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The Chinese University of Hong Kong EDD 5161R99 Group Project Chan Kwok Ping (S98118370) Seto Fung Mei (S98038260)
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Form 5 --- lecturing
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Learning 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:
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Content /History (1) Sign of inequality Sign of inequality (2) Godfrey Harold Hardy Godfrey Harold Hardy /Inequality & Coordinate PlaneInequality & Coordinate Plane /Solving Quadratic Inequality by Method of Graph SketchingSolving Quadratic Inequality by Method of Graph Sketching /ExerciseExercise
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Method of Graph sketching
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Solve the quadratic inequality x 2 – 5x + 6 > 0 graphically.
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Procedures: Step (2): we have y = (x – 2)(x – 3), i.e. y = 0, when x = 2 or x = 3. Factorize x 2 – 5x + 6, The corresponding quadratic function is y = x 2 – 5x + 6 Sketch the graph of y = x 2 – 5x + 6. Step (1): Step (3): Step (4): Find the solution from the graph.
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Sketch the graph y = x 2 – 5x + 6. x y 0 What is the solution of x 2 – 5x + 6 > 0 ? y = (x – 2)(x – 3), y = 0, when x = 2 or x = 3. 23
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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) 23
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x y 0 x < 2 When x < 2, the curve is above the x-axis i.e., y > 0 x 2 – 5x + 6 > 0 x > 3 When x > 3, the curve is above the x-axis i.e., y > 0 x 2 – 5x + 6 > 0 23
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From the sketch, we obtain the solution or
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Graphical Solution: 0 2 3
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Solve the quadratic inequality x 2 – 5x + 6 < 0 graphically. Same method as example 1 !!!
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x y 0 2 < x < 3 When 2 < x < 3, the curve is below the x-axis i.e., y < 0 x 2 – 5x + 6 < 0 23
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From the sketch, we obtain the solution 2 < x < 3
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0 2 3 Graphical Solution:
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Solve Exercise 1: x 1 Answer: x y 0 0–21 Find the x-intercepts of the curve: (x + 2)(x – 1)=0 x = –2 or x = 1 –2 1
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Solve Exercise 2: –3 < x < 4 Answer: x y 0 0–34 Find the x-intercepts of the curve: x 2 – x – 12 = 0 (x + 3)(x – 4)=0 x = –3 or x = 4 –3 4
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Solve Exercise 3: –7 < x < 5 Solution: x y 0 0–75 Find the x-intercepts of the curve: (x + 7)(x – 5)=0 x = –7 or x = 5 –7 5
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Solve Exercise 4: Solution: x y 0 Find the x-intercepts of the curve: (x + 3)(3x – 2)=0 x = –3 or x = 2/3 –3 2323 0 2323 x –3 or x 2/3
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y x origin O x - axis x P ( x, y ) The horizontal number line is 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.
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( 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? (+,+)
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5 4 3 2 1 -4-3-201234 -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? ( ,+)
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5 4 3 2 1 -4-3-201234 -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? (,)(,)
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5 4 3 2 1 -4-3-201234 -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 ) N P (+, )
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( ,+) (+,+) (,)(,) (+, ) y>0+ Shade the part that y>0 (i.e.”+”).
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5 4 3 2 1 -4-3-201234 -2 -3 -4 x y ( ,+) (+,+) (,)(,) (+, ) y<0 Shade the part that y<0 (i.e.” ”).
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5 4 3 2 1 -4-3-201234 -2 -3 -4 x y ( ,+) (+,+) (,)(,) (+, ) x>0+ Shade the part that x>0 (i.e.”+”).
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5 4 3 2 1 -4-3-201234 -2 -3 -4 x y ( ,+) (+,+) (,)(,) (+, ) x<0 Shade the part that x<0 (i.e.” ”).
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x< 1 Shade the part that x< 1.
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x<2 Shade the part that x<2.
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x>1 Shade the part that x>1.
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2<x<1 Shade the part that 2<x<1.
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3<x<2 Shade the part that 3<x<2.
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x 1 Shade the part that x 1.
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x 4 Shade the part that x 4.
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