LESSON 5:
_______
Date:
Composition of Functions
I. OBJECTIVES:
Students will:
1. know the process of simplifying the composition of functions
2. be able to explain the methods of simplifying the composition of functions
II. PRELIMINARIES:
Perform the indicated operations of the following polynomials.
1. (3x – 5 + 6x2 + x3 – x4) + (5 – 5x + 5x3 + 3x2 + 3x4)
2. ( 3x5 – 4x4 + 6x3 – 8x +4) – (5x2 – 2x5 – 4x4 – 7)
3. (2x2 + 4y) (2x2 + 6y)
4. 16m3n6 + 24m5n6 – 36m6n3
– 4m4n3
3
5. (x + 1) ÷ (x – 1)
How did you simplify the polynomial applying the operations of algebraic
expressions?
ARALINKS Forum for APK
III. LESSON Use DEVELOPMENT: Video Teaching on the discussion of the concepts
Concepts and key ideas
Functions or functional notation f(x) means the value of the function f at the number x.
If y is the value of f at x or y = f(x) , then x is called the independent variable since any element
of the domain can replace it and y is called the dependent variable because its value depends on
the value of independent variable x. All the values of x are called the domain and all the values
of y are called the range. The letters in the alphabet f, g, and h are the most commonly used
notations for functions. There are operations applied to obtain new functions these are addition,
subtraction, multiplication, division and composite functions. To find the sum of the functions
f and g use the equation, (f + g)(x) = f(x) + g(x). To find the difference of the functions f and g
use the equation, (f – g)(x) = f(x) – g(x). To find the product of the functions f and g use the
equation, (f • g)(x) = f(x) • g(x). To find the quotient of the functions f and g use the equation, (f ÷
g)(x) = f(x) ÷ g(x). For the composite functions f with g are functions as defined as (f ο g)(x) =
f[g(x)]. The domain of f ο g consists of all real numbers ‘x’ in the domain of g which g(x) is the
domain of f. The symbol (f ο g)(x) is read as ‘the function of f of g(x).
Example 1:
Given: f(x) = x2 – 1 , g(x) = x2 + 2x + 1 , h(x) = x + 1
Perform the operations of functions from the given functions.
1. (f + g)(x)
2. (f + g + h)(x)
3. (f – g)(x)
4. ( g– h)(x) 5 . (f • g)(x)
6. (f • h)(x)
7. (f ÷ g)(x)
8. (g ÷ h)(x)
9. (g ο f)(x) 10. (f ο h)(x)
Solution: 1. Use the equation in finding the sum of functions and substitute the functions and
then simplify the equation.
(f + g)(x) = f(x) + g(x) = (x2 – 1) + (x2 + 2x + 1) = 2 x2 + 2x
Answer: The sum of the functions is 2 x2 + 2x.
2. (f + g + h)(x) = f(x) + g(x) + h(x) = (x2 – 1) +(x2 + 2x + 1) +(x + 1) = 2 x2 + 3x + 1
Answer: The sum of the functions is 2 x2 + 3x + 1.
3. To solve for the difference use the equation of the difference of functions,
substitute the function and then simplify.
(f – g)(x) = f(x) – g(x) = (x2 – 1) – (x2 + 2x + 1) = x2 – 1 – x2 – 2x – 1 = – 2x – 2
Answer: The difference of the functions is – 2x – 2.
4. ( g – h)(x) = g(x) – h(x) = (x2 + 2x + 1) – (x + 1) = x2 + 2x + 1 – x – 1 = x2 + x
Answer: The difference of the functions is x2 + x.
5. (f • g)(x) = f(x) • g(x) = (x2 – 1) • (x2 + 2x + 1) =
To find the product use either distributive property or vertical method
Distributive property:
(x2 – 1) • (x2 + 2x + 1) = x4 + 2x3 + x2 – x2 – 2x – 1 = x4 + 2x3 – 2x – 1
Vertical Method:
x2 + 2x + 1
• x2 – 1
x4 + 2x3 + x2
– x2 – 2x – 1
+
4
3
x + 2x 0 – 2x – 1
Answer: The product of the functions is x4 + 2x3 – 2x – 1.
6. (f • h)(x) = f(x) • h(x) = (x2 – 1) • (x + 1) = x3 + x2 – x – 1
Answer: The product of the functions is x3 + x2 – x – 1
7. (f ÷ g)(x) = f(x) ÷ g(x) = f(x) = (x2 – 1)
= (x + 1) (x – 1) = (x – 1)
g(x) (x2 + 2x + 1) (x + 1) (x + 1) (x + 1)
Answer: The quotient of the functions is (x – 1)
(x + 1)
8. (g ÷ h)(x) = g(x) ÷ hg(x) = g(x) = (x2 + 2x + 1)
h(x)
(x + 1)
Answer: The quotient of the functions is (x + 1)
= (x + 1) (x + 1) = (x + 1)
(x + 1)
9. (g ο f)(x) = g[f(x)] for function f(x) = x2 – 1 , g(x) = x2 + 2x + 1
= (x2 – 1)2 + 2(x2 – 1) + 1 : Substitute the function f(x) = x2 – 1 to the variable
x in the function g(x) = x2 + 2x + 1
= (x4 – 2x2 + 1) + (2x2 – 2) + 1 Simplify the function
= x4 – 2x2 + 1 + 2x2 – 2 + 1
= x4
Answer: The composite function is x4.
10. (f ο h)(x) = f[h(x)] for function h(x) = x + 1 , f(x) = (x2 – 1)
= (x + 1)2 – 1
: Substitute the function h(x) = x + 1 to the variable
x in the function f(x) = (x2 – 1)
2
= (x + 2x + 1) – 1
Simplify the function
= x2 + 2x + 1 – 1
= x2 + 2x
Answer: The composite function is x2 + 2x
IV. INTEGRATION:
1. IGNACIAN CORE VALUE: Faith
RELATED VALUE: Strong Faith in God
How Jesus used his power to help all people?
2. SOCIAL ORIENTATION: Leaders: Responsibility
If you will be given a power to be the leader in your community, what will be your goals?
Do you agree with the saying “More power, more responsibility” .
V. EVALUATION/ ASSESSMENT:
Exercise 1:
For the given functions f, g, and h, solve the following functions and show the solution.
1. (f + g)(x)
2. (f + g + h)(x)
3. (f – g)(x)
4. ( g– h)(x) 5 . (f • g)(x)
6. (f • h)(x)
7. (f ÷ g)(x)
8. (g ÷ h)(x)
9. (g ο f)(x) 10. (f ο h)(x)
Given:
a. f(x) = x + 5 , g(x) = x2 + 13x + 40 ,
b. a(x) =2x – 3 , b(x) = 2x2 + 5x – 12 ,
Solution: a.
1. (f + g)(x)
h(x) = x + 8
c(x) = x + 4
6. (f + g + h)(x)
2. (f – g)(x)
7. ( g– h)(x)
3. (f • g)(x)
8. (f • h)(x)
4. (f ÷ g)(x)
9. (g ÷ h)(x)
5. (g ο f)(x)
10. (f ο h)(x)
Solution: b.
1. (a + b)(x)
6. ( b – c)(x)
2. (a + b + c)(x)
7. (a • b)(x)
3. (a – b)(x)
8. (b ÷ c)(x)
4. (a • c)(x)
6. (a ÷ c)(x)
5. (b ο a)(x)
10. (a ο c)(x)
.
Exercise 2:
Find the value of each composite function, if
f(x) = 3x + 5 , g(x) = x2 – 1 ,
h(x) = (x + 1) – 1
1. (g ο f)(1)
5. ( f ο g)(2)
2. (g ο h)( – 1 )
6. (h ο f)( – 2 )
3. (f ο f)(2)
7. (g ο g)(3)
4 . ( h ο h)(2)
8. (f ο h)(x)
LESSON 6:
_______
Polynomial Equations (Non -Linear, Non –Quadratic)
Date:
I. OBJECTIVES:
Students will:
3. know the components of polynomial functions which are not linear and quadratic
4. be able to explain the different components of polynomial functions which are not
linear and quadratic
II. PRELIMINARIES:
Identify the degree (n) of the given polynomial. Write the answer on the space provided.
________ 1. x2 – 1
________ 6. x3 + x2 – x + 6
4
________ 2. x – 1
________ 7. x4 + x3 – x2 + 4x – 2
________ 3. 46
________ 8. 2 x5 + 3x2 – 9x + 6
________ 4. x + 8
________ 9. 4x3 + 2 x5 – x + 8
2
________ 5.
x – 2x + 11
________ 10. 3x3 + 3x4 – x2 + 3
How did you identify the value of n or the degree of the function?
III. LESSON DEVELOPMENT: Google Meet with the students
Concepts and key ideas
Polynomial is an algebraic expression where the exponent of the literal coefficients
must be a positive whole numbers. A polynomial function is a function P in n that can be
written as
P(x) = anxn + an–1 xn–1 + an–2 xn–2 + . . . + a2x2 + a1x1 + a0 where an, an–1, an – 2 . . . a2 , a1 , a0 ,
are
real numbers, the value of the leading numerical coefficient an ≠ 0, and n is a positive integer.
Where n is the degree of the polynomial. In linear function n = 1, for quadratic function n = 2,
for cubic function n = 3, for quartic function n = 4, and for quintic/quantic function n = 5. It can
be written in descending order of n. For the graph of the polynomial functions the number of
turning point is equal to n – 1. The x-intercepts is the point where the graph intersect with the x
– axis and y-intercept is the point where the graph intersect with the y-axis.
Example 1.
Identify the polynomial function from the given values of a4 = 4 , a3 = 2, a1 = 3, a2 = –4, a0 =
8.
Solution: Since the highest degree is the value of n = 4, then the polynomial is a quartic
polynomial.
4x4 + 2x3 – 4x2 + 3x + 8 : the value of a is the numerical coefficient of the term and
the subscript of a is the exponent of the literal coefficient
x.
Note: In a0 = 8, the exponent of x is zero, x0 = 1, that’s why there is no x in last term
and 8 is called a constant.
Answer: The polynomial function is P(x) = 4x4 + 2x3 – 4x2 + 3x + 8 .
Example 2:
Identify the polynomial function from the given values of a1 = 4 , a5 = 2, a3 = –3, a0 = –6.
Solution: Since the highest degree is the value of n = 5, then the polynomial is quantic
polynomial.
2x5 + 0x4 – 3x3 + 0x2 + 4x – 6
The value of a is the numerical coefficient of the term and the subscript of a is the
exponent of the literal coefficient x. From the given values of an there are no values
for
a4 and a2, represent it 0x4 and 0x4 to complete the function.
Note: In a0 = – 6 , the exponent of x is zero, x0 = 1 then (– 6)( x0 ) = (– 6)( 1 ) = – 6 that’s why
there is no x in last term and – 6 is called constant.
Answer: The polynomial function is P(x) = 2x5 – 3x3 + 4x – 6 .
IV. INTEGRATION:
1. IGNACIAN CORE VALUE: Excellence
RELATED VALUE: Integrity
What are the works of Jesus related to His title “The Best Teacher”
2. SOCIAL ORIENTATION: Government Leaders
Why is it that some government officials do not have good reputation? If you were the
President what
will you do with that officials?
V. EVALUATION/ ASSESSMENT:
Exercise 1:
Find the polynomial functions from the given value of each numerical coefficient.
1. a0 = 4, a1 = 5, a2 = 1, a3 = 2 ;
n=
: P(x) =__________________________
2. a0 = 2, a1 = – 5, a2 = 3, a3 = 4, a4 = 4 ; n =
: P(x)
=_________________________
3. a0 = –6 , a2 = – 8, a4 = –3 , a5 = 4 ;
n=
: P(x)
=_________________________
4. a1 = –8 , a3 = 3, a4 = –8 , a5 = 1 ; n =
: P(x) =__________________________
5. a0 = 2, a2 = 1, a3 = –8, a4 = 1 ; n =
: P(x) =__________________________
Exercise 2:
Complete the table of polynomial function.
an
n
P(x)
a0 =3, a5=1, a3= –8, a4=6, a1= 6
P(x) = x4 + 5x3 – 3x2 + 3x + 2
a0 = 3, a1= 4, a3= 5, a4= 6
P(x) = 3x3 – x2 + 6x – 9
Types of Polynomial
Function
a0 =5, a5=6, a3= –7, a4=1, a1=3, a2= 9
LESSON 7:
Date: _______
Zeros of Polynomial Functions
I. OBJECTIVES:
Students will:
1. know the relevance of continuous synthetic division in finding the zeroes of
polynomial
functions
2. be able to perform continuous synthetic division in finding the zeroes of polynomial
functions
II. PRELIMINARIES:
Divide the given polynomial function using synthetic division. Show the complete solution.
1. (x3 + x2 – 7x – 3) ÷ (x + 3)
4. (5x3 + 4x2 – 31x + 6) ÷ (x –2)
2. (2x3 – 5x2 + 4x – 1) ÷ (x – 1)
5. (5x4 – 405) ÷ (x –3)
3. (2x7 –3x5 –6x4 +2x3 –x +6) ÷ (x – 1)
6. (2x6 – 4x5 +6x –12) ÷ (x –2)
Identify the steps in finding the quotient of polynomials using synthetic division.
Use ARALINKS Forum for the APK
III. LESSON DEVELOPMENT: Google Meet with the students
Concepts and Key Ideas:
A polynomial function has a degree of n ≥ 1 and has exactly n roots. Linear function has
only one root where n = 1, quadratic function has two roots where n = 2 and 1 turning point,
cubic function has three roots where n = 3 and its turning point is 2, quartic function has four
roots where n = 4 with 3 turning points, and quantic function has five roots where n = 5 and
there were 4 turning points. The degree (n) of the polynomial function determines the number
of roots and the zeroes of the functions. In the rational zeroes/roots theorem states that if the
𝐵
rational number 𝐶 in the lowest terms is a root of the polynomial function P(x) = anxn + an–1 xn–1
+ an–2 xn–2 + . . . + a2x2 + a1x1 + a0 where an, an–1, an – 2 . . . a2 , a1 , a0 , whose coefficients are
integers, then B is a factor of a0 and C is a factor of an. To find the roots or zeroes of the
function use the method of synthetic division or remainder theorem. (Note if the remainder is
zero then the divisor and quotient are factors.)
Example 1:
Determine the zeroes or roots of P(x) = x3 + 6x2 + 11x + 6.
Solution: Since the leading coefficient is 1 and the constant is 6 with factors of 1, 2 and 3 use
the
value of x – c = (x + 3), (x + 2), (x + 1) then use the method of synthetic division
or remainder theorem.
Using synthetic division:
In x – c = x + 3 , Then x = – 3 = c
–3
1 6 11 6
– 3 – 9 –6
1 3 2 0 remainder = 0 , Then – 3 is a root
In x – c = x + 2 , Then x = – 2 = c
–2
1 6 11 6
– 2 – 8 –6
1 4 3 0 remainder = 0 , Then – 2 is a root
In x – c = x + 1 , Then x = – 1 = c
–1
1 6 11 6
– 1 – 5 –6
1 5 6 0 remainder = 0 , Then – 1 is a root
Using Remainder Theorem:
In x – c = x + 3 , Then x = – 3 = c
P( – 3) = x3 + 6x2 + 11x + 6
P( – 3) = ( – 3)3 + 6( – 3)2 + 11( – 3) + 6
P( – 3) = – 27 + 54 – 33 + 6
P( – 3) = 0 ; remainder = 0 , Then – 3 is a root
In x – c = x + 2 , Then x = – 2 = c
P( – 2) = x3 + 6x2 + 11x + 6
P( – 2) = ( – 2)3 + 6( – 2)2 + 11( – 2) + 6
P( – 2) = – 8 + 24 –22 + 6
P( – 2) = 0 ; remainder = 0 , Then – 2 is a root
In x – c = x + 1 , Then x = – 1 = c
P( – 3) = x3 + 6x2 + 11x + 6
P( – 3) = ( – 1)3 + 6( – 1)2 + 11( – 1) + 6
P( – 3) = – 1 + 6 – 11 + 6
P( – 3) = 0 ; remainder = 0 , Then – 1 is a root
Using Factor Theorem:
P(x) = x3 + 6x2 + 11x + 6
P(x) = (x + 3)(x + 2)(x + 1)
In x + 3 = 0, then x = – 3 ; root = – 3
In x + 2 = 0, then x = – 2 ; root = – 2
In x + 1 = 0, then x = – 1 ; root = – 1
Answer: The roots of x3 + 6x2 + 11x + 6 are – 3, – 2, and – 1.
IV. INTEGRATION:
1. IGNACIAN CORE VALUE: Excellence
RELATED VALUE: Resourcefulness
Can you reach your goals without working it?
2. SOCIAL ORIENTATION: Environmental Issues
How can we take care of our environment with the changes that happens in our life?
V. EVALUATION/ ASSESSMENT:
Exercise 1:
Determine the polynomial function from the given zeroes or roots. Complete the table.
Zeroes of the
Function
1.
2, 4, 1
Factors
Polynomial Function
2.
–3,–5,–2
3. – 2, 2, – 3, – 1
4. – 1, 2, – 3, 3, 0
5.
2, 1, – 2 ,
2
3
Exercise 2:
Solve the zeroes of the given polynomial functions using any method. Show the complete
solution.
1. P(x) = x3 – 6x2 + 11x – 6
5. P(x) = x3 – 8
2. P(x) = x4 – 5x2 + 4
6. P(x) = x4 – 3x3 – 4x2 + 12x
3. P(x) = x3 + 4x2 + x + 4
7. P(x) = 2x4 + 9x3 + 11x2 – 4
4. P(x) = x5 + 15x3 – 8x2 + 24x
8. P(x) = x5 – 3x4 – 3x3 + x2 – 5
Use ARALINKS for the formative assessment.
LESSON 8:
Date: _______
Graphs of Polynomial Functions
I. OBJECTIVES:
Students will:
1. know the techniques in the computations of lower bound (GILB) and upper bound
(LIUB) of zeroes in the graph of polynomial functions
2. be able to compute the lower bound (GILB) and upper bound (LIUB) of the zeroes in
the
graph of polynomial function
II. PRELIMINARIES:
Graph the following functions and be able to make the table of values.
1.P(x) = x + 4
Table of values
X -3 -2 -1
Y
Graph:
0 1 2 3
2. P(x) = x2 + 4x + 4
Table of values
X -3 -2 -1 0
Y
Graph:
1
2
3
III. LESSON DEVELOPMENT: Google Meet with the students
Concepts and Key Ideas:
The graphs of polynomial function can be easily determined according to its x and y –
intercepts. The number of x intercepts determined the number of roots and the value of ‘n’. The
number of turning points depends on the degree on the polynomial n less than one or n – 1. The
numbers that is associated to the limits of bounds which are called the upper bond and the lower
bond. The upper bond of the zeroes when c in x – c is positive and the resulting numerical
coefficients of the quotient and the remainder is also positive, then the value of ‘c’ is the value
of the upper bound. While when c in x – c is negative and the resulting numerical coefficients
in the quotient and the remainder form an alternating signs, then the value of ‘c’ is the lower
bound of the zeroes. To solve for the values of x and y in the table of values using continuous
synthetic division, the value of ‘c’ will be ‘x’ and the remainder will be the value of ‘y’. The
LIUB or the least integral upper bound which means that there is no zero/root of the polynomial
function greater than the highest zero/root. The GILB is the greatest integral lower bound which
means that there is no zero/root of the function lesser than the smallest zero.
Example 1:
Construct the graph of the polynomial function x3 – 4x2 – 4x + 16 and make the table of
values, then determine the upper and lower boundary (GILB,LIUB),the number of
turning points, the zeroes of the function and y intercepts.
Solution: Compute the values of x and y for the table of values using synthetic division.
– 3 1 – 4 – 4 16
– 2 1 – 4 – 4 16
– 3 21– 51
– 2 12 –16
1 – 7 17 – 35 : GILB
1 – 6 8 0 root is – 2
–1
1 – 4 – 4 16
–4 8 –4
1 – 8 4 12
1
1 – 4 – 4 16
1 –3 –7
1 –3 –7 9
2
1 – 4 – 4 16
2 – 4 – 16
1 – 2 – 8 0 root is 2
3
1 – 4 – 4 16
3 – 3 – 21
1–1 –7 –5
4
1 – 4 – 4 16
4 0 –16
1 0 – 4 0 root is 4
5
Table of Values:
X –3
Y –35
–2
0
–1
12
0
16
1
9
2
0
1 – 4 – 4 16
5 5
5
1 1 1 21 : LIUB
3
–5
4
0
5
21
30
20
10
0
-10
-3
-2
-1
0
1
2
3
4
5
-20
-30
Graph:
-40
Answer: The zeroes of the polynomial function x3 – 4x2 – 4x + 16 are – 2, 2 and 4. Since the
degree n = 3 then there are three roots. The LIUB is 5 because all the numerical
coefficients of the quotients are all positive and the GILB is – 3 because the
numerical
coefficients of the quotients have alternate signs. The number of turning points is 2
and
the y – intercept is 16.
IV. INTEGRATION:
1. IGNACIAN CORE VALUE: Faith
RELATED VALUE: Communion
When we hear mass, why is it that we become one family in Christ?
2. LESSON ACROSS DISCIPLINE: Science
When graphing the speed of any moving object, what happen to the graph if the time elapse
decreases while the distance increases?
V. EVALUATION/ ASSESSMENT:
Exercise 1:
Construct the graph of the polynomial function x3 – 7x – 6 and make the table of
values, then determine the upper and lower boundary (GILB,LIUB),the number of
turning points, the zeroes of the function and y intercepts.
Table of Values:
X
Y
Solution:
Graph:
Exercise 2:
Construct the graph of the given polynomial function and make the table of
values, then determine the upper and lower boundary (GILB,LIUB),the number of
turning points, the zeroes of the function and y intercepts.
1. P(x) = x4 + 2x3 – 5x2 – 6x
Table of Values:
X
Y
Solution:
Graph:
2. P(x) = x5 + 32
Table of Values:
X
Y
Solution:
Graph:
LESSON 9:
Date: _______
Problems On Polynomial Functions
I. OBJECTIVES:
Students will:
1. know the steps in solving real – life problems applying the concepts of polynomial
functions
2. be able to use polynomial function to solve real – life problems
II. PRELIMINARIES:
Solve the given word problems applying the fundamental operations of polynomials.
1. What must be added to 3x2 – 5x2 + 1 to get the sum of x3 – x2 + 5?
2. What must be subtracted from the expression 6x5 – 2x4 + 6x3 – 9x + 9 to get a
difference of 2x5 – 4x2 + 8x3 – x + 4 ?
3. A student bought a board sheet that measures (4x4 – 2x2 + 6x + 8) cm long and a
width of (x3 – 3x) cm. What is the area of the board sheet? ( Note: A = LW)
4. Mother bought a round table for their dining room. The radius of the table measures
(2x4 – 3x2 + 4) meter, what is the area of the table? (Note: A = π r2)
5. The area of a parallelogram is (80a6c5 + 60 a4c4) square meters. What is the height of
the parallelogram if its base is (20 a4c4) meters.
III. LESSON DEVELOPMENT:
Concepts and Key Ideas:
In solving word problems involving polynomial functions we follow the given steps such as:
Step 1: Read the problem twice to analyze the concept of the problem
Step 2: Identify all the given information and the missing quantity or term/s. It can also be illustrated
in
diagram and then label the parts to identify the given information in the problem.
Step 3: Identify the equation of the polynomial function in terms of P(x)..
Step 4: Solve the missing term/s applying different laws in algebra in simplifying it.
Step 5: Get the lowest or simplified answer of the term/s
Example 1: In a certain clinic, the patient asked for a medicine for his headache and the doctor gave him
ibuprofen tablets with the given milligrams. From the polynomial equation
f(t)= 0.5t4+ 3.5 t3 – 96.66t2+ 347.7 t relates to the amount in milligrams of ibuprofen left in
the
blood after t hour when 400 mg was administered to the patient. How many milligrams of
ibuprofen were left in the blood after 2 hours?
Solution: Step 1: Read and analyze first the problem
Step 2: Identify the given information: mass of 1 tablet of ibuprofen 400 mg, t = 2 hours,
Step 3: Polynomial equation = f(t)= 0.5t4+ 3.5 t3 – 96.66t2+ 347.7 t
Step 4: Solve the value of t= 2 hours in the equation f(t)= 0.5t4+ 3.5 t3 – 96.66t2+ 347.7 t
f(2)= 0.5(24)+ 3.5 (2)3 – 96.66(22)+ 347.7 (2) then simplify:
Step 5: f(2)= 344.8 mg
Answer: There were 344.8 mg of ibuprofen left in the blood after 2 hours.
Example 2: Due to the erratic trend of the prices of garlic in the Philippine market today, an
analyst
came up with the function f(x) = – 3x3 + 8x2 + 9x – 72 to predict the income in
hundreds of
pesos from producing x kilograms of garlic. How many kilograms of garlic must be
sold to
have an income?
Solution: Step 1: Read and analyze first the problem
Step 2: Identify the given information: f(x) = – 3x3 + 8x2 + 9x – 72, x kilograms of
garlic
Step 3: Polynomial equation = f(x) = – 3x3 + 8x2 + 9x – 72 find the Zeros or the roots
of the
function using synthetic division:
r
–1
8
9 – 72
Since the remainder are all zero then the zeros
of the function are – 3, 3 and 8 and the values
– 3 – 1 11 24
0
x will make f(x) = 0.
So there will be no income if the harvested
garlic is
3 –1
5
24
0
0
9
between 3 and 8 kilograms.
If x = – 3 kilograms is irrelevant
because there
8
–1
0
is no negative measurement or negative
kilogram.
(Note: if 3, 8 and -3 will be substituted to the
function
The answer will all be zero.)
IV. INTEGRATION:
1. Ignacian Core Value: Faith
Related Value: Justice
Why did God gave us power to balance our life?
2. Social Orientation: Power of state and Church
How are the power of the state and church work for the benefit of the people?
3. Lesson Across Discipline: AP (Formulation of Laws)
How did the state formulate laws and policies of the country to solve all the problems of the
people?
V. EVALUATION/ ASSESSMENT: Google Meet with the students
Exercise 1.a: Group Activity
Solve the given word problems applying the addition and subtraction of polynomials.
1. What must be added to 3x2 – 5x2 + 1 to get the sum of x3 – x2 + 5?
2. What must be subtracted from the expression 6x5 – 2x4 + 6x3 – 9x + 9 to get a
difference of 2x5 – 4x2 + 8x3 – x + 4 ?
3. What is the perimeter of a square lot if each side measures 4x3 + 6x2 – 3x + 1 meters?
4. The base of an isosceles triangle measures 2x3 – 7x2 – 6 cm. and the congruent sides
3x3 – 2x2 + 9 cm. Find the perimeter of the triangle.
5. The length of a rectangle measures 6x2 – 4 cm. while the width measures 3x2 – x – 4
cm, what is the measure of its perimeter?
6. The newspaper boy travels for a distance of 6x3 – 4x2 + x km in going to barrio to sell
newspapers and then he move to another barrio for 2x3 – 5x2 + 6 km, how far did he
traveled?
7. Mother bought a square table for their dining room. If the side of the table measures
2x3 – 4x2 + x – 5 meters, what is the perimeter of the table?
8.
A student received her allowance of Php 3x3 – x2 + 2x + 8 for a week. She spend
Php 5x3 – 2x2 + 2x for lunch and snacks and Php 3x3 + 3x2 – 3x + 10 for her travel
expenses. How much was left to her after a week?
9. Ms. Santos received her salary for the month at an amount
of Php 10x4 – 6x3 + x2 – 8. If she spend Php 2x3 + 2x2 + 3x + 1
for her house rental, Php x4 – 4x2 + x3 – 4 for her grocery,
how much money was left to her?
10. Mr. Perez want to fence his rectangular lot with a perimeter
of 6x3 + 4x2 – 8x + 10 meters. If the width of the lot measures
2x3 + 2x2 + x + 8 meters, how many fencing materials will he
need for the length of the lot?
Exercise 1.b: Group Activity
B. Solve the following word problems applying multiplication of polynomials. Draw the
figure if possible then show the complete solution.
1. A student bought a board sheet that measures (4x4 – 2x2 + 6x + 8) cm long and a
width of (x3 – 3x) cm. What is the area of the board sheet? ( Note: A = LW)
2. Mother bought a round table for their dining room. The radius of the table measures
(2x4 – 3x2 + 4) meter, what is the area of the table? (Note: A = π r2)
3. What is the volume of a cubical water tank if its side measures (3x4 + 4x3) meters?
(Note: V = s3)
4.
A carpenter will construct the triangular roof of Mr. Bautista’s house. The height of
the roof is (2x2 – 3x + 2) meter and the base is (4x2 – 3x2 + 6) meter. What is the
area of the roof? (Note: A = ½ bh)
Exercise 1.c: Group Activity
Solve the following word problems applying the process of dividing polynomials. Show
the complete solution with diagram.
1. The area of a parallelogram is (80a6c5 + 60 a4c4) square meters. What is the height of
the parallelogram if its base is (20 a4c4) meters.
2. The area of a square paper is (36 a4c6) square inches. What is the measure of the
length of the side of the square?
3. The cost of x ball pens is (2x3 + 5x2 + 3x) pesos. How much do you pay for each ball
pen?
4. The area of a rectangular garden is (x2 + 2x – 35) square meters. If the width measures
(x – 5) meter, find the measures of its length.
5. What is the measures of the base of a triangular wood if its area is (2x3 +7x2 + 10x +
8) square cm and its height is (x + 2) cm?
6. Mother bought (10x3y4) pieces of candies for the birthday of her son that cost
(20x5y6 + 50x2y2 – 25x4y4) pesos. What is the cost of each candy?
7. The area of a rectangular table measures (2x3 – x2 + 4x – 5) square meters.
If the width of the table is (x + 2) meter, w
hat is the measure of its length?
8. Ms. Santos bought a rectangular carpet for their receiving room
with an area of (4x4 – 3x2 – x + 5) square meter and its length
measures (x – 3) meter. What is the measure of the width of the carpet?
Exercise 2:
Solve the following word problems involving polynomial functions.
1. The El Nino phenomenon has brought people’s attention to monitor their own water
consumption. A certain water pump supplies water to a farm field. The farmers want to
know how much water is consumed for the entire field. They come up with the
formula: r is the radius of throw of water of a certain sprinkler releasing pressure x. If
the relation can be described as r(x) = – 0.234x2+1.029x + 5.39
What is the radius of throw if the pressure is 2.5 kg/cm2?
2. The Philippines has been called the “Next Tiger Economy of Asia”,
because of the increasing trust of foreign investors in the country.
The function f(x) = 0.645x3 – 8.3453x2 + 33.435x + 73.345 models
an amount of investment in millions of dollars, where x is the
number of years after 1998. How much is the investment in 2006?
3. If the function of g(x) = 2x3 – 15x2 + 33x – 20 yields revenue
from selling handcrafted x bags, how many bags should be
sold to have revenue?
4. If 2 and – 1 are zeros of polynomial function g(x) = ax3 + bx2 – x + d, and 6 is its y –
intercept, find the values of a, b, and d.
5. The Cruz family is remodeling their square living room by tearing down one wall and
extending the room 3 meters more. The room will be rectangular with a floor area of
180 square meters. What are the original dimensions of the room?
Quarterly examination using ARALINKs
Conference with parents.
Prepared by:
Maribeth R. del Rosario
Teacher
Checked by:
S.A.C. – Date
Acad
Date
Coor.- Directress/
Date
Principal
-