Powerpoint_Presentation_in_General_Mathe.pdf

1. Wearing of Face Mask
2. Wearing of Face Shield
3. Social Distancing
4. Frequent Washing of Hands
MINIMUM HEALTH
STANDARDS

SAMSUDIN N. ABDULLAH, PhD
Master Teacher II
Esperanza National High School
Esperanza, Sultan Kudarat, Region XII, 9806 Philippines
Email Ad: samsudinabdullah42@yahoo.com

I. Functions and Relations (Review Lessons)
- Ordered Pairs
- Arrow Diagrams
- Tables of Values
- Equations
- Graphs
II. Basic Functions and Their Graphs
- Constant Function
- Linear Function
- Quadratic Function
- Cubic Function
- Identity Function
- Absolute Value Function
- Piecewise Function
- Representing Real-Life Situations using Functions
GENERAL MATHEMATICS (Advanced Algebra, Basic Business Math and Logic)
COURSE OUTLINE
IN GENERAL MATHEMATICS

III. Evaluating Functions
IV. Operations on Functions
- Addition of Functions
- Subtraction of Functions
- Multiplication of Functions
- Division of Functions
- Composition of Functions
- Applications of Functions
V. Inverse Functions
- Finding the Inverse of a Function
- Graphing Inverse Functions
- Finding the Domain and Range of an Inverse
Function
COURSE OUTLINE

VI. Rational Functions
- Rational Equations and Inequalities
- Graphing Rational Functions
- Finding the Domain and Range of Rational Functions
VII. Exponential and Logarithmic Functions
- Representation of Exponential Function through its
Table of Values, Graph and Equation
- Logarithmic Functions and Their Graphs
- Laws of Logarithm
- Exponential and Logarithmic Equations and
Inequalities
- Finding the Domain and Range of Logarithmic
and Exponential Functions
- Applications of Exponential and Logarithmic Functions
COURSE OUTLINE

VIII. Basic Business Mathematics
- Simple Interest
- Compound Interest
- Applications of Simple and Compound Interests
- Stocks, Bonds and General Annuities (Optional)
IX. Logic
- Propositions and Symbols
- Truth Values
- Forms of Conditional Propositions
- Tautologies and Fallacies
- Writing Poofs
COURSE OUTLINE

FUNCTIONS
AND
RELATIONS

OBJECTIVES:
At the end of this video presentation, you
are expected to:
1. determine whether or not an ordered pair or
an arrow diagram represents a function;
2. tell whether a table of values or an equation is
a function or a relation; and
3. use vertical line test to tell whether a graph is
a function or a mere relation.

FUNCTIONS AND RELATIONS REPRESENTED BY
ORDERED PAIRS
1. {(1, 2), (3, 5), (4, 6), (5, 5)} –
2. {(0, 5), (0, 4), (3, 0), (2, 0)} –
3. {(1, 2), (3, 4), (5, 6), (7, 8)} –
4. {(-1, -1), (2, 5), (-1, 0)} –
5. {(-1, -1), (0, 0), (2, 2), (3, 3)} –
6. {(-2, 5), (-2, 4), (2, 3), (2, 6)} –
7. {(1, 2), (2, 3), (3, 4), (4, 5)} –
8. {(0, 1), (0, 2), (0, 3), (0, 5)} –
9. {(4, 5), (5, 5), (6, 7), (7, 8)} –
10.{(5, 5), (6, 6), (5, 7), (8, 8), (9, 8)} –
Direction: Tell whether or not each of the following relations
describes a function. Write “Function” or “Not Function”.
Function
Not Function
Function
Not Function
Function
Not Function
Function
Not Function
Function
Not Function

Direction: Determine whether or not each arrow diagram
represents a function. Write “Function” or “Relation”.
ARROW DIAGRAMS
1. Domain Range 2. Domain Range
5
1
2
3
4
5
-2
1
3
5
7
0
3
5
8
Relation Function

Function
a
b
c
d
1
2
3
4
Function
6
3
0
-3
11
9
7
-3
-6

Function
-1
-5
-7
-9
4
-2
-3
-5
-6
-8
3
5
6
8
Relation

Function
-2
-3
-4
-5
6
7
8
-1
-2
-3
-4
-5
3
5
6
8
9
Function

Relation
10
11
13
14
6
7
8
-5
-4
-3
-2
-1
9
7
6
5
4
Relation

1. 2.
Function Mere Relation
x y
0 -5
1 -2
2 1
3 4
4 7
5 10
TABLES OF VALUES
Direction: Determine whether or not each of the following
tables of values is a function. Write “Function” or “Mere
Relation”.
x y
9 3
4 2
0 0
4 -2
9 -3
x y
-2 7
4 6
2 5
-2 4
5 3
6 2
-2 1
3.
Mere Relation

4. 5.
Function Mere Relation
x y
81 -3
16 -2
1 -1
0 0
1 1
16 2
81 3
x y
-5 7
4 4
2 -3
-2 2
5 1
6 -2
-5 7
6.
Function
x y
-5 -1
4 3
3 5
2 6
-5 -1
1 4

7.
8.
Function
x -2 -1 0 1 2
y -8 -1 0 1 8
x -2 -1 0 1 2
y 11 6 1 -4 -9
Function
9.
Mere
Relation
10.
Mere
Relation
x 2 -1 4 0 3 2
y -1 2 5 8 11 14
x -1 3 2 -4 3 5
y 4 -1 3 5 -2 4

EQUATIONS
Direction: Which of the following equations are functions?
Which are not? Write “Function” or “Not Function”.
1. y = 2x + 5
2. y = x2 + 5
3. y = 3𝑥 − 5
4. y = ± 5𝑥 − 3
5. y2 = x2 + 7
Function
Function
Function
Not Function
Not Function

6. y³ + x⁴ = 4
7. y⁵ = x⁴ − 1
8. y⁶ + x⁵ = 3
9. y⁷ − x2 = 16
10. x⁴ = 5 – y⁵
Function
Function
Not Function
Function
Function

Direction: Use the vertical line test to determine whether
or not each graph represents a function. Write “Function”
or “Not Function”.
GRAPHS
1.
y
x
y
x
The vertical line intersects the
graph at two points.
•
The vertical line intersects
the graph one point.
2.
Not Function
Function
•
•

3.
y
x
4.
x
y
x
5. 6.
y
x
Function Function
Not Function
Function
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

7. 8.
9. 10.
y
x
y
x
y
x
y
x
•
•
Not Function Function
Not Function
Function
•
•
•
•
•
•
•
•
•
•
•
•
•

ODD AND EVEN FUNCTIONS
Odd Functions and Even Functions
are functions that satisfy specific
symmetry relations.
If the function is odd, then its graph
is symmetric about the origin. f(x) = x³
is an odd function because f(-x) = -f(x)
for all x. f(x) = x is also odd function
because f(2) = 32 and f(-2) = -32 and 32
and -32 are additive inverses.

If the function is even, then
its graph is symmetric about
the y-axis. f(x) = x² is even
function because f(-x) = f(x).
f(x) = -x is also even function
because f(3) = -81 and
f(-3) = -81.

Is f(x) = -x³ + 1 an odd function? No,
it is not since f(1) = 0 and f(-1) = 2
wherein 0 and 2 are not additive
inverses. Is f(x) = (x + 1)² an even
function. No, it is not since f(2) = 9 and
f(-2) = 1 where 9 ≠ 1.
f(x) = -x³ + 1 and f(x) = (x + 1)² are
neither odd functions nor even
functions.

k
k
k
Odd Function Even Function
k k
k

DIRECTION: Tell whether each equation is odd,
even or neither from its equation. Write Odd
Function, Even Function or Neither.
5. y = 3 – x²
k k k
Odd Function
1. y = -x³
2. y = (x + 1)² Neither
3. y = 1 – x³ Neither
4. y = x + 2 Neither
Even Function

10. y = 2x² + 3
k k k
Odd Function
6. y = 3x
7. y = 3x Odd Function
8. y = (x – 2)³ Neither
9. y = -(x + 3)² Neither
Even Function
DIRECTION: Tell whether each equation is odd,
even or neither from its equation. Write Odd
Function, Even Function or Neither.

DIRECTION: Tell whether each graph is odd, even
or neither. Write Odd Function, Even Function or
Neither.
k k k
y
x
y
x
x
1.
Even Function
2.
Odd Function
3.
Even Function
4.
Even Function
y
x

k k k
y
x
y
x
5.
Even Function
6.
Odd Function
y
x
Odd Function
7.
8.
Neither
Neither.
y
x

k k k
9.
10.
Neither
Neither
x
y
x
y
11.
Odd Function
12.
Even Function
Neither.

k k k
DIRECTION: Show algebraically whether each
function is odd, even or neither.
1. f(x) = -3x³ + 2x
Solution:
f(x) = -3x³ + 2x
f(-x) = -3(-x)³ + 2(-x)
= -3(-1)³(x³) + 2(-1)(x)
f(-x) = 3x³ – 2x
-(3x³ – 2x) = -3x³ + 2x
f(-x) = -f(x)
Odd Function

k k k
2. f(x) = 𝟑𝒙𝟔 − 𝟖𝒙𝟒 + 𝟓𝒙𝟐 + 𝟔
Solution:
f(x) = 3𝑥6 − 8𝑥4 + 5𝑥2 + 6
f(-x) = 3(–𝑥)6 − 8(–𝑥)4 + 5(–𝑥)2 + 6
f(-x) = 𝟑𝒙𝟔 − 𝟖𝒙𝟒 + 𝟓𝒙𝟐 + 𝟔
f(-x) = f(x)
Even Function

k k k
3. f(x) = 𝟐𝒙𝟒 − 𝟕𝒙𝟑 + 𝟑𝒙𝟐 + 𝟏
Solution:
f(x) = 2𝑥4−7𝑥3 + 3𝑥2 + 1
f(-x) = 2(–𝑥)4−7(–𝑥)3 + 3(–𝑥)2 + 1
f(-x) = 𝟐𝒙𝟒 + 𝟕𝒙𝟑 + 𝟑𝒙𝟐 + 𝟏
f(-x) ≠ f(x)
Neither

k k k
4. f(x) =
𝑥
2𝑥5+𝑥3
Solution:
f(x) =
𝒙
𝟐𝒙𝟓+𝒙𝟑
f(-x) =
(−𝑥)
2(−𝑥)5+(−𝑥)3
=
−𝑥
−2𝑥5−𝑥3
=
−(𝑥)
−(2𝑥5+𝑥3)
f(-x) =
𝒙
𝟐𝒙𝟓+𝒙𝟑
f(-x) = f(x)
Even Function
y
x

k k k
5. f(x) =
1
𝑥7+ 4
Solution:
f(x) =
𝟏
𝒙𝟕+ 𝟒
f(-x) =
1
(−𝑥)7+ 4
=
1
−𝑥7+ 4
f(-x) =
𝟏
𝟒 − 𝒙𝟕
−(
𝟏
𝟒 − 𝒙𝟕) =
1
𝑥7− 4
f(-x) ≠ -f(x)
Neither
10 y
x
-

k k
DIRECTION: Mentally solve whether each
Neither
6. f(x) = 2𝑥8
− 5𝑥3
+ 3
7. f(x) = 3𝑥5
− 7𝑥3
+ 4𝑥2 Neither
8. f(x) =
3
𝑥4+1
Even Function
9. f(x) =
𝑥
𝑥3−2
Neither
10. f(x) =
2
𝑥3−𝑥5 Odd Function
11. f(x) =
5
2𝑥4−2𝑥2+5
Even Function
12. f(x) =
3
𝑥6−2𝑥3+5
Neither

k
DIRECTION: Mentally solve whether each function is odd,
even or neither.
Neither
13. f(x) = 3𝑥9
+ 2𝑥3
+ 3
14. f(x) = 𝑥7
− 2𝑥5
+ 3𝑥3 Odd Function
15. f(x) =
3𝑥
𝑥5−2𝑥
Even Function
16. f(x) =
2𝑥
3𝑥4−𝑥2
Odd Function
17. f(x) =
1
𝑥2−3
Even Function
18. f(x) =
5
𝑥5+3𝑥3−𝑥
Odd Function
19. f(x) =
3
𝑥4−2𝑥3+3𝑥 Neither
20. f(x) = /x/ + 5 Even Function

k
Domain and Range of a Relation
x
y
•
Range: {y|y ≥ k}
or [k, +∞)
Domain: {x|x ε R}
or (-∞, +∞)
(h, k)

k
DIRECTION: Find the domain and range of each relation.
1. {(-2, 5), (-3, 6), (-4, 6), (-5, 8), (-4, 10)}
Domain: {-2, -3, -4, -5, -4} = {-5, -4, -3, -2}
Range: {5, 6, 6, 8, 10} = {5, 6, 8, 10}
2. {(-2, -5), (5, 0), (6, 9), (-7, 9), (8, 9), (-9, -9), (9, 0)}
Domain: {-2, 5, 6, -7, 8, -9, 9} = {-9, -7, -2, 5, 6, 8, 9}
Range: {-5, 0, 9, 9, 9, -9, 0} = {-9, -5, 0, 9}
3.
Domain: {-10, -8, -7, -6, -5, -4, -3}
Range: {-8, -5, 0, 3, 7, 8}
X Y
-3
-4
-5
-6
-7
-8
-10
-8
7
3
8
-5
0

k
4.
Domain: {x|-4 ≤ x ≤ 4} or [-4, 4]
Range: {y|-2 ≤ y ≤ 2} or [-2, 2]
5.
Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ≥ -4} or [-4, +∞)
x
y
1
2
1
-1
-1
-2
2 3 4
-2
-3
x
y
1
2
1
-1
-4
-1
-2
2 3 4
-2
-3
x
y
-2 2
-4●
(0, -4)

k
6.
Range: {y|y ≤ 2} or (-∞, 2]
7.
Domain: {x|x ≥ -2} or [-2, +∞)
Range: {y|y ε R} or (-∞, +∞)
-2
x
y
-2 2
-4
●
2 (0, 2)
-2
x
y
-2 2
-4
●
2 (0, 2)
(-2, -3)●
-6

k
8.
Range: {y|y ≥ -6} or [-6, +∞)
9.
Range: {y|y ≤ -4, y ≥ -2} or (-∞, -4] ᴜ [-2, +∞)
y
x
-2
-4
-2 2
(-1, -4)
(-1, -2)
-6
-8
●
●
y
x
-2
-4
-2 2
(0, -6)
-6●

k
10.
Domain: {x|x < 5} or (-∞, 5)
Range: {y|y ≤ 4} or (-∞, 4]
11.
Range: {y|y ≤ −
15
4
, -2 ≤ y < 4, y ≥ 5} or
(-∞, −
15
4
] ᴜ [-2, 4) ᴜ [5, +∞)
2 4 6
2
4
6
y
x
(5, 3)
o
(4, 4)
2 4 6
2
4
6
y
x
(3, -1)●
(4, -2)
-2
o
(3, -5)
-4
-6
o
●
-8
-10
(2, −
𝟏𝟓
𝟒
)
●
●
(6, 4)
(6, 5)

k
12.
Domain: {x|x ≠ 2} or (-∞, 2) ᴜ (2, +∞)
Range: {y|y ≠ -4} or (-∞, -4) ᴜ (-4, +∞)
13.
Range: {y|y < -6, y = -3, y ≥ -1}
or (-∞, -6) ᴜ {-3} ᴜ [-1, +∞)
2 4 6
-2
-4
-6
-8
o(2, -4)
y
x
2 4 6
-2
-4
-6
-8
o(3, -6)
-2
y
x
o ●
-4
●
(3, -3)
(-2, -3)
(-2, -1)

k
14. y + 5 = 0 Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y = -5} or {-5}
15. y = 3x − 5 Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ε R} or (-∞, +∞)
16. y = -2(x + 1)² Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ≤ 0} or (-∞, 0]
17. y = (x − 3)² Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ≥ 0} or [0, +∞)
18. y = 3x² − 4 Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ≥ -4} or [-4, +∞)
19. y = 3 − 2x² Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ≤ 3} or (-∞, 3]

k
21. y = x² + 4x + 5
Solution:
h = −
b
2a
= −
4
2 1
= -2
k = (-2)² + 4(-2) + 5 = 1
Range: {y|y ≥ 1} or [1, +∞)
20. y = -(x + 2)² − 8 Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ≤ -8} or (-∞, -8]
22. y = -2x² − 8x + 3
Solution:
h = −
b
2a
= −
(−8)
2 −2
=
8
−4
= -2
k = -2(-2)² − 8(-2) + 3 = -8 + 16 + 3 = 11
Range: {y|y ≤ 11} or (-∞, 11]

k
24. y = 2𝑥 − 5
Solution:
2x − 5 ≥ 0
2x ≥ 5
x ≥
5
2
Domain: {x|x ≥
5
2
} or [
5
2
, +∞)
Range: {y|y ≥ 0} or [0, +∞)
23. y = 2x³ − 3x + 7 Domain: {x|x ε R} or (-∞, +∞)
Range: {y|y ε R} or (-∞, +∞)
25. y = - 3 − 4𝑥
Solution:
3 − 4x ≥ 0
-4x ≥ -3
x ≤
3
4
Domain: {x|x ≤
3
4
} or (-∞,
3
4
]
Range: {y|y ≤ 0} or (-∞, 0]

k
27. y = 𝑥2 − 4
Solution:
𝑥2
− 4 = 0
𝑥2 = 4
x = ± 4
x = ± 2
Domain: {x|x ≤ -2, x ≥ 2} or (-∞, -2] ᴜ [2, +∞)
Range: {y|y ≥ 0} or [0, +∞)
26. x² + y² − 64 = 0 Domain: {x|-8 ≤ x ≤ 8} or [-8, 8]
Range: {y|-8 ≤ y ≤ 8} or [-8, 8]
28. y = - 25 − 9𝑥2
Solution:
25 − 9𝑥2 = 0
- 9𝑥2
= -25
𝑥2
=
25
9
x = ±
25
9
= ±
5
3
Domain: {x|−
5
3
≤ x ≤
5
3
} or [-
5
3
,
5
3
]
25 = ± 5
Range: {y|-5 ≤ y ≤ 0} or [-5, 0]

ASSIGNMENT 1
Copy and answer the following
problems. Take a clear photo of
your answer and submit it online
through my email address
samsudinabdullah42@yahoo.com or
my messenger account Samsudin N.
Abdullah.

1. {(0, -2), (1, -1), (0, 0), (1, 1), (4, 2)} –
2. {(5, 2), (0, 0), (2, 5), (5, -3)} –
3. {(a, 1), (a, 2), (a, 3), (b, 2), (c, 4)} –
4. {(x, 5), (x, -6), (x, 7), (x, 8)} –
5. {(8, 0), (7, 0), (0, 0), (-2, 0)} –
6. {(3, 2), (3, 1), (3, 0), (3, -1)} –
7. {(-7, 5), (-7, 8), (6, 9), (6, 7)} –
8. {(5, 8), (6, 0), (-3, 9), (-3, 8)} –
9. {(2, 2), (4, 4), (5, 5), (6, 6)} –
10. {(3, 8), (2, 3), (8, 0), (0, 0)} –
A. Direction: Tell whether or not each of the
following relations represents a function. Write
“Function” or “Not Function”.

B. Direction: Identify whether or not each of the
following arrow diagrams represents a function.
Write “Function” or “Not Function”.
1. Domain Range 2. Domain Range 3. Domain Range
1
2 5 3 1 2
5 3 2 5
4 9 7 3 10
5 16 17
4. Domain Range 5. Domain Range 6. Domain Range
1 4 1 4 3
2 5 2 6 6 11
3 6 5 10 9

1. Domain Range 8.Domain Range 9.Domain Range 10.Domain Range
3 5 0 3
5 6 7 -5 5 8 15 -4
6 8 9 -7 10 -12
9 10 11 -10 15 20 30 - 15
10 13 -15 20
7.
1
2
3
4
-1
-2
-3
-4
-11
-21
-31
-41
10

1.
2.
3.
C. Direction: Determine whether or not each of
the following tables of values is a function. Write
“Function” or “Mere Relation”.
x -1 -2 -3 4 6
y 5 1 8 3 8
x 3 -4 0 -2 -4
y 2 3 -1 -5 6
x 6 4 2 -2 -4
y 1 3 -3 -4 6

4. x y
7 -1
3 3
-2 5
-4 2
7 3
8 -2
9 -3
5. x y
-4 2
5 4
6 6
-4 2
-2 1
-1 -2
0 -3
6.
x y
2 -4
3 -3
4 -3
5 -2
6 -1
7 0
8 -4
7. x y
-1 0
2 1
-3 2
4 3
-5 4
5 5
8. x y
5 2
5 0
5 -2
5 -4
5 -6
5 -8
9. x y
1 3
2 3
3 3
4 3
5 3
6 3
10. x y
1 2
2 5
3 4
4 5
3 -3
6 2

1. y = 4 – 2x
2. y = 5 – x³
3. y⁴ = 5x – 2
4. y⁵ + 3x = 1
5. 2y⁴ = 5x – 4
D. Direction: Which of the following are
functions? Which are not? Write “Function” or
“Not Function”.
6. x2 = y³+ 2
7. y⁶ = 2x – 5
8. y⁴ = 5x – 2
9. x⁵ = 3 – y⁵
10. y = – 3𝑥 − 5

E. Direction: Which of the following graphs are
functions? Which are not? Write “Function” or
“Not Function”.
1. y 2. y 3. y
x x x
4. y 5. y 6. y
x x x

7. y 8. y 9. y
x x x
10. y 11. y 12. y
x x x
13. y 14. y 15. y
x x x

Generalization:
FUNCTION is a rule or correspondence (relation) that no two
distinct ordered pairs have the same first element (abscissa). Let X & Y
be two nonempty sets of real numbers. A function from X into Y is a
relation that associates with each element of X a unique element of Y.
Set X is called the Domain of the function. For every element x in set X,
the corresponding element y in set Y is called the value of the function
at x, or the image of x. This set of values or images of the elements of
the domain is called the Range of the function. All functions are
relations but not all relations are functions.
Not all graphs represent functions. Vertical Line Test is
employed to determine whether a graph is a function or not. If the
vertical line intersects the graph at most one point, the graph is a
function. Otherwise, it is not.
Not all functions are one-to-one. To determine whether or not a
function is one-to-one, Horizontal Line Test is applied. If the horizontal
line intersects the graph at only one point, the graph is one-to-one.
Otherwise, it is not.

BASIC FUNCTIONS
AND THEIR
GRAPHS

OBJECTIVES:
are expected to:
1. demonstrate your understanding about a
constant function, a linear function and a
quadratic function; and
2. set up a table of values for each of these basic
functions to sketch the graph.

CONSTANT FUNCTION
CONSTANT FUNCTION is a linear function for which
the range does not change no matter which member of the
domain is used. Its graph is a horizontal line. It is denoted by
the equation y = c where c is a constant. The degree of a
constant function is 0. The slope (m) of a constant function is 0
and its y-intercept (b) is c.
Examples:
Set up a table of values for each constant function and
sketch the graph.
1. y = 3
x -2 2
y 3 3
x
y
• •
(2, 3)
(-2, 3)
y = 3
2
4
2
-2
-1
m = 0
b = 3

CONSTANT FUNCTION
2. y = -2
x 0 3
y -2 -2
x
y
• •
y = -2
1
-1
-3
-2 2 4
(3, -2)
(0, -2)
m = 0
b = -2

LINEAR FUNCTION
LINEAR FUNCTION is a function whose graph is
a non-vertical straight line. It is denoted by the equation
y = mx + b where m is the slope and b is the y-intercept.
The degree of a linear function is 1.
Examples:
Construct a table of values for each linear function
and sketch the graph.
1. y = 2x – 3
x -1 0
Y -5 -3
y = 2x – 3
-2
x
y
•
•
2
-2
-4
-6
(-1, -5)
(0, -3)
m = 2
b = -3

LINEAR FUNCTION
2. 𝒚 = −
𝟑
𝟐
𝒙 + 𝟒
x y
-2 7
4 -2
= − +
-2
x
•
2
(-2, 7)
y
2 4
4
6
8
-2 •(4, -2)
m = −
𝟑
𝟐
b = 4

QUADRATIC FUNCTION
Quadratic Function is a function whose graph is
a parabola that opens upward or downward. It is
denoted by the its vertex form y = a(x – h)² + k where
(h, k) is the vertex and x = h is the axis of symmetry. Its
degree is 2.
Examples:
Construct a table of values for each quadratic
function and draw the graph.
1. y = x²
x -2 -1 0 1 2
Y 4 1 0 1 4
V (0, 0)
Axis of Symmetry: x = 0 •
•
•
• •
-2 2
2
4
y
x
(2, 4)
(-2, 4)
(1, 1)
(-1, 1)
(0, 0)
y = x

QUADRATIC FUNCTION
2. y = x²− 2
x -2 -1 0 1 2
y 2 -1 -2 -1 2
V (0, -2)
Axis of Symmetry: x = 0
•
•
•
• •
2
y
x
(0, -2)
-1
-3
-2 2
(1, -1)
(-1, -1)
(-2, 2) (2, 2)
y = x − 2

QUADRATIC FUNCTION
3. y = -(x – 3)²
x 1 2 3 4 5
y -4 -1 0 -1 -4
•
•
•
•
•
y
x
(3, 0)
2 4 6
-2
-4
-6
(4,-1)
(2,-1)
(1, -4) (5, -4)
y = -(x – 3)²
V (3, 0)

QUADRATIC FUNCTION
4. y = 2(x + 2)² – 1
x -4 -3 -2 -1 0
y 7 1 -1 1 7
y
x
•
•
•
• •
2
4
6
8
-2
-2
-4
(0, 7)
(-4, 7)
(-3, 1) (-1, 1)
(-2, -1)
y = 2(x + 2)² – 1
V (-2, -1)
Axis of Symmetry: x = -2

OBJECTIVES:
are expected to:
cubic function, an identity function and an
absolute value function; and
2. set up a table of values for each of these basic
functions to sketch the graph.

CUBIC FUNCTION
Cubic Function is a function whose graph is a
curve with a point of symmetry. It is represented by the
equation y = a(x – h)³ + k where (h, k) is the point of
symmetry. The degree of a cubic function is 3.
Examples:
Set up a table of values for each cubic function
and sketch the graph.
1. y = x³
x -2 -1 0 1 2
Y -8 -1 0 1 8
(2, 8)
y
x
2
4
6
8
-2
-4
-6
-8
2
-2
••
•
•
•
(-2, -8)
(1, 1)
(-1, -1)
y = x³
Point of Symmetry: (0, 0)

CUBIC FUNCTION
1. y = x³
x -2 -1 0 1 2
Y -8 -1 0 1 8
Point of
Symmetry: (0, 0)
(2, 8)
y
x
2
4
6
8
-2
-4
-6
-8
2
-2
••
•
•
•
(-2, -8)
(1, 1)
(-1, -1)
y = x³
(0, 0)

CUBIC FUNCTION
2. y = -(x + 2)³ x -4 -3 -2 -1 0
y 8 1 0 -1 -8
y
(-4, 8) 8
-4
y = -(x + 2)³ 6
2
4
x
(-2, 1)
(-1, -1)
•
• -2
-4
-6
•
(0, -8) -8
•
•
2
Point of Symmetry: (-2, 0)
(-2, 0)
(-3, 1)

CUBIC FUNCTION
3. y = (x – 3)³ – 2 x 1 2 3 4 5
y -10 -3 -2 -1 6
Point of Symmetry: (3, -2)
y
y = (x – 3)³ – 2
2
4
x
(5, 6)
-2
-4
-6
•
-8
2
-11
6
4 6
•
•
•
•
(4, -1)
(3, -2)
(2, -3)
(1, -10)

IDENTITY FUNCTION
Identity Function is a function whose graph is a
straight line that passes through the origin and divides
the Cartesian plane into two equal parts. It is denoted by
the equation y = x. It is a type of a linear function.
Example:
Construct a table of values for the identity function
y = x. Then sketch the graph.
x -2 -1 0 1 2
y -2 -1 0 1 2
x
y
•
•
•
•
•
(2, 2)
(1, 1)
(0, 0)
(-1, -1)
(-2, -2)
1
2
3
-3 -2 -1
-1
-2
-3
y = x
1 2 3

ABSOLUTE VALUE FUNCTION
Absolute Value Function is a function that contains
an algebraic expression within absolute value symbols. It
is denoted by the equation y = a|x – h| + k where (h, k) is
the vertex and x = h is the axis of symmetry.
Examples:
Construct table of values for each absolute value
function. Then sketch the graph.
1. y = |x|
x -1 0 1
y 1 0 1
V (0, 0)
x
y
•
•
•
(0, 0)
1
2
3
-3 -2 -1 1
y = |x|
2 3
(1, 1)
(-1, 1)

2. y = -|x + 4|
x -5 -4 -3
y -1 0 -1
x
y
•
-2
y = -|x + 4|
(-3, -1)
-6
• •
(-5, -1)
(-4, 0)
-2
-4
V (-4, 0)
Axis of Symmetry: x = -4

3. y = 2|x – 3| – 4
x 2 3 4
y -2 -4 -2
x
y
4
(4, -2)
2
(3, -4)
6
•
•
-2
-4
(2, -2)
•
2
y = 2|x – 3| – 4
V (3, -4)

OBJECTIVES:
are expected to:
piecewise function;
2. set up a table of values for a piecewise
function to sketch the graph; and
3. represent real-life situations using functions.

PIECEWISE FUNCTION
Piecewise Function is a function defined by multiple
sub-functions or sequence of intervals. Its domain is
divided into parts and each part is defined by a different
function rule.
Examples:
Construct tables of values for each piecewise
1. y =
x if x ≥ 2
-2 if x < 2 •
•
•
2 4
-2
-4
-2
-4
2
4
y
x
y = x if x ≥ 2
y = -2 if x < 2
(3, 3)
(2, 2)
(2, -2)
(1, -2)
For y = x if x ≥ 2
x 2 3
y 2 3
For y = -2 if x < 2
x 1 2
y -2 -2

PIECEWISE FUNCTION
x² if x > -1
-1 if -4 < x ≤ -1
x if x ≤ -4
2. y =
For y = x2
if x >-1:
x -1 0 1 2
y 1 0 1 4
For y = -1 if -4 < x ≤ -1:
x -4 -1
y -1 -1
For y = x if x ≤ -4:
x -5 -4
y -5 -4
• 2 4
-2
-4
-2
-4
2
4
y
x
y = x² if x > -1
(2, 4)
-6
-6
-8
-8
•
•
•
•
•
(1, 1)
(-1, 1)
(-1, -1)
(-4, -1)
(-4, -4)
(-5, -5)
y = -1 if -4 < x ≤ -1
y = x if x ≤ -4

PIECEWISE FUNCTION
1
2
+ 6 if x > -4
2 if x = -4
-(x + 5)2 – 3 if x < -4
3. y =
For y =
1
2
𝑥 + 6 if x > -4
x -4 0
y 4 6
For y = -(x + 5)2 – 3
x -7 -6 -5 -4
y -7 -4 -3 -4
y
2 4
-2
-4
-2
-4
2
4
x
-6
-6
-8
-8
y =
𝟏
𝟐
𝒙 + 𝟔 if x > -4
(0, 6)
•
6
•
•
•
•
(-4, 4)
(-4, 2)
y = 2 if x =-4
y = -(x + 5)2
– 3 if x <-4
(-4, -4)
(-5, -3)
(-6, -4)
(-7, -7)

ASSIGNMENT 2
Copy and answer the
following problems. Take a clear
photo of your answer and submit
it online through my email address
samsudinabdullah42@yahoo.com
or my messenger account
Samsudin N. Abdullah.

Direction: Construct a table of values for each
1. y = 3
2. y = −
5
2
3. y = -3x + 2
4. y =
4
3
𝑥 − 2
5. y = (x − 2)²
6. y = -(x + 3)² − 2
7. y = (x – 2)² + 3
8. y = 4 – x³
9. y = -(x – 1)³
10. y = (x + 2)³ − 3
11. y = -|x − 4|
12. y = 2|x + 3| + 1
13. y =
14. y =
-x² + 2 if x < 1
-2 if 1 ≤ x < 3
x – 9 if x ≥ 3
DIRECTION: Construct a table of values for each
function. Then sketch the graph accurately.
x³ if x ≥ 1
3 if 1 < x ≤ 4
x if x > 4

REPRESENTING REAL-LIFE SITUATIONS
USING FUNCTIONS
Various types of relationships or real-life
situations apply the concept of functions.
Examples:
1. Distance is a function of time.
2. Height is a function of age.
3. A weekly salary is a function of the hourly pay
rate and the number of hours worked.
4. A circle’s circumference is a function of its
diameter.

REPRESENTING REAL-LIFE SITUATIONS
USING FUNCTIONS
5. Price of rice is a function of its weight.
6. Government employee’s salary is a function of a
month he has been working in his workplace.
7. Amount of sodas coming out of a vending
machine is a function of how much money is
being inserted.
8. Compound interest is a function of initial
investment, interest rate, and time.
9. Area of a square is a function of the square of its
side.
10. Volume of a rectangle is a function of its length,
weight and height.

OBJECTIVES:
are expected to:
1. represent real-life situations using functions;
and
2. solve word problems that apply the concept
of functions.

SOLVING PROBLEMS INVOLVING
FUNCTIONS
1. Mr. Maliga’s pizza costs Pph 50 with the first topping, and then an
additional Php 10 for each additional topping. What function represents
the cost of a pizza with at least one topping? If there are 3 toppings, how
much does the pizza cost? If there are 5 additional toppings, how much
does the pizza cost?
Solution:
Let x represent the number of toppings on a pizza and y
represent the amount of a pizza.
y = 10x + 50
y = 10(2) + 50 if x = 2
= 70
y = 10(5) + 50 if x = 5
= 100
Therefore, y = 10x + 50 represents the cost of a pizza with at least
one topping. If there are three toppings, the pizza costs Php 70. If there
are 5 additional toppings, the pizza costs Php 100.

FUNCTIONS
2. Mrs. Mohaima Ukom is in the business of repairing home computers. She
charges a base fee of Php 1,500 for each visit and Php 350 per hour for her labor.
What function represents the total cost for a home visit and hours of labor? How
much a customer will pay if he has 2 visits and 2 hours of labor per visit?
Solution:
Let x represent the hours of labor and y represent the amount a customer
will pay for 2 visits and 2 hours of labor per visit.
y = 350x + 3,000
y = 350(4) + 3,000
= 1,400 + 3,000
= 4,400
Therefore, y = 350x + 3,000 represents the amount a customer will pay
for home computer. A customer will pay a total of Php 4,400 for 2 visits and 2
hours of labor per visit.

FUNCTIONS
3. A computer shop salesman receives a weekly allowance of Php 3,500 and a 6% commission on
all sales. Write a function to represent the weekly earnings of salesman. Find the salesman’s
earnings for a week if he has Pph 50,000 total sales. What were the salesman’s total sales for a
week in which his earnings were Pph 9,500?
Solution:
Let x represent the salesman’s total sales and y represent his total earnings in a week.
y = 0.06x + 3,500
= 0.06(50,000) + 3,500
= 3,000 + 3,500
= 6,500
9,500 = 0.06x + 3,500
3,500 + 0.06x = 9,500
0.06x = 9,500 – 3,500
0.06x = 6,000
(
1
0.06
)(0.06x = 6,000)
x = 100,000
Therefore, y = 0.06x + 3,500
represents the weekly earnings of a
salesman. The salesman’s earnings for a
week is Pph 6,500. His total sales for a
week were Php 100,000.

ASSIGNMENT 3
Copy and answer the

DIRECTION: Apply the concept of basic functions to solve each
of the following problems. Choose only 5 out of 7 problems.
1. A rental company charges a flat fee of Php 150 and an additional Php 25
per mile to rent a moving van. Write an equation to represent the amount
of a rental fee. How much would a 85-mile trip cost? How many miles of
travel would cost Php 2,650?
2. School soccer team players are selling candles to raise money for an
upcoming field trip. Each player has 24 candles to sell. If a player sells 4
candles, a profit of Php 150 is made. Write an equation to represent the
profit. If each player is able to sell all the 24 candles, how much profit
does he receive? If one player makes a profit of Php 600, how many
candles does he sell? If another player sells 22 candles, how much profit
does he receive?
3. A game rental store charges Php 750 to rent the console and the game.
Php 150 is charged per additional hour. Determine the cost of renting for
15 hours. Determine the hours if the rental fee is Php 7,500.
4. A train leaves from a station and moves at a certain speed. After 2 hours,
another train leaves from the same station and moves in the same station
at a speed of 60 kph. If it catches up with the first train in 4 hours, what is
the speed of the first train?

DIRECTION: Apply the concept of basic functions to solve each
of the following problems. Choose only 5 out of 7 problems.
5. A GLOBE subscriber is charged Pph 300 monthly for a particular
mobile plan, which includes 100 free text messages. Text messages
in excess of 100 are charged Php 1 each. Represent the amount of a
subscriber he pays each month as a function of the number of
messages (m) sent in a month. If a subscriber is able to send 1,000
messages in a month, how much will be his bill in that specific
month?
6. An athlete begins the normal practice for the next marathon during
evening. At 6:00 pm, he starts to run and leaves his home. At 7:30
pm, he finishes the run at home and has run a total of 7.5 miles.
Represent his average speed over the course of run. How many
miles did he run after the first-half hour? If he kept running at the
same pace for a total of 3 hours, how many miles will he have run?
7. Initially, Trains A and B are 325 miles away from each other. Train
A is travelling towards B at 50 miles per hour and Train B is
travelling towards A at 80 miles per hour. At what time will the two
trains meet? At this time, how far did the trains travel?

EVALUATING A
FUNCTION

OBJECTIVES:
are expected to:
1. demonstrate your understanding about
evaluating a function; and
2. evaluate different types of function.

EVALUATING A FUNCTION
Evaluating a Function means replacing the
variable in the function with a given value. In
this lesson, the variable x is given a value from
the function’s domain and computing for the
result.
Function Machine
Input
(x)
Output
y = f(x)
Function Rule

Examples:
A. Evaluate the following functions at x = 1.5.
1. f(x) = 3x – 2
2. f(x) = 3x² – 4x
3. f(x) = 𝑥 + 4
4. f(x) =
2𝑥 + 1
𝑥 −1
5. f(x) = ⌊x⌋ + x
6. f(x) = -2⌊x⌋ – x
7. f(x) = 2x² – 5x + 3
8. f(x) =
3𝑥 − 8
5 − 𝑥
9. f(x) = 2x² – 5x + 3
10. f(x) =
3
2𝑥 − 11

Solutions:
1. f(x) = 3x – 2
f(1.5) = 3(1.5) – 2
= 4.5 – 2
f(1.5) = 2.5
2. f(x) = 3x² – 4x
f(1.5) = 3(1.5)² – 4(1.5)
= 3(2.25) – 6
= 6.75 – 6
f(1.5) = 0.75
3. f(x) = 𝑥 + 4
f(1.5) = 1.5 + 4
= 5.5
f(1.5) = 2.35
4. f(x) =
2𝑥 + 1
𝑥 −1
f(1.5) =
2(1.5) + 1
1.5 −1
=
3 + 1
0.5
=
4
0.5
f(1.5) = 8

Greatest Integer Function
0 1 2 3 4 5 6 7 8
-1
-2
-3
-4
-5
-6
-7
Evaluate y = ⌊x⌋ and graph.
1. ⌊-5.3⌋ = -6
2. ⌊-4.5⌋ = -5
3. ⌊-4⌋ = -4
4. ⌊-3.06⌋ = -4
5. ⌊-1.8⌋ = -2
6. ⌊1.6⌋ = 1
7. ⌊1⌋ = 1
8. ⌊2.8⌋ = 2
9. ⌊3.01⌋ = 3
10. ⌊4.8⌋ = 4
Greatest Integer Function is given by the equation
y = ⌊x⌋ where ⌊x⌋ is the greatest integer less than or equal
to x. It is also known as Floor Function.
2
2
4
4
6
6
-2
-2
-4
-4
-6
-6
y
x
• o
• o
• o
• o
• o
• o
• o
• o
• o
• o
• o
• o

Solutions:
5. f(x) = ⌊x⌋ + x
f(1.5) = ⌊1.5⌋ + 1.5
= 1 + 1.5
f(1.5) = 2.5
6. f(x) = -2⌊x⌋ – x
f(1.5) = -2 ⌊1.5⌋ – 1.5
= -2(1) – 1.5
= -2 – 1.5
f(1.5) = -3.5
0 1 2 3 4 5 6 7 8
-1
-2
-3
-4
-5
-6
-7

Solutions:
7. f(x) = 2x² – 5x + 3
f(1.5) = 2(1.5)² – 5(1.5) + 3
= 2(2.25) – 7.5 + 3
= 4.5 – 7.5 + 3
= 7.5 – 7.5
f(1.5) = 0
8. f(x) =
𝟑𝒙 − 𝟖
𝟓 − 𝒙
f(1.5) =
𝟑(𝟏.𝟓) − 𝟖
𝟓 −𝟏.𝟓
=
𝟑(𝟐.𝟐𝟓)− 𝟖
𝟑.𝟓
=
𝟔.𝟕𝟓 −𝟖
𝟑.𝟓
=
−𝟏.𝟐𝟓
𝟑.𝟓
f(1.5) = 0.36

Solutions:
9. f(x) = 2x² – 5x + 3
f(1.5) = 2(1.5)² – 5(1.5) + 3
= 2(2.25) – 7.5 + 3
= 4.5 – 7.5 + 3
= 0
10. f(x) =
𝟑
𝟐𝒙 − 𝟏𝟏
f(1.5) =
𝟑
𝟐(𝟏. 𝟓) − 𝟏𝟏
=
𝟑
𝟑 − 𝟏𝟏
=
𝟑
−𝟖
= -2

Evaluating a Function
B. Given: f(x) = 3x³ + 2x² + 3x – 3
Find: (1) f(-2)and (2) f(
𝟏
𝟑
)
Solution:
1) f(x) = 3x³ + 2x² + 3x – 3
f(-2) = 3(-2)³ + 2(-2)² + 3(-2) – 3
= 3(-8) + 2(4) – 6 – 3
= -24 + 8 – 9
= -33 + 8
= -25

Solution:
2) f(x) = 3x³ + 2x² + 3x – 3
f(
𝟏
𝟑
) = 3
𝟏
𝟑
³ + 2
𝟏
𝟑
+ 3
𝟏
𝟑
– 3
= 3
𝟏
𝟐𝟕
+ 2
𝟏
𝟗
+ 1 – 3
=
𝟏
𝟗
+
𝟐
𝟗
– 2
=
𝟏 + 𝟐 − 𝟏𝟖
𝟗
=
−𝟏𝟓
𝟗
= −
𝟓
𝟑

C. Given: f(x) =
𝟐𝒙𝟐 − 𝟓𝒙 + 𝟑
𝟑𝒙 −𝟓
Find: (1) f(-3) and 2) f(−
𝟏
𝟐
)
Solution:
1. f(x) =
𝟐𝒙𝟐 −𝟓𝒙 + 𝟑
𝟑𝒙 −𝟓
f(-3) =
𝟐(−𝟑)𝟐 −𝟓 −𝟑 + 𝟑
𝟑 −𝟑 −𝟓
=
𝟏𝟖 + 𝟏𝟓 + 𝟑
−𝟗 −𝟓
=
𝟑𝟔
−𝟏𝟒
= −
𝟏𝟖
𝟕

Solution:
2. f(x) =
𝟐𝒙𝟐 −𝟓𝒙 + 𝟑
𝟑𝒙 −𝟓
f(−
𝟏
𝟐
) =
𝟐(−
𝟏
𝟐
)𝟐 −𝟓 −
𝟏
𝟐
+ 𝟑
𝟑 −
𝟏
𝟐
−𝟓
=
𝟐
𝟏
𝟒
+
𝟓
𝟐
+ 𝟑
𝟑 −
𝟏
𝟐
−𝟓
=
𝟏
𝟐
+
𝟓
𝟐
+ 𝟑
−
𝟑
𝟐
− 𝟓
=
𝟏 + 𝟓 + 𝟔
𝟐
−𝟑 − 𝟏𝟎
𝟐
=
𝟏𝟐
𝟐
−𝟏𝟑
𝟐
= 6
𝟐
−𝟏𝟑
= −
𝟏𝟐
𝟏𝟑

D. Given: f(x) =
𝟑
𝟐𝒙 − 𝟓
Find: 1)f(16) and 2)f(
𝟏𝟐𝟕
𝟓𝟒
)
Solution:
1. f(x) =
𝟑
𝟐𝒙 − 𝟓
f(16) =
𝟑
𝟐(𝟏𝟔) − 𝟓
=
𝟑
𝟑𝟐 − 𝟓
=
𝟑
𝟐𝟕
= 3

Solution:
2. f(x) =
𝟑
𝟐𝒙 − 𝟓
f(
𝟏𝟐𝟕
𝟓𝟒
) =
𝟑
𝟐(
𝟏𝟐𝟕
𝟓𝟒
) − 𝟓
=
𝟑 𝟏𝟐𝟕
𝟐𝟕
− 𝟓
=
𝟑 𝟏𝟐𝟕 −𝟏𝟑𝟓
𝟐𝟕
=
𝟑 −𝟖
𝟐𝟕
= −
𝟐
𝟑
=
𝟑
(−
𝟐
𝟑
)³

ASSIGNMENT 4
Copy and answer the

A. Evaluate the following functions at x = -2.6. Round off your
answers in nearest hundredths if the answers are irrational
numbers.
1. f(x) = 3x³ − 8
2. f(x) =
𝟑
𝟓 − 𝟒𝒙
3. f(x) = -3⌊x⌋ + 2
B. If f(x) =
𝟑𝒙3−𝟖𝒙2+ 𝟓𝒙 + 𝟐
𝟓𝒙+𝟒
, then evaluate:
(1) f(-2), (2) f(3), (3) f(
𝟐
𝟑
) and 4)f(−
𝟐
𝟑
)
C. Given: f(x) =
𝟑
𝟑𝒙 − 𝟒
Find: (1) f(12), (2) f(
𝟔𝟖
𝟑
), (3) f(
𝟏𝟕𝟐
𝟖𝟏
) and (4) f(2x + 3)
4. f(x) =
𝟐𝒙𝟐 −𝟓𝒙 + 𝟐
𝒙 − 𝟓
5. f(x) =
𝟓𝒙 + 𝟖
2⌊x⌋−𝟑
6. f(x) = -3/2x – 5/ + 4

OBJECTIVES:
At the end of this video lesson, you are
expected to:
1. perform the following operations:
a. Addition of Functions;
b. Subtraction of Functions;
c. Multiplication of Functions;
d. Division of Functions; and
2. use the long process of multiplication,
distributive property and synthetic division to
find the product and quotient of complex
polynomial and rational functions.

Operations on Functions
Like numbers, functions can be added,
subtracted, multiplied or divided.
Examples:
A. Given the following functions, perform the
indicated operations:
Given: f(x) = 3x + 5
g(x) = 2x – 3
h(x) = 6x² + x – 15
k(x) =
2𝑥 −3
𝑥 + 4
Find:
1) (f + g)(x)
2) (g – f)(x)
3) (f – g)(x)
4) (f•g)(x)
5) (
𝑓
ℎ
)(x)
6) (
𝑔
𝑘
)(x)
7) (f – h)(-1)

Solutions:
1) (f + g)(x) = (3x + 5) + (2x – 3)
= 5x + 2
2) (g – f)(x) = (2x – 3) – (3x + 5)
= 2x – 3 – 3x – 5
= -x – 8 or -(x + 8)
3) (f – g)(x) = (3x + 5) – (2x – 3)
= 3x + 5 – 2x + 3
= x + 8
4) (f•g)(x) = (3x + 5)(2x – 3)
= 6x² + x – 15
= 6x² – 9x + 10x – 15
FOIL METHOD

Solutions:
5) (
𝑓
ℎ
)(x) =
3x + 5
6x² + x – 15
=
3x + 5
(3x + 5)(2x – 3) =
1
2x – 3
6) (
𝑔
𝑘
)(x) =
2x – 3
2x – 3
x + 4
= (2x – 3)(
x + 4
2x – 3
) = x + 4
7) (f – h)(-1) = 3(-1) + 5 – (6(-1)² + (-1) – 15)
= -3 + 5 – (6 –1 – 15)
= 2 – (–10)
= 2 + 10
(f – h)(-1) = 12

B. If f(x) =
2𝑥 + 1
𝑥 − 2
and h(x) =
3𝑥
𝑥 − 2
, find the
following:
1) (f + h)(x) 3) (f•h)(x) 5) (h – f)(-2)
2) (f – h)(x) 4) (
𝑓
ℎ
)(x) 6) (
ℎ
𝑓
)(x)
Solutions:
1) (f + h)(x) =
𝟓𝒙 + 𝟏
𝒙 − 𝟐
=
𝟏 − 𝒙
𝒙 − 𝟐
=
−𝑥 + 1
𝑥 − 2
3) (f•h)(x)=(
2𝑥 + 1
𝑥 − 2
)(
3𝑥
𝑥 − 2
) =
6𝑥2 + 3𝑥
(𝑥 − 2)
=
𝟔𝒙𝟐 + 𝟑𝒙
𝒙𝟐−𝟒𝒙 + 𝟒

Solutions:
4)(
𝑓
ℎ
)(x)=
2𝑥 + 1
𝑥 − 2
3𝑥
𝑥 − 2
= (
2𝑥 + 1
𝑥 − 2
)(
𝑥 − 2
3𝑥
) =
𝟐𝒙 + 𝟏
𝟑𝒙
5)(h – f)(-2) =
3(−2)
−2 − 2
−
2(−2) + 1
−2 − 2
=
−6
−4
−
−4 + 1
−4
=
3
2
−
−3
−4
=
3
2
−
3
4
=
6 −3
4
(h – f)(-2) =
𝟑
𝟒
6) (
ℎ
𝑓
)(x) =
3𝑥
𝑥 − 2
2𝑥 + 1
𝑥 − 2
= (
3𝑥
𝑥 −2
)(
𝑥 −2
2𝑥+1
) =
𝟑𝒙
𝟐𝒙 + 𝟏

Multiplication and Division of Functions
A.Given: f(x) = 3x – 8
g(x) = 2x² – 5
h(x) = 9x – 18x³ – 31x² + 52x – 32
Find:
1) (f•h)(x) = (3x – 8)(9x – 18x³ – 31x² + 52x – 32)
Solution:
9 -18 -31 52 -32
3 -8
27 -54 -93 156 -96
-72 144 248 -416 256
27 -126 51 404 -512 256
(f•h)(x) = 27x – 126x + 51x³ + 404x² – 512x + 256

A.Given: f(x) = 3x – 8
g(x) = 2x² – 5
h(x) = 9x – 18x³ – 31x² + 52x – 32
Find:
2) (g•h)(x) = (2x² – 5)(9x – 18x³ – 31x² + 52x – 32)
Solution:
9 -18 -31 52 -32
2 0 -5
18 -36 -62 104 -64
0 0 0 0 0
-45 90 155 260 160
18 -36 -107 194 91 260 160
(g•h)(x) = 𝟏𝟖𝒙𝟔
– 36x – 107x + 194x³ + 91x² + 260x + 160

Find:
3) (
ℎ
𝑓
)(x) =
9x – 18x³ – 31x² + 52x – 32
3x – 8
Solution:
3 2 -5 4
3 -8 9 -18 -31 52 -32
9 -24
6 -31
6 -16
-15 52
-15 40
12 -32
12 -32
0
Long Process of Division without Variable
(
ℎ
𝑓
)(x) =
9x – 18x³ – 31x² + 52x – 32
3x – 8
= 3x³ + 2x² – 5x + 4

Find:
4) (
ℎ
𝑓
)(x) =
9x – 18x³ – 31x² + 52x – 32
3x – 8
Solution:
8
3
9 -18 -31 52 -32
24 16 -40 32
9 6 -15 12 0
Synthetic Division
Then, divide the resulting coefficients by 3.
(
ℎ
𝑓
)(x) =
9x – 18x³ – 31x² + 52x – 32
3x – 8
= 3x³ + 2x² – 5x + 4

B. Given: f(x) =
5x + 3
10x² + 7x – 12
g(x) =
5x – 3
5x² – 32x – 21
Find: 1) (f•g)(x) and 2)(
𝑓
𝑔
)(x)
Solution:
1) (f•g)(x) = (
5x + 3
10x² + 7x – 12
)(
5x – 3
5x² – 32x – 21
)
(5x + 3)(5x – 3)= 25x² – 15x + 15x – 9 = 25x² – 9
(10x² + 7x – 12)(5x² – 32x – 21) = ?
10 7 -12
5 -32 -21
50 35 -60
-320 -224 384
-210 -147 252
50 -285 -494 237 252
(f•g)(x) = (
5x + 3
10x² + 7x – 12
)(
5x – 3
5x² – 32x – 21
) =
25x² – 9
𝟓𝟎𝒙𝟒−𝟐𝟖𝟓𝒙𝟑−𝟒𝟗𝟒𝒙𝟐+𝟐𝟑𝟕𝒙+𝟐𝟓𝟐

Solution:
2) (
𝑓
𝑔
)(x) =
5x + 3
10x²+ 7x – 12
5x – 3
5x² – 32x – 21
=
5𝑥+3
(2𝑥+3)(5𝑥−4)
•
(5𝑥+3)(𝑥 −7)
5𝑥−3
=
5𝑥+3
10𝑥2+7𝑥 −12
•
5𝑥2 −32𝑥 −21
5𝑥−3
(5x + 3)(5𝑥2
− 32𝑥 − 21) = 25𝑥3 − 160𝑥2 − 105𝑥 + 15𝑥2 − 96x − 63
= 25𝑥3 − 145𝑥2 − 201x − 63
(5x − 3)(10𝑥2
+ 7𝑥 − 12) = 50𝑥3
+ 35𝑥2
− 60𝑥 − 30𝑥2
− 21x + 36
= 50𝑥3 + 5𝑥2 − 81x + 36
(
𝑓
𝑔
)(x) =
5x + 3
10x²+ 7x – 12
5x – 3
5x² – 32x – 21
=
25𝑥3−145𝑥2−201𝑥−63
50𝑥3+5𝑥2−81𝑥+36

A. Given the following functions, perform the
indicated operations.
Given: f(x) = 2x + 9
g(x) = 3x – 11
h(x) = 6x² – 13x – 33
j(x) = 6x² + 5x – 99
Find:
1) (f + h)(x) 2) (g – j)(x) 3) (f•g)(x)
4) (f – h))(x) 5) (h•j)(x) 6) (
𝑔
ℎ
)(x)
7) (
𝑗
𝑓
)(x) 8) (f•h)(x) 9) (g – h)(-2) 10) (
ℎ
𝑗
)(x)

B. Given: f(x) =
𝑥 + 3
2𝑥2 + 𝑥 −15
and g(x) =
𝑥 + 4
2𝑥 − 5
Find: 1) (f + g)(x) 2) (f – g)(x)
3) (
𝑓
𝑔
)(x) 4) (f•g)(x)
Note: Reduce your answers in lowest terms.
C. If f(x) =
2𝑥 + 5
2𝑥2−3𝑥 −20
and h(x) =
3𝑥+5
6𝑥2+25𝑥+25
, then find each
of the following:
1) (f•h)(x) 2) (
𝑓
ℎ
)(x) 3) (
ℎ
𝑓
)(x)

OBJECTIVES:
At the end of this video lesson, you are
expected to:
composition of functions.
2. perform the composition of functions.

Composition of Functions
Composition of Function is an operation that takes
two functions f and h and produces a function f o h such
that (f o h)(x) = f(h(x)).
Input (x)
h(x)
h
f
f(h(x)
)

A. If f(x) = 2x + 1 and h(x) = 3x, evaluate the following:
1) (f o h)(x) = f(h(x))
= f(3x)
= 2(3x) + 1
(f o h)(x) = 6x + 1
2) (f o h)(1) = f(h(1))
= f(3(1))
= f(3)
= 2(3) + 1
(f o h)(1) = 7
3) (h o f)(1) = h(f(1))
= h(2(1) + 1)
= h(3)
= 3(3)
(h o f)(1) = 9

4) (f o f)(−
2
3
) = f(f(−
2
3
))
= f(2(−
2
3
) + 1)
= f(−
4
3
+ 1)
= f(−
4+3
3
)
= f(−
1
3
)
= 2(−
1
3
) + 1
= −
2
3
+ 1
= −
2+3
3
(f o f)(−
𝟐
𝟑
) =
𝟏
𝟑

B. If f(x) = x² + 2 and g(x) =
2
𝑥
, find the values of the following:
1) (f o g)(x) = f(g(x))
= f(
2
𝑥
)
=(
2
𝑥
)2
+ 2
=
4
𝑥
+ 2
=
4 + 2𝑥
𝑥
(f o g)(x) =
𝟐𝒙𝟐+𝟒
𝒙
2) (g o f)(x – 3) = g(f(x – 3))
= g((x – 3)² + 2)
= g(x² – 12x + 9 + 2)
= g(x² – 12x + 11)
(g o f)(x – 3) =
2
x² – 12x + 11
3) (f o g)(-3) = f(g(-3))
= f(
2
−3
)
= (−
2
3
)² + 2
=
4
9
+ 2
=
4 + 18
9
(f o g)(-3) = 𝟐𝟐
𝟗

C. If f(x) = 2x² – 3x + 5 and g(x) = 3x + 5, find:
1) (f o g)(x) = f(g(x))
= f(3x + 5)
= 2(3x + 5)² – 3(3x + 5) + 5
= 2(9x² + 30x + 25) – 9x – 15 + 5)
= 18x² + 60x + 50 – 9x – 10
(f o g)(x) = 18x² + 51x + 40
2) (g o f)(5) = g(f(5))
= g(2(5)² – 3(5) + 5)
= g(50 – 15 + 5)
= g(40)
= 3(40) + 5
(g o f)(5) = 125

D. If f(x) = 3𝑥 − 5 and h(x) = 5x² − 2, find:
1) (f o h)(-3) = f(h(-3))
= f(5(-3)² − 2))
= f(45 – 2)
= f(43)
= 3(43) − 5
= 124
= 4(31)
(f o h)(-3) = 2 𝟑𝟏
2) (h o f)(x + 3) = h(f(2x + 3))
= h( 3 2𝑥 + 3 − 5)
= h( 6𝑥 + 4)
= 5( 6𝑥 + 4)² − 2
= 5(6x + 4) − 2
(h o f)(x + 3) = 35x + 18

E. If f(x) = 3x³ − 2x² − 8x + 7, g(x) = 3x² − 5x + 6 and h(x) = 5x + 2,
find:
1) (f o h)(−
2
3
) = f(h(−
2
3
))
= f(5(−
2
3
) + 2)
= f(−
10
3
+ 2)
= f(−
10 + 6
3
)
= f(−
4
3
)
= 3(−
4
3
)³ − 2(−
4
3
)² − 8(−
4
3
) + 7
= 3(−
64
27
) − 2(
16
9
) +
32
3
+ 7
= −
64
9
−
32
9
+
32
3
+ 7
=
−64−32 + 96 + 63
9
=
63
9
(f o h)(−
𝟐
𝟑
) = 7

find:
2) (g o h)(3x − 5) = g(h(3x − 5))
= g(5(3x − 5) + 2)
= g(15x − 23)
= 3(15x − 23)² − 5(15x − 23) + 6
= 3(225x² − 690x + 529) − 75x + 115 + 6
= 675x² − 2070x + 1587 − 75x + 121
(g o h)(3x − 5) = 675x² − 2145x + 1708

find:
3) (f o h)(2x + 3) = f(h(2x + 3))
= f(5(2x + 3) + 2)
= f(10x + 17)
= 3(10x + 17)³ − 2(10x + 17)² − 8(10x + 17) + 7
= 3((10x)³ + 3(10x)²(17) + 3(10x)(17)² + (17)³) - 2(100x² + 340x + 289) – 80x – 136 + 7
= 3(1000x³ + 5100x² + 8670x + 4913) – 200x² – 680x – 578 – 80x – 129
= 3000x³ + 15300x² + 26010x + 14739 – 200x² – 760x – 707
= 3000x³ + 15100x² + 25250x 14032
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
.
.
.

F. 1) If g(x) = 2x − 7 and f(g(x)) = 6x − 13,
then find f(x).
Solution:
f(x) = 6(
𝑥
2
) − 13 + 21
= 3x + 8
Checking:
f(g(x)) = f(2x − 7 )
= 3(2x − 7) + 8
= 6x − 21 + 8
= 6x − 13

F. 2) If g(x) = 3x + 4 and f(g(x)) = 21x + 26,
then find f(x).
Solution:
f(x) = 21(
𝑥
3
) + 26 − 28
= 7x − 2
Checking:
f(g(x)) = f(3x + 4)
= 7(3x + 4) − 2
= 21x + 28 − 2
= 6x + 26

F. 3) If g(f(x)) = 12x + 39 and f(x) = 3x + 11,
then find g(x).
Solution:
g(x) = 12(
𝑥
3
) + 39 − 44
= 4x − 5
Checking:
g(f(x)) = 7(3x + 4) − 2
= 21x + 28 − 2
= 6x + 26

Try to answer!
A.If f(x) = 3x − 4 and h(x) = 2x,
evaluate the following:
1) (f o h)(x)
2) (h o f)(x)
3) (f o f)(x)
4) (f o h)(2x − 5)
5) (h o f)(x² − 2x + 3)

Try to answer!
B. If f(x) = x² − 2x + 3, g(x) = x + 4,
h(x) = x³ −3x² + 4x + 1, and
p(x) = 3𝑥² + 7 find the following:
1) (f o g)(x) 5) (h o g)(x)
2) (f o f)(-2) 6) (p o g)(x)
3) (g o f)(x) 7) (p o g)(5)
4) (g o g)(
2
3
) 8) (p o p)(x)

Try to answer!
C. 1) If g(x) = 3x − 2 and
f(g(x)) = 15x − 6, then find f(x).
2) If h(x) = 2x − 5 and
f(h(x)) = 8x − 17, then find f(x).
3) If g(h(x)) = 40x − 5 and
h(x) = 5x − 1, then find g(x).

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