1. Linear graphs

2. Graphs parallel to the y-axis
What do these coordinate pairs have in common:
(2, 3), (2, 1), (2, –2), (2, 4), (2, 0) and (2, –3)?
The x-coordinate in each pair is equal to 2.
Look what happens when these points are plotted on a graph. x y
All of the points lie on a straight line parallel to the y-axis.
Teacher notes
Explain that as long as the x-coordinate is 2, the y-coordinate can be any number: positive, negative or decimal.
Encourage students to be imaginative in their choice of points that lie on this line. For example, (2, ) (2, 43/78) or (2, – ).
Mathematical Practices
7) Look for and make use of structure.
Students should be able to see that all of the coordinate pairs share the x-coordinate 2 and realize that when plotted they make a straight line parallel to the y-axis. They should then be able to identify this line as x = 2.
Name five other points that will lie on this line.
x = 2
This line is called x = 2.

3. Graphs parallel to the y-axis
All graphs of the form
x = c,
where c is any real number, will be parallel to the y-axis and will intersect the x-axis at the point (c, 0). y x
Teacher notes
Stress that the graph of x = “a constant number” will always be parallel to the y-axis. In other words, it will always be vertical.
For each graph shown in the example, ask students to tell you the coordinate of the point where the line intersects the x-axis.
Ask students to tell you the equation of the line that coincides with the y-axis (x = 0).
x = –10
x = –3
x = 4
x = 9

4. Graphs parallel to the x-axis
What do these coordinate pairs have in common?
(0, 1), (4, 1), (–2, 1), (2, 1), (1, 1) and (–3, 1)?
The y-coordinate in each pair is equal to 1.
Look at what happens when these points are plotted on a graph. x y
All of the points lie on a straight line parallel to the x-axis.
y = 1
Teacher notes
Explain that as long as the y-coordinate is 1, the x-coordinate can be any number: positive, negative or decimal.
Encourage students to be imaginative in their choice of points that lie on this line. For example, ( , 1) (56/87, 1) or (– , 1).
Mathematical Practices
7) Look for and make use of structure.
Students should be able to see that all of the coordinate pairs share the y-coordinate 1 and realize that when plotted they make a straight line parallel to the x-axis. They should then be able to identify this line as y = 1.
Name five other points that will lie on this line.
This line is called y = 1.

5. Graphs parallel to the x-axis
All graphs of the form y = c,
where c is any real number, will be parallel to the x-axis and will intersect the y-axis at the point (0, c). x y
y = 5
y = 3
Teacher notes
Stress that the graph of y = “a constant number” will always be parallel to the x-axis. In other words, it will always be horizontal.
For each graph shown in the example, ask students to tell you the coordinate of the point where the line intersects the y-axis.
Ask students to tell you the equation of the line that coincides with the x-axis (y = 0).
y = –2
y = –5

6. Drawing graphs of linear functions
The x-coordinate and the y-coordinate in a coordinate pair can be linked by a function.
What do these coordinate pairs have in common?
(1, –1), (4, 2), (–2, –4), (0, –2), (–1, –3) and (3.5, 1.5)?
In each pair, the y-coordinate is 2 less than the x-coordinate.
These coordinates are linked by the function:
Teacher notes
Ask students if they can visualize the shape that the graph will have. This might be easier if they consider the points (0, 2), (1, 3), (2, 4) (3, 5), etc. Establish that the points will lie on a straight diagonal line. Stress that the graphs of all linear functions are straight lines. A function is linear if the variables are not raised to any power (other than 1).
Ask students to suggest the coordinates of any other points that will lie on this line. Look for imaginative answers.
Mathematical Practices
7) Look for and make use of structure.
Students should be able to see that for all of the coordinate pairs, the y-coordinate is 2 less than the x-coordinate. They should realize that when plotted they make a straight diagonal line, and be able to identify this line as y = x – 2.
y = x – 2
We can draw a graph of the function y = x – 2 by plotting points that obey this function.

7. Drawing graphs of linear functions
Given a function, we can find coordinate points that obey the function by constructing a table of values.
Suppose we want to plot points that obey the function
y = 2x + 5
We can use a table as follows: x
y = 2x + 5 –3 –2 –1 1 2 3
Teacher notes
Explain that when we construct a table of values, the value of y depends on the value of x. That means that we choose the values for x and substitute them into the equation to get the corresponding value for y.
The minimum number of points needed to draw a straight line is two. However, it is best to plot several extra points to ensure that no mistakes have been made.
The points given by the table can then be plotted to give the graph of the required function. –1 1 3 5 7 9 11
(–3, –1)
(–2, 1)
(–1, 3)
(0, 5)
(1, 7)
(2, 9)
(3, 11)

8. Drawing graphs of linear functions
For example, y
to draw a graph of y = 2x + 5:
1) Complete a table of values:
y = 2x + 5 x
y = 2x + 5 –3 –2 –1 1 2 3 5 7 9 11
2) Plot the points on a coordinate grid. x
3) Draw a line through the points.
Teacher notes
This slide summarizes the steps required to draw a graph using a table of values.
4) Label the line.
5) Check that other points on the line fit the rule.

9. Slopes of straight-line graphs
The slope of a line is a measure of how steep the line is.
The slope of a line can be positive, negative or zero.
an upwards slope
a horizontal line
a downwards slope y x y x y x
positive slope
zero slope
negative slope
If a line is vertical, its slope cannot be specified.
We often say the slope is “undefined”.

10. Calculating the slope
If we are given any two points (x1, y1) and (x2, y2) on a line we can calculate the slope of the line as follows: y x
(x2, y2)
change in y
change in x
rise
slope = =
run
y2 – y1
(x1, y1)
To help you find this more easily, you can draw a right triangle between the two points on the line.
x2 – x1
Teacher notes
Explain how drawing a right triangle on the line helps us calculate its slope. Explain too that since, for a straight line, the change in y is proportional to the corresponding change in x, the slope will be the same no matter which two points we choose on a line.
The formula for the slope is often also given as: slope = Δx/Δy. Explain that the delta, Δ, means “change in”.
y2 – y1
slope =
x2 – x1

11. The general equation of a straight line
The general equation of any straight line is:
This is called the slope-intercept form of a straight line equation.
y = mx + b
The value of m tells us the slope of the line.
The value of b tells us where the line crosses the y-axis.
This is called the y-intercept and it has the coordinates (0, b).
Teacher notes
Explain that the equation of a line can always be arranged into slope-intercept form: y = mx + b.
It is often useful to have the equation of a line in this form because it tells us the slope of the line and where it intersects the y-axis. These two facts alone can enable us to draw the line without having to set up a table of values.
Ask students what they can deduce about two graphs that have the same value for m. Establish that if they have the same value for m, they will have the same slope and will therefore be parallel.
For example, the line y = 3x + 4 has a slope of 3 and crosses the y-axis at the point (0, 4).

12. Rearranging to y = mx + b
Sometimes the equation of a straight line graph is not given in slope-intercept form, y = mx + b.
The equation of a straight line is 2y + x = 4. Find the slope and the y-intercept of the line.
Rearrange the equation by performing the same operations on both sides.
2y + x = 4
subtract x from both sides:
Teacher notes
Explain that if the equation of a line is linear (that is if x and y are not raised to any power except 1), it can be arranged to be in the form y = mx + b. Explain that the equation of a (straight) line can always be arranged to be in the form y = mx + b. (This is not true for lines parallel to the y-axis.)
It is often useful to have the equation of a line in this form because it tells us the slope of the line and where it cuts the y-axis. These two facts alone can enable us to draw the line without having to draw up a table of values.
Mathematical Practices
7) Look for and make use of structure.
Students should be able to look at the equation of a straight line in any form and, by rearranging it into the form y = mx + b, identify the slope and y-intercept.
2y = –x + 4
y =
–x + 4 2
divide both sides by 2:
y = – x + 2 1 2

13. Rearranging to y = mx + b
Once the equation is in slope-intercept form, y = mx + b, we can determine the value of the slope and the y-intercept.
y = – x + 2 1 2
m = so the slope of the line is 1 2 – 1 2 –
b = 2 and so the y-intercept is
(0, 2). x y
Mathematical Practices
7) Look for and make use of structure.
Students should be able to look at the equation of a straight line in any form and, by rearranging it into the form y = mx + b, identify the slope and y-intercept. 1
y = – x + 2 2

14. Substituting values into equations
A line with the equation y = mx + 5 passes through the point (3, 11). What is the value of m?
To solve this problem, we can substitute x = 3 and y = 11 into the equation y = mx + 5.
This gives us: = 3m + 5
subtract 5 from both sides:
6 = 3m
Teacher notes
Discuss ways to solve the problem. Some students may suggest plotting the point (3, 11) and drawing a straight line through this and the point (0, 5). The slope of the resulting line will give the value for m.
Ask students if they can suggest a method that does not involve drawing a graph.
Establish that if the line passes through the point (3, 11) then we can substitute x = 3 and y = 11 into the equation y = mx + 5. Reveal the equation 11 = 3m + 5 on the board and talk through the steps leading to the solution of this equation.
Mathematical Practices
1) Make sense of problems and persevere in solving them.
Students should be able to see that since they have been given the coordinates of a point that lies on the line, they have an x- and y-value that satisfy the equation of the line. This leaves one variable, m, which can be found using inverse operations.
7) Look for and make use of structure.
Students should be able to look at the equation of a straight line in any form and, by rearranging it into the form y = mx + b, identify the slope and y-intercept.
divide both sides by 3:
2 = m
m = 2
The equation of this line is therefore y = 2x + 5.

15. Parallel lines If two lines have the same slope, they are parallel.
Show that the lines 2y + 6x = 1 and y = –3x + 4 are parallel.
We can show this by rearranging the first equation so that it is in slope-intercept form, y = mx + b.
2y + 6x = 1
subtract 6x from both sides:
2y = –6x + 1
Mathematical Practices
7) Look for and make use of structure.
Students should realize that the slope of a line can be found by rearranging its equation into the form y = mx + b, and therefore see that parallel lines can be identified in this form. They should be able to rearrange the first given equation into this form and see that both lines have m = –3, concluding that they are parallel.
y =
–6x + 1 2
divide both sides by 2:
y = –3x + ½
The slope m is –3 for both lines, so they are parallel.

16. Perpendicular lines
If the slopes of two lines have a product of –1, then they are perpendicular.
if the slope of a line is m, then the slope of the line perpendicular to it is –
In general, 1 m
Write down the equation of the line that is perpendicular to y = –4x + 3 and passes through the point (0, –5).
The slope of the line y = –4x + 3 is
Mathematical Practices
7) Look for and make use of structure.
Students should realize that the slope of the given line is the value of m which is –4, and therefore a perpendicular line will have a slope of ¼. They should see that to complete the equation of this line in the form y = mx + b, the value of b, the y-intercept, is needed. Students should be aware that the y-intercept of a line is (0, b) and so realize that the given point on the line is in fact the y-intercept and b = –5.
–4. 1 4
The slope of the line perpendicular to it is therefore . 1 4
The equation of the line with slope and y-intercept –5 is:
y = x – 5 1 4

17. Investigate the cost of these music download services:
Music Download Prices
Investigate the cost of these music download services:
Yes Trax $9.99 per month $0.49 per song
Z-Tunes: No monthly charge.
$0.99 per song
hi-notes: $12.95 per month, $0.79 per song
MP3 House: $19.99 per month, no charge per song
Teacher notes
This exercise could be carried out in small groups. Students could be asked to display their graph and table on a poster.
The following slide shows the equations and graphs for each music downloads service. Use it to complete the questions with the entire class.
Z-Tunes: average $0.99 per song, no limit or subscription fee (y = 0.99x)
Yes Trax: $9.99 per month + $0.49 per song (y = 0.49x )
hi-notes: $12.95 per month + average $0.79 per song (y = 0.79x )
MP3 House: $19.99 per month, no charge per song (y = 19.99)
None of these music download services are real.
Mathematical Practices
2) Reason abstractly and quantitatively.
Students should be able to translate the given costs of each music download service into a function in terms of songs purchased and monthly cost.
4) Model with mathematics.
Students should be able to write a function to describe the cost of each music download service, given the cost per song and charge per month.
Write the monthly cost of each service as a function of the number of songs downloaded. Compare the services with a graph and a table. 17

18. Customs trap
A customs officer suspects that two boats will meet to transfer smuggled goods.
Using radar he writes down their coordinates at 5pm and 6pm.
The first boat is at (–32.64, –29.82) at 5pm, and at (–10.05, –9.64) at 6pm.
The second boat is at (105.64, 30.98) at 5pm and (77.18, 31.06) at 6pm.
Teacher notes
Use calculator values whenever possible.
To find the equation for the first boat’s graph, find the slope:
m = rise/run = (–9.64 – (–29.82)) / (–10.05 – (–32.64)) = 20.18/22.59 = 0.89 (nearest hundredth) Now substitute this and a known coordinate, (–32.64, –29.82), into y = mx + b to find b:
–29.82 = (0.89 × –32.64) + b
b = –0.77 (nearest hundredth)
So: y = 0.89x – 0.77
To find the equation for the second boat’s graph, find the slope:
m = rise/run = (31.06 – 30.98) / (77.18 – ) = 0.08/(–28.46) = –0.00 (2 d.p)
Now substitute this and a known coordinate, (105.64, 30.98), into y = mx + b to find b:
30.98 = (–0.00 × ) + b (Remember, use the exact value for m for an accurate result, not the rounded value of –0.00)
b = (nearest hundredth)
So: y = 31.28
To find where the boats meet, we need to find their intersection, so put the equations equal to each other:
31.28 = 0.89x – 0.77
x = 36.01
Their paths should cross at approximately (35.89, 31.28).
The graph is shown on the next slide, as well as the calculations needed to find the meeting time.
Mathematical Practices
1) Make sense of problems and persevere in solving them.
Students should think about how they can use linear graphs to find where and when the boats meet.
4) Model with mathematics.
Students should model the paths of the boats using straight lines.
Photo credit: © Alexandre Boavida, Shutterstock.com 2012
Sketch their paths on a graph, and write equations for these lines.
Where should the officer expect the boats to meet?

19. Customs trap graph The equations for the two boats’ paths are:
First boat: y = 0.89x – 0.77
Second boat: y = 31.28
Here is a graph of their paths. y
(35.89, 31.28) 60 50
Their meeting occurs at the intersection of the lines. 40
y = 31.28 30 20 10
y = 0.89x – 0.77
At approximately what time should the officer expect the boats to meet?
–30
–20
Teacher notes
To find the time at which the boats will meet, calculate how much time is represented by how many units of x.
The known x-coordinates for the first boat are:
5pm: x = –32.64
6pm: x = –10.05
meeting: x = 35.89
So, 1 hour on the x-axis is: –10.05 – (–32.64) = units
The meeting was – (–10.05) = units after 6pm.
In terms of the number of hours, this is: / = 2.03 hours after 6pm, i.e. 8pm.
Do the same for the second boat:
6pm: x = 77.18
Meeting: x = 35.89
The meeting was – = units after 6pm.
For the second boat, units = / = 1.82 hours. i.e. around 7.45pm.
The smugglers probably organised to meet at about 8pm.
–10 10 20 30 40 50 60 70 80 90
100
–10 x
–20
–30
–40