the chinese university of hong kong edd 5161r99 group project chan kwok ping (s98118370) seto fung...
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The Chinese University of Hong Kong
EDD 5161R99EDD 5161R99
Group ProjectGroup Project
Chan Kwok Ping (S98118370)Seto Fung Mei (S98038260)
Form 5 --- lecturing
Learning PrerequisitesLearning Prerequisites::Sketching the graph of the corresponding
quadratic expressions.Method of factorization.Basic knowledge of Rectangular coordinate plane.
Students will be able to solve the quadratic inequalities by graphical method.
Represent the solutions graphically.
Aims and Objectives:Aims and Objectives:
ContentContent History (1) Sign of inequality
(2) Godfrey Harold Hardy
Inequality & Coordinate Plane
Solving Quadratic Inequality by Method of Graph Sketching
Exercise
Method ofGraph
sketching
Solve the quadratic inequality Solve the quadratic inequality xx2 2 – 5– 5x x + 6 > 0 graphically.+ 6 > 0 graphically.
Procedures:
Step (2): we have y = (x – 2)(x – 3) ,i.e. y = 0, when x = 2 or x = 3.
Factorize x2 – 5x + 6,
The corresponding quadratic function is y = x2 – 5x + 6
Sketch the graph of y = x2 – 5x + 6.
Step (1):
Step (3):
Step (4): Find the solution from the graph.
Sketch the graph Sketch the graph y =y = xx2 2 – 5– 5x x + 6 .+ 6 .
x
y
06 5
2 x x y
What is the solution of What is the solution of xx2 2 – 5– 5x x + 6 > + 6 > 0 0 ??
y = (x – 2)(x – 3) , y = 0, when x = 2 or x = 3.
2 3
above the x-axis.so we choose the portion
x
y
0
We need to solve x 2 – 5x + 6 > 0,
The portion of the graph above the x-axis represents y > 0 (i.e. x 2 – 5x + 6 > 0)
The portion of the graph below the x-axis represents y < 0 (i.e. x 2 – 5x + 6 < 0)
2 3
x
y
0
6 52
x x y
When x < 2x < 2,the curve is
above the x-axisi.e., y > 0
x2 – 5x + 6 > 0
When x > 3x > 3,the curve is
above the x-axisi.e., y > 0
x2 – 5x + 6 > 0
2 3
From the sketch, we obtain the solution
3xor2x
Graphical Solution:
0 2 3
Solve the quadratic inequality Solve the quadratic inequality xx2 2 – 5– 5xx + 6 < 0 graphically. + 6 < 0 graphically.
Same method as example 1 !!!Same method as example 1 !!!
x
y
0
6 52
x x yWhen 2 < x < 32 < x < 3,
the curve isbelow the x-axis
i.e., y < 0x2 – 5x + 6 < 0
2 3
From the sketch, we obtain the solution
2 < x < 3
0 2 3
Graphical Solution:
Solve
Exercise 1:
.012 xx
x < –2 or x > 1
Answer:
x
y
0
1 2 x x y
0–2 1
Find the x-intercepts of the Find the x-intercepts of the curve:curve:
(x + 2)(x – 1)=0(x + 2)(x – 1)=0
x = –2 or x = 1x = –2 or x = 1
–2 1
Solve
Exercise 2:
.0122 xx
–3 < x < 4
Answer:
x
y
0
122
x x y
0–3 4
Find the x-intercepts of the curve:Find the x-intercepts of the curve:
xx22 – x – 12 = 0 – x – 12 = 0
(x + 3)(x – 4)=0(x + 3)(x – 4)=0
x = –3 or x = 4x = –3 or x = 4
–3 4
Solve
Exercise 3:
.107
22
xx
–7 < x < 5
Solution:
x
y
0
35 22
x x y
0–7 5
Find the x-intercepts of the Find the x-intercepts of the curve:curve:
(x + 7)(x – 5)=0(x + 7)(x – 5)=0
x = –7 or x = 5x = –7 or x = 5
10
7
22
xx
271022 xx
03522 xx
057 xx–7 5
Solve
Exercise 4:
.3233 xxx
Solution:
x
y
0
35 22
x x y
Find the x-intercepts of the Find the x-intercepts of the curve:curve:
(x + 3)(3x – 2)=0(x + 3)(3x – 2)=0
x = –3 or x = 2/3x = –3 or x = 2/3
3233 xxx
03233 xxx
0233 xx
–3 23
0–3 23
x –3 or x 2/3
y
xoriginO x - axis
x P( x , y )
The horizontal number lineis called the x-axis.
The vertical number line is called the y-axis.
y - axis
The point of intersection of the axes is called the origin O.
y
Any point P on the plane is described by two numbers x and y called coordinates.
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
( 3, 1 )A
( 4, 3 )B( 1, 2 )C
( 2, 5 )D
What are the sign of the x- and y-coordinates of A,B,C and D?
(+,+)
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
(+,+)
E (4, 4 )
(3, 1 )F
(1, 2 )G
(2, 3 )H
What are the sign of the x- and y-coordinates of E,F,G and H?
(,+)
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
(,+) (+,+)
I (4,2 )
(3, 5 )L
(1, 3)K
(2,1)J
What are the sign of the x- and y-coordinates of I,J,K and L?
(,)
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
(,+) (+,+)
(,)
What are the sign of the x- and y-coordinates of M,N,P and Q?
( 3,5 )Q
( 4,3 )
( 1,4 )M
( 2,1 )NP (+,)
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
(,+) (+,+)
(,) (+,)
Shade the part that y>0y>0 (i.e.”++”).
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
(,+) (+,+)
(,) (+,)
Shade the part that y<0y<0 (i.e.””).
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
(,+) (+,+)
(,) (+,)
Shade the part that x>0x>0 (i.e.”++”).
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
(,+) (+,+)
(,) (+,)
Shade the part that x<0x<0 (i.e.””).
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
Shade the part that x<x<11.
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
Shade the part that x<2x<2.
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
Shade the part that x>1x>1.
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
Shade the part that 2<x<12<x<1.
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
Shade the part that 3<x<23<x<2.
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
Shade the part that xx<<2 or x>12 or x>1.
5
4
3
2
1
-4 -3 -2 -1 0 1 2 3 4
-1
-2
-3
-4
x
y
Shade the part that xx<<1 or x>41 or x>4.