D. none of these
III. Give four examples of quadratic equations.
9. ____________________
11. ____________________
10. ____________________
12. ____________________
IV.
Write each of these quadratic equations in standard form and determine a, b, and c.
Quadratic Equation
13.𝑥 2 + 7 + 2𝑥 = 0
14.5 − 2𝑥 2 = 6
15.2𝑥(𝑥 − 3) = 15
16.(2𝑥 + 7)(𝑥 − 1) = 0
17.(𝑥 − 4)2 + 8 = 0
Standard Form
a
b
c
J. Additional activities for application or remediation
Remediation 1. Identify whether each of the following equations is quadratic or not. Write
YES if it is quadratic and NO if it is not quadratic.
1.
2.
3.
4.
5.
2𝑥 − 3𝑦 = 7
2𝑥 2 = 8 − 3𝑥
𝑥 3 + 2𝑥 2 + 1 = 0
𝑥 2 = 16
𝑥 + (2𝑥 − 2) = 4
_____
_____
_____
_____
_____
6. 𝑡 = 5
7. 2𝑦 2 = 84
8. 3𝑥(𝑥 + 4) = 8
9. 3𝑥 + 4𝑦 = 12
10. (𝑦 − 3)2 − 12 = 5
_____
_____
_____
_____
_____
You may check your answers by turning to Page 10 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Remediation 2. Write each quadratic equation in standard form, ax2 + bx + c = 0 then
identify the values of a, b, & c.
Standard Form
a
b
c
1. 8x = x2 -7
_________________
____
____
2. 6x( x – 2) = 4
_________________
____
____
2
3. ½ x = 3x
_________________
____
____
4. x2 + 8 – 2x = 0
_________________
____
____
2
5. 3 – 4x – 6x = 0
_________________
____
____
You may check your answers by turning to Page 10 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
____
____
____
____
____
Page 8 of 148
KEY TO CORRECTION
Activity 1: LINEAR OR NOT?
On the space provided, write LINEAR if the given is a linear equation. Otherwise, write
NOT LINEAR.
1. x2 – 5x + 3 = 0
NOT LINEAR
2. 2s + 3t = -7
LINEAR
3. 6x( x – 2) = 8
NOT LINEAR
4. 8k – 3 = 12
LINEAR
5. 9r2 –25 = 0
NOT LINEAR
6. c = 12n – 5
LINEAR
2
7. ½ x = 3x
NOT LINEAR
8. ¾ (h + 6) = 0
LINEAR
9. (x + 9) (x – 2) = 0
NOT LINEAR
10. 9 – 4x = 15
LINEAR
11. ( x + 7)2 = 1
NOT LINEAR
12. 4m2 + 4m +1 = 0
NOT LINEAR
13. 10( x – y) = 3
LINEAR
14. 7 + 3x2 = - 6x
NOT LINEAR
15. x2 – 15x = 7
NOT LINEAR
Activity 2: GUESS MY ANAGRAM
Think of other words you can form out of the following anagrams. Write your answers on
the space provided.
1. ACID QUART
QUADRATIC
2. CON DES
SECOND
3. ERG EDE
DEGREE
4. QUAT IONE
EQUATION
5. DARTS AND
STANDARD
6. MORF
FORM
Activity 3: COLOR ME
B
l
u
e
G
r
e
e
n
O
r
a
n
g
e
G
r
e
e
n
B
l
u
e
Page 9 of 148
Activity 4: IT’S YOUR TURN!
Write each quadratic equation in standard form, ax2 + bx + c = 0 then identify the values of
a, b, & c.
Standard Form
a
b
c
1. x2 + 4x = 8
x2 + 4x – 8 = 0
1
4
-8
2
2
2. (x + 3) = 0
x + 6x + 9 = 0
1
6
9
3. 9 = -11x2 –25x
11x2 + 25x + 9 = 0
11
25
9
2
4. (x+2)(x–2)= - 2x
x + 2x – 4 = 0
1
2
-4
2
5. x (x + 2) = 0
x + 2x = 0
1
2
0
Remediation 1. Identify whether each of the following equations is quadratic or not. Write
YES if it is quadratic and NO if it is not quadratic.
1.
NO
6. NO
2. YES
7. YES
3. NO
8. YES
4. YES
9. NO
5. NO
10. YES
Remediation 2. Write each quadratic equation in standard form, ax2 + bx + c = 0 then
identify the values of a, b, & c.
Standard Form
a
b
c
1. 8x = x2 -7
x2– 8x – 7 = 0
1
-8
-7
2. 6x( x – 2) = 4
6x2– 12x – 4 = 0
6
-12
-4
3. ½ x2 = 3x
½ x2 – 3x = 0
½
-3
```0
4. x + 8 – 2x = 0
x – 2x+ 8 = 0
1
-2
8
5. -3 – 4x – 6x2 = 0
6x2+ 4x + 3= 0
6
4
2
2
3
If you have questions regarding the lessons, you may contact me at # 09434634247.
“What the mind of man can achieve and believe, it can achieve.”
Page 10 of 148
SELF-INSTRUCTIONAL PACKETS (SIPacks)
MATHEMATICS Grade 9 – SSC
School
PAMPANGA HIGH SCHOOL
Teacher
Teaching Dates/ Week
IVAN T. SALAS
Quarter
September 1 - 4, 2020 (Week 2)
FIRST
I. OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations,
inequalities and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various
situations, formulate real-life problems involving quadratic equations, inequalities and
functions, and rational algebraic equations and solve them using a variety of strategies.
C. Learning Competencies
The Learners
➢ Solves quadratic equations by; (a) extracting square roots; (b) factoring;
(c) completing the square; (d) quadratic formula. (M9AL-Ia-b-1)
D. Objectives
At the end of the period, the learners will be able to:
➢ Solves quadratic equations by; (a) extracting square roots; (b) factoring; (c)
completing the square; (d) quadratic formula. (M9AL-Ia-b-1)
II. CONTENT
Topic/Lessons: Solving quadratic equations by; (a) extracting square roots;
(b) factoring; (c)completing the square; (d) quadratic formula
Learning Resources:
References: Learner’s Material for Mathematics Grade 9 pp. 18-55
Teacher’s Guide for Mathematics pp. 19-38
Patterns and Practicalities
III.
PROCEDURE:
DAY 1
A. Reviewing previous lesson or presenting the new lesson
CONGRATULATIONS! You have successfully accomplished the maiden lesson in
Grade 9 Math. And with that, I can say “Ang Galing Mo!”
Before moving on to the next lesson, let’s have a short review of the previous lesson.
Complete the definition below by supplying the missing word/number, expression or
equation in each blank.
Review 1:
Quadratic Equationin one variable is a __________________ sentence of degree _____.
It can be written in the standard form__________________, where a, b,&c are ________
numbers and _____can not be equal to _______.
In the standard form of the quadratic equation, ax2,is called the ______term, bx is the
____________ term, while c is the _____________term.
Page 11 of 148
Next! Try this one.
Review 2: Match each quadratic equation in Column A with their corresponding
standard form in Column B.
Write the letter of your answers in your notebook.
Answer
Column A
Column B
2
1. x - 36 = 2x
A. x2 + 2x + 38 = 0
2
2. 4x - 5 = 3x + 15
B. 3x2 - 4x + 5 =0
3. 2x + x2 = 36
C. x2 + 2x – 36 = 0
4. 3x2 - x + 5= 3x
D. x2- 2x – 36 = 0
5. x2 + 2x + 2 = -36
E. 3x2 + 4x+ 20 = 0
Now, check your work by turning to Page 32
for the key to correction.
Give yourself 1 point for every correct answer.
How many correct answers did you get?
Score
12-14
9-11
5-8
0-4
Description
Very Good
Good
Fair
Turn to DLP1 for
Week 1 for review
Rate yourself by using the table to the right.
B. Establishing a purpose for the lesson
Now that you have learned to illustrate quadratic equations, let us proceed to next
lesson - the different methods of solving quadratic equations.
There are several methods for solving quadratic equations. These methods include
extracting the square root, factoring, completing the square, and using the quadratic
formula
Each method has its advantages and disadvantages.
As you go through this lesson, think of this important question: “How does finding
solutions of quadratic equations facilitate in solving real life problems and in making
decisions?”
Are you ready and excited to learn all these?
C. Presenting examples/instances of the lesson
The first method to be discussed is the extracting square root method or square root
method.
In the previous discussion, quadratic equation may come in different forms. One is the
form
ax2 + c = 0 or ax2 = c.
Examples are: x2 = 100, 4x2 = 64, t2 – 81 = 0, and (x – 4)2 = 169
Any quadratic equation that can be written in the form ax2 = c can be solved best using
the square root methodor extracting the square root.
Page 12 of 148
Learn about the properties of square root:
1. If c > 0 ( meaning c is positive), the x2 = c has two real solutions or roots: x = √𝑐
2. If c = 0, the x2 = c has one solution or root: x = 0
3. If c  0 ( meaning c is negative), the x2 = c has no real solutions or roots
The solutions to quadratic equations are called roots.
Let us have some examples on how to solve quadratic equations using the square root
method.
Read each example carefully and note the steps followed in solving each equation.
Solving quadratic equations by extracting square root or square root method
Example 1. Solve x2 – 100 = 0
Solution:
a. Write the equation in
x2 – 100 = 0
the form ax2 = c
Since c = 100 and its greater than 0 then you can have 2
roots.See property 1
b. If a is not equal to1,
make it equal to one by
dividing both sides of
equation by the value
of a.
c. Extract the root to find
the value of the
variable
Since a = 1 so this step is not applicable in this particular case
d. Check by substituting
each root in the
originalequation.
For x = 10: x2 – 100 = 0For x = -10:x2 – 100 = 0
(10)2 – 100 = 0
(-10)2 – 100 = 0
100 – 100 = 0
100 – 100 = 0
0 = 0 TRUE
0 = 0 TRUE
Both roots satisfy the equation.
Therefore, the roots of x2 – 100 = 0 are10 or -10
√x² = √100
x = 10 “ read as plus or minus 4”
x = 10can be written as x = 10 or x = -10
It’s your turn, try this. Solve x2 – 81 = 0
You should get x =  9. If you get the correct answer you may proceed to example 2.
If not, review example 1 and try again.
Example 2. Solve x2 = 0
Solution:
a. Write the equation in
the form ax2 = c
b. If a is not equal to1,
make it equal to one by
dividing both sides of
equation by the value
of a.
c. Extract the root to find
the value of the
variable
x2 = 0
this is in the form ax2 = c already
Since c = 0, then you can have 1 root, x = 0, see property 2
Since a = 1, skip step b
√x² = √0
x=0
Page 13 of 148
d. Check by substituting
each root in the original
equation.
Example 3. Solve 2x2 + 9 =1
Solution:
a. Write the equation in
the form ax2 = c
b. If a is not equal to1,
make it equal to one by
dividing both sides of
equation by the value
of a.
If x =0 : x2 = 0
(0)2= 0
0 = 0 TRUE
Since 0 satisfy the equation; therefore, the roots of x2 = 0is 0
2x2 + 9 =1
2x2 + 9 -9 =1 – 9
2x2 = -8
Add -9 to both sides of the equation
2x 2 −8
=
2
2
x2 = -4
Since c = -4 and is less than 0, then the equation has no real
roots because negative numbers has no real square roots,
See property 3.
There is no real number when squared gives -4.
Therefore 2x2 + 9 =1 has no real roots or solutions
You try this. Solve 4x2 + 81 = 60
You should get x2 = -7. This equation has no real solutions
If you get the correct answer you may proceed to example 4.
If not, review example 3 and try again.
Example 4. Solve 4x2 = 64
Solution:
a. Write the equation in
the form ax2 = c
4x2 = 64
this is already in ax2 = c form
b. If a is not equal to1,
make it equal to one by
dividing both sides of
equation by the value
of a.
c. Extract the root to find
the value of the
variable
4𝑥 2 64
=
4
4
d. Check by substituting
each root in the original
equation.
For x = 4: x2 – 16 = 0
(4)2 – 16 = 0
16 = 0
0 = 0 TRUE
𝑥 2 = 16
√x² = √16
x=4
x=4
x2 – 16 = 0
(-4)2 – 16 = 016 –
16 – 16 = 0
0 = 0 TRUE
For x = -4:
Both values satisfy the equation.
Therefore, the roots or solutions of 2x2= 64 are 4 or -4
Try this. Solve 3x2 = 27
You should get x =  3 as solutions or roots.
If you get the correct answer you may proceed to example 5.
If not, review example 4 and try again.
Page 14 of 148
Example 5. Solve 2( x – 5)2 =32
Solution:
a. Write the equation in
2( x – 5)2 =32 this is already in ax2 = c
the form ax2 = c
b. If a is not equal to1,
make it equal to one by
dividing both sides of
equation by the value
of a.
c. Extract the root to find
the value of the
variable
d. Check by substituting
each root in the original
equation.
2( x − 5)2 32
=
2
2
( x – 5)2 = 16
√(𝑥 − 5)2 = √16
x–5=4
x–5+5=54
x=54
when x = 4:
x=5+4
x=9
when x = - 4:
x=5-4
x=1
For x = 9: 2( x – 5)2 =32 For x = 1: 2( x – 5)2 =32
2(9 – 5)2 =32 2(1 – 5)2 =32
2(4)2 =32 2(- 4)2 = 32
.
2(16) = 32
2(16) = 32
32 = 32
32 = 32
0 = 0 TRUE
0 = 0 TRUE
Both values satisfy the equation.
Therefore, the roots of x2 – 100 = 0 are 10 or -10
Try this. Solve 5(x – 3)2 = 125
You should get x = 8 or x = -2 as roots or solutions.
If you get the correct answer you may proceed example 6.
If not, review example 5 and try again.
Example 6. Solve (2x – 3)² = 18
Solution:
a. Write the equation in the form
ax2 = c
b. If a is not equal to1, make it
equal to one by dividing both
sides of equation by the value
of a.
c. Extract the root to find the
value of the variable
(2x – 3)² = 18 this is in the form ax2 = c already
Since a = 1, skip step b
√(2𝑥 − 3)² =√18
2x – 3=√18
simplify the right hand side
2x – 3 =(√9)(√2
2x – 3 =  3 √2
2x – 3 + 3 = 3 √2 + 3Add 3 to both side
2x = 3  3 √2Divide both sides by 2
3 ± 3√2
x=
2
The roots are x =
3+3√2
2
or x =
3−3√2
2
These kinds of roots are called irrational numbers.
Page 15 of 148
d. Check by substituting each
root in the original equation.
Note: Irrational numbers will be discussed thoroughly
under radicals.
Again, it’s your turn. Solve (4x – 5)2 = 12.
You should get 𝑥 =
5 ±2√3
as
4
roots.
If you get the correct answer you may answer the given exercises below before we
proceed to the next part of the discussion.
If not, review example 5 and try again.
Comprehension Check 1:
Apply the given steps in each given examples above,
solve each quadraticequation by square root method.
Do this in your notebook.
1) x2 -144 = 0
2) 4x2 – 144 = 0
3)( x- 4)2 = 225
4) ( 2x2 – 1)2= 225
Now, check your work by turning to Page 32 for the
key to correction. How many correct answers did you
get? Rate your result using the table above.
Score
7-8
4 -6
2-3
0-1
Description
Very Good
Good
Fair
Turn to page31
sec. J and work on
the enrichment
activities
If you would like to learn more about square root method you may visit;
https://www.youtube.com/watch?v=NnjVQRwAaMg
D. Discussing new concepts and practicing new skills # 1
DAY 2
How did you find solving quadratic equations using the square root method?
Are you ready for the next method of solving quadratic equations?
I hear you right! Without much ado, let’s go to the next part of our discussion
This part of the lesson will explain how to solve quadratic equations by the method of
factoring.
Solving quadratic equations by Factoring
This method requires knowledge of a property that involves the number zero. To elicit this
property, consider this equation.
x • y = 0
What x-values and y-values make this equation true?
According to the multiplication property of zero, the equation is true when either...
x = 0 or y = 0
The Zero Product Property
For any real numbers x and y, if xy = 0, then x = 0 or y = 0; and if either x = 0 or y = 0, then xy = 0.
We will use this property to solve quadratic equations in the examples below.
We start by Solving Quadratic Equations of the form ax2 + bx = 0
When the constant is 0, the quadratic equation will be of the form ax2 + bx = 0.
Page 16 of 148
Let’s do some examples.
Example 1:
Solve 3x2 + 18 x = 0.
Notice that 3x2 and 18x has a common factor which is 3x.
Solution:
Rewirte in standard form
3x2 + 18 x = 0.
Factor
3x ( x + 6 ) = 0
Apply Zero Product Property 3x = 0 ;
x+6=0
Solve for x
x=0
x = -6
The roots are 0 and -6.
Check:
If x = 0:
3(0)2 + 18(0) = 0
0=0
3(-6)2 + 18(-6) = 0
0=0
Both roots satisfy the original equation.
If x = -6
Will you try solving for the roots of 4x2 – 2x = 0.
You should get 0 and ½ as roots.
If you get the correct answer you may proceed with Example 2.
If not, I am sorry but you have to go back to example number 1 and try all over
again.
Example 2:
Solve 11x2 – 13x = 8x – 3x2
Solution:
Rewrite in standard form
11x2 + 3x2 – 13x – 8x = 0
Combine like terms
14x2 – 21x = 0
Factor
7x (2x – 3) = 0
Apply Zero Product Property
7x = 0 2x – 3 = 0
Solve for x
x = 0 ; x = 3/2
The roots are 0 and 3/2.
Check:
If x = 0:
11(0)2 – 13(0) = 8(0) – 3(0)2
0=0
If x = 3/2:
11(3/2)2 – 13(3/2) = 8(3/2) – 3(3/2)2
11(9/4) – 13(3/2) = 8(3/2) – 3(9/4)
21/4 = 21/4
Both roots satisfy the original equation.
Now it’s your turn. Try solving for the roots of 7x2 + 18x = 10x2 + 12x.
You should get 0 and 2 as roots.
If you get the correct answer you may proceed with the next part of the discussion.
If not, I am sorry but you have to go back to example number 1 and try all over again.
Have you not noticed it? That in examples 1 and 2, one of the roots is 0 and the other
are nonzero numbers. Is this true to all quadratic equations? Let’s find out.
Take a look at this.
Given the quadratic equation of the form ax2 + bx = 0
Factor
x (ax +b) = 0
Set factors to 0
x=0
ax + b = 0
Solve for x
x=0
x =- b/a
Page 17 of 148
Did you see it? So always remember;
The roots of the quadratic equation of the form ax2 + bx = 0 are x = 0 and x = -b/a
In fact, you can obtain the roots without factoring just by looking the coefficients.
Let’s try to solve the same equations as in examples 1 and 2 without factoring.
Example 3. Solve 3x2 + 18 x = 0.
Solution.
Rewrite in standard form
Identify a and b
Apply the formula
3x2 + 18 x = 0.
a=3 ;
b = 18
x=0
x = -b/a
x = -18/3
x = -6
The roots are x = 0 and x = -6
Compare these roots in example 1. What can you say?
Example 4:
Solve 11x2 – 13x = 8x – 3x2
Solution:
Rewrite in standard for
Combine like terms
Determine a and b
Apply the formula
11x2 + 3x2 – 13x – 8x = 0
14x2 – 21x = 0
a = 14, b = -21
x=0
x = -b/a
x = -(-21/14)
x =3/2
Compare these roots in example number 2. Are you convinced?
Now it’s your turn. Try solving for the roots of 4x2 – 2x = 0 and 7x2 + 18x = 10x2 + 12x
without factoring.
You should get 0 and ½ , 0 and 2 as roots respectively.
If you get the correct answer you may answer given exercises below before you proceed
to the next part of the discussion
If not, I am sorry but you have to go back to examples 3 and 4 and try all over again.
Comprehension Check 2:
Determine the roots of these quadratic equations either by applying The Zero Product
Property or without factoring. Write your answers in your notebook.
1. 2x2 – 5x = 0
Ans. __________
Score
Description
2
2. x –6x = 0
Ans. __________
7-8
Very Good
2
4
-6
Good
3. 8x – 12x = 0
Ans. __________
2
2
2
3
Fair
4. 8x + 5x = 2x - 15xAns. __________
Now, check your work by turning to Page 33
for the key to correction. How many correct
answers did you get?
0-1
Turn to page 31
sec. J and work
on the enrichment
activities
Rate your result using the table to the right.
Page 18 of 148
Let us now have Solving Quadratic Equations of the Form ax2 + bx + c = 0
We can also use factoring in solving quadratic equations when none of the constants a, b, or
c are missing.
Example 5. Solve x2-6x+8=0
Solution:
Factor the left hand side (LHS):
Apply Zero Prodct Property:
Solve each resulting equation:
The roots or solutions are 4 and 2
Check the values of the variable
obtained
by substitutingeach in the equation :
(x-4)(x-2)=0
x-4=0 ; x-2=0
x=4
; x =2
If x = 4
If x = 2
42 -6(4) + 8 =0
16 – 24 + 8 =0
0 =0
22 -6(2) + 8 =0
4 – 12 + 8 =0
0=0
Both roots satisfy the original equations.
Now, you try solving for the roots of x2+17x - 18 = 0.
You should get -18 and 1 as roots or solutions to the equation.
If your answer is correct, you may proceed to example number 6.
If not, I am sorry but you have to go back to example number 5 and try all over again.
Example 6: Solve 2x2 – 5x – 3 = 0
Solution
Factor LHS:
(2x + 1) (x – 3) = 0
Apply Zero Prodct Property:
2x + 1 = 0
x–3=0
Solve for x:
x = -1/2
x=3
The roots or solutions to the equation are: -1/2 and 3.
Check:
If x = -1/2:
2(-1/2)2 – 5(-1/2) – 3 = 0
If x = 3:
2(3)2 – 5(3) – 3 = 0
2 (1/4) + 5/2 – 3 = 0
18 – 15 – 3 = 0
½ + 5/2 – 3 = 0
0=0
3 – 3 = 0 (since ½ + 5/2 = 6/2 = 3)
0=0
Both roots satisfy the original equation.
Now, you try solving for the roots of 3x2 - 7x - 6 = 0.
You should get 3 and -2/3 as roots or solutions to the equation.
If your answer is correct, you may proceed to example number 7.
If not, I am sorry but you have to go back to example number 6 and try all over again.
Example 7: Find the roots of 𝑥 2 + 9𝑥 = −8.
Solution
Write the equation in standard form:
Factor the quadratic expression:
Apply Zero Prodct Property:
Solve for x:
The roots are:
x2 + 9x + 8 = 0
(x + 1) (x + 8) = 0
x+1=0
x+8=0
x = -1 x = -8
-1 and -8
Page 19 of 148
Check:
If x =-1: (-1)2 + 9(-1) = -8
1 – 9 = -8
-8 = -8
If x = -8: (-8)2 + 9(-8) = -8
64 - 72 = -8
-8 = -8
Both roots satisfy the original equation.
Solve for the roots of 2x2 – 5x = -2.
You should get ½ and 2 as roots or solutions to the equation.
If your answer is correct, you may answer given exercises below before we proceed to t
he next part of the discussion
If not, I am sorry but you have to go back to example number 7 and try all over again.
Comprehension Check 3:
Solve for the roots of the given quadratic equations. Write your answers in your
notebook.
1. (3y – 1) (y + 2) = 0
Ans. ___________
2. m2 – 9m + 14 = 0
Ans. ___________
3. 2x2+13x = 7
Ans. ___________
4. 4x2 – 12x + 9 = 0
Ans. __________
Now, check your work by turning to page 33for the
key to correction.
How many correct answers did you get?
Rate your result using the table above.
Score
7-8
4 -6
2-3
0-1
Description
Very Good
Good
Fair
Turn to page 31
sec. J and workon
the enrichment
activities
For more examplws about factoring you may visit:
https://www.youtube.com/watch?v=6PTt9BoIfHg
E. Discussing new concepts and practicing skills #2
DAY 3
Not all quadratic equations can be solved by the square root method and the factoring
method.
Now, let us look at another method of solving quadratic equations;
Solving quadratic equations by completing the square
What do we mean when we say: “Complete the Square”?
To complete the square of the expression x2 + bx, add the square of the half the
b 2
coefficient of x to make x2+ bx + (2) .
Page 20 of 148
Example 1. Complete the square
Quick Review:
a. x2 + 10x
10
2
=5
Taking half the x-coefficient
(5)2 = 25
Squaring
x2 + 10x + 25
Adding 25 to complete the square
The trinomial x2 + 10x + 25 is the square of x +5 or (x +5)2
b. x2 - 3x
3
2
=
3
Taking half the x-coefficient
2
3
( 2 )2 =
x2 - 3x +
A trinomial is a perfect
square trinomial if it can
be factored into
a binomial multiplied to
itself. (This is the part
where you are moving the
other way). In a perfect
square trinomial, two of
your terms will
be perfect squares.
9
Squaring
4
9
Adding
4
The trinomial x2 - 3x +
𝟗
𝟒
9
4
to complete the square
is the square of x -
𝟑
𝟐
𝟑
or (x - )2
𝟐
Now try this. Complete the square
a. x2 + 12x
b. x2 - 7x
You should get a. x2 + 12x + 36 and b. x2 - 7x +
49
4
If your answer is correct, you may answer given exercises below before we proceed to
the next part of the discussion
If not, I am sorry but you have to go back to example number 1 a and b.
Now that you are ready, let us use the technique of completing the square to solve
quadratic equations.
Example 2. Solve by completing the square
a) 2x2 + 8x - 10 = 0
STEPS
1) Write the equation in the form
ax2 + bx= c
2) if a is not equal to one, make it equal to
one by dividing both sides of equation
by the value of a.
APPLICATION
2x2 + 8x = 10
2x2 + 8x = 10
2
2
x + 4x = 5
Page 21 of 148
3) Get 1/2 of b and square it. Add this to
both sides of the equation. The left side
of the equation becomes a perfect
square trinomial.
x2 + 4x + (4/2)2 = 5 +(4/2)2
x2 +4x + 4 = 5 + 4
x2 + 4x +4 = 9
4) Express the left member of the
equation as a square of a binomial.
5) Solve for the roots by using extracting
the roots method .
6) Check by substituting each root in the
original equation.
(x + 2)2 =9
( x + 2)2
= 9
x+2= 3
x = 1 , x = -5
When x=1
→ 2(12) + 8(1) – 10 = 0
→ 2 +8 – 10 = 0
→0 = 0
When x = -5
→ 2(-5)2 + 8(-5) – 10 = 0
→ 50 - 40 – 10 = 0
→0 = 0
Now, you try solving for the roots of 3x2 - 6x - 18 = 0.
You should get 3 and -1 as roots or solutions to the equation.
If your answer is correct, you may proceed to example number 2b.
If not, I am sorry but you have to go back to example number 2a and try all over again.
b) x2 + 5x + 4 = 0
APPLICATION
STEPS
1)Write the equation in the form ax2 +
bx= c
x2 + 5x = -4
2)if a is not equal to one, make it equal
to one by dividing both sides of
equation by the value of a.
Sincethe coefficient of x2 is 1 you can
skip this step
3)Get 1/2 of b and square it. Add this to
both sides of the equation. The left side
of the equation becomes a perfect
square trinomial.
x2 + 5x + (5/2)2 = - 4 +(5/2)2
x2 + 5x + (25/4) = - 4 + (25/4)
x2 + 5x + (25/4) = (9/4)
4)Express the left member of the
equation as a square of a binomial.
(x + (5/2))2 = (9/4)
5) Solve for the roots by using
extracting the roots method .
√(𝑥 + 5) = √9
2
2
4
x + (5/2) =  (3/2)
x = -1 , x = - 4
Page 22 of 148
6) Check by substituting each root in
the original equation.
When x=-1
→ (-12) + 5(-1) + 4 =0
→1 – 5 + 4 = 0
→ 0=0
When x = - 4
→ (- 4)2 + 5(- 4) + 4 = 0
→ 16 - 20 + 4 = 0
→ 0=0
Now, you try solving for the roots of x2 + 7x + 6 = 0.
You should get -6 and -1 as roots or solutions to the equation.
If your answer is correct, you may answer the given exercises below before you proceed to
the next part of the discussion
If not, I am sorry but you have to go back to example number 2b and try all over again.
Comprehension Check 4.
Use completing the square to find the solutions of each equation. Write your answers in your
notebook.
a) x2 + 8x + 12 = 0
b) x2 + 10x = 3
You should get for a) -6 and -2 and for b) 𝑥 = −5 ± √28 as roots or solutions to the
equations.
If your answer is correct, you may proceed to the next part of the discussion
If not, I am sorry but you have to go back to example number 2a and 2b and try all over
again.
How are you so far?
We`ve been solving quadratic equations in 3 ways (extracting square roots, factoring, and
completing the square). Why do we need to learn all these ways? What is its implication to
life?
FINALLY, we come to the last method of solving quadratic equation.
The method using: The Quadratic Formula .
To begin with, I’ll show you how the quadratic formula was derived using the completing the
square method of solving quadratic equations.
From the standard form of quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
𝑎𝑥 2 + 𝑏𝑥 = −𝑐
𝑎𝑥 2
𝑎
+
𝑏𝑥
𝑎
𝑏
=−
𝑥2 + 𝑥 = −
𝑎
𝑏
𝑥2 + 𝑥 + (
𝑎
𝑏2
Rearrange the terms in general form
𝑐
Divide both sides by 𝑎. This is possible since 𝑎 ≠ 0
𝑎
𝑐
Simplify
𝑎
𝑏2
𝑐
) = (4𝑎2) − 𝑎
4𝑎2
𝑎𝑥 2
𝑎
.
Complete the square on the left side by the adding
the square of half the coefficient of x. That is,
1 𝑏 2
2 𝑎
𝑏 2
2𝑎
( . ) =( ) =
𝑏2
4𝑎 2
Page 23 of 148
2
𝑏
(𝑥 + 2𝑎) =
𝑏2 −4𝑎𝑐
Factor the left side of the equation and simplify the
4𝑎2
right side.
𝑥+
𝑏
2𝑎
= ±√
𝑏2 −4𝑎𝑐
Get the square roots of each side provided
4𝑎2
𝑏 2 − 4𝑎𝑐 ≥ 0
𝑥+
𝑏
2𝑎
𝑥=−
𝑥=
=±
𝑏
2𝑎
±
√𝑏2 −4𝑎𝑐
Simplify √4𝑎2 .
2𝑎
√𝑏2 −4𝑎𝑐
𝑏
Transpose 2𝑎 to the right side to solve for x.
2𝑎
−𝑏±√𝑏2 −4𝑎𝑐
Write the right side as a single fraction.
2𝑎
The Quadratic Formula:
If ax2 + bx + c = 0, then
𝑥=
−𝑏±√𝑏 2 −4𝑎𝑐
2𝑎
.
Here is how we use the formula in solving any quadratic equations.
Solving quadratic equations using the Quadratic Formula.
Example 1. Solve x2 -6x = -8
Solution:
Write in standard form
Determine the values of a,b,and
Substitute the values of a,b, and c in the
x2 -6x + 8 = 0
a =1, b=-6 , c= 8
𝑥=
−(−6)±√(−6)2 −4(1)8
2(1)
quadratic formula.
Solve to find the solutions.
𝑥=
𝑥=
𝑥=
6±√36+32
2
6±√4
2
6+2
2
𝑥=4
Check by substituting each solution in
in the original equation
,𝑥 =
6−2
2
,𝑥=2
42 – 6(4) = - 8 ;
16 – 24 = - 8 ;
-8 = -8 ,
22 – 6(2) = - 8
4 - 12 = - 8
-8 = -8
Both values satisfy the original equation.
Therefore, the roots are 4 and 2.
Example 2. Solve (x + 2)2 = 9
Solution:
Write in standard form
Determine the values of a,b,and c
Substitute the values of a,b, and c in the
x2 + 4x – 5 = 0
a=1, b=4 c= -5
𝑥=
−(4)±√(4)2 −4(1)(−5)
2(1)
Page 24 of 148
Solve to find the solutions.
𝑥=
−4±√16+20
2
−4±√36
2
𝑥=
𝑥=
−4+6
2
,
𝑥=1
Check by substituting each solution in
in the original equation
𝑥=
−4−6
2
𝑥 = −5
(x + 2)2 = 9
(1 + 2 )2 =9
( 3 )2 = 9
9=9
(x + 2)2 = 9
5 + 2)2 = 9
( -3 )2 = 9
9=9
Both values of x make the equation true.
Therefore, the roots or solutions are 1 and –5.
Now, try these. Solve using the quadratic formula
a. x2 + 10x + 9 = 0.
b. 3x2 – -4x = 0
You should get for a) -1 and -9 and for b) 0 and 4/3 as roots or solutions to the equations.
If your answer is correct, you may proceed to the next part of the lesson.
If not, I am sorry but you have to go back to example numbers 1 and 2 and try all over again.
For more about quadratic formula and completing the square you may visit
:https://www.youtube.com/watch?v=jgZG010wvAo&t=72s
F. Developing mastery
DAY 4
Activity 1: I Believe!
To unlock a beautiful quotation, solve each equation by extracting the square roots.
Find the solution set in the bottom block. Transfer the word from the top box to the
bottom box where the answers are found. Write your answers in your notebook.
1
=0
49
your journey.
3𝑥 2 = 12
Believing you
4𝑥 2 − 225 = 0
Believing in
3𝑥 2 − 147 = 0
an endless
(𝑥 − 4)2 = 169
is the end of
𝑥2 − 7 = 0
yourself is
(𝑥 + 7)2 = 289
destination.
𝑥 2 = 81
have failed
________________
15 15
( ,− )
2
2
________________
(2, −2)
_________________
(√7, −√7)
_________________
(7, −7)
_________________
(10, −24)
_________________
(9, −9)
_________________
(17,9)
_________________
1 1
( ,− )
7 7
𝑥2 −
Page 25 of 148
You may check your answers by turning to Page 33 for the key
to correction.
Score
7-8
4-6
2-3
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
0-3
Rate the result using the table.
Description
Very Good
Good
Fair
Turn to page
31 sec. J and
work on the
enrichment
activities
Activity 2: Mathematics Meaning
We get the word mathematics from the Greek word mathematics. What does it mean?
6
11 7
2
3
7
1
6
4
3
9
1
12 5
10 11 10 8
To find the term, find the solutions of the given quadratic equation on the blank above
the numbers corresponding to the solutions. Write your answers in your notebook.
QUADRATIC EQUATION
SOLUTIONS
1) x 2 - 4x + 4 = 0
L
(1,-1)
2) x 2 +11x + 28 = 0
N
(4, -4)
3) 4x 2 + 11x + 6 = 0
G
(-3, 2/3)
4) -10x 2 - 22x - 4 = 0
S
(7, - 4)
5) 2x 2 +9x + 4 = 0
A
(2, -3/2)
6) 7x 2 --3x - 4 = 0
I
(5/3, -5/3)
7) x 2 -3x - 28 = 0
T
(-2, -1/5)
8) 3x 2 +11x + 6 = 0
O
(-1/2,- ¾)
9) 1- x 2 = 0
E
(2)
10) x 2 - 16 = 0
P
(7, 4)
11) 9x 2 - 25= 0
D
(1, -4/7)
12) 2x 2 + x - 6 = 0
R
(-1/2, -4)
You may check your answers by turning to Page 33 for the key to
correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Rate the result using the table.
Score
11-12
9-10
6-8
0-5
Description
Very Good
Good
Fair
Turn to page
31 sec. J
and work on
the
enrichment
activities
Page 26 of 148
Activity 3. Read the direction carefully and write your answers on the space provided.
I am beautiful, up in the sky.
I am magical, yet I cannot fly.
To people I bring luck, to some people, riches.
The boy at my end does whatever he wishes.
What am I?
To answer the riddle, match the quadratic equation in Column A with its roots in
Column B. Write the letter of your answer in the box. Put your answers in your
notebook.
COLUMN A
COLUMN B
1. x2 + 12x + 32 = 0
I. x = -6 and x = 2
2. x2 – 10x = -16
N. x = 12 and x = 6
3. x2 + 4x – 12 = 0
R. x = -4 and x = -8
4. x2 – 18x = -72
W. x = 4 and x = -6
5. x2 – 6x = 16
A. x = 8 and x = 2
6. x2 – 4x = 12
O. x = 6 and x = -2
7. x2 + 2x = 24
B. x = 8 and x = -2
1
2
3
4
5
You may check your answers by turning to Page 33 for the key
tocorrection.
Give yourself one (1) point for every correct answer.
How many correct answers did you get? Rate the result using
the table.
.
6
7
Score
Description
6-7
Very Good
4–5
Good
2–3
Fair
0-1
Turn to page
32 sec. J and
work on the
enrichment
activities
Page 27 of 148
Activity 4: See the Difference!
Direction: Use the quadratic formula to solve the following quadratic equation. Then look for
your answer from the code key and write the letter corresponding to your answer on the
boxes above the question number in the tables that follow. Put your answers in your
notebook.
No.
Given
Code key
Answer
2x2+3x+1=0
M
3,4
3x2- x- 1=0
0
−1 ± √−47
4
3
2x2+7x+3=0
I
1 ± √21
10
4
x2- 7x +12=0
T
1 ± √13
6
5
3x2-14x-5=0
N
5
2
6
2x2+x+6=0
A
5, - 3
7
x2- 16=0
S
8
2x2+2x-2=0
L
1, -
9
2x2+3x-20=0
E
-1√−2
10
5x2-x-1=0
R
4 , -2
11
x2-2x-8 =0
W
-4 4
12
2x2+9x+9=0
H
-3, - 2
13
2x2+4x+6=0
D
-3, - 2
1
2
, -4
1
−1 ± √5
2
1
2
1
3
Question :
What is the difference between a train conductor and a math teacher?
1)
2)
2)
3)
4)
6 9 13
4 10 9 12 8
2 3 13
2 11 5 10 9
7 3 10 1 13
6)
7)
8)
9)
10)
2 3 13
6 2 3 13 11
2 11 5 10 9 8
2 3 13
4 10 3 12
Page 28 of 148
You may check your answers by turning to Page 33
for the key tocorrection.
Give yourself one (1) point for every correct answer.
Score
Description
9 - 10
Very Good
6–8
Good
3–4
Fair
0-2
Turn to page
32 sec. J and
work on the
enrichment
activities
How many correct answers did you get?
Rate the result using the table.
G. Finding practical applications of concepts and skills in daily living
Problem: A cement walk of constant width is built around a 10m by 20m rectangular pool.
The total area of the walk and the pool is 504 m2. Find the width of the walk.
.
(expected answer :width=4 m)
H. Making generalizations and abstractions about the lesson
Let us wrap up with the different steps in solving quadratic equations for each method
Steps in solving quadratic equations by extracting square roots.
1. Write the equation in the form ax2 = c
2. If a is not equal to1, make it equal to one by dividing both sides of equation by the value
of a.
3. Extract the root to find the value of the variable
4. Check by substituting each root in the original equation.
Steps in solving Quadratic Equations by Factoring
1. Write the equation in the form ax2 + bx + c = 0.
2. Factor the left-hand side of the equation.
3. Set each factor equal to zero using the Principle of Zero Products.
4. Solve each resulting linear equation.
5. Check the results in the original equation.
Steps in solving Quadratic Equations by Completing the Square
1. Divide both sides of the equation by a then simplify.
2. Write the equation such that the terms with variables are on the left side of the equation
and the constant term is on the right side.
3. Add the square of one-half of the coefficient of x on both sides of the resulting equation.
The left side of the equation becomes a perfect square trinomial.
4. Express the perfect square trinomial on the left side of the equation as a square of a
binomial.
5. Solve the resulting quadratic equation by extracting the square root.
6. Solve the resulting linear equations.
7. Check the solutions obtained against the original equation.
8. check the results in the original equation.
Steps in solving quadratic equation using the quadratic formula.
1. Write the equation in standard form
2. Determine the values of a, b, c and and substitute them in the quadratic formula.
3. Solve to find the solutions
4. Check by substituting each solution in the original equation.
Page 29 of 148
Advantages and disadvantages of the different methods of solving quadratic equations.
Method
Disadvantages
Solves equation of the form
x2 = c very quickly
Can be used to solve any
quadratic equation.
May take a long time when the
origInal equation is not written in
the form x2 = c
Factoring
May solve quadratic
equations very quickly
Applicable only on some
equations; many quadratic
equations are difficult to solve
using factoring
Completing the Square
Good to use on equations of
the form x2 + bx = -c, when b
is even
Can be used to solve any
quadratic equation
May be difficult to use when a ≠
1 or b is not even in x2 + bx = -c
Can be used to solve any
quadratic equation.
May Take more time than
factoring or the extracting square
root method for certain
equations.
Extracting Square Roots
The Quadratic Formula
I.
Advantages
Evaluating learning
Direction: Use yellow papers to answer the evaluation. Provide necessary solutions for
your answers. Use the format shown below to label your paper and it is to be
submitted to your Math teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 2 Parent’s Signature: __________________
EVALUATION 2.2
I.
Choose the letter of the best answer.
1. What value of c will make x2 + 5x + c a perfect square trinomial?
A. 1/4
C.2/5
B. 5/2
D.25/4
2. Which of the following has 4 and -9 as roots?
A. x2 + 5x - 36 =0
C. x2 + 5x + 36 = 0
2
B. x - 5x - 36 =0
D. x2 - 5x + 36 =0
3. Which of the following is a solution of (x – 6)(x + 5) = 12?
A. 5
C. 7
B. 6
D. 8
1
4. What are the roots of 2 𝑥 2 + 4 = −5?
A. ±2√3
C. ±2√5
B. ±3√3
D. ±5√2
5. If -2 is one of the roots of 2x2 – 3x – 14 = 0, what is the other root?
A. 3/2
C. 7/2
B. -5/2
D. -9/2
Page 30 of 148
II.
Given the quadratic equations below, find the roots using the appropriate method. Use
only one method for each item. No duplication of method is allowed for all the items.
6. 2𝑥 2 − 7 = 25
7. 3𝑥 2 + 11𝑥 = 4
8. 2𝑥 2 − 2𝑥 = 12
9. 5𝑥 2 − 4𝑥 + 2 = 0
10. x2 + 5x = 0
J. Additional activities for application or remediation
PRACTICE 1: Extracting Square Roots
Solve the following equations by extracting square roots.
1. t2 = 81
Answer: _________________
2. 2s2 = 50
Answer: _________________
3. (k + 7)² = 289
Answer: _________________
4. (2m – 1)² = 225
Answer: _________________
5. 3h² - 147 = 0
Answer: _________________
You may check your answers by turning to Page 31 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
PRACTICE 2 : Factoring
A. Find the roots of the following quadratic equation.( 𝑎𝑥 2 + 𝑏𝑥 = 0 )
1.
𝑥 2 − 6𝑥 = 0
2. 8𝑥 2 + 12𝑥 = 0
Answer: _________________
Answer: _________________
B. Find the roots of the following quadratic equation. (𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0)
1. 𝑥 2 − 2𝑥 − 15 = 0
Answer: _________________
2. 5 𝑥 2 − 19𝑥 + 12 = 0
Answer: _________________
You may check your answers by turning to Page 31 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 31 of 148
PRACTICE 3: Completing the Square
Put
if the given roots are the solutions of each quadratic equation, otherwise
_____1. x – 18x = -72
x = 12 and x = 6
_____2. 2x2 – 3x + 2 = 0
x = -1/2 and x = 2
_____3. x2 – 6x + 16 = 0
x = 8 and x = -2
_____4. 4x2 – 5x - 9 = 0
x = 9/4 and x = -1
_____5. 7x – 6x - 5 = 0
x = -1 and x = -5
2
2
.
You may check your answers by turning to Page 32 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
PRACTICE 4 : The Quadratic Formula
Solve the following using the Quadratic Formula.
1. 𝑥 2 + 10𝑥 = 3
2. 𝑥 2 − 6𝑥 − 10 = 0
3. 2𝑥 2 + 8𝑥 − 10 = 0
4. 2𝑥 2 − 3𝑥 − 9 = 0
5. −3𝑥 2 − 𝑥 + 2 = 0
You may check your answers by turning to Page 23 for the key to correction.
You may check your answers by turning to Page 32 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
KEY TO CORRECTION:
Review 1:
Quadratic Equationin one variable is a _________________
sentence of degree _____.
mathematical
2
2
It can be written in the standard form__________________,
where a, b,&c are _______
real
𝑎𝑥 + 𝑏𝑥 = 0
a
zero
numbers and ___can
not be equal to _______.
quadratic
In the standard form of the quadratic equation, ax2,is called the _____________term,
bx is
linear
constant
the ____________ term, while c is the _____________term.
Review 2:1. D
2. E
3. C
Comprehension Check 1:
1) x=12
3) x=17,x=-9
4. B
4)
5. A
2)
x=15
x=8,x=-7
Page 32 of 148
Comprehension Check 2:
1. x = 0 ; x = 5/2
2. x = 0 ; x = 6
3. x = 0 ; x = 3/2
4. x = 0 ; x = -10/3
Comprehension Check 3:
1. y = 1/3 ; y = - 2
2. m = 7 ; m = 2
3. x = -7 ; x = 1/2
4. x = 3/2
Activity 2: Mathematics Meaning
We get the word mathematics from the Greek word mathematics. What does it mean?
D
I
S
P
O
S
E
D
T
O
L
E
A
R
N
I
N
G
6
11
7
2
3
7
1
6
4
3
9
1
12
5
10
11
10
8
Activity 3. RIDDLE
R
A
I
N
B
O
W
Activity 4: See the Difference!
Question :What is the difference between a train conductor and a math teacher?
1)
2)
3)
4)
5)
𝑶𝑵 𝑬
𝟔 𝟗 𝟏𝟑
𝑴 𝑰 𝑵 𝑫 𝑺
𝟒 𝟏𝟎 𝟗 𝟏𝟐 𝟖
𝑻𝑯 𝑬
𝟐 𝟑 𝟏𝟑
𝑻 𝑹 𝑨 𝑰 𝑵
𝟐 𝟏𝟏 𝟓 𝟏𝟎 𝟗
𝑾𝑯 𝑰 𝑳 𝑬
𝟕 𝟑 𝟏𝟎 𝟏 𝟏𝟑
6)
7)
8)
9)
10)
𝑻𝑯 𝑬
𝟐 𝟑 𝟏𝟑
𝑶𝑻𝑯 𝑬 𝑹
𝟔 𝟐 𝟑 𝟏𝟑 𝟏𝟏
𝑻 𝑹 𝑨 𝑰 𝑵𝑺
𝟐 𝟏𝟏 𝟓 𝟏𝟎 𝟗 𝟖
𝑻𝑯 𝑬
𝟐 𝟑 𝟏𝟑
𝑴 𝑰 𝑵 𝑫
𝟒 𝟏𝟎 𝟗 𝟏𝟐
PRACTICE 1: Extracting Square Roots
Solve the following equations by extracting square roots.
1. t =  9
2. s =  5
3. k = 10 ; k = - 24
4. m = 3 ; m = - 2
5. h =  7
Page 33 of 148
PRACTICE 2 : Factoring
A.
1.
x=0 ; x=6
2. x = 0 ; x = - 3/2
B.
1. x = 5 ; x = -3
2. x = 3 ; x = 4/5
PRACTICE 3: Completing the Square
1.
2.
3.
4.
5.
PRACTICE 4: The Quadratic Formula
1. x = −5 + 2√7 and x = −5 − 2√7
2. x = 3 + √19 and x = 3 − √19
3. : x = −5 andx = 1
3
4. x = − 2 andx = 3
2
5. x = −1 andx = 3
If you have questions regarding this lesson, you may contact me at #09434634247.
Page 34 of 148
SELF-INSTRUCTIONAL PACKETS (SIPacks)
MATHEMATICS Grade 9 – Regular
PAMPANGA HIGH SCHOOL
School
IVAN T.SALAS
Teacher
I.
Teaching Dates/ Week
September 7 - 11, 2020 (Week 3)
Quarter
FIRST
OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations,
inequalities and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various
situations, formulate real-life problems involving quadratic equations, inequalities and
functions, and rational algebraic equations and solve them using a variety of strategies.
C. Learning competencies:
The learner:
1. characterizes the roots of a quadratic equation using the discriminant (M9AL-Ic-1);
2. describes the relationship between the coefficients and the roots of quadratic equation
(M9AL-Ic-2);
3. solves equations transformable to quadratic equations (including rational algebraic
equations) (M9AL-Ic-d1).
D. Objectives:
At the end of the lesson, the learners should be able to:
1.) solve the discriminant;
2.) describe the nature of the roots of the quadratic equation using the value of the
discriminant;
3.) solve for the sum and product of roots of quadratic equations;
4.) describe the relationship between coefficients and the roots of a quadratic
equations;
5.) determine the equation given the sum and products of its roots; and
6.) solve equations transformable to quadratic equations (including rational
algebraic equations).
II.
CONTENT
The Nature of the Roots of a quadratic Equation
The Sum and the Product of Roots of Quadratic Equations
Equations Transformable to Quadratic Equations
Learning Resources:
References:
Learner’s Material for Mathematics Grade 9 pp. 56-95
Teacher’s Guide for Mathematics pp. 39-53
Grade 9 Mathematics: Patterns and Practicalities
Other Learning Resources
Page 35 of 148
DAY 1 (Setember 7)
PROCEDURES
A. Reviewing previous lesson or presenting the new lesson
Hi! How are you?
Congrats! You were able to accomplish the lessons for the second
week given to you. You did it well and with that, I salute you.
Now before you go to another topic, let us have a review of your
knowledge and understanding of the number system. Do you still
remember it?
Activity 1: COMPLETE THE DIAGRAM
COMPLEX NUMBERS
ZERO
INTEGERS
REAL NUMBERS
RATIONAL NUMBERS
NATURAL NUMBERS
FRACTIONS(Non-Integers)
WHOLE NUMBERS
NEGATIVE NUMBERS
IMAGINARY NUMBERS
IRRATIONAL NUMBERS
Complete the diagram for the Complex Number System above by writing the correct
answer on each box. Choose the correct answer from inside the box to the right. Write
your answers in your notebook.
You may check your answers by turning to Page 55.
Give yourself one (1) point for every correct answer. I hope you got them all correctly!
B. Establishing a purpose for the lesson
Let us have another short activity!
Activity 2: WHICH ARE REAL? WHICH ARE NOT?
Page 36 of 148
Questions:
1. Which of the numbers are real? Which are not real?
2. Which of the numbers are rational? irrational?
3. Which of the numbers are perfect squares? not perfect squares?
You may check your answers by turning to Page 55.
C. Presenting examples/instances of the lesson
Your knowledge and understanding of the number system, coupled with your skills in
finding
solutions to quadratic equations will help you through to our discussion in this
SIPack.
You are now ready to level up to the next lesson about determining the nature of the
roots of a quadratic equation using the discriminant: b2 – 4ac.
−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
Remember the quadratic formula:
The expression under the radical, b2 – 4ac is called the discriminant of the quadratic
equation ax2 + bx + c = 0.
The value of b2 – 4ac is used to describe the nature of the roots of a quadratic equation
- the number and the types of solutions. See table below.
Table 1. Table of Discriminants
Value of Discriminant
b2 – 4ac
Nature of the Roots
Number of Solutions
Types of Roots
one real solution
Real, Rational, Equal
(RRE)
b2 – 4ac > 0 and a
Perfect Square
two different real solution
Real, Rational, Unequal
(RRU)
b2 – 4ac > 0 and Not a
Perfect Square
two different real solution
Real, Irrational, Unequal
(RIU)
b2 – 4ac  0 ( Negative)
two different imaginary
solution
No Real Roots
b2 – 4ac = 0
(Zero)
Examples: Use the discriminant to determine the nature of the roots of each quadratic
equation.
Example 1: Solve x2 - 4x + 4 = 0.
Solution:
Write in standard form
Identify a, b, and c:
Solve for b2 - 4ac:
:
x2 - 4x + 4 = 0.
a = 1, b = -4, c = 4
(- 4)2 – 4 (1) (4) = 0
Since the value of b2 - 4ac = 0, this quadratic equation has one real, rational
solution.
Page 37 of 148
To check: Solve x2 - 4x + 4 = 0.
By factoring:
(x – 2) (x – 2) = 0
x–2=0 ; x-2=0
x=2
x=2
The roots of the quadratic equation x2 - 4x +4 = 0 are real, rational and equal. In terms of
numbers of solutions, since the roots are the same then it has one real solution.
Now, try this…
Describe the roots of x2 + 6x + 9 = 0.
You should get 0 as the value of the discriminant.
If you got the correct answer, you may proceed to example 2. If not, I am sorry
but you have to try it again.
Example 2: Solve x2 + 7x + 10 = 0
Solution:
Write in standard form
:
Identify a, b, and c:
Solve for b2 - 4ac:
x2 + 7x + 10 = 0
a =1, b = 7, c = 10
2
(7) – 4 (1) (10) = 9
Since the value of b2 - 4ac > 0, is positive and a perfect square, this quadratic equation
has two, real, rational and unequal solutions.
.
To check: Solve x2 + 7x + 10 = 0
By factoring:
(x + 2) (x + 5) = 0
x+2=0 ; x+5=0
x = -2
x = -5
The roots of the quadratic equation x2 + 7x +10 = 0 are real, rational but not equal.
Now, try this…
Describe the roots of x2 + 9x + 20 = 0. You should get 1 as the value of the
discriminant.
If you got the correct answer, you may move on to example 3. If not, you need
to go back to example number 2 and then try to answer it again.
Example 3: Solve x2 + 6x = - 3.
Solution:
Write in standard form
Identify a, b, and c:
Solve for b2 - 4ac:
:
x2 + 6x + 3 = 0
a =1, b = 6, c = 3
(6)2 – 4 (1) (3) = 24
Since the value of b2 - 4ac > 0, is positive and not a perfect square, this quadratic
equation has two, real, irrational and unequal solutions.
Page 38 of 148
To check: Solve x2 + 6x = - 3 using the quadratic formula.
𝑥=
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
−6 ±
𝑥=
√(6)2
− 4(1)(3)
2(1)
−6 ± √36 − 12
2
𝑥=
−6 ± √24
2
𝑥=
−6 + √24
2
𝑥=
−6 − √24
2
𝑥 = −3 + √6 and 𝑥 = −3 − √6
−6 ± √24
2
The roots of the quadratic equation x2 +6x +3 = 0 are real, irrational and unequal.
𝑥=
Now, try this…
Describe the roots of x2 + 8x + 5 = 0.
You should get 44 as the value of the discriminant.
If you got the correct answer, you may proceed to example 4. If not, I am
sorry but you need to try it all over again.
Example 4: Solve x2 - 2x + 5 = 0.
Solution:
Write in standard form:
Identify a, b, and c:
Solve for b2 - 4ac:
x2 + 2x + 5 = 0
a =1, b = 2, c = 5
(2)2 – 4 (1) (5) = -16
Since the value of b2 - 4ac < 0, is negative, this quadratic equation has no real roots or
solutions.
To check: Solve x2 + 2x + 5 = 0 using the quadratic formula.
𝑥=
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
−(−2) ±
√(−2)2
− 4(1)(5)
𝑥=
2 + √−16
2
𝑥=
2 − √−16
2
2(1)
𝑥=
2 ± √4 − 20
2
𝑥=
2 ± √−16
2
The roots of the quadratic equation x2 - 2x +5 = 0 are not real numbers.
Now, try this…
Describe the roots of x2 - 5x + 7 = 0.
You should get -3 as the value of the discriminant.
If you got the correct answer, you may proceed to the next lesson. If
not, I am sorry but you need to try it all over again.
Page 39 of 148
For a quick review of the four examples;
Equation
1. x2 - 4x + 4 = 0
Discriminant
(- 4)2 – 4 (1) (4) = 0
b2 - 4ac is zero
Solutions
Nature of Roots
x=2
one real, rational
x = -2 ; x = -5
two, real, rational
and unequal
(7)2 – 4 (1) (10) = 9
2
2. x + 7x + 10 = 0
b2
- 4ac is positive
perfect square
(6)2 – 4 (1) (3) = 24
3. x2 + 6x + 3 = 0
b2 - 4ac is positive not
perfect square
no real roots or
solutions.
2 ± √−16
𝑥=
2
(2) – 4 (1) (5) = -16
2
4. x2 + 2x + 5 = 0
two, real,
irrational and
unequal
−6 ± √24
𝑥=
2
b2 - 4ac is negative
D. Discussing new concepts and practicing new skills # 1
DAY 2 (September 8)
You have just seen that the coefficients a, b, and c of a quadratic equation can be used to
tell the nature of the roots of a quadratic equation by the expression b2 – 4ac called
disciminant.
The next part of the discussion will answer the question:
How are the coeffecient of the terms of quadratic equation related to the roots of quadratic
equation?
Study the table below.
Table 2: The Sum and Product of the Roots
Coefficients
Roots
Sum
Product
x1 + x2
x1• x2
Equation
a
b
c
x1
x2
1. x2 + 7x + 12 = 0
1
7
12
-3
-4
-7
12
2. 2x2 – 3x – 20 = 0
2
-3
-20
5/2
4
3/2
-10
What do you notice to the relationship of the sum and products of the roots to the value
of a, b, and c of the given equations?
Yes, I can hear you!
If you let x1 and x2 be the roots of a quadratic equation:
The sum of the roots of the quadratic equation x1 + x2 =
𝒄
and the product x1 • x2 =
.
𝒂
−
𝒃
𝒂
Page 40 of 148
I’ll show you how these relationships were derived.
From the quadratic formula, let 𝑥1 =
−𝑏+√𝑏2 −4𝑎𝑐
2𝑎
and 𝑥2 =
−𝑏−√𝑏2 −4𝑎𝑐
be the roots.
2𝑎
We can derive the sum and the product of the roots of quadratic formula.
Sum of the Roots of Quadratic Formula
𝑥1 + 𝑥2 =
−𝑏 +
√𝑏 2
− 4𝑎𝑐
2𝑎
−𝑏 − √𝑏 2 − 4𝑎𝑐
+
2𝑎
𝑥1 + 𝑥2
=
(−𝑏 + √𝑏 2 − 4𝑎𝑐) + (−𝑏 − √𝑏 2 − 4𝑎𝑐)
2𝑎
𝑥1 + 𝑥2 =
−2𝑏
2𝑎
−𝑏
𝑥1 + 𝑥2 =
𝑎
Product of the Roots of Quadratic
Formula
𝑥1 + 𝑥2
−𝑏 + √𝑏 2 − 4𝑎𝑐 −𝑏 − √𝑏 2 − 4𝑎𝑐
=(
)(
)
2𝑎
2𝑎
(−𝑏)2 − (√𝑏 2 − 4𝑎𝑐)2
𝑥1 + 𝑥2 =
(2𝑎)2
𝑥1 + 𝑥2 =
𝑏 2 − 𝑏 2 + 4𝑎𝑐
4𝑎2
4𝑎𝑐
𝑐
→ 𝑥1 + 𝑥2 =
2
4𝑎
𝑎
𝑥1 + 𝑥2 =
Let’s have some examples.
Example 1: Find the sum and the product of the roots of 2x2 + 8x – 10 = 0.
Solution:
Write in standard form:
2x2 + 8x – 10 = 0.
Identify a, b, and c:
a = 2, b = 8, c = -10
Sum of the roots:
Product of the roots:
−𝑏
𝑎
𝑐
𝑎
=
=
−(8)
2
−10
2
= −4
= −5
To check: Find the roots of 2x2 + 8x – 10 = 0 using any method
The roots are:
1 and -5.
Let:
x1 = 1 and x2 = –5.
Sum of the roots:
x1 + x2 = 1 + (–5) = –4
Product of the roots:
x1 • x2 = (1)(–5) = –5
Therefore, the sum and the product of the roots of 2x2 + 8x – 10 = 0 are –4 and –5,
respectively.
Now, identify the roots, the sum and products of roots of x2 – 5x + 6 =0
You should get 2 and 3 as roots, 5 is the sum and 6 is the product.
If you get the correct answer, you may proceed with Example 2.
If not, I am sorry but you have to go back to example 1 and try all over again.
Page 41 of 148
Example 2: Find the sum and the product of the roots of x2 + 7x – 18 = 0.
Solution:
Write in standard form:
x2 + 7x – 18 = 0
Identify a, b, and c:
a = 1, b = 7, c = -18
−b
a
Sum of the roots:
c
=
=
a
Product of the roots:
−(7)
1
−18
1
= −7
= −18
To check:
Find the roots of x2 + 7x – 18 = 0 using any method
The roots are:
-9 and 2.
Let:
x1 = -9 and x2 = 2.
Sum of the roots:
x1 + x2 = (-9) + 2 = –7
Product of the roots:
x1 • x2 = (-9)(2) = –18
Now, let us try solving the roots, the sum and product of the roots of given equations in
Activity 3, Write your answers in your notebook.
Activity 3: Sum and Product of the Roots.
Use the values of a, b, and c of each of the following quadratic equations in determining
the sum and the product of its roots. Verify your answers by obtaining the roots of the
equation.
1. x2 + 3x + 2 = 0
roots: ___, ____ sum: _____ product: _____
2
2. 2w – 3w – 20 = 0
roots: ___, ____ sum: _____ product: _____
3. 15h2 – 7h – 2 = 0
roots: ___, ____ sum: _____ product: _____
Now, check your work by turning to Page 55 for the key to
correction.
Give yourself 1 point for every correct answer. How many
correct answers did you get?
Score
11-12
8-10
6-7
0-5
Rate your result using the table.
Description
Very Good
Good
Fair
Turn to page 18
sec. J and work on
the enrichment
activities
If I may ask you: Is it possible to write or determine the quadratic equation given its
roots?
The answer is yes.There are two ways of finding the quadratic equation given its roots:
First method:
We use the equations describing the roots to come up with two binomials whose product is
zero.If the resulting equation is simplified ,it becomes a quadratic equation in the form ax²
+bx + c=0
Second method:
We get the sum and product of the roots and substitute these in the equation
x² + 𝑏𝑎 x +𝑎𝑐 = 0 ,where 𝑎𝑏 is the sum of the roots and
𝑐
𝑎
is the product of the roots.
Page 42 of 148
Let’s have some examples.
Writing the quadratic equation in the form ax².+bx+c=0 given the roots -1 and -3.
Method 1: Let x1 = -1 and x2 = -3
Getting 0’s on one side:
Applying Zero product Property:
Multiplying using FOIL:
Combine like terms:
x + 1 = 0 and x + 3 = 0
(x+1) (x+3) = 0
x2 + 3x + x
+3=0
x² + 4x + 3 = 0
The quadratic equation:
x² + 4x + 3 = 0
To check if the equation is correct , substitute the values of x in the equation obtained
for x = -1:
(-1)2 + 4(-1) + 3 = 0
for x = -3:
(-3)2 + 4(-3) + 4 = 0
1–4+3=0
9 – 12 + 4 = 0
0=0
0=0
Both roots satisfy the equation
Now, write the quadratic equation whose roots are -4 and – 5 using method 1.
You should get x2 + 9x + 20 = 0 as the equation.
If you get the correct answer, you may proceed with Method 2.
If not, I am sorry but you have to go back to example Method 1 and try all over
again.
Method 2: Let x₁ = -1 or x₂ = -3
Sum of the roots:
x₁ + x₂ = -1 + -3 = -4
𝑏
x₁ + x₂ = - 𝑎 = -(4) = -4
Thus,
𝑏
𝑎
Product of the roots:
( x₁ )( x₂) = (1)(3) =3
( x₁ )( x₂) =
=4
Thus
𝑐
𝑎
𝑐
𝑎
=3
= 3
ax².+ bx + c = 0 → x² + 𝑎𝑏 x + 𝑎𝑐 = 0
x² + (4)x + (3) =0
x²+ 4x +3 =0
Substitute the values in the equation:
The quadratic equation is:
Now, write the quadratic equation whose roots are -4 and – 5 using Method 2
You should get x2 + 9x + 20 = 0 as the equation.
If you get the correct answer, you may proceed to the next part of the lesson.
If not, I am sorry but you have to go back to example Method 2 and try all over
again.
Now, let us try writing the equationt of each given roots in Activity 4, write your answers
in your notebook.
Activity 4: Write the quadratic equation in the form ax².+ bx + c = 0, given the
following roots.
1.) -1 and -2
3.) -4 and 2
2.) 2 and 3
4.) −
5
2
and −
3
2
Page 43 of 148
Now, check your work by turning to Page 57
for the key to correction. Give yourself 1 point for every
correct answer.
Score
4
3
2
How many correct answers did you get?
Rate your result using the table.
0-1
Description
Very Good
Good
Fair
Turn to page 18
sec. J and work on
the enrichment
activities
E. Discussing new concepts and practicing new skills # 2
Remember in the previous discussion that there are quadratic equations which are not in
simplified form.
This part of the lesson is focus on solving equations transformable to quadratic
equations.
Let us have the following examples.
Example 1: Solve 𝑥(𝑥 − 5) = 36.
Write the quadratic equation in
standard form :
𝑥(𝑥 − 5) = 36
𝑥 2 − 5𝑥 = 36
𝑥 2 − 5𝑥 − 36 = 0
(𝑥 − 9)(𝑥 + 4) = 0
Solve the equation using any method.
By factoring:
𝑥−9=0 ; 𝑥+4=0
𝑥=9
; 𝑥 = −4
Therefore, the roots of the quadratic equation 𝑥(𝑥 − 5) = 36 are 9 and -4.
To check: For x = 9: 𝑥 (𝑥 − 5) = 36 For x = 4:
9 (9 -5) = 36
9 (4) = 36
36 = 36 True
𝑥(𝑥 − 5) = 36
-4(-4 -5) = 36
-4 (-9) = 36
36 = 36 True
Find the roots of x(x -10) = - 21
You should get x= 7 or x = 3
If you got the correct answer you may proceed to the next example
If not, I am sorry but you need to go back to example 1 and try it again.
Example 2: Find the roots of the equation (𝑥 + 5)2 + (𝑥 − 2)2 = 37.
𝑥 2 + 10𝑥 + 25 + 𝑥 2 − 4𝑥 + 4 = 37
𝑥 2 + 10𝑥 + 25 + 𝑥 2 − 4𝑥 + 4 = 37
2𝑥 2 + 6𝑥 − 8 = 0
(2𝑥 − 2)(𝑥 + 4) = 0
Solve the equation by factoring:
2𝑥 − 2 = 0
𝑥+4=0
𝑥=1
𝑥 = −4
Therefore, the roots of the quadratic equation (𝑥 + 5)2 + (𝑥 − 2)2 = 37 are 1 and -4.
To check:For x = 1: (𝑥 + 5)2 + (𝑥 − 2)2 = 37
For x= 4: (𝑥 + 5)2 + (𝑥 − 2)2 = 3
(1 + 5)2 + (1 − 2)2 = 37
(−4 + 5)2 + (−4 − 2)2 = 3
(6)2 + (−1)2 = 37
(1)2 + (−6)2 = 37
37 = 37 True
37 = 37 True
Write the quadratic equation in
standard form:
Page 44 of 148
Try This :
Find the roots of (x+1)2 + (x - 3)2 = 15
You should get
2+√14
2
or
2−√14
2
If you got the correct answer, you may proceed to the next example. If not, I
am sorry but you need to go back to example 2 and try it again.
Example 3. Solve the rational algebraic equation
6
𝑥
+
𝑥−3
4
= 2.
The given rational algebraic equation can be transformed to a quadratic equation. To
solve the equation, the following procedure can be followed.
6
𝑥
+
𝑥−3
4
6
= 2→4𝑥 (𝑥 +
𝑥−3
4
) = 4𝑥 (2)
Multiply both sides of the equation by the
24 + x – 3x = 8x
24+x2 – 3x = 8x
Least Common Multiple (LCM) of all denominators.
In the given equation, the LCM is 4x
x2 – 11x + 24 = 0
Write the resulting quadratic equation in standard
form
(x – 3) (x – 8) = 0
x – 3 = 0 or x – 8 = 0
x = 3 or x = 8
Find the roots of the resulting equation using any of
the methods of solving quadratic equations. Try
factoring in finding the roots of the equation.
2
6
Check whether the obtained values of x make the equation 𝑥 +
6
If the obtained values of x make the equation 𝑥 +
𝑥− 3
4
𝑥−3
4
= 2 true.
= 2 true, then the solutions of the
equation are x = 3 or x = 8.
Try This:
1
Find the roots of 3𝑥 +
4𝑥
6
=1
1
You should get x = 1 or x = 2
If you got the correct answer, you may proceed. If not, I am sorry but you
need to go back to example 3 and try it again.
Page 45 of 148
F. Developing mastery
DAY 3 (September 9)
Answer Activity 1 below. Read the direction carefully and write your answers on the space
provided.
Activity 1:
Give the values of the discriminant of the following quadratic equations to get the code to be
used in completing the message on the next page.Write your answers in your notebook.
QUADRATIC EQUATIONS
1) 1) x2 + 4x – 21 = 0
2) x2 + 7x + 12 = 0
3) 2x2 – x – 10 = 0
4) 15 x2 – 10x + 9 = 0
5) 2x2 + 5x + 2 = 0
6) 3x2 + 8 = x2 – 8x
7) 4x2 + 3x + 3 = 0
8) 18x2 + 8x + 5 = 0
9) 9x2 – 4x + 3 = 0
10) 5x2 = – 15
11) x2 + 7x - 4 = 0
12) x2 - 12x + 35 = 0
13) x2 + 10x + 9 = 0
14) x2 - 4x + 12 = 0
𝒃𝟐 − 𝟒𝒂𝒄
F. 81
R. 0
S. -300
Y. 1
I. -39
E. 100
N. -296
W. 9
T. -92
G. -440
V. 64
D. -32
H. 65
O. 4
Page 46 of 148
Activity 1:
Message
5 6 12 8 4
1 13 1 8
7 10
7 3
7 3
8 12
5 6 12 8 4
1 13 1 6 2 12 8 1
14 12 7 8 4
6 7 4 11 9
1 13 1 8
7 10
7 10
7 9
6 7 4 11 9
12 8 1
7 10
14 12 7 8 4
7 9
Describe the nature of the roots of the given quadratic equations above.
QE 1. __________________________
QE 8 ________________________________
QE 2. __________________________
QE 9 ________________________________
QE 3. __________________________
QE 10 ________________________________
QE 4. __________________________
QE 11 ________________________________
QE 5. __________________________
QE 12 ________________________________
QE 6. __________________________
QE 13 ________________________________
QE 7. __________________________
QE 14 ________________________________
You may check your answers by turning to Page 56 for the
key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get? Rate the result
using the table.
Score
11 - 14
7 - 10
3-6
0-2
Description
Victory
Nice One
Notbad you’re
almost there.
Turn to page 52
sec. J and work
on the
enrichment
activities
Answer Activity 2 below. Read the direction carefully and write your amswers on the space
provided. Write your answers inyour notebook.
Activity 2:
A. Given the equation, determine the sum and the products of its roots.
1. x2 – 12x – 45 = 0
Sum: ________
Product: ________
2. 12x2 – x – 6 = 0
Sum: ________
Product: ________
3. x2 + 9 = 0
Sum: ________
Product: ________
Score
7-8
Description
Very Good
Page 47 of 148
You may check your answers by turning to Page 56 for the
key to correction.
5-6
3-4
Give yourself one (1) point for every correct answer.
0-2
How many correct answers did you get? Rate the result
using the table.
Good
Fair
Turn to page
53 sec. J and
work on the
enrichment
activities
Answer Activity 3 below. Read the direction carefully and write your answers on the space
provided. Write your answers in your notebook.
Activity 3: Here are the Roots. Where is the Trunk
Write the quadratic equation in the form ax2 + bx + c = 0 given the following roots.
1. 5 and 9
Answer: ___________________
2. 8 and 10
Answer: ___________________
3. 6 and 3
Answer: ___________________
4. -2/3 and ¾
Answer: ___________________
5. -3 and 15
Answer: ___________________
You may check your answers by turning to Page 56 for the
key to correction. Give yourself one (1) point for every correct
answer.
Score
5
4
3
How many correct answers did you get?
Rate the result using the table.
1-2
Description
Very Good
Good
Fair
Turn to page 53
sec. J and work
on the
enrichment
activities
Answer Activity 4 below. Read the direction carefully and write your answer in your
notebook.
Page 48 of 148
Activity 4: Find my Food Maze!
Help the insects find their food by transforming the algebraic expressions into quadratic
equations first, and then trace their path using a crayon or any coloring material.
You may check your answers by turning to Page 57 for
the key to correction. The rubric on the right side is your
guide in checking your work.
If you got a score below 6, you may proceed to section J
for enrichment activities.
Rubrics
Accuracy→ 7 points
7 → got 3 correct
answers
4 → got 2 correct
answers
1 → got 1 correct answer
Correct Path→ 3 points
3 → traced 3 correct path
2 → traced 2 correct path
1 → traced 1 correct path
G. Finding practical application of the concepts and skill in daily living
DAY 4 (September 10)
1. Since the nature of the roots is determined by the kind of discriminant, once this is
identified we can still use the concept of quadratic equation when going on a river cruise.
A 3-hour river cruise goes 15 km upstream and then back again. The river has a current
of 2 km an hour. What is the boat's speed and how long was the upstream journey?
We can represent the problem by the equation 3x² − 30x − 12 = 0.
By using the Quadratic Formula, the answer would be:
Boat's Speed = 10.39 km/h (to 2 decimal places)
And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min
And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min
Page 49 of 148
2. For the Sum and Product of the Roots of Quadratic Equations:
The concept of the sum and product of the roots of quadratic equations will help you
determine not only the roots but also the equation. In doing so, you can apply this in
Aeronautics by finding how high above sea level can a rocket reach and what time will it
splash down into the ocean or on land.
3. Equations Transformable to Quadratic Equations
In Electronics,
Two resistors are in parallel, like in this diagram:
The total resistance has been measured at 2 Ohms, and one of the resistors is known to
be 3 ohms more than the other. What are the values of the two resistors?
1
1
1
The formula to work out total resistance "RT" is:𝑅 = 𝑅 + 𝑅
𝑇
1
2
In this case, we have 𝑅𝑇 = 2 and 𝑅2 = 𝑅1 + 3
Substituting, we have
1
2
1
=𝑅 +𝑅
1
1
, and the equation would be 𝑅12 − 𝑅1 - 6 = 0. The roots
1 +3
are -2 and 3, and since 𝑅1 cannot be negative, so 𝑅1 = 3 ohms. The two resistors are 3
ohms and 6 ohms.
H. Making generalization and abstraction about the lesson
As for our summary, in determining the nature of the roots and the number
of solutions in a quadratic equation you have to:
1. Check first if the equation is expressed in the standard form ax2 + bx + c =
0 if not you need to write it in its standard form.
2. Identify the values of a, b and c.
3. Find the value of the discriminant using the expression b2 – 4ac.
4. Determine the nature of the roots using the Table of Discriminant as your
guide.
Page 50 of 148
There are two ways of finding the quadratic equation given its roots:
First method:
We use the equations describing the roots to come up with two binomials whose
product is zero. If the resulting equation is simplified, it becomes a quadratic
equation in the form ax² + bx + c=0
Second method:
We get the sum and product of the roots and substitute these in the equation
x² +
𝑏
𝑐
x+
𝑎
𝑎
To transform rational algebraic equations into quadratic equations:
1. Write the equation into standard form
2. Solve the equations using any method
3. Check the roots if it satisfy or make the equation true
Page 51 of 148
I. Evaluating Learning
Direction: Use yellow papers to answer the evaluation. Provide necessary solutions for
your answers. Use the format shown below to label your paper and it is to be
submitted to your Math teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 3 Parent’s Signature: __________________
EVALUATION 3
I. Determine the nature of the roots and number of solutions of the following quadratic
equations using the discriminant. The first one was done for you.
Quadratic equation
1. x2 + 7x + 12 = 0
Discriminant ( b2 – 4ac)
No. of Solutions
Nature of the Roots
b2 – 4ac = (7)2 – 4(1)(12)
= 49 - 48
=1
( positive perfect square)
Two different real
solutions
Real, rational,
unequal
2. 1. x2 + 5x + 10 = 0
3. 2. x2 - 10x + 8 = 0
4. 3. x2 + 3x + 2 = 0
5. 4. x2 + 8x + 16 = 0
6. 5. 3x2 - 5x = -4
II. Find the sum and the product of the roots of the equations from the given problem.Match
the quadratic equation in column A with its sum and product in column B, then answer
the question below.
This man is responsible for reshaping the way early man believed the solar system
worked. He proposed that the Earth was not the center of the universe, and that the
sun was instead at the center of our solar system. Who was he?
Column A
Column B
3
1. x2 + 4x + 3 = 0
N: Sum = -2, Product = 4
2. 6x2 + 12x – 18 = 0
C: Sum = -4, Product = 3
3.x2 + 4x – 21 = 0
I: Sum = 3, Product = − 9
4. 2x2 + 3x – 2 = 0
O: Sum = -2, Product = -3
5.3x2 – 10x – 8 = 0
C: Sum = 4, Product = − 8
6.4x2 + 8x + 3 = 0
P: Sum = -4, Product = -21
7.9x2 – 6x = 8
U: Sum = 10, Product = 5
8.8x2 = 6x + 9
E: Sum = − 2, Product = -1
9.10x2 – 19x + 6 = 0
S: Sum = 2, Product = 0
10.2x2 – 3x = 0
R: Sum =
2
8
3
9
19
3
3
3
10
,
3
8
Product = − 3
Page 52 of 148
EVALUATION CONT…
Answer
1
2
3
4
5
6
7
8
9
10
III. Write the quadratic equation in the form ax2 + bx + c = 0 given the following roots.
1. -9 and 0
2. 2.5 and 4.5
3. - 3 and - 3
4. - 8 and -10
5. - 5/6 and - 1/6
IV.
Transform each into quadratic equation and then solve using any method.
1. x(x + 3)= 28
2. 3s(s-2) = 12s
3. (t-1)² +( t-8)² =45
4. (3r+1)² + (r+2)² = 65
5.
(𝑥+2)²
5
(𝑥−2)²
+
3
=
16
3
J. Additional activities for application of remediation
DAY 5 (September 11)
Remediation 1: Nature of the roots
Write in your notebook the correct nature of the roots of the following quadratic
equations.
1. x2 +10x + 9 = 0
RRE
RRU
RIU
no real roots
2. x2 + 7x = 4
RRE
RRU
RIU
no real roots
3. x2 +10x + 25 = 0
RRE
RRU
RIU
no real roots
4. x2 -4x = -9
RRE
RRU
RIU
no real roots
5. 3x2 -2x -5 = 0
RRE
RRU
RIU
no real
roots
You
may check your answers by turning to Page 57 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 53 of 148
Remediation 2: Sum and Product of the Roots
Find the roots, the sum and the product of roots of the given equations below. Write
your answers in your notebook.
Quadratic Equation
Roots
Sum of Roots
Product of roots
1. x2 - 5x + 6 =0
2. x2 + 4x - 12 = 0
3. 2x2 + 4x – 6 = 0
4. x2 + 7x + 10 = 0
5. 3x2 +7x – 6 = 0
You may check your answers by turning to Page 57 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Remediation 3: Writing Quadratic Equations Given the Roots.
Write the quadratic equation in the form ax2 + bx + c = 0 given the following roots.
Write your answers in your notebook.
Roots
Equation
6. 3 and -4
7. 2 and 3/5
8. -7 and 4
9. -7 and -7
10. 1 and -9
You may check your answers by turning to Page 58 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 54 of 148
Remediation 4: Transform the following into the form ax² + bx + c = 0
What do you call a horizontal line placed over an expression to show that everything
below the line belongs to one group? This symbol is used in division too.
You can find the answer to the question by transforming the ff. rational algebraic
expressions into quadratic equations of the form ax² + bx + c = 0.
An answer box is provided below. Write the letter of the correct answer on the space
provided. Write your answers in your notebook.
1.
2.
3.
3
𝑥
+
𝑥−2
−4𝑥
𝑥−1
6.
7.
8.
1
1
𝑥−5
5
𝑥−2
𝑥+3
𝑥−4
4
𝑥−3
=1
𝑥+4
3𝑥+2
4. 1 =
5.
5
1
−2
𝑥
𝑥 2 −2𝑥
+ =
+
4
𝑥−4
=
+
𝑥+7
𝑥
=7=
+
𝑥−5
𝑥+4
𝑥
𝑥+2
2
3x2 – x – 4 =0
I
3x² + 3x =0
𝑥 2 −1
C
7x² - 64x+192=0
7𝑥−42
V
x²-4x – 12=0
U
2x²-4x – 35=0
L
7x²-15x – 18=0
N
x² - 1=0
U
2x²-16x – 22=0
=
𝑥+1
𝑥+2
M
−8
𝑥+4
+1
10
𝑥+2
+2
= -2
3
4
5
6
7
8
You may check your answers by turning to Page 58 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 55 of 148
KEY TO CORRECTION
Activity 1: COMPLETE THE DIAGRAM
COMPLEX NUMBERS
IMAGINARY NUMBERS
Activity 2: WHICH ARE REAL? WHICH ARE NOT?
7
5
√15
9
1. .Real numbers: 24.5, 8 ,12, 289,√25 ,
,√35
√−21
9
Not real numbers: √−15 ,
7 5
8 12
2. Rational numbers: 24.5 , , , 289,√25
Irrational numbers:
√15
9
,√35
3. . Perfect square numbers: 289
Not perfect square numbers:
7 5
, , √25
8 12
√15
,
9
,
√35
Activity 3: Sum and Product of the Roots.
1.
x2 + 3x + 2 = 0
roots: -1, -2
2.
2w2 – 3w – 20 = 0
roots: 4,
3.
15h2 – 7h – 2 = 0
roots:
−1
5
−5
2
,
2
3
sum: -3
3
sum: 2
sum:
7
15
product: 2
product: -10
product:
−2
3
Page 56 of 148
DEVELOPING MASTERY (KEY TO CORRECTIONS)
Activity 1:
“Wrong is wrong even if everyone is doing it. Right is right even if no one is doing it.”
1.
2.
3.
4.
5.
6.
7.
RRU
RRU
RRU
No real roots
RRU
RRE
No real roots
Activity 2:
Sum and Product
1. Sum = -3,
2. Sum =
2
− 3,
3. Sum = -9,
8.No real roots
9. No real roots
10. No real roots
11. RIU
12. RRU
13. RRU
14. No real roots
of the Roots.
Product = 15
3
Product = − 4
Product = 0
Activity 3: Here are the Roots. Where is the Trunk
Write the quadratic equation in the form ax2 + bx + c = 0 given the following roots.
1.
2.
3.
4.
5.
x2 -14x + 45 = 0
x2 - 18x + 80 = 0
x2 – 9x + 18 = 0
12x2 + x - 6 = 0
x2 – 12x - 45 = 0
Page 57 of 148
Activity 4 :
1. x(x-9) + 35 = 27
2
2. (x + 6)² = 15
6
3. 𝑥+ 𝑥+5= 1
x² - 9x + 35 = 27
x² + 12x + 36 = 15
x² - 9x + 8 = 0
x² + 12x + 21 = 0
2
x(x + 5){𝑥 +
6
𝑥+5
= 1}
2(x + 5) + 6x = x² + 5x
2x + 10 + 6x = x² + 5x
x² - 3x – 10 = 0
ADDITIONAL ACTIVITIES/REMEDIATION (KEY TO CORRECTION)
Remediation 1:
1. RRU
2. RIU
3. RRE
4. no real roots
5. RRU
Remediation 2:
Quadratic Equation
Roots
Sum of Roots
Product of roots
1. x2 - 5x + 6 =0
2, 3
5
6
2. x2 + 4x - 12 = 0
-6, 2
-4
-12
3. 2x2 + 4x – 6 = 0
-3, 1
-2
-3
4. x2 + 7x + 10 = 0
-5, -2
-7
-10
5. 3x2 +7x – 6 = 0
-3,
−7
3
-2
2
3
Page 58 of 148
Remediation 3:
1. x2 + x -12 = 0
2. 5x2 -13x + 6 = 0
3. x2 + 3x - 28 = 0
4. x2 -49 = 0
5. x2 + 8x -9 = 0
Remediation 4:
VINCULUM
If you have any questions regarding the lesson, feel free to contact me at # 09434634247.
“It does not matter how slow you go
as long as you do not stop.” - Confucius
Page 59 of 148
SELF-INSTRUCTIONAL PACKETS (SIPacks)
MATHEMATICS Grade 9 – SSC
School
PAMPANGA HIGH SCHOOL
Teaching Dates/
Week
September 14 - 18, 2020 (Week 4)
Teacher
IVAN T. SALAS
Quarter
FIRST
I.
OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations,
inequalities and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various situations,
formulate real-life problems involving quadratic equations, inequalities and functions, and
rational algebraic equations and solve them using a variety of strategies.
C. Learning competencies:
The learnersolves problems involving quadratic equations and rational algebraic equations.
(M9AL-Ie-1).
D. Objectives:
At the end of the lesson, the learners should be able to solve problems involving quadratic
equations and rational algebraic equations.
II.
CONTENT
Solving Problems Involving Quadratic Equations
Learning Resources:
References: Learner’s Material for Mathematics Grade 9 pp. 88-95
Teacher’s Guide for Mathematics pp. 49-53
Grade 9 mathematics (Patterns and Practicalities)
III.
DAY 1 (September 14)
PROCEDURES
A. Reviewing previous lesson or presenting the new lesson
Welcome to another week of home schooling!
How was your home schoolingfor the previous week? I’m certain it went
well with the help of your parents and other persons around you.
In this detailed lesson plan, you will be learning about the wide
applications of quadratic equations as we are about to apply the
concepts, understanding and skills you gained from last week’s lesson.
You will be thinking like a mathematician, asking lots of questions,
gaining insights from the previous lessons and connecting it in real life.
But before that, let us answer first Activity 1 by solving the given
quadratic equations.
Page 60 of 148
Activity 1: REVIEW
Solve 𝑥 2 + 3𝑥 − 10 = 0 using:
a. Factoring
b. Completing the Square
c. The Quadratic Formula
You may check your answers by turning to Page 69
Give yourself one (1) point for every correct answer. I hope you got them all correctly!
B. Establishing a purpose for the lesson (Paghabi ng layunin ng aralin)
Let us have another short activity!
This activity is entitled “Turn the Fish.”This is a test of your problem-solving skills.
Activity 2: TURN THE FISH.
You may check your answers by turning to Page 69
I hope you got the moves correctly!
If I may ask, “What mathematical concepts and thinking skills did you apply in solving the two
activities?”
C. Presenting examples/instances of the lesson
Now, we will solve various real-life applications of quadratic equations. You may use whatever
is the most convenient method for you in solving the equations.
Let’s try the following examples!
Example 1: Geometry Problem
An amusement park wants to place a new rectangular billboard to inform visitors of
their new attractions. Suppose the length of the billboard to be placed is 4m longer than its
width and the area is 96 m2. What will be the length and the width of the billboard?
Page 61 of 148
Solution:
Let x- be width of the billboard.
Let x + 4 -be the length of the billboard.
If we represent the width, in meters, of the billboard by x, then its length is x + 4. Since the area
of the billboard is 96 m2, then (𝑥)(𝑥 + 4) = 96.
The equation (𝑥)(𝑥 + 4) = 96 is a quadratic equation that can be written in the form 𝑎𝑥 2 + 𝑏𝑥 +
(𝒙)(𝒙 + 𝟒) = 𝟗𝟔
𝑐 = 0.
𝒙𝟐 + 𝟒𝒙 = 𝟗𝟔
𝒙𝟐 + 𝟒𝒙 − 𝟗𝟔 = 𝟎
Solve the resulting equation using any method of solving quadratic equation. In this example,
factoring is used.
𝒙𝟐 + 𝟒𝒙 − 𝟗𝟔 = 𝟎
(𝒙 − 𝟖)(𝒙 + 𝟏𝟐) = 𝟎
𝒙−𝟖=𝟎
𝒙 + 𝟏𝟐 = 𝟎
𝒙=𝟖
𝒙 = −𝟏𝟐
The equation has two solutions: 𝒙 = 𝟖
or 𝒙 = −𝟏𝟐 .
However, we only consider the positive value of x since the situation involves measure of
length. The width of the billboard is 8.
To find the length of the billboard, use 𝑥 + 4,
L = 𝑥 + 4,
L=8+4
L = 12.
Hence, the width of the billboard is 8m and its length is 12m.
Now, you try this problem.
The length of the rectangular lot is 4 m longer than the width. If the area of the lot is 12
m , what will be the length and the width of the rectangular lot?
2
You should get, width = 2 m and length = 6 m.
If your answer is correct, you may proceed to Example 2. If not, I am sorry but you have to go
back to Example 1 and try all over again.
Here is another application.
Example 2: Number Problem
The product of two consecutive positive numbers is 240. Find the numbers.
Solution:
Let:
x – be the first number
x + 1 – be the second number (because the numbers are consecutive,
the other number is one more than the first)
Page 62 of 148
Their product equals 240:
x (x + 1) = 240
Multiplying:
x2 + x = 240
Write in standard form:
x2 + x – 240 = 0
Solve by factoring:
(x + 16) (x – 15) = 0
x + 16 = 0
x = -16
(the product results to a quadratic equation)
x - 15 = 0
x = 15
Since we are only looking for positive numbers, disregard x = -16
Therefore, the positive consecutive numbers are 15 and 16.
(This problem could have been set up with x representing the first number and x – 1
representing the second number.)
Note: For consecutive even or odd numbers, represent the first number as x and the other
number as x + 2 since consecutive even or odd numbers differ by 2.
Now try this.
The product of two consecutive positive numbers is 650. Find the numbers.
You should get, 25 and 26 as the two consecutive positive numbers.
If your answer is correct, you may proceed to the next part of the discussion. If not, I am sorry,
but you have to go back to Example 2 and try all over again.
D. Discussing new concepts and practicing new skills # 1
DAY 2 (September 15)
Revenue Problem
Another application of quadratic equation is in business. A common business application of
quadratic equations occurs when raising a price results in lower sales or lowering a price
results in higher sales. The question is what to charge to bring in the most revenue.
The revenue formula is R = P ● Q where R represents the revenue, P represents the price,
and Q represents the number sold.
If the price is increased, then P will equal the current price plus the variable times the
increment. If the price is decreased, then P will equal the current price minus the variable
times the increment.
If the sales level is increased, then Q will equal the current sales level plus the variable
times the increment gain. If the sales level is decreased, then Q will equal the current sales
level minus the variable times the increment loss.
Example 3: Revenue Problem
A gadget store sells 20 units of earphones per week at Php80. The store owner believes that
for each decrease of Php5 in the price, six more earphones are sold. What price should be charged
if the revenue needs to be Php2,240?
Let x be the number of Php5 decreases in the price
Then: P = 80 – 5x,
Q = 20 + 6x, R = (80 – 5x) (20 + 6x)
If the revenue needs to be Php2240:
2240 = (80 – 5x) (20 + 6x)
Page 63 of 148
2240 = 1600 + 380x -30x2
Simplifying:
30x2 – 380x + 640 = 0
1
1
(10) (30x2 – 380x + 640) = 0 (10)
3x2 - 38x + 64 = 0
(3x – 32) (x – 2) = 0
Solve the equation: (by factoring)
3x – 32 = 0
x=
If x =
𝟑𝟐
𝟑
x–2=0
32
x=2
3
32
, the price for each earphone will be P = 80 – 5(
3
) = Php 26.67
If x = 2, the price for each earphone will be P = 80 – 5 (2) = Php 70
Try this.
A beach resort manages 16 cottages. Each cottage can be rented if the daily rent is Php1000.
For each Php200 increased in the rent, one renter will be lost. What should be the daily rent if the
rental company needs Php20,800 each day in revenue.
You should get Php1,600 or Php2600.
If your answer is correct, you may proceed to the next part of the discussion. If not, I am
sorry, but you have to go back to Example 3 and try all over again.
Work Problem
Another application of quadratic equations is in work problems.
In the work formula Q = r ● t, where Q = quantity, r = rate, and t = time. Q is usually 1.
The equation to solve is: W1 rate + W2 rate = Together rate
The given information in the problem is usually the time one or both workers need to complete
𝑄
the job. We want the rates not the times. We can solve for r in Q = r ● t to be r = but Q is
𝑡
1
usually 1, thus r = .
𝑡
The equation to solve is usually:
1
𝑊1 𝑡𝑖𝑚𝑒
+
Solve work problems by filling in the table below
Worker
Quantity (Q)
W1
1
W2
1
Together
1
1
𝑊2
=
1
𝑇𝑜𝑔𝑒𝑡ℎ𝑒𝑟 𝑡𝑖𝑚𝑒
:
Rate (r)
1
𝑊1 𝑡𝑖𝑚𝑒
1
𝑊2 𝑡𝑖𝑚𝑒
1
𝑇𝑜𝑔𝑒𝑡ℎ𝑒𝑟 𝑡𝑖𝑚𝑒
Time (t)
W1 time
W2 time
Together time
Page 64 of 148
Example 4: Work Problem
Together Jack and Poy can paint a wall in 18 minutes. Alone, Jack needs 15
minutes longer to paint the wall than Poy needs. How much time does Jack and Poy
each need to paint the wall by himself?
Solution:
Let t - be the number of minutes Poy needs to paint the wall; and
t + 15 – be the number of minutes John needs to paint the wall
Using the table:
Worker
Quantity (Q)
Poy
1
Jack
1
Together
1
The equation to solve is:
1
𝑡
+
Rate (r)
1
𝑡
1
𝑡 + 15
1
18
1
𝑡+15
=
1
18
Time (t)
t
t +15
18
The LCD is 18t (t + 15)
1
𝑡
1
)
𝑡+15
Multiply both sides by the LCD
18𝑡(𝑡 + 15) ( ) + 18𝑡(𝑡 + 15) (
and simplify
18(𝑡 + 15) + 18𝑡 = 𝑡(𝑡 + 15)
1
18
= 18𝑡(𝑡 + 15) ( )
18𝑡 + 270 + 18𝑡 = 𝑡 2 + 15𝑡
0 = 𝑡 2 − 21𝑡 − 270
you have now quadratic equation
Solve the quadratic equation:
0 = (𝑡 − 30)(𝑡 + 9)
(by factoring)
(Note: you can use any method)
𝑡 − 30 = 0
𝑡+9=0
𝑡 = 30
𝑡 = −9 (this does not lead to a solution)
Thus,
Poy needs 30 minutes to paint the wall by himself and John needs 30 + 15 = 45 minutes.
It is your turn. Try this.
Jack and Jill working together can clean their room six minutes. By herself Jill
needs five minutes more than Jack to clean the room. How long would each need to
clean the room if he or she were to work alone?
You should get 10 minutes for Jack and 15 minutes for Jill.
If your answer is correct, you may proceed to the next part of the discussion. If not, I am sorry,
but you have to go back to Example 4 and try all over again.
Page 65 of 148
E. Continuation of the discussion of new concepts
DAY 3 (September 16)
Projectile Problem (Height of Falling Objects)
The height of an object dropped, thrown or fired can be computed using quadratic equations.
The general formula is:
h = -16t2 + v0t + ho
h = object’s height in feet (h = 0 when object reaches the ground)
t = time in seconds
Where:
vo = object’s initial velocity (speed when t = 0) in feet per second,
If the object is thrown or fired upward, vo is positive and negative
if thrown downward. If the object is dropped vo = 0
ho = initial height (height when t = 0) in feet
Example 5:
Height of Falling Object
An object is dropped from a height of 1600 ft. How long will it take for the object to
hit the ground?
Given:
h0 = 1600 ft
v0 = 0
initial height
initial velocity is zero since the object is dropped
Required:
t – time in seconds when the object hits the ground
Solution:
Apply the formula:
h = -16t2 + v0t + ho
h = -16t2 + (0)t + 1600
h = -16t2 + 1600
when the object hits the ground h = 0
Thus,
0 = -16t2 + 1600
results in a quadratic equation
Write in standard form:
16t2 – 1600 = 0
Solve the equation:
16t2 = 1600
16𝑡 2
16
=
(by square root method)
1600
16
divide both sides by a = 16
𝑡 2 = 100
√𝑡 2 = √100
extract the square roots
𝑡 = ±10
𝑡 = 10
𝑡 = −10 (this is not a solution)
Therefore, the object will hit the ground 10 seconds after it is dropped.
Now it is your turn. Following the example above, try this.
A ball is dropped from the top of a four-story building. The building is 48 ft. tall.
How long will it take the ball to reach the ground?
Page 66 of 148
You should get 𝒕 = √𝟑 ≈ 𝟏. 𝟕𝟑 seconds.
If your answer is correct, you may proceed to the next part of the discussion. If not, I am sorry,
but you have to go back to Example 5 and try all over again.
Distance Problem
There are several distance problems that quadratic equations can solve. But in all of these
types, the formula d = r ● t is the key. Where, d = distance, r = rate, and t = time
“Stream” distance problems usually involve boats (travelling upstream or downstream) and
planes (traveling against a headwind or a tailwind).
Example 6: Distance Problem
A small motorboat traveled 15 miles downstream then turned around and traveled 15
miles back. The total trip took two hours. The stream’s speed is 4 mph. How fast
would the boat have traveled in still water?
Solution:
Let:
r represents the boat’s speed in still water
r + 4 represents the average speed downstream
r – 4 represents the average speed upstream
Since the boat was in the water a total of 2 hours and the distance traveled in each direction is
15 miles then,
15
𝑟+4
15
𝑟−4
hours, the time the boat traveled downstream
hours, the time the boat traveled downstream
The time the boat traveled upstream plus the time it traveled downstream equals 2 hours.
15
15
So, the equation I to solve is:
+ 𝑟−4 = 2
𝑟+4
Multiply both sides by the LCD
(𝑟 − 4)15 + (𝑟 + 4)15 = 2(𝑟 + 4)(𝑟 − 4)
and simplify:
15𝑟 − 60 + 15𝑟 + 60 = 2(𝑟 2 − 16)
The LCD is (𝑟 + 4)(𝑟 − 4)
30𝑟 = 2𝑟 2 − 32
0 = 2𝑟 2 − 30𝑟 − 32
1
(2) 0 =
2𝑟 2 −30𝑟−32
2
0 = 𝑟2 − 15𝑟 − 16
Write in standard form:
𝑟 2 − 15𝑟 − 16 = 0
Solve the equation:
(𝑟 − 16)(𝑟 + 1) = 0
results into a quadratic equation
by factoring
𝑟 − 16 = 0
𝑟 + 1 = 0)
𝑟 = 16
𝑟 = −1
(this is not a solution)
Therefore, the boat average speed in still water r = 16 mph.
Page 67 of 148
Now your turn. Try this.
Calvin rides his power boat up and down a drainage ditch. The water in the
drainage ditch flows at 6 miles per hour. Calvin takes 5 hours longer to travel 360
miles upstream than he does to travel 360 miles downstream. What is the speed of
Calvin's boat in still water?
You should get r = 30 mph.
If your answer is correct, you may proceed to the next part of the discussion. If not, I am sorry,
but you have to go back to Example 6 and try all over again.
F. Developing mastery
DAY 4 (September 17)
IT’S YOUR TURN!
Solve the following problems involving quadratic equations (2 pts. each).
Write your complete solutions in your notebook.
1. A rectangular garden has an area of 14 m2 and a perimeter of 18 m. Find the
dimensions of the rectangular garden
2. The product of two consecutive odd numbers is 399. Find the numbers.
3. A sari-sari store owner sells 60 AAA sardines per week when the cost is Php12 per
sardines. For each Php1.50 decrease in the price, the store sells an additional 16
sardines per week. What should be the price if the storeowner needs revenue of
Php810 per week for the sale of these sardines. How many will be sold at this/these
price/prices?
4. Together Tom and Jerry can wash a car in 16 minutes. Working alone Tom needs 24
minutes longer than Jerry does to wash the car. How long would it take for each Tom
and jerry to wash the car?
5. An object is dropped from a 56-foot bridge over a bay. How long will it take for the
object to reach the water?
6. Airport X and Airport Y are 1000 miles apart. A plane flew into a 50-mph headwind
from Airport X to Airport Y. on the return flight the 50-mphwind became a tailwind.
The plane was in the air a total of 4.5 hours for the round trip. What would have been
the plane’s average speed without the wind?
You may check your answers by turning to Page 70for the key to
correction. Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Rate the result using the table.
Score
11-12
9-10
5-8
0-4
Description
Very Good
Good
Fair
Turn to page 69
sec. J and work
on the enrichment
activities
Page 68 of 148
G. Finding practical application of the concepts and skill in daily living
You have just seen how wide the applications of quadratic equations is. They are scattered
everywhere in everyday life. You cannot disassociate yourself from it.
Here are some illustrations of quadratic applications.
Can you think of another real-life applications of quadratic equation other than these?
G. Making generalization and abstraction about the lesson
STEPS IN SOLVING WORD PROBLEMS
To solve problems involving quadratic equations:
1. Read the problem carefully and thoroughly.
2. Make illustrations if necessary and label it properly.
3. Identify the given data and the required.
4. Find an appropriate equation or formula that relate the given and the required.
5. Solve and check.
I.
Evaluating learning
DAY 5 (September 18)
Direction: Use yellow papers to answer the evaluation. Provide necessary solutions for
your answers. Use the format shown below to label your paper and it is to be
submitted to your Math teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 4 Parent’s Signature: __________________
EVALUATION 4
1. The width of a rectangle is 4 cm less than the length. The area is 320 cm 2. Find the
length and width.
2. The numbers differ by 7. Their product is 228. What are the numbers?
3. A shoe store sells a certain imported athletic shoe for $40 per pair. The store averages
sales of 80 pairs per week. The store owner past’s experience leads him to believe that
for each $2 increase in the price of the shoe, one less pair would be sold each week.
What price would result in $3,648 weekly sales?
4. Michael and Jackson together can unload a truck in 1 hour and 20 minutes. Working
alone Michael needs 36 minutes more to unload the truck than Jackson needs. How
long would each Michael and Jackson need to unload the truck by himself?
5. A ball is dropped from a sixth-floor window at a height of 70 ft. When will the ball hit the
ground?
Page 69 of 148
6. The current in a stream moves at a speed of 3-mph. A boat travels 40 miles upstream
and 40 miles downstream in a total of 14 hours. What is the speed of the boat in still
water?
J. Additional activities for application of remediation
Solve the following problems involving quadratic equations.
1. The length of a rectangle is 6 inches more than its width. The area of a rectangle is 91
square inches. Find the dimensions of the rectangle.
2. The product of two consecutive even numbers is 288. What are the numbers?
3. A Duty-Free Shop sells 4000 gallons of milk per week when the price is $2.80 per gallon.
Customer research indicates that for each decrease in the price, 200 more gallons of
milk will be sold. What does the price need to be so that weekly milk sales reach
$11,475?
4. Two printing presses working together can print a magazine order in 6 hours. Printing
Press I can complete the job alone in five fewer hours than Printing Press II. How long
would each press need to print the run by itself?
5. A small object falls from a height of 200 ft. How long will it take to reach the height of
88 ft.?
6. A plane on a flight from Point A to Point B flew with a 20-mph tailwind. On the return
flight, the plane flew into a 20-mph headwind. The distance between Point A and Point
Bis 1000 miles and the plane was in the air a total of 5.5 hours. What would have been
the plane’s average speed without the wind?
You may check your answers by turning to Page 70 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
KEY TO CORRECTIONS
Activity 1: REVIEW
a. x = -5 and x = 2
b. x = -5 and x = 2
c. x = -5 and x = 2
Activity 2: TURN IN THE FISH
1
2
7
4
3
5
6
5
8
7
2
1
Page 70 of 148
DEVELOPING MASTERY (KEY TO CORRECTIONS)
IT’S YOUR TURN!
1. Length = 13 m, Width = 7 m
2. The two numbers are19 and 21
3. When x = 1.25, P = $10.13, and Q = 80
When x = 3, P = $7.50 and Q = 108.
4. Jerry needs 24 minutes, Tom needs 48 minutes
5. t = 1.87 seconds
6. r = 450 mph
ADDITIONAL ACTIVITIES/REMEDIATION (KEY TO CORRECTION)
REMEDIATION:
1. L = 13 inches and W = 7 inches
2. The consecutive even numbers are 16 and 18 or -16 and -18
3. If x =2.50, the price the price should be 2.80-0.10(2.5)=$2.55.
If x= 5.5, the price should be 2.80-0.10(5.5)=$2.25.
4. Printing Press II can print the run alone in 15 hours and Printing Press I 10 hours.
5. The object will reach the height of 88 ft. after about 2.65 seconds
6. The plane would have averaged about 365 mph without the wind.
If you have any questions regarding the lesson, feel free to contact me at #09434634247
Page 71 of 148
SELF-INSTRUCTIONAL PACKETS (SIPacks)
MATHEMATICS Grade 9 - SSC
School
PAMPANGA HIGH SCHOOL
Teacher
IVAN T. SALAS
I.
Teaching Dates/
Week
September 21-25, 2020 (Week 5)
Quarter
FIRST
OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations, inequalities
and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various situations,
formulate real-life problems involving quadratic equations, inequalities and functions, and
rational algebraic equations and solve them using a variety of strategies.
C. Learning Competencies
The Learner
1. Illustrates quadratic inequalities. (M9AL-If-1)
2. Solves quadratic inequalities. (M9AL-If-2)
3. Solves problems involving quadratic Inequalities (M9AL-If-g-1)
D. Objectives
At the end of the lesson, the learners should be able to:
1) illustrates quadratic inequalities in one variable;
2) solves quadratic inequalities;
3) solves problems involving quadratic inequalities.
II.
CONTENT
Illustrations of Quadratic Inequalities
Solving Quadratic inequalities
Learning Resources
A. References:
Learner’s Material for Mathematics Grade 9, pp. 96-113
Teacher’s Guide for Mathematics, pp. 59-66
Grade 9 Patterns and Practicalities, pp 59-66 by Nivera, Gladys C.
and Lapinid, Minie Rose C.
B. Other Learning Resources
III.
PROCEDURES
DAY 1 (September 21)
A. Reviewing previous lesson or presenting the new lesson
Hi! I hope you enjoyed last week’s lesson.
This week, you are going to expand your knowledge of the different
mathematics concepts and your skills in performing mathematical
operations in dealing with the new topic.
To start with, let us answer the first activity
Page 72 of 148
Activity 1: Find the solution set of each of the following mathematical sentences by
matching column A with its graph in column B. Write your answers in your notebook.
Activity 1: MATCH ME WITH MY GRAPH?
A
B
1. 4x < 12
a.
2. x + 5 > 3
b.
2
3. x ≤ 2
3
c.
4. -5x ≤ 10
d.
5. -2x + 3 > 7
e.
3
6. x – 5 ≤ -2
4
f.
You may check your answers by turning to Page 85
Give yourself one (1) point for every correct answer. I hope you got them all correctly!
B. Establishing a purpose for the lesson
Let us have another short activity!
This activity is entitled “Am I Equal or Not.”
Write the word“equal” if the picture shows equality. Otherwise, write “unequal”
Write your answers in your notebook.
Activity 2: Am I Equal or Not?
Activity 2: AM I EQUAL OR NOT
1.
__________________
2.
__________________
Page 73 of 148
3.
__________________
4.
___________________
5.
____________________
You may check your answers by turning to Page 85.
Give yourself one (1) point for every correct answer. I hope you got them all correctly!
You may now continue with our new lessons.
In the previous activity, were you able to recall your past lessons? Have you observed the
symbols: <, > , ≤, ≥ that were used in activity 1 and the words that describes the pictures in
activity 2? All of these tell us about inequality.
This DLP focuses on quadratic inequality in one variable.
C. Presenting examples/instances of the lesson
What is quadratic inequality?
Definition
A quadratic inequality in one variable is an inequality that contains a polynomial of
degree 2 and can be written in any of the following forms:
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 > 0
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 < 0
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 ≥ 𝟎
𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 ≤ 𝟎
where 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 are real numbers and 𝑎 ≠ 0.
Here are examples of quadratic inequalities in one variable.
1. 3𝑥 2 + 10𝑥 + 3 > 0
3. 3𝑟 2 + 𝑟 − 5 ≥ 0
2. 𝑠 2 − 9 < 2𝑠
4. 𝑡 2 − 2𝑡 ≤ 15
Page 74 of 148
D. Discussing new concepts and practicing new skills #1
In this part of the lesson, you will learn, how to solve quadratic inequalities in the form
ax2 + bx + c < 0
ax2 + bx + c ≤ 0
ax2 + bx + c > 0
ax2 + bx + c ≥ 0
A solution of a quadratic inequality in one variable is a value of the variable that makes the
inequality a true statement.
Solving Quadratic Inequalities
Example 1 : Solve x2 – 2x – 3 < 0.
Solution :
Write the inequality into its
corresponding equality/equation:
x2 – 2x − 3 = 0
solve the equation by factoring:
(Note: you can use other method)
(x - 3) (x + 1) = 0
x-3 =0
x+1 = 0
x=3
x = -1
Plot these two numbers (points) in a number line:
A
B
C
Note that these two numbers divide the number line into three intervals.
Choose one representative number (test point) from each interval and substitute these
values in the original inequality. If the resulting inequality is true, then the interval containng
the test point is a solution interval.
Interval
Test Point
Inequality
(-3) – 2(-3) – 3 < 0
9+6–3<0
15 < 0
(0)2 – 2(0) – 3 < 0
0–0–3<0
-3 < 0
(5)2 – 2(5) – 3 < 0
25 - 10 – 3 < 0
12 < 0
Result
2
For A: 𝑥 < −1
-3
For B: −1 < 𝑥 < 3
0
For C: 𝑥 > 3
5
False
True
False
The inequality 𝑥 2 − 2𝑥 − 3 < 0 is true for any value of x in interval B, −1 < 𝑥 < 3
Since the inequality symbol is <, the numbers -1 and 3 are not included in the solution set.
The solution set of the inequality 𝑥 2 − 2𝑥 − 3 < 0 is x = {𝑥: −1 < 𝑥 < 3} “read as values of x
such that x is greater than -1 but less than 3”.
Another way of writing the solution set is(−1, 3).
The graph is shown below.
Page 75 of 148
Note: We use hollow circle ( ) and ( ) for < and >, solid circle ( ) and [ ] for ≤ and ≥.
The hollow circles (unshaded circles) are used in the graph to show that -1 and 3 are
not part of the solution set.
Now, on your own, will you try to find the solution set of 𝑥 2 − 6𝑥 − 7 < 0 ?
You should get the solution set {𝑥: −1 < 𝑥 < 7 } .
The graph is shown below.
If your answer is correct, you may proceed to the next part of our
discussion. If not, I am sorry but you have to go back to example
number 1 and try all over again.
Let us have another example.
Example 2. Solve 𝑥 2 + 𝑥 − 2 ≥ 0.
Solution:
Write the corresponding equation
Solve by factoring
:
𝑥2 + 𝑥 − 2 = 0
(𝑥 + 2)(𝑥 − 1) = 0
𝑥+2=0
𝑥−1=0
𝑥 = −2
𝑥=1
Plot the points 1 𝑎𝑛𝑑 − 2 on the number line.
A
B
C
These two numbers 1 𝑎𝑛𝑑 − 2 divide the number line into three intervals/regions.
Test a number from each interval/region against the inequality.
Interval/Region
For A:
𝑥 ≤ −2
For B: −2 ≤ 𝑥 ≤ 1
For C:
𝑥≥1
Test Point
−3
0
2
𝑥 2 + 𝑥 − 2 ≥ 0.
(−3)2 + (−3) − 2 ≥ 0
9−6−2≥0
1≥0
(0)2 + 0 − 2 ≥ 0
0−0−2≥0
−2 ≥ 0
(2)2 + 2 − 2 ≥ 0
4+2−2≥0
4≥0
Results
True
False
True
Therefore, the inequality 𝑥 2 + 𝑥 − 2 ≥ 0 is true for any value of x in the intervals A and B
𝑥 ≤ −2 or 𝑥 ≥ 1 . Since the inequality symbol is ≥, the numbers −2 𝑎𝑛𝑑 1 are included in the
solution set.
The solution set of 𝑥 2 + 𝑥 − 2 ≥ 0 is {𝑥: 𝑥 ≤ −2 𝑜𝑟 𝑥 ≥ 1}.
Its graph is shown below.
Page 76 of 148
Note: The solid circles (shaded circles) are used in the graph to show that -2 and 1 are part of
the solution set.
Now, on your own, will you try to find the solution set of 𝑥 2 − 2𝑥 − 8 ≥ 0 ?
You should get the solution set {𝑥: 𝑥 ≤ −2 𝑜𝑟 𝑥 ≥ 4} .
The graph is shown below.
If your answer is correct, you may proceed to the next part of our
discussion. If not, I am sorry but you have to go back to example number
2 and try all over again.
E. Discussing new concepts and practicing new skills #2
DAY 2 (September22)
In our previous lesson you’ve learned to find the solution set qudratic inequalities. This
time you will work on a situation involving quadratic inequalities. But before we proceed, you
need to know and familiarize the following phrases which will guide you solve problems that
applies the concept of quadratic inequalities.
INEQUALITY VOCABULARY
> (greater than)
< (less than)
more than
above
over
greater/larger than
exceeds/increased
longer than
is higher than
bigger
is less than
is under
is below
shorter/smaller
than
fewer than
is lower than
beneath
a better deal
≥ (greater than or
equal to)
at least
minimum
top
is no less than
≤ (less than or
equal to)
at most
maximum
bottom
is no more than
at or above
at or below
Now, we will solve real-life applications of quadratic inequalities.
Let’s try the following examples!
Problem Solving Involving Quadratic Inequalities
Example 1.
The profit P that a company earns for selling x number of toy cars can be modeled
by P(x) = -25x² + 1000x – 3000. How many toy cars must be sold for a profit of at
least Php 5000?
Solution:
Since the profit P(x) has to be at least Php 5000, then we write the quadratic inequality as
-25x² + 1000x – 3000 ≥ 5000
Write the inequality into its related equation:
Simplify
Divide both sides by -25 to simplify:
Write in standard form
-25x² + 1000x – 3000 = 5000
-25x² + 1000x – 8000 = 0
-x² + 25x - 320 = 0
x² - 25x + 320 = 0
Page 77 of 148
−(−25)±√(−25)2 −4(1)(320
Solving using quadratic formula:
𝑥=
We get:
x = 28.94 or
2(1)
x = 11.06
Thus,
A
B
C
Region
Test Point Value
-25x² + 1000x –
3000 ≥ 5000
Result
A
5
1375 ≥ 5000
False
B
15
6375 ≥ 5000
True
C
30
4500 ≥ 5000
False
The range of possible values of x is 11.06< x < 28.94.
Since we are talking of number of toy cars, the solution set is at least[12, 28].
To make a profit of at least Php 5000, at least 12 and at most 28 toy cars must be sold by the
company.
Now, you try this problem!
The total profit function P(x) for a company producing x thousands of
pesos given by P(x) = -2x² + 26x - 44. Find the value of x for which the
company makes a profit.
[𝐻𝐼𝑁𝑇: 𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑎𝑛𝑦 𝑚𝑎𝑘𝑒𝑠 𝑎 𝑝𝑟𝑜𝑓𝑖𝑡 𝑤ℎ𝑒𝑛 𝑃(𝑥) > 0].
You should get, −1.52 < 𝑥 < 14.52
If your answer is correct, you may proceed to Example 3. If not, I am
sorry but you have to go back to Example 1 and try all over again.
Example 2.
The length of a rectangle is 3m greater than the width. Find the possible widths (w)
such that the area of the rectangle will be greater than 15m².
Solution:
We need to consider the formula in getting the area of a rectangle which is A = l x w.
Representation:
Let x be the width (w) of the rectangle
Let x + 3 be the length (l) of the rectangle
(𝑥)(𝑥 + 3) > 15.
Since the area of the rectangle
2
is greater than 15 m , then:
𝑥 2 + 3𝑥 > 15
Write the corresponding equality
Write in standard form:
x² + 3x = 15
x² + 3x – 15 = 0
Solve using quadratic formula:
x=
We get
x = 3.40
Thus,
The width is:
−3±√32 −4(1)(−15)
2(1)
x = -6.40 (Not solution
because length cannot be
negative)
w = x = 3.40 ≈3 meters
Page 78 of 148
The length is:
l = x + 3 = 3 + 3 = 6 meters
To check:
x(x + 3) > 15
3(3 + 3) > 15
3(6) > 15
18 > 15
Therefore, the possible widths:
w > 3 meters
Now, you try this problem!
The perimeter of a rectangle is 20 cm. Find the possible dimensions if
the area of the rectangle must be less than 78 cm².
You should get, x = 15.15 and x = -5.15.
If your answer is correct, you may proceed to Example 3. If not, I am
sorry but you have to go back to Example 2 and try all over again
Example 3.
An object is launched from ground level directly upward at a rate of 144 feet per
second. The equation of the object’s height is y = -16x² + 144x. What values of x is
the object at or above a height of 288 feet?
Solution:
Since the phrase at or above is used, we’ll make use of ≥.
So, we will have -16x² + 144x ≥ 288.
-16x² + 144x ≥ 288
-16x² + 144x = 288.
Write the inequality in its corresponding equality.
-16x² + 144x - 288 = 288 – 288
Add -288 on both sides of the equation.
-16x² + 144x – 288 = 0
Standard form
-16(x² - 9x + 18) = 0
Factor out -16.
x² - 9x + 18 = 0
Divide both sides by -16
You may use any of the methods in solving quadratic equation. In this example, factoring is
used.
x² - 9x + 18 = 0
(x – 6) (x – 3) = 0
x–6=0
x -3 = 0
x=6
x=3
With the obtained roots of the quadratic equation, the object is at or above a height of
288
feet in 3 to 6 seconds.
Now, you try this problem!
A projectile is fired straight up from the ground with an initial velocity of
80 feet per second. Its height s(t) = -16t² + 80t. Find the interval of time
for which the height of the projectile is greater than 96 feet.
You should get, t = 6 and t = -1.
If your answer is correct, you may proceed to the next level of our study.
Congratulations!
Page 79 of 148
F. Developing mastery
DAY 3 (September 23)
Answer Activity 1 below. Read the direction carefully and write your answers in your
notebook.
Activity 1: Identify whether each of the following a quadratic inequality or not. Write QI if it is
quadratic inequality and NQI if it is not.
1. 𝑥 2 + 9𝑥 + 20 = 0
Answer:
_______
2
2. 2𝑡 < 21 − 9𝑡
Answer:
_______
2
3. 𝑟 + 10𝑟 ≤ −16
Answer:
_______
4. 𝑚2 = 6𝑚 − 7
Answer:
_______
5. 𝑏 − 7 ≥ 0
Answer:
_______
2
6. 17 − 2𝑥 > 5𝑥
Answer:
_______
Now, check your work by turning to Page 85 for the key to
correction. How many correct answers did you get? Rate your
result using the table.
If your score is at least 4 out of 7, you may now proceed to next
activity.
Score
6
5
3–4
0-2
Description
Very Good
Good
Fair
Turn to page
83 sec. J and
work on the
enrichment
activities
Answer Activity 2 below. Read the direction carefully and write your answers in your
notebook.
Activity 2: Solve and graph each quadratic inequality.
1. (𝑥 + 7)(𝑥 + 1) < 0
Solutions:_______________ Graph: ____________________
2. 𝑥 2 − 6𝑥 + 5 > 0
Solutions:________________Graph: ____________________
3. 2𝑥 2 − 4𝑥 ≥ 16
Solutions:________________Graph: ____________________
4. 2𝑥 2 − 5𝑥 ≤ 3
Solutions:_______________ Graph: ____________________
5. 3𝑥 2 + 2𝑥 ≥ 5
Solutions:_______________ Graph: ____________________
Now, check your work by turning to Page 85 for the key to
correction.
How many correct answers did you get?
Score
9-10
7-8
5-6
Rate your result using the table.
0-4
If your score is at least 3 out of 4, you may now proceed to next
activity.
Description
Very Good
Good
Fair
Turn to page
83 sec. J and
work on the
enrichment
activities
Page 80 of 148
Answer Activity 3 below. Read the direction carefully and write your answers in your
notebook.
Activity 3
Consider the given below. The steps in solving the problem are expressed in different
orders. You need to number from 1 to 10 the bar pieces in proper order to make a correct
sequence of the solution.
The difference of two numbers is 9. What are the two numbers if their product is at least 90?
Now, check your work by turning to Page 86 for the key to correction. How many correct
answers did you get? Rate your result using the table.
If your score is at least 5 out of 10, you may now proceed to
Score
Description
9–10
Very Good
next activity.
7–8
5–6
0-4
Good
Fair
Turn to page
84 sec. J and
work on the
enrichment
activities
G. Finding practical application of the concepts and skill in daily living
DAY 4 (September 24)
Solving Quadratic Inequalitiescan be applied in real world.
Inequalities are used more often in “real life” than equalities.
Businesses are use inequalities to control inventory,plan production lines,and produce pricing.
Other examples of inequality in society today are gender inequality, health care wherein some
individuals receive better and more professional care compared to other and social class
Page 81 of 148
H. Making Generalization and abstraction about the lesson
Quadratic Inequalities –These mathematical sentences contain a polynomial of
degree 2 and can be written in any of the following forms:
ax2 + bx + c > 0
ax2 + bx + c ≥ 0
ax2 + bx + c < 0
ax2 + bx + c ≤ 0
where 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 are real numbers and 𝑎 ≠ 0.
A solution of quadratic inequality in one variable is a value of the variable that
makes the inequality a true statement.
To solve quadratic inequalities, follow these steps:
1. Solve the inequality as though it were an equation.
The real solutions to the equation become boundary points for the solution to the
inequality.
2. Make the boundary points solid circles if the original inequality includes equality;
otherwise, make the boundary points open circles.
3. Select points from each of the regions created by the boundary points. Replace
these “test points” in the original inequality.
4. If a test point satisfies the original inequality, then the region that contains that test
point is part of the solution.
5. Represent the solution in graphic form and in solution set form.
Page 82 of 148
I. Evaluating learning
Direction: Use yellow papers to answer the evaluation. Provide necessary solutions for
your answers. Use the format shown below to label your paper and it is to be
submitted to your Math teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 5 Parent’s Signature: __________________
EVALUATION 5
I.
Identify whether each mathematical sentence is a quadratic inequality or not.
Write TIK if it is quadratic inequality, and TOK if it is not.
1. 2𝑥 2 − 22 ≤ 6𝑥
Answer: _______
2. 4𝑥 ≥ 23 − 𝑥
Answer: _______
3. (3𝑥 − 7)(𝑥 + 3) > 21
Answer: _______
2
4. 9𝑟 + 4𝑟 = 12
Answer: _______
5. 𝑥 2 − 4𝑥 ≥ 15 + 𝑥 2 − 2𝑥
Answer: _______
II.
Match each quadratic inequality in Column A with its solution set in Column B.
Answer
III.
Column A
6. 𝑥 2 − 4 < 0
A. 1 < 𝑥 < 3
7. 𝑥 2 ≥ 16
B. 0 < 𝑥 < 9
8. 𝑥 2 − 9𝑥 < 0
C. −2 < 𝑥 < 2
9. 𝑥 2 − 4𝑥 + 3 < 0
D. 𝑥 ≤ 2 and 𝑥 ≥ 3
10. 𝑥 2 − 𝑥 − 6 ≥ 0
E. 𝑥 ≤ −4 and 𝑥 ≥ 4
Find the solution set of each of the following quadratic inequalities, then graph.
Inequality
11. 𝑥 2 + 9𝑥 + 14 < 0
12. 𝑥 2 − 10𝑥 + 16 > 0
13. 2𝑥 2 + 16𝑥 ≤ −14
IV.
Column B
Solution set
Graph
Solve. Show your complete solutions.
14. A rectangular tile has length 4x cm and width (x + 3) cm. The area of the rectangle is
less than 112 cm2. By writing down and solving an inequality, determine the set of
possible values of x.
15. A tennis ball is hit into the air. The height of the ball, h(t), in feet, at time t, in seconds, is
modeled by the equation h(t) = -16t2 + 64t + 4. Determine the interval of time when the
height of the ball is greater than or equal to 52 feet?
Page 83 of 148
J. Additional activities for application of remediation
DAY 5 (September 25)
Remediation 1: Determine whether each mathematical sentence is a quadratic inequality
or not. Write your answers in your notebook.
1. 𝑥 2 + 9𝑥 + 14 > 0
6. 3𝑚 + 20 ≥ 0
2. 3𝑠 2 − 5𝑠 = 1
7. 12 − 5𝑥 + 𝑥 2 = 0
3. (2𝑟 − 5)(𝑟 + 4) > 0
8. 𝑥 2 − 1 < 𝑥 + 1
4. 4𝑡 2 − 7𝑡 + 2 ≤ 0
9. (4ℎ2 − 9) + (2ℎ + 3) ≥ 0
5. 𝑥 2 < 10𝑥 − 3
10. 15 − 2𝑥 = 3𝑥 2
You may check your answers by turning to Page 86 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Remediation 2: Find the solution set of each of the following quadratic inequalities, then
graph. Write your answers in your notebook.
1. (𝑥 − 2)(𝑥 + 5) > 0
2. 𝑥 2 + 𝑥 < 20
3. 2𝑥 2 − 18𝑥 ≤ 0
4. 3𝑥 2 + 12𝑥 + 9 ≥ 0
You may check your answers by turning to Page 86 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 84 of 148
Remediation 3: LET’S INTERPRET!
Match the keys to their corresponding padlock by interpreting the phrases to their
equivalent inequalities. Cut out the keys and paste them to their correct padlocks.
1
x is at most
most 7 7
xxxxxx
2
x isless
no less
than
no
than
7 7
y <77
y≤7
oooooo7
33
7rrrrrrrr
y is beneath 7
x>
x > 77
Remediation 3: LET’S INTERPRET!
You may opt to just match the each item in the keys to their corresponding item in the padlocks.
Write your answers in your notebook.
You may check your answers by turning to Page 87 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 85 of 148
KEY TO CORRECTION
Activity 1: MATCH ME WITH MY GRAPH
Activity 2: AM I EQUAL OR NOT
1. e
4. d
1. Unequal
4. Equal
2. a
5. f
2. Unequal
5. Unequal
3. c
6. b
3. Unequal
(DEVELOPING MASTERY)
Activity 1:
1. NQI
2. QI
3. QI
4. NQI
5. NQI
6. QI
Activity 2:
1. Solutions: {-7 < x <-1}
2. Solutions:{x < 1 or x > 5}
3. Solutions: {x ≤ -2 or x ≥ 4}
𝟏
4. Solutions: {𝒙: − 𝟐 ≤ 𝒙 ≤ 𝟑}
𝟓
5. Solutions: {𝒙: 𝒙 ≤ − 𝟑 𝒐𝒓 𝒙 ≥ 𝟏}
Page 86 of 148
(REMEDIAL)
Remedial 1:
1.
2.
3.
4.
5.
Quadratic Inequality
Not Quadratic Inequality
Quadratic Inequality
Quadratic Inequality
Quadratic Inequality
6. Not Quadartic Inequality
7. Quadraticv Inequality
8. Quadratic Inequality
9. Quadratic Inequality
10. Not Quadratic Inequality
Remediation 2:
1. Solutions: {x : x < -5 or x > 2}
2. Solutions: {x : -5 < x < 4}
3. Solutions:{x : 0 ≤ x ≤ 9}
4. Solutions: {x : x ≤ -3 or x ≥ -1}
Page 87 of 148
Remedial3: LET’S INTERPRET!
If you have any questions regarding this lesson, feel free to contact me at #09434634247.
Page 88 of 148
SELF-INSTRUCTIONAL PACKETS (SIPacks)
MATHEMATICS Grade 9 – SSC
School
PAMPANGA HIGH SCHOOL
Teaching Dates/
Week
Sept. 28 – Oct. 2, 2020 (Week 6)
Teacher
IVAN T. SALAS
Quarter
FIRST
I. OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations, inequalities
and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various situations,
formulate real-life problems involving quadratic equations, inequalities and functions, and
rational algebraic equations and solve them using a variety of strategies.
C. Learning competencies:
The learner models real-life situations using quadratic functions and represents a quadratic
function using: (a) table of values; (b) graph; and (c) equation. (M9AL-Ig-2, M9AL-Ig-3)
D. Objectives:
At the end of the lesson, the learners should be able to:
1) recall the definition of a function;
2) differentiate a quadratic function from a linear function in terms of equation, graph, and table
of values;
3) determine whether the given equations, tables of values and graphs represent a quadratic
function or not.
II.
CONTENT
Introduction to Quadratic Functions
Learning Resources:
References:
III.
PROCEDURES
Learner’s Material for Mathematics Grade 9 pp. 125-129
Teacher’s Guide for Mathematics pp. 84-90.
Grade 9 mathematics (Patterns and Practicalities) pp. 67-72
DAY 1 (September 28)
A. Reviewing previous lesson or presenting the new lesson
HELLO! How are you today?
Do you miss going to school? The situation may not permit us
as of this time. But don’t you worry everything will be okay. You
will soon be meeting your friends, classmates and teachers in
God’s perfect timing.
For the meantime, let us continue discussing new lessons.
For this week, we will focus on quadratic functions. If you’re
excited to learn about this topic, let’s learn Math together!
Page 89 of 148
Before we proceed with our actual topic, let us first have a review of relations and functions.
Recall, that the correspondence between the elements of two sets which determine a set of
ordered pairs is a relation.
A relation is a function when no two ordered pairs (x, y) have the same first component
(x-coordinate). The set of x-coordinates is called the domain, while the set of y-coordinates
is called the range.
A function may be represented as ordered pairs, mapping diagram, table of values and
graphs.
Let us have some examples of relations that are functions and not functions.
FUNCTIONS
NOT FUNCTIONS
A. Ordered Pair
{(1, 2), (2, 2), (3, 4), (4, 4), (5, 6), (6, 6)}
{(1, 3), (2, 5), (2, 7), (3, 7), (3, 9), (4, 9)}
Domain: { 1, 2, 3, 4, 5, 6 } set of x-coordinates
Range: { 2, 4, 6 } set of y-coordinates
B. Mapping Diagram
X
Y
1
2
2
3
4
4
5
6
6
Domain: { 1, 2, 3, 4, 5, 6 }
Range: { 2, 4, 6 }
C. Table of Values
X
1
2
3
Y
2
2
4
4
4
5
6
6
6
Domain: { 1, 2, 3, 4, 5, 6 }
Range: { 2, 4, 6 }
Domain and Range are the same for A, B, and
C since we used only one function in three
different representations.
X
1
2
3
4
X
Y
1
3
Y
3
5
7
9
2
5
2
7
3
7
3
9
4
9
The relation is not a function because we
have the same first component or xcoordinates - 2 and 3
A graph represents a function if no vertical line can be drawn intersecting the graph in no
more than one point. This method is called the vertical line test.
Page 90 of 148
In figure (c) above, a vertical line can be drawn intersecting the graph in two points. The
graph does not represent a function. Figures (a) and (b) are graphs of a function.
Now that you already have an idea what a function is, proceed to the next part.
B. Establishing a purpose for the lesson
Let us play, Mobile Legends: Bang Bang!
If you are familiar with the game Mobile Legends: Bang Bang (MLBB), you know that two
opposing teams fight to reach and destroy the enemy's base while defending their own
base. In each team, there are five players who each control a character, known as a "hero".
For this activity, you are given ten (10) MLBB heroes wherein each hero represents a
function. You have to form two opposing teams consisting of five (5) heroes by grouping the
MLBB heroes and functions into Linear and Quadratic. Please be guided that each MLBB
team must have the following heroes: Marksman, Assassin or Fighter, Mage, Support and
Tank.
MOBILE LEGENDS:
BANG BANG
Form two opposing teams consisting of five (5) heroes each by
grouping the MLBB heroes and given functions into Linear and Quadratic. In answering, write
each given function inside the empty boxes under your chosen team. Write your answers in
your notebook.
GIVEN FUNCTIONS
LINEAR
FUNCTIONS
Did you enjoy the game?
Now, you may check your answers by turning to Page 96 for
the key to corrections. Give yourself one (1) point for every
correct answer. How many correct answers did you get?
Rate the result using the table.
Once done, you can proceed to the next part of the
discussion.
QUADRATIC
FUNCTIONS
Score
9 - 10
6-8
3-5
0-2
Description
Very Good
Good
Fair
Turn to page96
sec. J and work
on the
enrichment
activities
Page 91 of 148
DAY 2 (September 29)
C. Presenting examples/instances of the lesson
Let us now differentiate linear functions and quadratic functions in terms of their definitions,
forms, equations, table of values and their graphs.
COMPARISONS BETWEEN LINEAR FUNCTION AND QUADRATIC FUNCTION
LINEAR FUNCTION
QUADRATIC FUNCTION
DEFINITION:
Linear functions are first degree functions. The
highest exponent of the variable𝑥 is one (1).
Quadratic functions are second degree
functions. The highest exponent of the variable
𝑥 is two (2)
GENERAL FORMS:
𝑦 = 𝑚𝑥 + 𝑏
𝑓(𝑥) = 𝑚𝑥 + 𝑏
𝑔(𝑥) = 𝑚𝑥 + 𝑏
EQUATION:
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑔(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑦 = 𝑥2
𝑦 = 3𝑥
Example:
TABLE OF VALUES:
Assign values for x (inputs) and substitute values in the equation to solve for y (outputs).
x
-3
-2
-1
0
1
2
3
3x
3 (-3)
3 (-2)
3 (-1)
3 (0)
3 (1)
3 (2)
3 (3)
THUS,
x
-3
y
-9
First
Difference
In y
-2
-6
-1
-3
y
-9
-6
-3
0
3
6
9
0
0
1
3
2
6
3
9
V V
V V V
V
-3
-3
-3
-3
-3
-3
x
y
-3
9
First
Difference
in y
Second
Difference
In y
Constant or equal first difference in y
x2
(-3)2
(-2)2
(-1)2
(0)2
(1)2
(2)2
(3)2
x
-3
-2
-1
0
1
2
3
-2
4
-1
1
y
9
4
1
0
1
4
9
0
0
1
1
2
4
3
9
V V V V
V V
5
-3
3
1
-1
-5
V
V V V V
2
2
2
2
2
Constant or equal second differencein y
Note: the differences between the adjacent 𝑦 – values was obtained by subtracting from the
left to the right.
GRAPH:
Using each table of values above, plot the points and then connect.
Page 92 of 148
y
y
14
8
12
6
10
4
8
2
−4 −3 −2 −1
O
1
2
3
4
6
x
−2
4
−4
2
−6
−4 −3 −2 −1
O
−8
−2
−10
−4
STRAIGHT LINE
1
2
3
4
x
PARABOLA
D. Discussing new concepts and practicing new skills # 1
DAY 3 (September 30)
From the given comparison above, the function 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, where a, b, and c are real
numbers and 𝑎 ≠ 0 is a quadratic function. Quadratic function in this form is said to be in
general form.
The following are examples of quadratic functions:
𝑦 = 𝑥2
a=1
b=0
c=0
2
𝑓(𝑥) = 𝑥 + 3𝑥 − 5
a=1
b=3
c = -5
2
𝑔(𝑥) = 2𝑥 − 6𝑥
a=2
b = -6
c=0
Can you give 3 examples of quadratic functions and write a, b, and c?
Quadratic Function
a
b
c
1.
2.
3.
E. Continuation of the discussion of new concepts
If for every equal differences in x-values of a given table, produces constant or equal second
differences in y-values, then the given table of values of x and y represents a quadratic
function.
The differences between the adjacent y–values can be obtained by subtracting from the left
to the right.
Okay, try this.
Determine whether the given tables of values represent a linear function or a quadratic
function.
Page 93 of 148
A.
B.
x
-2
-1
0
1
2
3
x
-2
y
-3
-1
1
3
5
7
y
9
Answer: ___________________________
-1
3
0
1
2
1
3
9
3
19
Answer: ___________________________
Try to graph the the given table of values A and B above.
What is the graph of Table A? Table B?
You may check your answer by turning to Page96.
If you got the correct answer, you may proceed to next part of the discussion. Otherwise,
review the “comparisons between linear functions and quadratic functions” above and try all
over again.
F. Developing mastery
TRY THIS!
Tell whether each of the following is a linear function or quadratic function. On the
space provided before the number, draw a smiley (☺) if your answer is linear function, and
draw a heart (♡) if your answer is quadratic function. Write your answers in your notebook.
__________ 1.𝑦 = 𝑥
________________11.
__________ 2.𝑦 = 𝑥 2 − 1
__________ 3.𝑦 = 9𝑥 2 − 𝑥 − 1
__________ 4.𝑦 = 22 + 𝑥
__________ 5.g(𝑥) = 𝑥(𝑥 + 3)
__________ 6.𝑦 = 2𝑥 − 10
__________ 7.𝑦 = (𝑥 − 3)(3𝑥 + 2)
__________ 8.𝑓(𝑥) = 3𝑥 − 1
__________ 9.
𝒙
𝒚
__________ 10. 𝒙
𝒚
__________12.
−3 −2 −1 0
1
5 10 15 20 25
−2 −1
6
3
0
2
1
3
2
6
You may check your answers by turning to Page 96 for the key
to corrections. Give yourself one (1) point for every correct
answer.
How many correct answers did you get?
Rate the result using the table.
Score
11 -12
8-10
5-7
0-4
Description
Very Good
Good
Fair
Turn to page 96
sec. J and work
on the enrichment
activities
Once done, you can proceed to the next part.
Page 94 of 148
G. Finding practical application of the concepts and skill in daily living
Quadratic functions are more than algebraic curiosities—they are widely used in
science, business, and engineering. The U-shape of a parabola can describe the trajectories
of water jets in a fountain and a bouncing ball, or be incorporated into structures like the
parabolic reflectors that form the base of satellite dishes and car headlights. Quadratic
functions help forecast business profit and loss, plot the course of moving objects, and
assist in determining minimum and maximum values. Most of the objects we use every day,
from cars to clocks, would not exist if someone, somewhere hadn't applied quadratic
functions to their design.
H. Making generalization and abstraction about the lesson (Paglalahatngaralin)
I know how the difference between a linear function and a quadratic
function. First, in terms of their equation, linear functions are first degree
functions where the highest exponent of the variable x is one (1), while
quadratic functions are second degree functions where the highest exponent
of the variable x is two (2).Then, in terms of their table of values, linear
functions produces constant first difference in y, while quadratic functions
produces constant second difference in y.Finally, in terms of their graphs,
linear functions form a line while quadratic functions form a parabola.
I. Evaluating learning
DAY 4 (October 1)
EVALUATION 6
Direction: Copy the Quadratic Function Maze below on a separate sheet to answer the
evaluation. Provide necessary solutions for your answers. Use the format
shown below to label your paper and it is to be submitted to your Math
teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 6 Parent’s Signature: __________________
The grade for your output will be based on the criteria below. Please be guided accordingly.
Accuracy (12 Points)
12 points Got 11 – 12 correct answers
10 points Got 8 – 10 correct answers
7 points Got 4 – 7 correct answers
3 points Got 1 – 3 correct answers
No point Got no correct answers
Neatness of Work (5 Points)
5 points
Free from erasures
3 points
Has minimal erasures
1 point
Not neatly done
Page 95 of 148
−𝑥 + 5 =
(
𝑔 𝑥 ) = 3𝑥 + 22
y = −5𝑥 + 11
4𝑥 + 𝑥 2 = 𝑦
(
y = 2 𝑥 2 + 3)
S
T
A
R
T
y
1
2
3
4
5
y
-1
0
1
2
3
y
-3
-2
-1
0
1
y
-2
-1
0
1
2
x
7
14
21
28
35
x
-4
0
-4
-16
-36
x
-9
-6
-3
0
3
x
2
-1
-2
-1
2
(
𝑔 𝑥 ) = (𝑥 − 2)(𝑥
+ 4)
1
(
𝑓 𝑥) = 𝑥 + 2
2
(
𝑓 𝑥 ) = 𝑥(𝑥 − 2)
𝑦 = 3𝑥 2 − 5
y
2
3
4
5
6
x
-3
-4
-3
0
5
F
I
N
I
S
H
(
𝑓 𝑥 ) = 3 + 2𝑥 2
𝑦 = 3𝑥 2 + 𝑥 + 2
(
y = 3 𝑥 + 6)
(
𝑓 𝑥 ) = 25 + 5𝑥
𝑦 = 2𝑥 + 2
QUADRATIC FUNCTION MAZE
Draw a path from Captain Ri to South Korea by colouring the boxes containing quadratic
functions.
Page 96 of 148
J. Additional activities for application or remediation.
DAY 5 (October 2)
In the oval callout, describe the ways of recognizing a quadratic function. Write your
answers in your notebook.
QUADRATIC
FUNCTIONS
GRAPH
EQUATION
TABLE OF VALUES
You may check your answers by turning to Page 97 belowfor the key to correction.
Give yourself two (2) points for every correct answer.
How many correct answers did you get?
KEY TO CORRECTIONS
MOBILE LEGENDS:
BANG BANG
GIVEN FUNCTIONS
LINEAR
FUNCTIONS
QUADRATIC
FUNCTIONS
(DISCUSSING NEW CONCEPTS AND PRACTICING SKILLS #2)
Table A:
Table B:
Linear Function
Quadratic Function
Graph: Line
Graph: Parabola
Page 97 of 148
(DEVELOPING MASTERY)
TRY THIS!
☺
__________
1.
♡
__________
3.
♡
__________ 2.
☺
______________
11.
♡
_______________ 12.
__________
4.
☺
♡
__________ 5.
__________
6.
☺
♡
__________ 7.
__________
8.
☺
__________
9
☺
♡
__________10.
(REMEDIAL ACTIVITY)
QUADRATIC
FUNCTIONS
Quadratic functions
produce constant second
difference in y.
TABLE OF VALUES
GRAPH
The graph of a
quadratic function
is a parabola.
Quadratic functions
aresecond degree functions.
The highest exponent of the
variable 𝑥 is two (2).
EQUATION
If you have any problem regarding this lesson, you may contact me at #09434634247.
Page 98 of 148
SELF-INSTRUCTIONAL PACKETS (SIPacks)
MATHEMATICS Grade 9 –SSC
School
PAMPANGA HIGH SCHOOL
Teacher
IVAN T. SALAS
Teaching Dates/
Week
October 6-8, 2020 (Week 7)
October 5 -Holiday
Quarter
FIRST
A. OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations, inequalities
and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various situations,
formulate real-life problems involving quadratic equations, inequalities and functions, and
rational algebraic equations and solve them using a variety of strategies.
C. Learning competencies:
The learner
1) Transforms the quadratic function defined by y = ax2 + bx + c into the form y = a(x –
h)2 + k. (M9AL-Ih-1)
2) Graphs a quadratic function: (a) domain; (b) range; (c) intercepts ;(d) axis of
symmetry; (e) vertex; (f) direction of the opening of the parabola. (M9AL-Ig-h-i-1)
3) Analyzes the effects of changing the values of a, h and k in the equation y = a(x – h)2
+ k of a quadratic function on its graph. (M9AL-Ii-2)
D. Objectives:
At the end of the lesson, the learners should be able to:
1. transforms quadratic functions from general form 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 to vertex form
𝑎(𝑥 − ℎ)2 + 𝑘.
2. graphs a quadratic function: (a) domain; (b) range; (c) intercept; (d) axis of
symmetry; (e) vertex; (f) direction of the opening of the parabola.
3. analyzes the effects of changing the values of a, h ad k in the function ation 𝑎(𝑥 −
ℎ)2 + 𝑘 of a quadratic function on its graph.
II.
CONTENT
Graphs of Quadratic Function
Learning Resources:
A. References:
Learners’ Materials Grade 9 Mathematics, pp. 140-150
Grade 9 Mathematics Patterns and Practicalities by Nivera,
Gladys C. and Lapinid, Minie Rose C., pp. 70-72
Advanced Algebra Textbook by Bernabe, Julieta G., et.al.,
pp.56 – 61
B. Other Learning Resources
Page 99 of 148
III.
PROCEDURES
DAY 1 (October 6)
A. Reviewing previous lesson or presenting the new lesson
Hi! How well have you performed in the previous weeks of
homeschooling? I just hope that you are able to cope with the new
normal of learning.
For this detailed lesson plan, we will focus on the graph of quadratic
functions. If you’re excited to learn about this topic, let’s start in
reviewing the Cartesian Plane. Do you remember its parts? What
are they?
Before we proceed with our actual topic, let us first have a review of the Cartesian Plane and its
parts
.
DID YOU KNOW?
The Cartesian Plane, also known as the Coordinate
Plane, is named after the French mathematician Rene
Descartes. He was born in La Haye, France on March
31, 1596. Descartes was taught at home at the age of
eight. At that point he entered the Jesuit College of La
Fleche, where he continued his schooling until
graduating at the age of eighteen.
While at La Fleche, Descartes suffered health problems
and because of this, his teachers allowed him to stay in
bed for most of the morning. Even though he missed
almost of his morning classes, he was still be able to
keep up with all of his studies. It has been rumored that
Descartes’ inspiration for the coordinate plane came
about because of all the time he spent in bed. The story
goes one day when Descartes was in bed, he noticed a
fly crawling around the ceiling. He tried to think of wats to
describe where the fly was located and realized that he
could do so by describing the fly’s position by its distance
from each wall. Then he tried to relate the fly’s position to
appoint, and, well, one thing leads to another, and voila!
He came up with the coordinate plane and Cartesian
coordinate!
Did you understand the story?
Proceed with the next part of the review, which is, the parts of a cartesian plane.
For this part, you choose from the words on the box below the Cartesian Plane and put
them in their proper places or parts.
Page 100 of 148
Check the correct answer by turning to page 122
Now that you already have an idea what a Cartesian plane is, proceed to the next part.
B. Establishing a purpose for the lesson
For the next part, let’s have the activity entitled “Let’s See Philippines”, in this activity
pictures of some tourist spots and wonderful places in the Philippines are presented below.
You need to identify the name and location of the given pictures.
LET’S SEE
PHILIPPINES!
Name: __________________________
Name: __________________________
Location: ________________________
Location: ________________________
Page 101 of 148
Name: __________________________
Name: _________________________
Location: ________________________
Location: _______________________
Name: __________________________
Name: _________________________
Location: ________________________
Location: _______________________
Did you enjoy the activity?
Just like in Cartesian plane, the points must be in proper
location so that you will make the graph of quadratic function
properly! Now, you may check your answers by turning to
Page 122 for the key to corrections. Give yourself one (1)
point for every correctanswer. How many correct answers did
you get?
Rate the result using the table.
Score
10 – 12
7–9
4–6
0–3
Description
Very Good
Good
Fair
Review you
Philippine
Geography
Once done, you can proceed to the next part of the discussion.
Page 102 of 148
C. Presenting examples/instances of the lesson
Study how the vertex form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘 of a quadratic function is derived from the
form 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 and vice versa.
Given: 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑏
= 𝑎 (𝑥 2 + 𝑎 𝑥) + 𝑐
factor out 𝑎 in 𝑎𝑥 2 + 𝑏𝑥
𝑏2
𝑏
𝑏2
= 𝑎 (𝑥 2 + 𝑎 𝑥 + 4𝑎2 ) + 𝑐 − 4𝑎
= 𝑎 (𝑥 +
𝑏 2
)
2𝑎
+
complete the square
4𝑎𝑐−𝑏2
4𝑎
simplify
𝑏
𝒇(𝒙) = 𝒂(𝒙 − 𝒉)𝟐 + 𝒌
let ℎ = − 2𝑎 and 𝑘 =
Given: 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘
= 𝑎(𝑥 2 − 2ℎ𝑥 + ℎ2 ) + 𝑘
= 𝑎𝑥 2 − 2𝑎ℎ𝑥 + 𝑎ℎ2 + 𝑘
𝑏
)𝑥 +
2𝑎
𝑏2
+𝑘
4𝑎
= 𝑎𝑥 2 − 2𝑎 (−
2
= 𝑎𝑥 + 𝑏𝑥 +
𝑎 (−
𝒇(𝒙) = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
4𝑎𝑐−𝑏2
4𝑎
square the binomial
simplify
𝑏 2
)
2𝑎
+𝑘
let ℎ = −
let 𝑐 =
𝑏2
4𝑎
𝑏
2𝑎
+𝑘
Writing quadratic function 𝒇(𝒙) = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 in the vertex form 𝒇(𝒙) = 𝒂(𝒙 − 𝒉)𝟐 + 𝒌.
Example 1. Express 𝑓(𝑥) = 𝑥 2 − 4𝑥 − 10 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘 .
Solution.
𝑓(𝑥) = 𝑥 2 − 4𝑥 − 10
given
𝑓(𝑥) = (𝑥 2 − 4𝑥) − 10
group together the terms containing x.
𝑓(𝑥) = (𝑥 2 − 4𝑥) − 10
factor out a (in this case a = 1)
𝑓(𝑥) = (𝑥 2 − 4𝑥 + 4) − 10 − 4
complete the expression in parenthesis to
make it a perfect square trinomial by adding
a constant term then subtract the same
value from the constant term.
𝑓(𝑥) = (𝑥 − 2)2 − 14
express the perfect square trinomial as a
square of binomial and simplify the constant
term.
Thus, the vertex form of 𝒇(𝒙) = 𝒙𝟐 − 𝟒𝒙 − 𝟏𝟎 is 𝒇(𝒙) = (𝒙 − 𝟐)𝟐 − 𝟏𝟒
Now, try this.
Express 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 3 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘.
You should get 𝑓(𝑥) = (𝑥 − 1)2 − 4 .
If you get the correct answer, you may proceed to Example 2.
If not, I am sorry but you have to go back to Example 1 and try to solve it again.
Page 103 of 148
Example 2. Express 𝑓(𝑥) = 3𝑥 2 + 12𝑥 + 13 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘 .
Solution.
𝑓(𝑥) = 3𝑥 2 + 12𝑥 + 13
given
𝑓(𝑥) = (3𝑥 2 + 12𝑥) + 13
group together the terms containing x.
𝑓(𝑥) = 3(𝑥 2 + 4𝑥) + 13
factor out a (in this case a = 3)
𝑓(𝑥) = 3(𝑥 2 + 4𝑥 + 4) + 13 − (3)4
𝑓(𝑥) = 3(𝑥 2 + 4𝑥 + 4) + 13 − 12
𝑓(𝑥) = 3(𝑥 2 + 4𝑥 + 4) + 1
complete the expression in parenthesis to
make it a perfect square trinomial by adding
a constant term then subtract the same
value from the constant term.
𝑓(𝑥) = 3(𝑥 + 2)2 + 1
express the perfect square trinomial as a
square of binomial and simplify the constant
term.
Thus the vertex form of 𝒇(𝒙) = 𝟑𝒙𝟐 + 𝟏𝟐𝒙 + 𝟏𝟑 is 𝒇(𝒙) = 𝟑(𝒙 + 𝟐)𝟐 + 𝟏
Now, try this.
Express 𝑓(𝑥) = 2𝑥 2 + 4𝑥 + 3 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘.
You should get 𝑓(𝑥) = 2(𝑥 + 1)2 + 5.
If you get the correct answer, you may proceed to Example 3.If not, I am sorry but you have
to go back to Example 2 and try to solve it again.
Example 3. Express 𝑓(𝑥) = −3𝑥 2 + 3𝑥 + 2 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘 .
Solution.
𝑓(𝑥) = −3𝑥 2 + 3𝑥 + 2
given
𝑓(𝑥) = (−3𝑥 2 + 3𝑥) + 2
group together the terms containing x.
𝑓(𝑥) = −3(𝑥 2 − 𝑥) + 2
factor out a (in this case a = -3)
1
1
𝑓(𝑥) = −3 (𝑥 2 − 𝑥 + 4) + 2 − (−3) (4)
1
3
𝑓(𝑥) = −3 (𝑥 2 − 𝑥 + 4) + 2 + (4)
1
11
𝑓(𝑥) = −3 (𝑥 2 − 𝑥 + 4) + ( 4 )
complete the expression in parenthesis to
make it a perfect square trinomial by adding
a constant term then subtract
the same value from the constant term.
1 2
11
𝑓(𝑥) = −3 (𝑥 − 2) + ( 4 )
express the perfect square trinomial as a
square of binomial and simplify the constant
term.
𝟏 𝟐
𝟐
Thus the vertex form of 𝒇(𝒙) = −𝟑𝒙𝟐 + 𝟑𝒙 + 𝟐 is 𝒇(𝒙) = −𝟑 (𝒙 − ) +
𝟏𝟏
𝟒
Now, try this. Express 𝑓(𝑥) = −𝑥 2 + 2𝑥 − 4 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘.
You should get 𝑓(𝑥) = −(𝑥 − 1)2 − 3.
If you get the correct answer, you may proceed to the next discussion.If not, I am sorry but
you have to go back to Example 3 and try to solve it again.
Page 104 of 148
D. Discussing new concepts and practicing new skills # 1
DAY 2 (October 7)
For this part, let us look at the graph of a quadratic function and observe its
characteristics.
The graph of a quadratic function which is a smooth curve is called a parabola.
Parabolas have certain common characteristics:
1. It has a turning point called the vertex which is either the lowest point (minimum point) or
the highest point(maximum point) of the graph of a quadratic function. The vertex is the
point (ℎ, 𝑘).
2. It has a line called the axis of symmetry which divides the graph into two parts such that
one-half of the graph is a reflection of the other half. The line 𝑥 = ℎis the axis of
symmetry.
3. If 𝑎 > 0, the parabola opens upward. If 𝑎 < 0, the parabola opens downward.
4. If the parabola opens upward, the vertex of the parabola is the lowest point. If its opens
downward, the vertex is the highest point. 𝑘 is the minimum or the maximum value of
the function.
5. It has a minimum point when the graph opens upward and a maximum point when the
graph opens downward.
6. The x-intercept(s) of the graph are the zeroes of the quadratic function and the roots of
the corresponding/related quadratic equation. The y-intercept is when 𝑥 = 0 in the
quadratic function.
7. The domain of quadratic function is the set of all possible values of x. The range
depends on whether the parabola opens upward or downward. If it opens upward, the
range is the set {𝑦: 𝑦 ≥ 𝑘} ; if it opens downward, then the range is the set {𝑦: 𝑦 ≤ 𝑘}.
These characteristics will be discussed further in the different forms of the quadratic
function.
The graph of a quadratic function can be drawn using the vertex, the axis of symmetry, and
some symmetric points.
Follow these steps in drawing the graph of a quadratic function.
1. Find the coordinates of the vertex (h, k).
2. Identify the axis of symmetry (x = h)
3. Determine the direction of the opening and the minimum/maximum value.
4. Form a table of values for 𝑥 and 𝑓(𝑥).
5. Draw the graph.
Now, let us have the graph of the following quadratic function.
Example 1. Graph the function 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7.
Step 1: Find the coordinates of the vertex.
Method 1: Express 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘.
Let 𝑓(𝑥) = 𝑦:
𝑦 = 𝑥 2 − 8𝑥 + 7
Add -7 to both sides:
𝑦 − 7 = 𝑥 2 − 8𝑥
Complete the square:
𝑦 − 7 + 16 = 𝑥 2 − 8𝑥 + 16
Express left side as square of
𝑦 + 9 = (𝑥 − 4)2
binomial:
Add -9 to both sides:
𝑦 = (𝑥 − 4)2 − 9
Therefore the vertex (h, k) is ( 4, -9 )
Page 105 of 148
Method 2: Using ℎ = −
𝑏
2𝑎
and 𝑘 =
4𝑎𝑐−𝑏 2
4𝑎
𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7
Identify a, b, and c:
a=1
Substitute values in the formula:
b = -8 c = 7
𝑏
ℎ = − 2𝑎
𝑘=
−8
ℎ = − 2(1)
𝑘=
8
4𝑎𝑐−𝑏2
4𝑎
4(1)(7) − (−8)2
4(1)
28 − 64
4
ℎ=2
𝑘=
ℎ=4
𝑘 = −9
The vertex (h,k) is at (𝟒, −𝟗).
Step 2: Identify the axis of symmetry.
the axis of symmetry is obtained using 𝑥 = −
𝑏
2𝑎
or 𝑥 = ℎ.
The axis of symmetry is 𝑥 = 4 since ℎ = 4.
Step 3: Determine the direction of the opening and the minimum/maximum value.
Notice that the coefficient of 𝑥 2 , which is a = 1, is positive.
Therefore, the graph opens upward.
The vertex will be the lowest point. The minimum value is −9.
Step 4: Form a table of values for 𝑥 and 𝑓(𝑥).
Choose values for x that are less than and greater than h = 4.
vertex
𝑥
1
2
3
4
5
6
7
𝑓(𝑥)
0
- 5
-8
-9
-8
-5
0
x
1
2
3
4
5
6
7
𝒙𝟐 − 𝟖𝒙 + 𝟕
(1)2 − 8(1) + 7
(2)2 − 8(2) + 7
(3)2 − 8(3) + 7
(4)2 − 8(4) + 7
(5)2 − 8(5) + 7
(6)2 − 8(6) + 7
(7)2 − 8(7) + 7
𝒇(𝒙)
0
-5
-8
-9
-8
-5
0
Page 106 of 148
Step 5: Draw the graph.
In the Cartesian Plane, plot the
Points (1,0), (2, −5), (3, −8),
(4, −9), (5, −8), (6, −5), 𝑎𝑛𝑑 (7,0).
and connect the points with a smooth line
x=4
Axis of Symmetry: x = h
Vertex: (h,k)
(4,-9)
➢ Solve for the x-intercepts and y-intercept of 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7.
For x-intercepts,
Solution:
𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7
𝑥 2 − 8𝑥 + 7 = 0
(𝑥 − 1)(𝑥 − 7) = 0
𝑥−1=0
𝑥−7=0
𝑥=1
𝑥=7
Set 𝑦 = 𝑓(𝑥) = 0
Solve 𝑥 2 − 8𝑥 + 7 = 0 using any method.
Solve for 𝑥.
The x-intercepts of the graph of 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7 are 𝟏 𝒂𝒏𝒅 𝟕.
This means that the graph intersects the x-axis at 1 and 7
For y-intercept,
Solution:
𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7
𝑥 2 − 8𝑥 + 7 = 0
𝑓(0) = (0)2 − 8(0) + 7
𝑓(0) = 7 or 𝑦 = 7
Set 𝑥 = 0 in 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7.
The y-intercept of the graph of 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7 is 7.
This means that the graph intersects the y-axis at 7
*Notice that the y-intercept of 𝑓(𝑥) = 𝑥 2 + 𝑏𝑥 + 𝑐 is the constant c.
➢ Determine the domain and range.
The domain (𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥) of 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7 is the set of all real numbers.
The range (𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑦) of 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 7 is the set {𝑦: 𝑦 ≥ −9}
Let us have another example in graphing a quadratic function.
Example 2. Draw the graph of the quadratic function 𝑓(𝑥) = −𝑥 2 + 2𝑥 + 3.
Step 1: Find the coordinates of the vertex.
Method 1: Express 𝑓(𝑥) = −𝑥 2 + 2𝑥 + 3 in the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘.
Let 𝑓(𝑥) = 𝑦:
𝑦 = −𝑥 2 + 2𝑥 + 3
Page 107 of 148
Add -3 to both sides:
Factor out a:
Complete the square:
Express left side as square of
binomial:
𝑦 − 3 = −𝑥 2 + 2𝑥
𝑦 − 3 = −(𝑥 2 − 2𝑥)
𝑦 − 3 + (−1) = −(𝑥 2 − 2𝑥 + 1)
𝑦 − 3 − 1 = −(𝑥 − 1)2
𝑦 − 4 = −(𝑥 − 1)2
Add 4 to both sides:
𝑦 = (𝑥 − 1)2 + 4
Therefore the vertex (h, k) is (1, 4)
Method 2: Using ℎ = −
𝑏
2𝑎
and 𝑘 =
4𝑎𝑐−𝑏 2
4𝑎
𝑓(𝑥) = −𝑥 2 + 2𝑥 + 3
Identify a, b, and c:
a = -1
b=2
𝑏
ℎ = − 2𝑎
Substitute values in the formula:
𝑘=
2
ℎ = − 2(−1)
𝑘=
2
c=3
4𝑎𝑐−𝑏2
4𝑎
4(−1)(3) − (2)2
4(−1)
−12 − 4
ℎ=2
𝑘=
ℎ=1
𝑘=4
−4
The vertex (h,k) is at (𝟏, 𝟒).
Another way in solving for 𝑘 is to evaluate the function 𝑦 = −𝑥 2 + 2𝑥 + 3 when
ℎ = 1 𝑜𝑟 𝑥 = 1.
Solve for 𝑘.
𝑦 = −𝑥 2 + 2𝑥 + 3
𝑦 = −(1)2 + 2(1) + 3
Since ℎ = 1, the x-coordinate of the vertex is 1.
𝑦 = −1 + 2 + 3
𝑦 = 4 or 𝑘 = 4
The vertex is at (1, 4).
Step 2: Identify the axis of symmetry.
The axis of symmetry is obtained using 𝑥 = −
𝑏
2𝑎
or 𝑥 = ℎ.
The axis of symmetry is 𝑥 = 1 since ℎ = 1
Step 3: Determine the direction of the opening and the minimum/maximum value.
The coefficient of 𝑥 2 is a = −1.
Therefore, the graph opens downward.
The vertex will be the highest point. The maximum value is 4.
Step 4: Form a table of values for 𝑥 and 𝑓(𝑥).
vertex
𝑥
-2
-1
0
1
2
3
4
𝑓(𝑥)
-5
0
3
4
3
0
-5
Page 108 of 148
−𝒙𝟐 + 𝟐𝒙 + 𝟑
−(−2)2 + 2(−2) + 3
−(−1)2 + 2(−1) + 3
−(0)2 + 2(0) + 3
−(1)2 + 2(1) + 3
−(2)2 + 2(2) + 3
−(3)2 + 2(3) + 3
−(4)2 + 2(4) + 3
x
-2
-1
0
1
2
3
4
𝒇(𝒙)
-5
0
3
4
3
0
-5
Step 5: Draw the graph.
In Cartesian plane, plot the points (−2, 5), (−1, 0), (0, 3), (1, 4), (2, 3), (3, 0), 𝑎𝑛𝑑 (4, −5). and
connect the points with a smooth line.
y
6
𝑓(𝑥) = −𝑥 2 + 2𝑥 + 3
Vertex: (h, k)
( 1, 4)
4
2
−4
−3
−2
−1
O
1
2
3
4
5
6
x
−2
−4
Axis of Symmetry: x = h
x=1
−6
➢ Solve for the x-intercepts and y-intercept of 𝑓(𝑥) = −𝑥 2 + 2𝑥 + 3.
For x-intercepts,
Solution:
𝑓(𝑥) = −𝑥 2 + 2𝑥 + 3
−𝑥 2 + 2𝑥 + 3 = 0
Set 𝑓(𝑥) = 0
2
−1(−𝑥 + 2𝑥 + 3) = (0) − 1
Multiply both sides of the equation by -1 to
2
make −𝑥 positive.
𝑥 2 − 2𝑥 − 3 = 0
(𝑥 + 1)(𝑥 − 3) = 0
by factoring method
𝑥+1=0
𝑥−3=0
Solve for 𝑥.
𝑥 = −1
𝑥=3
The x-intercepts of the graph of 𝑦 = −𝑥 2 + 2𝑥 + 3 are −𝟏 𝐚𝐧𝐝 𝟑.
The graph intersects the x-axis at -1 and 3
For y-intercept,
Solution:
𝑦 = −𝑥 2 + 2𝑥 + 3
𝑦 = −(0)2 + 2(0) + 3
𝑓(0) = 0 − 0 + 3
𝑓(0) = 3 or 𝑦 = 3
Set 𝑥 = 0 in 𝑦 = −𝑥 2 + 2𝑥 + 3
The y-intercept of the graph of 𝑦 = −𝑥 2 + 2𝑥 + 3 is 3.
The graph intersects the y-axis at 3
Page 109 of 148
➢ Determine the domain and range.
The domain (𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑥) of 𝑦 = −𝑥 2 + 2𝑥 + 3 is the set of all real numbers.
The range (𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑦)of 𝑦 = −𝑥 2 + 2𝑥 + 3 is the set {𝑦: 𝑦 ≤ 4}
Now, it’s your turn. Will you try to draw the graph of the quadratic function
𝑔(𝑥) = 𝑥 2 + 6𝑥 + 5? Then, identify the following:
a. vertex
b. opening of the graph
c. minimum/maximum value
d. axis of symmetry,
e. x-intercepts and y-intercept
f. domain and range
You should have the graph of 𝑔(𝑥) = 𝑥 2 + 6𝑥 + 5 as shown below.
y
8
6
4
2
−7
−6
−5
−4
−3
−2
−1
O
1
2
x
−2
(
- 3, - 4)
−4
−6
You should get the following answers.
a. The vertex is at (−3, −4).
b. The opening of the graph is upward since 𝑎 is positive.
c. It has a minimum value of −4.
d. The axis of symmetry is 𝑥 = −3.
e. The x-intercepts are – 1 and – 5. While the y-intercept is 5.
f. The domain is the set of all real numbers and the range is the set {𝑦: 𝑦 ≥ −4}
If you get the graph and answers correctly, you may proceed to the next part of our lesson.
If not, I am sorry but you have to go back to example number 1 and try all over again.
E. Continuation of the discussion of new concepts
Now you have learned how to graph a quadratic function, let us now study in detail the graphs
of the various forms of the quadratic functions and the behavior of the graph in the coordinate
plane.
THE GRAPH OF 𝒇(𝒙) = 𝒂𝒙𝟐
Given on the figure below are graphs of some quadratic functions of the form 𝑓(𝑥) = 𝑎𝑥 2 for
|𝑎| > 1 and |𝑎| < 1 compared with the parent quadratic function 𝑓(𝑥) = 𝑥 2 where 𝑎 = 1. Also
given are the graphs of 𝑓(𝑥) = 𝑎𝑥 2 where 𝑎 < 0.
Page 110 of 148
𝑓(𝑥) = 2𝑥 2
y
6
𝑓(𝑥) = 𝑥 2
4
𝑓(𝑥) =
2
−6
−4
−2
O
2
4
6
1 2
𝑥
2
x
1
𝑓(𝑥) = − 𝑥 2
2
−2
−4
𝑓(𝑥) = −𝑥 2
−6
𝑓(𝑥) = 2𝑥 2
Properties of the graph of the function 𝑓(𝑥) = 𝑎𝑥 2 .
1. The vertex is at the origin (0,0).
2. The line of symmetry is the y-axis, x = 0
3. If a is positive, the graph opens upward and the vertex is the minimum point.
4. If a is negative, the graph opens downwar and the vertex is the maximumpoint.
5. If |𝑎| > 1 the graph is narrower than the graph of 𝑓(𝑥) = 𝑥 2
6. If |𝑎| > 1 the graph is narrower than the graph of 𝑓(𝑥) = 𝑥 2
Thus, we conclude that the coefficient “a” 𝑓(𝑥) = 𝑎𝑥 2 makes the graph narrower or wider.
This behavior of the graph is called stretching.
Examples: For each quadatic function, determine the coordinates of the vertex, the equaton of
the axis of symmetry, identify whether the graph opens upward or downward, identify whether
the vertex is a minimum or a maximum point and tell whether it is narrower or wider than the
graph of 𝑓(𝑥) = 𝑥 2 .if it opens upward and 𝑓(𝑥) = −𝑥 2 if it opens downward.
vertex
Opening
Max/min
point
Axis of
symmetry
( 0,0 )
upward
x=0
2. 𝑓(𝑥) = 3 𝑥 2
( 0,0 )
upward
x=0
3. 𝑓(𝑥) =
−5𝑥 2
( 0,0 )
downward
x=0
Function
1. 𝑓(𝑥) = 4𝑥 2
2
Wider/Narrower
than 𝑓(𝑥) =
±𝑥 2 .
Narrower than
𝑓(𝑥) = 𝑥 2 .
Wider than
𝑓(𝑥) = 𝑥 2
Narrower than
𝑓(𝑥) = −𝑥 2 .
1
4. 𝑓(𝑥) = 4 𝑥 2
5. 𝑓(𝑥) = 3𝑥 2
Now try to answer nos. 4 and 5 in the table above.
Check your work by turning to page 122 for the key to correction.
Page 111 of 148
THE GRAPH OF 𝒇(𝒙) = 𝒂𝒙𝟐 + 𝒌
Study and analyze the following graphs of quadratic function of the form 𝑓(𝑥) = 𝑎𝑥 2 + 𝑘
y
10
8
𝑓(𝑥) = 𝑥 2 + 2
6
4
𝑓(𝑥) = 𝑥 2
2
−6
−4
−2
O
2
4
6
x
𝑓(𝑥) = 𝑥 2 − 2
−2
−4
Properties of the graph of the function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑘.
1. The graph of 𝑓(𝑥) = 𝑎𝑥 2 + 𝑘 is similar to the graph of 𝑓(𝑥) = 𝑎𝑥 2 except that it is
translated/shifted |𝑘| units vertically. If k is positive, the translation is upward. If k is
negative, the translation is downward.
2. The vertex is at (0, 𝑘)
3. The line of symmetry is the y-axis, x = 0
4. If a is positive, the vertex is the minimum point.
5. If a is negative, the vertex is the maximum point.
Example1: For each quadatic function, determine: the coordinates of the vertex, the equation of
the axis of symmetry, identify whether the graph opens upward or downward, tell whether the
vertex is a minimum or a maximum point. The first two items are done for you.
Function
vertex
Axis of
symmetry
Opening
min/max
point
1. 𝑦 = 4𝑥 2 + 3
( 0,3 )
x=0
upward
minimum
2. 𝑓(𝑥) = −𝑥 2 − 7
( 0,-7 )
x=0
downward
maximum
3. 𝑔(𝑥) = −5𝑥 2 − 1
1
4. 𝑝(𝑥) = 2𝑥 2 + 2
Now try to answer nos. 3 and 4 in the table above.
Check your work by turning to page 122 for the key to correction.
Example 2: Write the resulting function in each of the following translation. The first two items
are done for you.
1. 𝑓(𝑥) = 𝑥 2 is translated 3 units downward
Answer: 𝑓(𝑥) = 𝑥 2 − 3
2
2. 𝑔(𝑥) = −2𝑥 is translated 4 units upward
Answer: 𝑔(𝑥) = −2𝑥 2 + 4
Page 112 of 148
3. ℎ(𝑥) = 4𝑥 2 is translated 6 units upward
4. 𝑝(𝑥) = −3𝑥 2 is translated 2 units downward
Answer: _____________
Answer: _____________
Now try to answer numbers 3 and 4 above.
Check your work by turning to page 122 for the key to correction.
THE GRAPH OF 𝒇(𝒙) = 𝒂(𝒙 − 𝒉)𝟐
Another form of quadratic function is 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 .
Let us study the following graphs of the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 below.
y
𝑓(𝑥) = (𝑥 + 2)2
10
8
𝑓(𝑥) = 𝑥 2
6
4
𝑓(𝑥) = (𝑥 − 2)2
2
−6
−4
−2
O
2
4
6
x
−2
−4
Properties of the graph of the function 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2.
1. The graph 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2.of is similar to the graph of 𝑓(𝑥) = 𝑎𝑥 2 except that it is
translated/shifted |ℎ| units horizontally. If h is positive, the translation is to the right. If
h is negative, the translation is to the left.
2. The vertex is at (ℎ, 0)
3. The line of symmetry is x = h
4. If a is positive, the vertex is the minimum point.
5. If a is negative, the vertex is the maximum point.
Example1: For each quadatic function, determine the coordinates of the vertex, the equation of
the axis of symmetry, identify whether the graph opens upward or downward, tell whether the
vertex is a minimum or a maximum point. The first two items are done for you.
1.
2.
3.
4.
Function
vertex
𝑦 = (𝑥 + 3)2
𝑓(𝑥) = −2(𝑥 − 7)2
𝑔(𝑥) = (𝑥 + 7)2
𝑝(𝑥) = −5(𝑥 + 7)2
( -3,0 )
( 7,0 )
Axis of
symmetry
x = -3
x=7
Opening
upward
downward
min/max point
minimum
maximum
Now try to answer nos. 3 and 4 in the table above.
Check your work by turning to page 122 for the key to correction.
Page 113 of 148
Example 2: Write the resulting function in each of the following translation. The first two items
are done for you.
1. 𝑓(𝑥) = 𝑥 2 is translated 3 units to the left
Answer: 𝑓(𝑥) = (𝑥 + 3)2
2. 𝑔(𝑥) = −2𝑥 2 is translated 1 unit to the right
Answer: 𝑔(𝑥) = −2(𝑥_ − 1)2
3. ℎ(𝑥) = 4𝑥 2 is translated 9 units to the right
Answer: _______________
2
4. 𝑝(𝑥) = −3𝑥 is translated 2 units to the left
Answer: _______________
Now try to answer nos. 3 and 4above.
Check your work by turning to page 122 for the key to correction.
THE GRAPH OF 𝐟(𝐱) = 𝐚(𝐱 − 𝐡)𝟐 + 𝐤
Shown below are grphs of the form 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘
y
12
10
𝑓(𝑥) = (𝑥 − 4)2 + 4
8
6
𝑓(𝑥) = 𝑥 2
4
2
−8
𝑓(𝑥) = 𝑥 2
−6
−4
−2
O
2
4
6
8
x
−2
−4
−6
𝑓(𝑥) = −(𝑥 + 4)2 − 4
−8
−10
−12
Take note of the following characteristics of the graphs.
𝑓(𝑥) = (𝑥 − 4)2 + 4: vertex (4,4) ; axis of symmetry: x = 4
𝑓(𝑥) = −(𝑥 + 4)2 − 4: vertex (-4,-4) ; axis of symmetry: x = -4
Te graph of the function 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘 has the same characteristics as the graph of
𝑓(𝑥) = 𝑎𝑥 2 . Only, it is the translation of the graph of 𝑓(𝑥) = 𝑎𝑥 2 |ℎ| units horizontally and |𝑘|
units vertically.
The vertex is at ( h,k ) and the axis of symmery is x = h.
Example1: For each quadatic function, determine the coordinates of the vertex, the equation of
the axis of symmetry, identify whether the graph opens upward or downward, tell whether the
vertex is a minimum or a maximum point. The first two items are done for you.
Axis of
min/max
Function
vertex
Opening
symmetry
point
2
( 5,3 )
x=5
upward
minimum
1. 𝑓(𝑥) = 3(𝑥 − 5) + 3
2
x=3
downward
maximum
2. 𝑓(𝑥) = −5(𝑥 − 3) − 4 ( 3,-4 )
3. 𝑔(𝑥) = −(𝑥 + 7)2 + 1
4. 𝑝(𝑥) = (𝑥 − 8)2 − 1
Now try to answer nos. 3 and 4 in the table above.
Check your work by turning to page 122 for the key to correction.
Page 114 of 148
Example 2: Write the resulting function in each of the following translation. The first two items
are done for you.
1. The graph of 𝑓(𝑥) = 2𝑥 2 is translated 3 units to the right and 4 units upward
Answer: 𝑓(𝑥) = 2(𝑥 + 3)2 + 4
2. The graph of 𝑓(𝑥) = −2(𝑥 − 3)2 is translated 5 units to the right and 7 units upward
Answer: 𝑓(𝑥) = −2(𝑥 + 2)2 + 7
3. The graph of 𝑓(𝑥) = −3𝑥 2 is translated 3 units to the right and 4 units upward
Answer: ________________
4. The graph of 𝑓(𝑥) = −4(𝑥 + 1)2 is translated 3 units to the right and 4 units downard
Answer: ________________
Now try to answer nos. 3 and 4 above.
Check your work by turning to page 122 for the key to correction.
F. Developing mastery
A. Match the given quadratic function 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 to its equivalent vertex form
𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘. Write your answers in your notebook.
𝒚 = 𝒙𝟐 + 𝟒𝒙 + 𝟓
𝒚 = −(𝒙 − 𝟏)𝟐 + 𝟐.
𝒚 = 𝒙𝟐 − 𝟒𝒙 + 𝟏
𝒚 = 𝟐(𝒙 − 𝟏)𝟐 + 𝟐.
𝒚 = 𝟐𝒙𝟐 − 𝟒𝒙 + 𝟒
𝒚 = −𝟓(𝒙 − 𝟏)𝟐 + 𝟑
𝒚 = −𝒙𝟐 − 𝟐𝒙 + 𝟏
𝒚 = (𝒙 − 𝟐)𝟐 − 𝟑
𝒚 = −𝟓𝒙𝟐 + 𝟏𝟎𝒙 − 𝟐
𝒚 = (𝒙 + 𝟐)𝟐 + 𝟏.
B. Choose me!
Answer each of the following questions. Write your answers in your notebook.
1. What is the vertex of f(x) = x2 + 4x + 7?
a. V (-2, 3)
b. V (3, -2)
2. What is the opening of the graph of y = -3x2 – 10x + 4?
a. Upward
b. Downward
3. What is the axis of symmetry of g(x) = -5x2 – 5?
a. x = 0
b. x = 5
Page 115 of 148
4. What is the minimum value of f(x) = ½ x2 + 3?
a. 0
b. 3
5. Which has a maximum point?
a. y = x2 – 3x + 5
b. y = -x2 + 3x – 5
6. Which has a lower vertex?
a. y = x2 + 2x + 3
b. y = x2 – 4x + 7
7. Which has the higher vertex?
a. f(x) = -x2 – 6x + 15
b. f(x) = -x2 + 6
Now, check your work by turning to page 123 for the key to
correction. How many correct answers did you get? Rate your
result using the table above.
If your score is at least 4 out of 7, you may now proceed to next
part of the discussion.
Score
Description
6-7
Very Good
4–5
Good
2–3
Fair
0-1
Turn to page
120-121sec. J
and work on
the enrichment
activities
C. MATCH OR MISMATCH!
Decide whether the given graph is a match or mismatch with the indicated equation of quadratic
function. Write match if the graph corresponds with the correct equation. Otherwise, indicate the
correct equation of the quadratic function. Write your answers in your notebook.
1.
y
10
9
8
𝑦 = (𝑥 + 4)2
7
6
5
4
3
__________________
2
1
−4
−3
−2
−1
O
1
2
3
4 x
__________________
−1
y
2.
4
3
𝑦 = 2𝑥 2 − 3
2
1
−4
−3
−2
−1
O
−1
1
2
3
4 x
____________________
−2
−3
____________________
−4
Page 116 of 148
y
3.
6
𝑦 = 2𝑥 2 − 12𝑥 + 36
5
4
3
___________________
2
1
−1
O
1
2
3
4
5
6 x
___________________
−1
4.
y
4
𝑦 = 𝑥 2 − 12𝑥 + 36
3
2
1
−1 O
___________________
1
2
3
4
5
6
7
8
x
___________________
−1
5.
y
5
4
𝑦 = −𝑥 2 + 4
3
2
1
−5 −4 −3 −2 −1 O
−1
1
2
3
4
x
__________________
−2
−3
__________________
−4
−5
Now, check your work by turning to page 123 for the key to correction. How many correct
answers did you get?
G. Finding practical application of the concepts and skill in daily living
Throwing a ball, shooting a cannon, diving from a platform and hitting a golf ball are all
examples of situations that can be modeled by quadratic functions. In many of these
situations you will want to know the highest or lowest point of the parabola, which is known
as the vertex.
H. Making generalization and abstraction about the lesson
DAY 3 (October 8)
To transform 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 in the form 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 using completing the square,
a. Group together the terms containing x.
b. Complete the expression in parenthesis to make it a perfect square trinomial by adding
a constant term then subtract the same value from the constant term.
Note : To determine what to add to complete the square, take half of coefficient of x and
square it then add on both sides of the equation.
If the a >1, to find the real value added to the entire equation, be sure to multiply the
constant term by the value of a when you subtract it to the constant term since it has been
factored out in the beginning.
c. Express the perfect square trinomial as a square of binomial and simplify the constant
term.
Page 117 of 148
To transform 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 in the form 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 using the vertex h and k,
a. Identify the values of a, b and c in 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
−𝑏
2𝑎
and 𝑘 =
4𝑎𝑐−𝑏 2
,
4𝑎
b.
Ffind the vertex using ℎ =
c.
Substitute 𝑎, ℎ and 𝑘 in the form 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 .
Characteristics of a parabola.
✓ The turning point of the parabola is called the vertex which is either the lowest point
(minimum point) or the highest point(maximum point) of the graph of a quadratic
function. The vertex is the point (ℎ, 𝑘).
✓ The axis of symmetry is the vertical line which divides the graph into two parts such that
one-half of the graph is a reflection of the other half. The line 𝑥 = ℎis the axis of
symmetry.
✓ If 𝑎 > 0, the parabola opens upward. If 𝑎 < 0, the parabola opens downward.
✓ If the parabola opens upward, the vertex of the parabola is the lowest point. If its opens
downward, the vertex is the highest point. 𝑘 is the minimum or the maximum value of
the function.
✓ The x-intercept(s) of the graph are the zeroes of the quadratic function and the roots of
the corresponding/related quadratic equation.
✓ The y-intercept is when 𝑥 = 0 in the quadratic function.
✓ The domain of quadratic function is the set of all possible values of x. The range
depends on whether the parabola opens upward or downward. If it opens upward, the
range is the set {𝑦: 𝑦 ≥ 𝑘} ; if it opens downward, then the range is the set {𝑦: 𝑦 ≤ 𝑘}.
✓ The graph of a quadratic function can be drawn using the vertex, the axis of symmetry,
and some points (table of values).
❖ Different forms of the Quadratic Function
1. f(x) = ax²
2. f(x) = ax² + k
3. f(x) = a(x – h)²
4. f(x) = a(x – h)² + k
➢ In the graph of f(x) = ax², if |a| >1, the graph of the parabola is narrower than the
parent function. If the |a|<1, the graph of the parabola is wider than the parent graph.
This behavior of the graph is called stretching.
➢ In the graph of f(x) = x2 + k which has the same shape as the graph of f(x) = x2,
if k is positive, the graph of f(x) = x2 shifts “k” units upward. If k is negative, the graph of
f(x) = x2 shifts “k” units downward.
➢ In the graph of f(x) = (x - h)2 which has the same shape as the graph of f(x) = x2,
if “h” is positive, the graph of f(x) = (x - h)2 is the graph of f(x) = x2 shifted “h” units to
theright. If “h” is negative, the graph of f(x) = (x - h)2 is the graph of f(x) = x2 shifted “h” units
to the left.
➢ In the graph of f(x) = a(x - h)2 + k with vertex (h, k), the longitudinal and vertical
movement of the graph of f(x) is called translation. If a > 0, the parabola opens
upward and the minimum function value is k. If a < 0, the parabola opens
downward and the maximum function value is k.
Note:
•
“a” determines the opening of the size of the parabola
•
“h” indicates shifting/translating/moving the graph horizontally
Page 118 of 148
I. Evaluating learning
Direction: Use yellow paper and graphing paper to answer the evaluation. Provide
necessary solutions for your answers. Use the format shown below to label
your paper and it is to be submitted to your Math teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 7 Parent’s Signature: __________________
EVALUATION 7
I.
Write the indicated letter of the quadratic function in the form y = a(x-h)2 + k into the
box that corresponds to its equivalent general form y = ax2 + bx+ c
I
y = (x-1)2 – 4
5
4
S y = 2(x + )2 –
49
8
2
E y = (x- 3)2 + 2
A y = 3(x + 2)2 –
1
2
N y = (x - 0)2 – 36
T y = (x - 1)2 – 16
F y = (x - 3)2 + 5
1
3
M y = (x - 2)2 + 2
U y = -2(x - 3)2 + 1
H y = 2(x + 1)2 - 2
Dialog Box:
7
y = x2 – x + 4
y = 3x2 + 12x +
23
2
y = x2 - 2x – 15
y = 2x2 + 4x
y = x2 – 2x – 3
y = 2x2 + 5x – 3
y = x2 – 6x + 14
y = -2x2 + 12x – 17
y = x2 – 36
Page 119 of 148
II.
DRAW AND DESCRIBE ME!
Sketch the graph of each quadratic function and identify the vertex, domain, range, and the
opening of the graph. State whether the vertex is a minimum or a maximum point and write
the equation of its axis of symmetry.
1. f(x) = x2 + 3x – 2
Vertex _____________
Opening of the graph _____________
Vertex is a _____________ point
Equation of the axis of symmetry ______
Domain: _____
Range: _____
2. f(x) = -x2 - 2x + 1
Vertex _____________
Opening of the graph _____________
Vertex is a _____________ point
Equation of the axis of symmetry ______
Domain: _____
Range: _____
3. f(x) = 2x2 + 4
Vertex _____________
Opening of the graph _____________
Vertex is a _____________ point
Equation of the axis of symmetry ______
Domain: _____
Range: _____
4. f(x) = -3x2 + 2x – 5
Vertex _____________
Opening of the graph _____________
Vertex is a _____________ point
Equation of the axis of symmetry ______
Domain: _____
Range: _____
5. f(x) = x2 – 2x
Vertex _____________
Opening of the graph _____________
Vertex is a _____________ point
Equation of the axis of symmetry ______
Domain: _____
Range: _____
Page 120 of 148
III.
In each of the following quadratic functions, give the coordinates of the vertex,
equation of the axis of symmetry, identify whether the graphs opens or downward
and tell whether the vertex is a maximum or minimum point.opening and movement
of the vertex.
Quadratic function
Vertex
Axis of
Symmetry
Max/Min
point
Opening
1. f(x) = -3x2
2. f(x) = (x - 4)2
3. f(x) = 2x2 - 3
4. f(x) = -2(x - 3)2 + 6
5. f(x) = ½ (x + 2)2 + 3
IV.
Write the resulting function in each of the following translations.
______________
______________
______________
______________
______________
1. The graph of 𝑓(𝑥) = 3𝑥 2 is translated 7 units upward.
2. The graph of 𝑓(𝑥) = −𝑥 2 − 1 is translated 2 units upward.
3. The graph of 𝑓(𝑥) = 2𝑥 2 is translated 8 units to the right.
4. The graph of 𝑓(𝑥) = −3(𝑥 + 1)2 is translated 2 units to the left
5. The graph of 𝑓(𝑥) = −3(𝑥 − 3)2 + 4 is translated 3 units to the
right and 2 units upward..
J. Additional activities for application of remediation
Write all your answers in your notebook.
I.
Transform the following quadratic functions from general to vertex form.
1. y = x2 + 8x +21
2. y = x2 – 4x + 8
II.
Graph the quadratic function 𝑔(𝑥) = 2𝑥 2 + 4𝑥 + 5 and identify the following:
a. vertex
b. opening of the graph
c. minimum/maximum value
d. axis of symmetry,
e. domain and range
Page 121 of 148
III.
Loop the words given below either horizontally, vertically or diagonally.
Turn to page 123 for the key to corection for this additional activity.
Did you get the correct answers?
Page 122 of 148
KEY TO CORRECTIONS
NAME
LOCATION
1. Banaue Rice Terraces
Ifugao
2. Hundred Islands
Pangasinan
3. Chocolate Hills
Bohol
4. Mayon Volcano
Albay
5. Taal Volcano
(Discussion of new concepts)
6. Boracay
Batangas
Aklan
Graph of 𝑓(𝑥) = 𝑎𝑥 2
Example 4: vertex: (0,0), upward, x = 0, graph is wider than 𝑓(𝑥) = 𝑥 2
5.vertex (0,0), upward, x = 0, graph is narrower than 𝑓(𝑥) = 𝑥 2
Graph of 𝑓(𝑥) = 𝑎𝑥 2 + 𝑘
Example 1.
3. Vertex: (0,-1), x = 0, downward. Maximum
1
4.Vertex: (0, 2), x = 0, upward, minimum
Example 2:
3. ℎ(𝑥) = 4𝑥 2 + 6
4.𝑝(𝑥) = −3𝑥 2 − 2
Graph of 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2
Example 1:
3. Vertex: (-7,0), x = -7, upward, minimum
4.vertex (-7,0), x = -7, downward, maximum
3. ℎ(𝑥) = 4(𝑥 − 9)2
4. 𝑝(𝑥) = −3(𝑥 + 2)2
Graph of 𝑓(𝑥) = 𝑎(𝑥 − ℎ)2 + 𝑘
Example 1:
3. Vertex: (-7,1), x = -7, downward , maximum
4.vertex: (8,-1), x = 8, upward, minimum
Example 2:
3. 𝑓(𝑥) = −3(𝑥 − 3)2 + 4
4. 𝑓(𝑥) = −4(𝑥 − 3)2 − 4
Example 2:
Page 123 of 148
(Developing Mastery)
A. Match the given quadratic function 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 to its equivalent vertex form 𝑦 =
𝑎(𝑥 − ℎ)2 + 𝑘.
B.
1.
2.
3.
4.
5.
6.
7.
Choose Me!
A
B
A
B
B
A
A
C. Match or Mismatch!
1. Mismatch
𝑦 = 𝑥2 + 4
2. Match
3. Mismatch
𝑦 = 2(𝑥 − 3)2 + 1
4. Match
5. Match
(Additional Activities)
Key to Enrichment
1. y = (x+ 4)2 + 5
2. y = (x - 2)2 + 4
Graph of 𝑔(𝑥) = 2𝑥 2 + 4𝑥 + 5
y
12
11
10
9
8
7
6
5
4
3
2
1
−5
−4
−3
−2
−1
O
1
2
3
4
5 x
−1
−2
a.
b.
c.
d.
g.
Vertex: (-1,3)
opening of the graph: Upward
minimum/maximum value: 3
axis of symmetry: x = 1
The domain is the set of all real numbers and the range is the set {𝑦: 𝑦 ≥ 3}
Page 124 of 148
Page 125 of 148
SELF-INSTRUCTIONAL PACKETS
MATHEMATICS Grade 9 – SSC
School
PAMPANGA HIGH SCHOOL
Teacher
IVAN T. SALAS
Teaching Dates/
Week
Quarter
October 9 and 12, 2020
(Week 8)
FIRST
I. OBJECTIVES
A. Content standards:
The learner demonstrates understanding of key concepts of quadratic equations,
inequalities and functions, and rational algebraic equations.
B. Performance standards:
The learner is able to investigate thoroughly mathematical relationships in various
situations, formulate real-life problems involving quadratic equations, inequalities and
functions, and rational algebraic equations and solve them using a variety of strategies.
C. Learning competencies:
The learner
1. Determines the equation of a quadratic function given: (a) a table of values; (b)
graph; (c) zeros(M9AL-Ij-1)
2. Solves problems involving quadratic functions. (M9AL-Ii-j-2)
D. Objectives:
At the end of the lesson, the learners should be able to:
1. Find the equation of a quadratic function given: (a) a table of values; (b) graph;
(c) zeros
2. Solve real-world problems that involves quadratic function
II.
CONTENT
APPLICATION OF QUADRATIC FUNCTIONS
Learning Resources:
A. References:
Learners’ Materials Grade 9 Mathematics, pp. 140-150
B. Other Learning Resources:
(Math Trivia): www.mceducation.us
https://sciencing.com/everyday-examples-situations-apply-quadraticequations-10200.html
III.
PROCEDURES
DAY 1 (October 9)
A. Reviewing previous lesson or presenting the new lesson
Hi! How are you? I pray for your wellness. We are now down to the last week of the
quarter. Hooray!
Yesterday is World Teacher’s Day. Did you greet your teachers and your adviser?
We should also greet our parents and guardians as they are our first teacher in our lives
and they are helping you in your studies today while we are experiencing this turmoil
because of the pandemic.
We are going to explore more on quadratic function and this time, we we will focus on
zeros of quadratic function.
Page 126 of 148
Before we dig deeper in our lesson, let us have a review.
REVIEW:
Consider the given function:
f (x) = -2x2 + 5x + 3 for the following questions:
1. Does the parabola open upward or downward?
b
2. Determine the equation of the axis of symmetry. x = 2a
3. Determine the coordinates of the vertex. (h,k)
4. Determine the x and y intercepts.
x-intercepts: set f(x) = 0
y-intercepts: set x=0
5. Graph the function draw your own Cartesian plane.
6. What is the domain and range?
You may check your answers by turning to Page (+14).
Give yourself one (1) point for every correct answer. I hope you got them all correctly!
B. Establishing a purpose for the lesson
In this lesson, we will use the zeros in determining quadratic function. However, before
that, let us have a Math trivia about zero.
Different names for
the number ‘0’ include
zero, nought, naught,
nil, zilch, and zip.
0
Zero is the only number
that cannot be
represented in Roman
Numerals.The Latin word
‘nulla’ would have been
used instead.
Interesting facts, right? Do you know any trivia about zeros?
C. Presenting examples/instances of the lesson
In this lesson, we will learn how to find the quadratic function given its roots. Zeros are
the roots or x-intercept/s of a function.
Example 1. Find a quadratic function whose zeros are -1 and 4.
Solution:
If the zeros are -1 and 4, then
x = -1 and x = 4
x + 1 = 0 and x – 4 = 0
Applying the Zero Product Property
(x + 1)(x – 4) = 0
Apply the FOIL Method
𝑥 2 − 3𝑥 − 4 = 0
Equation of the quadratic function
2
𝑦 = 𝑥 − 3𝑥 − 4
Quadratic function
Now, you try to find the quadratic function whose zeros are -1 and 5.
You should get 𝑦 = 𝑥 2 + 4𝑥 − 5as function.
If your answer is correct, you may proceed to next example.If not, I am sorry but you
have to go back to example number 1 and try all over again.
Page 127 of 148
Example 2. Find an equation of a quadratic function with zeros -2 and -1.
Solution:
If the zeros are -2 and -1, then
x = -2 and x = -1
x + 2 = 0 and x + 1 = 0
Applying the Zero Product Property
(x + 2)(x + 1) = 0
Apply the FOIL Method
2
𝑥 + 3𝑥 + 20 = 0
Equation of the quadratic function
𝑦 = 𝑥 2 + 3𝑥 + 20
Quadratic function
Now, you try to find the function whose zeros are -9 and 5.
You should get 𝑦 = 𝑥 2 + 4𝑥 − 45as function.
If your answer is correct, you may proceed to next part.If not, I am sorry but you have
to go back to example number 1 and try all over again.
The equation of the quadratic function can be determined using table of values.
Example 1. Determine the equation of the quadratic function represented by the table of values
below.
3
x
-3
-2
-1
0
1
2
y
24
16
10
6
4
4
6
Solution:
Step 1. Notice that you can’t find any zeros from the given table of values. In this
case, take any three ordered pairs from the table, and use these as the values
of x and y in the equation 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
using point ( 1, 4)
using point (-1, 10)
using point (2, 4)
4 = 𝑎(1)2 + 𝑏(1) + 𝑐
𝟒= 𝒂+𝒃+𝒄
equation 1
2
10 = 𝑎(−1) + 𝑏(−1) + 𝑐
𝟏𝟎 = 𝒂 − 𝒃 + 𝒄
equation 2
4 = 𝑎(2)2 + 𝑏(2) + 𝑐
𝟒 = 𝟒𝒂 + 𝟐𝒃 + 𝒄
equation 3
We obtain a system of equation in a, b, and c. Using elimination method;
Step 2. Add corresponding terms in eq. 1 and eq. 2 to eliminate b.
eq.1 + eq. 2
We have
4=𝑎+𝑏+𝑐
(+) 10 = 𝑎 − 𝑏 + 𝑐
𝟏𝟒 = 𝟐𝒂 + 𝟐𝒄
equation 4
Step 3. Multiply the terms in eq.2 by 2 and add the corresponding terms in eq.3
to eliminate b.
2(eq.2) + eq.3
20 = 2𝑎 − 2𝑏 + 2𝑐2
2(10) = (𝑎 − 𝑏 + 𝑐)2
+) 4 = 4𝑎 + 2𝑏 + 𝑐
We have
𝟐𝟒 = 𝟔𝒂 + 𝟑𝒄
equation 5
Notice that equation 4 and equation 5 constitute a system of linear equations in
two variables.
Page 128 of 148
Step 4. To solve for c, multiply the terms in equation 4 by -3 and add
corresponding terms in equation 5.
3(eq.4) – equation 5
(−3)(14) = 2𝑎 + 2)(−3)
42 = 6𝑎 + 6𝑐
24 = 6𝑎 + 3𝑐
18 = 3𝑐
𝒄=𝟔
Step 5. To solve for a, substitute the value of c in equation 4 or 5.
14 = 2𝑎 + 2𝑐
14 = 2𝑎 + 2(6)
2𝑎 = 14 − 12
2𝑎
2
Using equation 4:
=
2
2
𝒂=𝟏
Step 6. To solve for b, substitute the value of c and a in equation 1, 2, or 3
4=𝑎+𝑏+𝑐
4=1+𝑏+6
4=7+𝑏
𝑏 =4−7
𝒃 = −𝟑
Using equaton 1:
Step 7. Substitute the values of a, b and c in 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
a =1, b = -3, c = 6 →
𝑦 = 𝑥 2 − 3𝑥 + 6
Thus, the equation of the quadratic function is 𝒚 = 𝒙𝟐 − 𝟑𝒙 + 𝟔 .
Your turn! Determine the equation of the quadratic function represented by the table
of values below.
6
x
1
2
3
4
5
y
5
11
19
29
41
55
You should get𝑦 = 𝑥 2 + 3𝑥 + 1.
If you get the correct answer, you may proceed with the next discussion. If not, I am
sorry but you have to go back to example number 1 and try all over again.
D. Discussing new concepts and practicing new skills # 1
When the vertex and any point on the parabola are clearly seen, the equation of the
quadratic function can easily be determined by using the form of a quadratic function 𝑦 =
𝑎(𝑥 − ℎ)2 + 𝑘.
Page 129 of 148
Example 1. Find the equation of the quadratic function determined from the graph
below.
Solution:
Step 1. Identify the vertex of the graph and a point that passes through the graph.
(You may use the x or y intercepts).
The vertex of the graph of quadratic function is (2, -3).
The graph passes through the point (5,0).
Step 2. Find the value of a.
By replacing x and y with 5 and 0,
and h and k with 2 and -3, respectively,
𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘
0 = 𝑎(5 − 2)2 + (−3)
0 = 𝑎(3)2 − 3
3 = 9𝑎
3
1
𝑎=9=3
Step 3. Write the equation in vertex form and in the form y = ax2 + bx +c.
1
3
1
(𝑥 2 − 4𝑥 + 4) − 3
3
1
1
1
(𝑥 2 ) − 3 (4𝑥) + 3 (4) −
3
1 2
4
5
𝑥 − 3𝑥 − 3
3
𝑦 = (𝑥 − 2)2 − 3
𝑦=
𝑦=
𝑦=
1
(vertex form)
3
4
(since 3 − 3 =
1
4−9
3
4
5
= −3)
5
Thus, the quadratic equation is 𝑦 = 3 (𝑥 − 2)2 − 3 or 𝑦 = 3 𝑥 2 − 3 𝑥 − 3.
Your turn!
Determine the equation of the quadratic function whose
graph is given. The vertex (1, -4) and a point on the graph
(3, 0).
You should get 𝑦 = 𝑥 2 − 2𝑥 − 3.
If you get the correct answer, you may proceed with the next
example. If not, I am sorry but you have to go back to
example number 1 and try all over again
Page 130 of 148
Example 2. Determine the equation of the quadratic function whose graph is given
below.
Solution.
Step 1. Identify the vertex of the graph and a point that passes through the graph
(you may use the x or y intercepts).
The vertex of the graph of quadratic function is (1, -3).
The graph passes through the point (0,-4).
Step 2. Find the value of a.
By replacing x and y with 0 and -4
and h and k with 1 and -3, respectively,
𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘
−4 = 𝑎(0 − 1)2 + (−3)
−4 = 𝑎(−1)2 − 3
3 − 4 = 1𝑎
𝑎=
−1
1
→
𝑎 = −1
2
Step 3. Write the equation in vertex form and in the form y = ax + bx +c.
𝑦 = −1(𝑥 − 1)2 − 3or𝑦 = −(𝑥 − 1)2 − 3 (vertex form)
𝑦 = −(𝑥 2 − 2𝑥 + 1) − 3
𝑦 = −𝑥 2 + 2𝑥 − 1 − 3
𝑦 = −𝑥 2 + 2𝑥 − 4
Thus, the quadratic equation is𝑦 = −(𝑥 − 1)2 − 3or 𝑦 = −𝑥 2 + 2𝑥 − 4.
Try this!
Find the equation of the quadratic function determined from the
graph below.
The vertex (-1, 9) and a point on the graph (0, 8).
You should get 𝑦 = −𝑥 2 − 2𝑥 + 8.
If you get the correct answer, you may proceed with the next
discussion. If not, I am sorry but you have to go back to example
number 2 and try all over again.
E. Continuation of the discussion of new concepts
The application of quadratic function can be seen in many different fields like physics,
industry, business, and in variety of mathematical problems. In this section, you will
explore situations that can be modeled by quadratic functions.
Page 131 of 148
Consider the problem below.
If the perimeter of the rectangle is 100 m, find its dimensions of the rectangle if its
area is maximum.
Complete the table below for the possible dimensions of the rectangle and their
corresponding areas. The first column has been completed for you.
Width (W)
5
10
15
20
25
30
35
40
45
Length
45
(l)
Area
225
(A)
Answer:
45
Width (W)
5
10
15
20
25
30
35
40
Length (L)
45
40
Area (A)
225
400
35
525
30
25
20
15
10
5
600
625
600
525
400
225
a. What is the largest area that you obtained? A = 625 m²
b. What are the dimensions of a rectangle with the largest area? W = 25 m and L=25 m
c. The perimeter P of a given rectangle is 100 m. Make a mathematical statement for the
perimeter of the rectangle. 2L + 2W = 100
d. Simplify the obtained equation and solve for the length l of the rectangle in terms of its
width w. L = 50 – W
e. Express the area A of a rectangle as a function of width w. A = 50W - W²
f. What kind of equation is the result? The function is a quadratic.
g. Express the function in standard form. What is the vertex? In standard form, the area is
A = -(W – 25)² + 625. The vertex is (25, 625).
h. Graph the data from the tablein a, showing the relationship between the width and the
area.
i. What have you observed about the vertex of the graph in relation to the dimensions and
the largest area? The coordinates of the vertex is related to the width and the largest area.
How did you find the activity? Can you still recall the properties of quadratic function? Did
you use them in solving the given problem?
Page 132 of 148
To understand more on how the concepts of the quadratic function can be applied to
solve geometry problems, study the illustrative example presented below.
Example 1.
What are the dimensions of the largest rectangular field that can be enclosed by
80 meters of fencing wire?
Solution:
Let L and W be the length and width of the rectangle.
The perimeter of the rectangle is given by:
P = 2L + 2W
Since P = 80 m, thus,
2L + 2W = 80
Substitution in the formula for the perimeter of the
rectangle
L + W = 40
Factor out 2.
L = 40 – W
Add –W on both sides of the equation to express
the length as a function of W.
A (w) = W × L
Substituting in the formula for the area A of a
A (w) = W (40 – W)
rectangle gives you,
A (w) = -W² + 40W
A (w) = - (W – 20)² + 400.
By completing the square
The vertex of the graph of the function A(w) is (20, 400). This point indicates a maximum
value of 400 for A(w) that occurs when W = 20. Thus, the maximum area is 400 m² when the
W = 20 m. If the width is 20 m, then L= (40 – 20) m = 20 m also. The field with maximum
area is a square.
Try this problem!
What are the dimensions of the largest rectangular field that can be enclosed
with 60 m of wire?
You should get 15 m by 15 m
If your answer is correct, you may proceed to Example 2. If not, I am sorry but you
have to go back to Example 1 and try all over again.
Example 2.
A garments store sells about 40 t-shirts per week at a price of Php 100 each. For
each Php 10 decrease in price, the sales lady found out that 5 more t-shirts per
week were sold.
a. Write a quadratic function in standard form that models the revenue from t-shirts
sales.
b. What price produces the maximum revenue?
Solution:
You should know that Revenue R(x) = (price per unit) × (number of units produced or sold).
Therefore, Revenue R(x) = (Number of t-shirts sold) ×(Price per t-shirt)
Let 40 + 5x
→ represents the t-shirts sold plus 5 more
100 – 10x → indicates the price per week with 10 decreased in price
Page 133 of 148
For a:
Substituting, we’ll have:
Revenue R(x) = (40 + 5x) (100 – 10x)
R(x) = -50x²+ 100x + 4000
For b:
If we transform the function into the form y = a(x – h)² + k, it will be:
R(x) = -50(x – 1)² + 4050
The vertex is (1, 4050).
Thus, the maximum revenue is Php 4050
The price of the t-shirt to produce a maximum revenue can be determined by
P(x) = 100 – 10x
P(x) = 100 – 10(1)
P(x) = 90
Thus, Php 90.00 is the price of the t-shirt that produces maximum revenue.
You try this!
A company of cellular phones can sell 200 units per month at Php 2000
each. Then they found out that they can sell 50 more cell phone units every
month for each Php 100 decrease in price.
a. Write a quadratic function in standard form that models the revenue from
cell phone sales.
b. What price per cell phone unit gives them the maximum monthly sales?
You should get:
a. R(x) = -5 000x² + 80 000x + 400 000
b. Php 1 200.00
If your answer is correct, you may proceed to Example 3. If not, I am sorry but you
have to go back to Example 2 and try all over again.
Example 3.
From a 96-foot building, an object is thrown straight up into the air then follows a
trajectory. The height S(t) of the ball above the building after t seconds is given by
the function S(t) = 80t – 16t².
1. What maximum height will the object reach?
2. How long will it take the object to reach the maximum height?
3. Find the time at which the object is on the ground?
Solution:
1. The maximum height reached by the object is the ordinate of vertex of the parabola of
the function S(t) = 80t – 16t².
Transform S(t) = 80t – 16t²
using completing the square.
S(t) = 80t – 16t²
S(t) = -16t² + 80t
S(t) = -16(t² - 5t)
S(t) = -16(t² - 5t +
25
)
4
+ 100
5
S(t) = -16(t - 2)2 + 100
5
The vertex is ( 2, 100).
Page 134 of 148
Thus, the maximum height reached by the object is 100 ft. from the top of the building. This
is 196 ft. from the ground.
2. The time for an object to reach the maximum height is the abscissa of the vertex of the
parabola or the value of h.
S(t) = 80t – 16t²
5
2
S(t) = -16(t - )2 + 100
5
Since the value of h is 2 or 2.5, then the object is at its maximum height after 2.5 seconds.
3. To find the time it will take the object to hit the ground,
Let S(t) = -96, since the height of the building is 96 ft.
The problem requires us to solve for t.
h(t) = 80t – 16t²
given
-96 = 80t – 16t²
substitute -96 for h(t)
16t² -80t – 96 = 0
express in standard form
16(t² - 5t – 6) = 0
factor out 16
t² - 5t – 6 = 0
divide both sides of the equation by 16
(t – 6)(t + 1) = 0
use factoring
t – 6 = 0, t + 1 = 0
use zero product property
t = 6, t = -1
Thus, it will take 6 seconds before the object hits the ground.
You try this!
An object is thrown vertically upward with a velocity of 96 m/s. The distance S(t)
above the ground after t seconds is given by the formula S(t) = 96t – 5t².
1. How high will it be at the end of 3 seconds?
2. How much time will it take the object to be 172 m above the ground?
3. How long will it take the object to reach the ground?
You should get:
a. 243 ft. b. 2 seconds
c. 1915 seconds
If your answer is correct, I must say that you understood the lesson well.
Page 135 of 148
DAY 2 (October 12)
F. Developing mastery
Answer Activity 1 below. Read the direction carefully and write your answers in your
notebook.
Activity 1: IT’S YOUR TURN!
Find the equation of quadratic functions given its zeros.
ZEROS
EQUATION
1. -3 and 8
____________________
2. -2 and 4
____________________
3. -6 and 1
____________________
4. -7 and -6
____________________
5. -6 and -1
____________________
You may check your answers by turning to Page 143 for the
key to correction.
Give yourself one (1) point for every correct answer.
Rate the result using the table.
Score
5
4
3
2
1
Description
Very Good
Good
Fair
Turn to page
141 section J
and work on
the enrichment
activities
Answer Activity 2 below. Read the direction carefully and write your answers in your
notebook.
Activity 2:
Determine the equation of the quadratic function represented by the table of values below.
1.
2.
You may check your answers by turning to Page 143 for the key to correction. Give
yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 136 of 148
Answer Activity 3 below. Read the direction carefully and write your answers in your
notebook.
Activity 3:
Find the equation of the quadratic function presented by each of the following graphs.
Graph
Vertex form
y=ax2 + bx + c
1
V(1,1)
P(0,2)
2
V(-2, 8) P(-4,0)
3
V(1, -3)
P(2,0)
You may check your answers by turning to Page 143 for the key to correction. Give
yourself one (1) point for every correct answer.
Score
Description
How many correct answers did you get? Rate the result
6
Very Good
using the table.
5
Good
3-4
Answer Activity 4 below. Read the direction carefully and write
your answers on the space provided
0-2
Fair
Turn to page
142 section J
and work on
the enrichment
activities
Page 137 of 148
Activity 4:Make the Heart Whole Again.
Heal the broken heart by matching the functions to their corresponding maximum area,
maximum height and maximum revenue. Write your answers in your notebook.
You may check your answers by turning to Page 144 for the
key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get? Rate the result
using the table.
G. Finding practical application of the concepts and
skill in daily living
Score
3
2
0-1
Description
Very Good
Good
Turn to page
142 section J
and work on
the enrichment
activities
Many real-world situations deal with quadratics and parabolas:
Throwing a ball, shooting cannon, diving from a platform and hitting a golf ball are
all examples of situations that can be modeled by quadratic functions.
Sometimes calculating a business profit requires using a quadratic function. If you want
to sell something – even something as simple as lemonade – you need to decide how
many items to produce so that you will make a profit.
People frequently need to calculate the area of rooms, boxes or plots of land. An
example might involve building a rectangular box where one side must be twice the
length of the other side. For example, if you have only 4 square feet of wood to use for
Page 138 of 148
the bottom of the box, with this information, you can create an equation for the area of
the box using the ratio of the two sides.
H. Making generalization and abstraction about the lesson
To find the equation of the quadratic function using graph,
a. Identify the vertex of the graph and a point that passes through the graph.
b. Find the value of a, and
c. Write the equation in vertex form and general/standard form.
Deriving quadratic functions from:
• Zeros
To find the equation of the quadratic function if the zeros (𝑟1 𝑎𝑛𝑑 𝑟2 ) are given,
multiply the two binomials such that:
𝑓(𝑥) = (𝑥 − 𝑟1 )(𝑥 − 𝑟2 )
• Table of Values
To find the equation of the quadratic function if the table of values is given:
a. take any three ordered pairs (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ), and (𝑥3 , 𝑦3 ), from the table;
b. use these values of 𝑥 𝑎𝑛𝑑 𝑦 in the equation 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 to write three
equations
𝑦1 = 𝑎(𝑥1 )2 + 𝑏(𝑥1 ) + 𝑐
𝑦2 = 𝑎(𝑥2 )2 + 𝑏(𝑥2 ) + 𝑐
𝑦3 = 𝑎(𝑥3 )2 + 𝑏(𝑥3 ) + 𝑐
c. using the three derived equations, solve for 𝑎, 𝑏 𝑎𝑛𝑑 𝑐; and
d. substitute the values obtained for 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 in 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
•
Graph
When the vertex (ℎ, 𝑘) and any point (𝑥, 𝑦) on the parabola are clearly seen, the
equation of the quadratic function can be determined by using the form of a
quadratic function 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘.
To solve problems involving quadratic functions:
✓ Read and understand the problem.
✓ Identify the given and use a representation for the unknown.
✓ Write the equation for the problem.
✓ Solve the equation.
✓ Graph, if necessary.
Page 139 of 148
I.
Evaluating learning
Evaluation 8
Direction: Use yellow papers to answer the evaluation. Provide necessary solutions for
your answers. Use the format shown below to label your paper and it is to be
submitted to your Math teacher.
Name:_______________________
Section:_______________________
Subject: Mathematics 9 Week Number: 8 Parent’s Signature: __________________
I.
Find the equation of quadratic functions given its zeros.
ZEROS
II.
EQUATION
1. -1 and 5
____________________
2. 3 and 8
____________________
3. -7 and 5
____________________
4. 3 and -6
____________________
5. -5 and -2
____________________
Determine the equation of the quadratic function represented by the table of values
below.
6.
7.
Page 140 of 148
Continuation EVALUATION 8
III.
Match each graph with its correct equation. Write the letter of your answer.
Column A
1. v( -2, -6) p ( 0, -2)
Column B
A. 𝑦 = 𝑥 2 − 4𝑥 + 7
2. v ( 2, 3 ) p ( 0, 7 )
B. 𝑦 = −2𝑥 2 − 4𝑥 + 1
3. v ( 2, -2 ) p ( 4, 0 )
C. 𝑦 = 𝑥 2 + 4𝑥 − 2
4. v ( 3, 3 ) p ( 0, -6 )
1
D. 𝑦 = 2 𝑥 2 − 2𝑥
5. v ( -1, 3 ) p ( 0, 1)
E. 𝑦 = −𝑥 2 + 6𝑥 − 6
Page 141 of 148
Continuation EVALUATION 8
IV.
V.
Analyze the graph below representing the relationship between the time in seconds
when a projectile is launched vertically into the air, and its height at that time.
1.
What is the maximum height reached by the
projectile?
2.
How long does it take for the projectile to reach its
maximum height?
3.
How long does it take for the projectile to return to the
ground?
4.
What is the axis of symmetry of the graph?
Solve the following problems
1. Find the maximum rectangular area that can be enclosed by a fence that is 362
meters long.
2. The height in meters of an object projected upward after t seconds given by ℎ(𝑡) =
192𝑡 − 16𝑡 2 .
a. After how many seconds does the object reach its maximum height?
b. After how many seconds will the object agin touch the ground?
J. Additional activities for application of remediation
Write your answers in your notebook.
REMEDIATION 1: Find the equation of quadratic functions given its zeros.
ZEROS
1. 9 and 12
2. -7 and -8
3. -14 and -12
EQUATION
____________________
____________________
____________________
You may check your answers by turning to Page 144 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
REMEDIATION 2:
Determine the equation of the quadratic function represented by the table of values below.
You may check your answers by turning to Page 144 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
Page 142 of 148
REMEDIATION 3: Determine the equation of the quadratic function whose graph is given
below.
1. V(-1, -2) P(0,-1)
2. V(-2, 1) P(0,13)
You may check your answers by turning to Page 144 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
REMEDATION 3:Arrange the jigsaw puzzle to form the image and see the application of
quadratic function in real life. Cut out the puzzle piece and form the image.
Page 143 of 148
You may check your answers by turning to Page 144 for the key to correction.
Give yourself one (1) point for every correct answer.
How many correct answers did you get?
KEY TO CORRECTIONS
REVIEW: f(x) = -2x2 + 5x + 3
1. The parabola opens down.
2.
3.
4.
5
x= 4 or 1.25
5 49
(4 , 8 )or (1.25,6.125)
1
x -intercepts: (− 2,0) and
Graph:
(3,0)
y - intercepts: (0,3)
5. (See the graph on the side)
6. Domain: all real numbers
49
Range: (-∞, 8 ) 𝑜𝑟 𝑦 ≤
49
8
KEY TO CORRECTIONS (DEVELOPING MASTERY)
Activity 1:IT’S YOUR TURN!
Find the equation of quadratic functions given its zeros.
ZEROS
EQUATION
1. -3 and 8
𝑦 = 𝑥 2 − 5𝑥 − 24
2. -2 and 4
𝑦 = 𝑥 2 − 2𝑥 − 8
3. -6 and 1
𝑦 = 𝑥 2 + 5𝑥 − 6
𝑦 = 𝑥 2 + 13𝑥 + 42
4. -7 and -6
5. -6 and -1
𝑦 = 𝑥 2 + 7𝑥 + 6
Activity 2:
Determine the equation of the quadratic function represented by the table of values.
1. y = 2x2 – 8x – 4
2. y = 3x2 – 6x – 1
Activity 3:
Find the equation of the quadratic function presented by each of the following graphs.
Graph
Vertex form
y = ax2 + bx + c
1
V(1,1)
P(0,2)
2
V(-2, 8) P(-4,0)
3
V(1, -3)
P(2,0)
𝑦 = (𝑥 − 1)2 + 1
𝑦 = 𝑥 2 − 2𝑥 + 2
𝑦 = −2(𝑥 + 2)2 + 8
𝑦 = −2𝑥 2 − 8𝑥
𝑦 = 3(𝑥 − 1)2 − 3
𝑦 = 3𝑥 2 − 6𝑥
Page 144 of 148
Activity 4: Make the Heart Whole Again
KEY TO CORRECTION (REMEDIATION)
REMEDIATION 1: Find the equation of quadratic functions given its zeros.
ZEROS
EQUATION
1. 9 and 12
𝑦 = 𝑥 2 − 21 + 108
2. -7 and -8
𝑦 = 𝑥 2 + 15𝑥 + 56
3. -14 and -12
𝑦 = 𝑥 2 + 2𝑥 − 168
REMEDIATION 2:
𝒚 = 𝒙𝟐 − 𝟒
REMEDIATION 3:
1. 𝑦 = 𝑥 2 + 2𝑥 − 1
2. 𝑦 = 3𝑥 2 + 12𝑥 + 13
3.
REMEDIATION 4:
If you need help with this lesson, you may CONTACT ME AT # 09434634247.
Page 145 of 148
Department of Education
Region III
Division of City of San Fernando
PAMPANGA HIGH SCHOOL
PHS Blvd., Brgy. Lourdes, City of San Fernando, Pampanga
Tel. No. (045) 961-4261 website: www.pampangahigh.school
Parent’s Signature: ___________
Date: ______________________
Name: ______________________________
Grade and Section: ____________________
SUMMATIVE TEST IN GRADE 9 MATH - SSC
FIRST QUARTER
SY 2020-2021
A. Match each item in column A with items in column B.
Column A
Column B
𝑏
𝑎
___
1. axis of symmetry
a. −
___
2. vertex
b. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
___
3. discriminant
c.
___
4. x- intercept
d. 𝑏 2 − 4𝑎𝑐
___
5. vertex form of quadratic function
e. (ℎ, 𝑘)
___
6. standard form of quadratic equation
f.
___
7. quadratic formula
g. √−1
___
8. sum and product of the roots
of 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
If 𝑎𝑏 = 0 𝑡ℎ𝑒𝑛 𝑎 = 0 𝑜𝑟 𝑏 = 0
𝑥=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
h. 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘
___
9. zero product property
i. 𝑥 = ℎ
___
10. not real roots/imaginary number i
j.
𝑥 𝑖𝑛 (𝑥, 0)
B. Choose the letter of the best answer and write it on the space provided before the
number.
___11. What are the roots of 2𝑥 2 + 𝑥 − 21 = 0?
7
A. − 2, 𝑎𝑛𝑑 3
B.
7
2,
𝑎𝑛𝑑 − 3
C.
7
2,
D. −
𝑎𝑛𝑑 3
7
2,
𝑎𝑛𝑑 − 3
___12. Which is the vertex of the parabola 𝑦 = 3𝑥 2 − 12𝑥 + 14?
A. (0, 0)
C. (2, 2)
B. (1, 0)
D. (1, −1)
___13. What is the vertex form of the quadratic function 𝑓(𝑥) = −𝑥 2 + 2𝑥 + 2?
A. 𝑦 = (𝑥 − 1)2 + 3
C. 𝑦 = (𝑥 − 1)2 − 3
B. B. 𝑦 = −(𝑥 − 1)2 + 3
D. 𝑦 = (𝑥 − 1)2 − 3
2
___14. What is the axis of symmetry of 𝑦 = 3𝑥 − 6𝑥 + 12?
A. 𝑥 = 0
C. 𝑥 = 4
B. 𝑥 = 1
D. 𝑥 = 9
___15. Which quadratic equation has −3 ± 2√3 as roots?
A. 𝑥 2 + 3𝑥 − 6 = 0
C. 𝑥 2 + 9𝑥 − 6 = 0
B. 𝑥 2 + 9𝑥 + 6 = 0
D. 𝑥 2 + 6𝑥 − 3 = 0
2
___16. What value of k will make 2 a root of𝑥 + 𝑘𝑥 + 8 = 0?
A. -6
C. 6
B. 0
D. 12
Page 146 of 148
___17. What is the sum of the roots of 𝑥 2 + 2𝑥 − 6 = 0?
A. -6
C. 0
B. -2
D. 2
___18. What is the product of the root of 2𝑥 2 − 𝑥 + 12 = 0?
A. 0
C. 4
B. 2
D. 6
2
___19. What are the roots of 𝑥 − 4 = 0?
A. ±2
C. ±2√2
B. ±√2
D. 2 ± √2
___ 20. Which quadratic function has a graph that opens downward?
A. 𝑓(𝑥) = 𝑥 2
C. 𝑓(𝑥) = −(𝑥 + 1)2 + 4
1
B. 𝑓(𝑥) = 𝑥 2 − 10𝑥 − 25
D. 𝑓(𝑥) = 4 (𝑥 + 2)2 − 9
___21. Which of the following has the widest graph?
1
A. 𝑓(𝑥) = 2 𝑥 2
C.
𝑓(𝑥) = 2𝑥 2
B. 𝑓(𝑥) = 𝑥 2
D. 𝑓(𝑥) = 4𝑥 2
___22. Which table of values represents a quadratic function?
A. x
____
-2
-1
0
1
2
3
C. x
y
1
2
3
4
5
6
y
B. x
y
-2
-2
-1
-1
0
0
1
1
2
2
3
3
D. x
y
-3
-2
-1
0
1
2
9
4
1
1
1
4
-2
-1
0
1
2
3
2
-1
-2
-1
2
7
23. What is the resulting function if 𝑓(𝑥) = 𝑥 2 is translated 5 units to the right?
A. 𝑓(𝑥) = 𝑥 2 + 5
C. 𝑓(𝑥) = 5𝑥 2
C. 𝑓(𝑥) = 𝑥 2 − 5
D. 𝑓(𝑥) = (𝑥 − 5)2
C. For numbers 24-29, consider the graph to the right.
y
Determine the following:
7
24. vertex of the graph: ________________________
6
25. axis of symmetry of the graph: _______________
5
4
26. number of zeros: __________________________
3
27. equation of the function: ___________________
2
28. domain: __________________________________
1
29. range: ___________________________________
−5 −4 −3 −2 −1
O
1
2
3
x
−1
−2
D. Solve each completely. (2 points each)
30-31).To entertain PBA patrons during halftime, T-shirts are hurled into the crowds using a
slingshot. The function ℎ(𝑡) = −16𝑡 2 + 80𝑡 + 4 , gives the height h in feet of the shirt after t
secons. How long will it take for the T-shirt to reach its maximum height? What is its
maximum height?
32-33)
3𝑥 2 − 𝑥 − 24 = 0
34-35) 2𝑥 2 − 5𝑥 − 18 ≥ 0
Page 147 of 148
E. Graph 𝒇(𝒙) = −𝒙𝟐 + 𝟐𝒙 + 𝟐.
36. Determine the opening __________________;
37. Find the coordinates of the vertex _____________;
38. Give the equation of the axis of symmetry ______________;
39. Set up a table of values; and
x
y
40. Plot the points and connect them with a smooth curve.
Page 148 of 148