1314mathy10W27shadeinequal

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Year 10 wk27 maths

Plotting st line graphs shading

inequalities


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

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5-6

-7

-8

-9

-10

Straight Line Graphs

m gives the gradientof the line and cgives the y intercept.

All straight line graphsare of the form y = mx +c

The gradient is definedas the rise/run

rise = 2run = 1

y intercept = 3

y = 2x + 3


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

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5-6

-7

-8

-9

-10





rise = 3

run = 1

y intercept = -4

y = 3x - 4


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

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5-6

-7

-8

-9

-10





rise = 1run = 2

y intercept = 1

y = x + 1


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

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5-6

-7

-8

-9

-10




The gradient is definedas the rise/run rise = 1

run = 3

y intercept = -5

1y = x -5

3


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

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5-6

-7

-8

-9

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rise = 3

run = 4

y intercept = 2

3y = x +2

4


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rise = 4

run = 5

y intercept = 3

4y =- x +3

5

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

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10





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

y

1

2

3

4

5

6

7

89

10

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

Determine the equations of the following lines

1

2

3

4

5

6


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Objective- To graph inequalities on

the coordinate plane.

Recall

Graph n


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Graph y >3 on the coordinate plane.

x

y


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Graph x -2 on the coordinate plane.

x

y


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Graph y -3x +2 on the coordinate plane.

x

y

Boundary Line

y =-3x +2

m =

-3

b =2

3

1

Test a point not on the line

test (0,0)

0 -3(0) +2

Not true!


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A short review on graphing inequalities.

In order to graph the inequality y > 3

x first graph the equationy = 3x. This line will be the borderline between the points that

make y > 3x and the points that make y < 3 x.

In y = mx + b formwe have y = -x + 3.

In this case we have

a line whose slope is

1 and whose y-

intercept is 3.

4.0 2.0 2.0 4.0

4.0

2.0

2.0

4.0

x

y

Now we have to decide which side

of the line satisfies y > 3 x.


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A short review on graphing inequalities.

All we have to do is to choose one point that is off the line and

test it in the original inequality. If the point satisfies theinequality then we are on the correct side of the line and we

shade that side. If the point does not satisfy the line, we shade

the other side.

The most popularpoint to use in the

shading test is (0, 0).

4.0 2.0 2.0 4.0

4.0

2.0

2.0

4.0

x

y

THE TEST: substitute(0, 0) into y > 3 x

and see if you get a

true statement.

0 > 3 - 0 0 > 3, which is false.


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Since (0, 0) did not satisfy the inequality y > 3x we conclude

that (0, 0) is on the wrong side of the tracks and we shade the other

side. Our conclusion is that every point in the shaded area is part

of the solution set for y > 3

x.

You can reinforce this idea

by testing several points in

the shaded area.

(2, 2) 2 > 32 2 > 1

(0, 3) 3 > 3

0 3 > 3

(4, 1) 1 > 34 1 > -1

Each point that we pick in the shaded

area generates a true statement.

4.0 2.0 2.0 4.0

4.0

2.0

2.0

4.0

x

y


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Hwk wk27 shading inequalities


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1314mathy10W27shadeinequal

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