Educational Communication and
Technology (EDD5161A)
PowerPoint Presentation
Topic: Concept and Graphical Representation of
Functions
Group 21
Member:
Lam King Ho (98003010)
Lau Shu Fat (98003430)
Li Chi Chung (98002690)
Lam Wing Yin (98002850)
Poon Hing Sheng (98249700)
GO
Concept of function:
Trigonometric function
Logarithmic function
Quadratic function
Design Objectives:
Enhance students’ interest in understanding concept of
functions and graphical representation of functions
Provide a time-saving package for student/teacher in drawing
graphs of different functions
Enable a quick understanding of graphical variations upon
changes of the coefficients
Applicable to varies level of students ( Form 3 to 5)
Guidelines to use:
Designed for students to use before or after the class but
NOT in the class
Designed as assisting program besides the classroom
teaching
Target Student :
Age: Form two to Form four
Ability: Band 3
Attitude: self-motivated, eager to learn
Authors’ part of design:
Concept of function ( Lam king ho & Poon hing sheng )
Trigonometric function ( Li chi chung & Lau shu fat )
Logarithmic function ( Lam wing yin & Lam king ho )
Quadratic function ( Lau shu fat & Lam king ho )
Function: f(x)=x+5
Function: f(x)=x+5
x
Function: f(x)=x+5
x
Value
Function: f(x)=x+5
x= 1
Value= 6
Function: f(x)=x+5
x= -3
Value= 2
Function: f(x)=x+5
x= 4/9
Value= 49/9
Let f(x) = x + 5
when x = 1 ,
f(1) = 1 + 5
=6
Let f(x) = x + 5
when x = -3,
f(-3) = -3 + 5
=2
Let f(x) = x + 5
find f(4/9).
y=
2
ax
+ bx + c
y
x
Use this
to plot the following graphs
Then comparing their shapes
y = x2
1.
2.
y = -x2
y = x2 - x - 6
y = -x2 - x - 6
y = 2x2 - x + 5
3.
y = -2x2 - x + 5
y
Eg.1
y = ax2 + bx + c
For a > 0
Curve open
upwards
x
y
Eg.2
y = ax2 + bx + c
For a > 0
Curve open
upwards
x
y
y = ax2 + bx + c
For a > 0
Curve open
upwards
x
y = ax2 + bx + c
y
For a < 0
Curve open
downwards
x
Eg.1
y = ax2 + bx + c
y
For a < 0
Curve open
downwards
x
Eg.2
y = ax2 + bx + c
y
For a < 0
Curve open
downwards
x
y
y = ax2 + bx + c
For c > 0
y-intercept is
positive
x
Eg.1
y
(0, c)
Eg.2
y = ax2 + bx + c
For c > 0
y-intercept is
positive
x
y
y = ax2 + bx + c
(0, c)
(0, c)
For c > 0
y-intercept is
positive
x
y
x
y = ax2 + bx + c
Eg.1
For c < 0
y-intercept is
negative
y
Eg.2
x
(0, c)
y = ax2 + bx + c
For c < 0
y-intercept is
negative
y
(0, c)
x
(0, c)
y = ax2 + bx + c
For c < 0
y-intercept is
negative
y
y = ax2 + bx + c
Eg.1
For c = 0
x
y
y = ax2 + bx + c
For c = 0
x
Eg.2
y
y = ax2 + bx + c
For c = 0
x
Click the corrected answer
y
a > 0,c > 0
a < 0,c > 0
a < 0,c < 0
a > 0,c < 0
x
y = ax2 + bx + c
y = sinx, y = cosx, y = tanx
y
x
Use this
to plot the following graphs
Then comparing their shapes
1. sinx
2. cosx
3. tanx
y
Eg.1
1. y = sin(x)
2. y = sin(2x)
x
y
Eg.2
1. y = sin(x)
2. y = sin(2x)
x
y = sin(x)
y
1. y = sin(x)
2. y = sin(2x)
y = sin(2x)
x
y = sin(x)
y
Eg.1
1. y = cos(x)
2. y = cos(2x)
x
y
Eg.2
1. y = cos(x)
2. y = cos(2x)
x
y = cos(x)
y
y = sin(2x)
1. y = cos(x)
2. y = cos(2x)
x
y = cos(x)
Eg.1
y
1. y = tan(x)
2. y = tan(2x)
x
Eg.2
y
1. y = tan(x)
2. y = tan(2x)
x
y = tan(x)
y
1. y = tan(x)
2. y = tan(2x)
x
y = tan(x)
y = tan(2x)
Match the corrected order:
Sin ,Cos, Tan
y
A,B,C
B,C,A
C,B,A
C,A,B
A
B
x
C
Meaning of Common
Logarithm
Graph of y=10x
y
10
9
8
7
6
5
4
3
2
1
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
x
1.0
Using the graph of log y = x to find
the value of the following:
Graph of y=10x
y
10
a) log 2 = ( 0.3)
9
8
7
b) log 5 = (0.7 )
6
5
4
3
c) log 10 = ( 1 )
2
1
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
x
1.0
For y = 10x
When x = 0.85, value of y is :
a:
b:
c:
d:
Graph of y=10x
y
y
y
y
=
=
=
=
y
10
9
8
7
6
5
4
3
2
1
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
x
1.0
6
7
8
9