straight-line-graphs11111111111111111.ppt

Straight Line Graphs
Sections
1) Horizontal, Vertical and Diagonal Lines
(Exercises)
2) y = mx + c
(Exercises : Naming a Straight Line
Sketching a Straight Line)
3) Plotting a Straight Line - Table Method
(Exercises)
4) Plotting a Straight Line – X = 0, Y = 0 Method
(Exercises)
5) Supporting Exercises
Co-ordinates Negative Numbers Substitution

x
y
1
-5
-4
-3
-2
-1
4
3
2
1
-5 -4 -3 -2 0 2 3 5
4
-1
Naming horizontal and vertical lines
(-4,-2) (0,-2) (-4,-2)
y = -2
(3,4)
(3,1)
(3,-5)
x = 3
(x,y)
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1
-5
-4
-3
-2
-1
4
3
2
1
-5 -4 -3 -2 0 2 3 5
4
-1
Now try these lines
(-4,2) (0,2) (-4,2)
y = 2
(-2,4)
(-2,1)
(-2,-5)
x = -2
(x,y)
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y
x

-5
-4
-3
-2
-1
4
3
2
1
1
-5 -4 -3 -2 0 2 3 5
4
-1
See if you can name lines 1 to 5
(x,y)
1
5 3
4
2 Back to Main Page
y
x
y = 1
x = 1 x = 5
y = -4
x = -4

1
-5
-4
-3
-2
-1
4
3
2
1
-5 -4 -3 -2 0 2 3 5
4
-1
Diagonal Lines
(-4,-3) (0,1) (2,3)
(3,3)
(1,1)
(-3,-3)
y = -x
(x,y)
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(2,-2)
(-1,1)
(-3,3)
y = x
y = x + 1
y
x

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1 2
1
-5
-4
-3
-2
-1
4
3
2
1
-5 -4 -3 -2 0 2 3 5
4
-1
3
4
Now see if you can identify these diagonal lines
x
y
y = x - 1
y = x + 1
y = - x - 2
y = -x + 2

y = mx + c
Every straight line can be written in this form. To do this the
values for m and c must be found.
y = mx + c
c is known as the intercept
m is known as the gradient
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y
x
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
–
7 –
6 –
5 –
4 –
3 –
2 –
1 -
1
-
2
-
3
-
4
-
5
-
6
Find the Value of c
This is the point at
which the line crosses
the y-axis.
Find the Value of m
The gradient means
the rate at which the
line is climbing.
Each time the lines
moves 1 place to the
right, it climbs up by 2
places.
Finding m and c
y = 2x +3
y = mx +c
So c = 3
So m = 2
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y
x
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
–
7 –
6 –
5 –
4 –
3 –
2 –
1 -
1
-
2
-
3
-
4
-
5
-
6
Find the Value of c
This is the point at
which the line crosses
the y-axis.
Find the Value of m
The gradient means
the rate at which the
line is climbing.
Each time the line
moves 1 place to the
right, it moves down
by 1 place.
Finding m and c
y = 2x +3
y = mx +c
So c = 2
So m = -1
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y
x
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
–
7 –
6 –
5 –
4 –
3 –
2 –
1 -
1
-
2
-
3
-
4
-
5
-
6
Line 1
m =
c =
Equation:
Some Lines to Identify
Line 2
m =
c =
Equation:
1
2
y = x + 2
Line 3
m =
c =
Equation:
1
-1
y = x - 1
-2
1
y = -2x + 1
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y
x
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
–
7 –
6 –
5 –
4 –
3 –
2 –
1 -
1
-
2
-
3
-
4
-
5
-
6
Exercise
Back to Main Page
Click for Answers
1
2
3
5
4
1) y = x - 2
2) y = -x + 3
3) y = 2x + 2
4) y = -2x - 1
5) y = -2x - 1
2

Further Exercise
Sketch the following graphs by using y=mx + c
1) y = x + 4
2) y = x - 2
3) y = 2x + 1
4) y = 2x – 3
5) y = 3x – 2
6) y = 1 – x
7) y = 3 – 2x
8) y = 3x
9) y = x + 2
2
10) y = - x + 1
2
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The Table Method
We can use an equation of a line to plot a graph by
substituting values of x into it.
Example
y = 2x + 1
x = 0 y = 2(0) +1 y = 1
x = 1 y = 2(1) +1 y = 3
x = 2 y = 2(2) +1 y = 5
Now you just have to plot the points on to a graph!
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x 0 1 2
y 1 3 5

The Table Method
0 1
-1 4
3
2
-2
-3
-4
-1
-2
-3
-4
1
2
3
4
y = 2x + 1
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x 0 1 2
y 1 3 5

The Table Method
Use the table method to plot the following lines:
1) y = x + 3
2) y = 2x – 3
3) y = 2 – x
4) y = 3 – 2x
Click to reveal plotted lines
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x 0 1 2
y

The Table Method
0 1
-1 4
3
2
-2
-3
-4
-1
-2
-3
-4
1
2
3
4
4
3
1
2 Back to Main Page
Click for further
exercises

Further Exercise
Using the table method, plot the following graphs.
1) y = x + 2
2) y = x – 3
3) y = 2x + 4
4) y = 2x – 3
5) y = 3x + 1
6) y = 3x – 2
7) y = 1 – x
8) y = 1 – 2x
9) y = 2 – 3x
10) y = x + 1
2
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2

This method is used when x and y are on the same side.
Example: x + 2y = 4
The x = 0, y = 0 Method
To draw a straight line we only need 2 points to join
together.
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These points are where x = 0 (anywhere along the y
axis) and y = 0 (anywhere along the x axis).
If we find the 2 points where the graph cuts the
axes then we can plot the line.
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y
x
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
-6 -5 -4 -3 -2 -1-
1
-
2
-
3
-
4
-
5
-
6
This is where the graph
cuts the y – axis (x=0)
This is where the graph
cuts the x – axis (y=0)
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By substituting these values into the equation we
can find the other half of the co-ordinates.
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Example
Question: Draw the graph of 2x + y = 4
Solution
x = 0
2(0) + y = 4
y = 4
1st
Co-ordinate = (0,4)
y = 0
2x + 0 = 4
2x = 4
x = 2
2nd
Co-ordinate = (2,0)
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So the graph will look like this.
y
x
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
–
7 –
6 –
5 –
4 –
3 –
2 –
1 -
1
-
2
-
3
-
4
-
5
-
6
2x + y = 4
Back to Main Page

Exercise
Plot the following graphs using the x=0, y=0 method.
1) x + y = 5
2) x + 2y = 2
3) 2x + 3y = 6
4) x + 3y = 3
Click to reveal plotted lines
Back to Main Page

Answers
y
x
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
–
7 –
6 –
5 –
4 –
3 –
2 –
1 -
1
-
2
-
3
-
4
-
5
-
6
1. 3x + 2y = 6
2. x + 2y = 2
3. 2x + 3y = 6
4. x - 3y = 3
Click for further
exercises
Back to Main Page

Exercise
1) x + y = 4
2) 2x + y = 2
3) x + 2y = 2
4) x + 3y = 6
5) 2x + 5y = 10
6) x – y = 3
7) 2x – y = 2
8) 2x – 3y = 6
9) x + 2y = 1
10) 2x – y = 3
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Using the x = 0, y = 0 method plot the following graphs:

What are the Co-ordinates of these points?
-1
1
-5
-4
-3
-2
5
4
3
2
1
-5 -4 -3 -2 0 2 3 5
4
-1
(x,y)
Back to Main Page

Negative Numbers
(1) 2 + 3 (2) 6 - 5 (3) 3 - 7 (4) -2 + 6
(5) -1 - 2 (6) -4 + 5 (7) -2 - 2 (8) 0 – 4
(9) -3 + 6(10) -4 - 1 (11) 6 - 8 (12) -5 - 2
(13) -8 + 4 (14) -5 - (- 2) (15) 0 - (- 1)
(16) 7 - 12 + 9 (17) -4 - 9 + -2 (18) 14 - (- 2)
(19) -45 + 17 (20) 4 - 5½
Addition and Subtraction
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Negative Numbers
(1) 4 x -3 (2) -7 x -2
(3) -5 x 4 (4) 28 ÷ -7
(5) -21 ÷ -3 (6) -20 ÷ 5
(7) -2 x 3 x 2 (8) -18 ÷ -3 x 2
(9) -2 x -2 x -2 (10) 2.5 x -10
Multiplication and Division
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Substituting Numbers into Formulae
Exercise
Substitute x = 4 into the following formulae:
1) x – 2
2) 2x
3) 3x + 2
4) 1 – x
5) 3 – 2x
6) 4 - 2x
7) x - 3
2
8) 3 - x
2
9) 2x – 6
Click forward to reveal answers
2
8
14
-3
-5
-4
-1
1
2
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Substituting Negative Numbers into Formulae
Exercise
Substitute x = -1 into the following formulae:
1) x – 2
2) 2x
3) 3x + 2
4) 1 – x
5) 3 – 2x
6) 4 - 2x
7) x - 3
2
8) 3 - x
2
9) 2x – 6
Click forward to reveal answers
-3
-2
-1
2
5
6
-3½
3½
-8
Back to Main Page

straight-line-graphs11111111111111111.ppt

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