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Chapter 6: Linear graphs and models

Ex 6A: Drawing straight-line graphs A linear equation connecting y and x is one that results in a straight line when you graph it. Equations such as y = 2x + 1; 3x – 2y = 12; y = x – 8 are examples of linear equations (since the powers on the pronumerals are 1). Hence, when graphed, they will produce a straight line. The Cartesian plane is a set of axes on which graphs can be drawn/plotted/sketched. It is made up of 2 lines:

• The x-axis, which is the horizontal axis ( )↔

• The y-axis, which is the vertical axis ( )b

e.g.

One way to plot a graph is to produce a table of values (as you have previously done in Ex 3B). We can plot the table of values on the Cartesian plane. This will produce a series of points like point A seen above. A ruler should be used to draw a line passing through all of the points.

Important: coordinate points or points on a line must be written in the form (x, y). That is, the x value must be written first, followed by the y value. They must be written in round brackets. The coordinates of point A on the graph above are (2, –3)

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Example 1 Plot the graph of y = 4x – 3 by forming a table of values of y using x = 0, 1, 2, 3, 4. Solution Produce a table of values.

x 0 1 2 3 4

y

Use substitution to find the y values, or use a CAS calculator.

And so the table of values is:

x 0 1 2 3 4

y –3 1 5 9 13

(x, y) (0, –3) (1, 1) (2, 5) (3, 9) (4, 13)

Important: Notice how the y values all increase/decrease by the same number ( 4± for this example). This will always happen with linear equations. A table of values can quickly be generated if you can see the pattern.

Plot the points and use a ruler to draw a line through all points. This is the graph of y = 4x – 3.

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Using CAS: Generating a table of values from a graph

Example Sketch the graph of y = 8 – 2x for [ 3,10] and [ 10,10]x y∈ − ∈ − and generate a table of values

Solution

1. Press /N and Add Graphs 2. Type 8 – 2x into f1(x) = followed by ·

3. To change the viewing Window press b41

4. Change the X/Y Min/Max values to the desired values then press · on OK. 5. To generate a table of values press /T

6. You can scroll up and down to view more values in the table.

7. Note: to remove the table of values press /Z

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Ex 6B: Determining the slope of a straight line A unique feature of a straight line is its steepness or slope. The slope of a straight line is a numerical measure of how steep a line is. The larger the magnitude of the slope, the steeper the line is. e.g. a line with a slope of 3 is steeper than a line with a slope of 1.5 (3 is larger than 1.5) a line with a slope of –2 is steeper than a line with a slope of –0.7 (2 is larger than 0.7) Another name for slope is gradient. The slope of a line can be calculated using the formula:

( )( )→

↑=

run horizontal

rise vetricalslope

Depending on which way a line slopes, the slope of a straight line could be:

• A positive number (think of walking up hill)

• A negative number (think of walking down hill)

• Zero (walking along flat horizontal ground)

• Undefined (trying to walk at a 90o angle – can’t be done!) Graphically these would look like:

Positive slope Negative slope

• Any line moving upwards from left to right will have a positive slope.

• Any line moving downwards from left to right will have a negative slope.

Zero slope Undefined slope

• All horizontal lines have a slope equal to zero.

• All vertical lines have an undefined slope (that is, it cannot be calculated).

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Example 1 Calculate the slope of the following lines. a.

b.

Solution

a.

Line is sloping upwards, so the answer should be a positive number.

22

4

run

riseslope ===

b.

Line is sloping downwards, so the answer should be a negative number.

)5.0or ( 2

1

6

3

run

riseslope −−=−==

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An alternative method to calculate the slope of a straight line. Given information about any two points on a line, it is possible to calculate the slope (gradient). Each point will have a set of coordinates.

e.g. point A has coordinates (x1, y1) and point B has coordinates (x2, y2)

The following formula can be used to calculate the slope of a straight line given any two points.

slope = y2 – y1

x2 – x1

Example 2

Find the slope of the line joining point A(–1, 3) to B(5, –5).

Solution

Let (x1, y1) = (–1, 3) and

(x2, y2) = (5, – 5)

3

4

6

8

15

8

)1(5

35slope

12

12

−=

−=

+

−=

−−

−−=

−

−=∴

xx

yy

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Ex 6C: The intercept-slope form of the equation of a straight line

The intercept-slope form (or general form) of a straight line is given by y = a + bx. where a is the y-intercept on the graph – where the graph cuts the y axis (vertical axis intercept) b is the slope of the graph

That is,

• The y-intercept corresponds to the first (constant) term in the equation.

• The slope is given by the coefficient of x in the equation.

Example 1 Determine the slope and the y-intercept of the lines below.

a. y = –3 + 4x b. xy3

24 −= c. y – 2x = 6 d. 4x + 3y = 9

Solution

a. constant = –3 coefficient of x = 4 y-intercept = –3 slope = 4

b. constant = coefficient of x = y-intercept = slope =

c. Transpose the equation for y.

xy 26 +=∴

constant = coefficient of x = y-intercept = slope =

d. Transpose the equation for y. constant = coefficient of x = y-intercept = slope =

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Sketching straight lines Only two coordinate points are needed in order to draw a straight line. If the equation of a straight line is written in intercept-slope form then one of the points is immediately available: the y-intercept. A second point can be easily calculated by substituting a suitable value of x into the equation.

Example 2 Sketch the graph of y = – 1 + 2x. Solution

Constant term = –1 ∴ y-intercept is –1 (mark this on a graph) Find a second point: Let’s choose x = 3 (you can pick any value except 0)

561321 =+−=×+−=y

Mark the point (3, 5) on the graph. Use a ruler to draw a straight line through the two points.

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Example 3 Sketch the graph of y = –5x

Solution Constant term = 0 (since it is missing) ∴ y-intercept is 0 (mark this on a graph) Find a second point: Let’s choose x = 2 (you can pick any value except 0)

1025 −=×−=y

Mark the point (2, –10) on the graph. Use a ruler to draw a straight line through the two points.

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Ex 6D: Finding the equation of a straight-line graph from its intercept and slope Now that we have learnt how to draw a straight line graph from its equation, we will now discover how to determine the equation from a graph. If a graph shows the y-intercept and another point it is a straightforward procedure to determine the equation of the line.

Procedure: 1. Read the value of the y-intercept on the graph. This gives the value of a 2. Use two points on the graph to find the slope (refer to formulas given in Ex 6B). This gives the value of b. 3. Substitute these two values into the general equation y = a + bx

Example 1 Determine the equation of the straight line graph shown below.

Solution The y-intercept is 5. ∴ a = 5

33

9

run

rise slope −=

−== or 3

3

9

03

54 slope

12

12 −=−

=−

−−=

−

−=

xx

yy 3−=∴b

Recall: the equation of a straight line is y = a + bx

xy 35 −=∴

The equation of the line is y = 5 – 3x.

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Handout: Sketching straight lines using x and y intercepts

As seen earlier, a sketch of a straight line can be made by finding and plotting any two

points. The two special points that are usually found are the x-intercept and the y-intercept.

• The x-intercept is the point where the line cuts the x-axis and occurs when y = 0.

• The y-intercept is the point where the line cuts the y-axis and occurs when x = 0. For example,

To calculate the intercepts, do the following: To find the x-intercept, put y = 0 into the equation and solve for x. To find the y-intercept, put x = 0 into the equation and solve for y.

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Example 1 Sketch a graph of the following, showing the intercepts. a. 2x + 8y = 20 b. y = – 6 + 3x

Solution

a. 2x + 8y = 20 x-intercept: put y = 0

( )0,1010

202

20082

⇒=∴

=

=×+

x

x

x

y-intercept: put x = 0

( )25,0

2

5

8

20

208

20802

⇒==∴

=

=+×

y

y

y

Mark these two points on a graph and use a ruler to draw a straight line through them.

b. y = – 6 + 3x x-intercept: put y = 0

( )0,22

36

360

⇒=∴

=

+−=

x

x

x

y-intercept: put x = 0

( )6,06

06

036

−⇒−=∴

+−=

×+−=

y

y

y

Mark these two points on a graph and use a ruler to draw a straight line through them.

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Graphing Straight lines Using X-Y intercepts – Worksheet

Sketch the following lines showing the x and y intercepts. Q1 y = x + 2 Q2 y = x – 4 Q3 y = 2x – 5 Q4 y = 3 – x Q5 y = 5 – x Q6 x + y = 6 Q7 x – y = 7 Q8 y = –2x

Q9 y = 2

x

Q10 x + 2y + 6 = 0

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Ex 6E: Finding the equation of a straight-line graph using two points on the graph

Unfortunately, not all straight line graphs show the y-intercept. When this happens, we have

to use the two-point method for finding the equation of the line.

The two-point method for finding the equation of a straight line is:

y – y1 = slope ×××× (x – x1)

where 12

12slopexx

yy

−

−= and the two points ( )11, yx and ( )22 , yx lie on the line.

Example 1

Determine the equation of the straight line shown below.

Solution

Let ( ) )4.8,6(, 11 =yx

( ) )2.25,30(, 22 =yx

Find the slope: 7.024

8.16

630

4.82.25slope

12

12 ==−

−=

−

−=

xx

yy

Using the two-point method, we have;

y – y1 = slope ×××× (x – x1)

y – 8.4 = 0.7 × (x – 6) (expand brackets)

y – 8.4 = 0.7 × x + 0.7 × –6

y – 8.4 = 0.7x – 4.2 (add 8.4 to both sides )

∴∴∴∴ y = 0.7x + 4.2

The equation of the line is y = 4.2 + 0.7x

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Example 2

Determine the equation of the line that passes through the points (–2, 4) and (4, –5).

Solution

Let ( ) )4,2(, 11 −=yx

( ) )5,4(, 22 −=yx

Find the slope: 5.12

3

6

9

24

9

)2(4

45slope

12

12 −=−=−

=+

−=

−−

−−=

−

−=

xx

yy

Using the two-point method, we have;

y – y1 = slope ×××× (x – x1)

y – 4 = –1.5 × (x – (–2))

y – 4 = –1.5 × (x + 2) (expand brackets)

y – 4 = –1.5 × x – 1.5 × 2

y – 4 = –1.5x – 3 (add 4 to both sides )

∴∴∴∴ y = –1.5x + 1

The equation of the line is y = 1 – 1.5x

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Ex 6F: Finding the equation of a straight-line graph from two points using a CAS calculator Finding the equation of a straight line, or drawing a straight on a graph is also known as

linear regression. It is a concept heavily used in Statistics. Your CAS calculator has line-

fitting (or linear regression) facilities on it. We will explore 2 different ways to determine the

equation of a straight line.

Example

Determine the equation of the line that passes through the points (–2, 4) and (4, –5)

Method 1 – no graph drawn Method 2 – graph is drawn

1. Start a new document (/N) and Add

Lists & Spreadsheet

2. Enter the coordinate values into lists

named x and y.

3. Complete a linear regression

(b414)

Type the name of the lists into X List

and Y List as shown below and press ·

4. The general equation is y = a + bx.

∴∴∴∴ y = 1 – 1.5x

Complete steps 1 and 2 as per method 1

3. Press /I and Add Data & Statistics

To plot the 2 points:

4. Add the variable x to the horizontal axis

by clicking on the words “Click to add

variable” along the horizontal.

5. Add the variable y to the vertical axis

by clicking on the words “Click to add

variable” along the vertical.

6. Complete the linear regression

(b462) to draw the line and

display the equation of the line.

∴∴∴∴ y = 1 – 1.5x

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Ex 6G: Linear modelling Linear modelling refers to the many situations in real life where two variables can be described by a linear relationship.

Important skills:

1. ( )

( ) 12

12

Run

RiseSlope

xx

yy

−

−=

→

↑=

2. The equation of a line (or a linear model) is y = a + bx or y – y1 = slope ×××× (x – x1)

where a represents: the y-intercept / an initial value / a once off flat fee/cost

b represents: the slope / gradient / rate / avg. speed / the cost per item 3. Use the variables given in the question (instead of y and x)

Interpretation of slope and y-intercept: Once you have a linear regression line, the slope and intercept can give important information

about the variables.

The slope, b, indicates the rate at which the variables are increasing or decreasing.

The y-intercept, a, indicates the approximate value of the y variable when x = 0.

For example,

Consider the regression line: Number of Bacteria = 203 + 226 × Day of experiment

• The y-intercept (203) tells us that there were 203 bacteria present in the beginning.

• The slope (+226) tells us that the number of bacteria are increasing by 226 each day of

the experiment.

Note: if the slope was a negative number we would use the word ‘decreasing’.

Using a generic sentence to interpret the slope and intercept:

To interpret the y-intercept (a), we generally state “The y variable is “a” units when the x

variable is zero units.”

To interpret the slope (b), we generally state “The y variable increases/decreases by “b”

units for every 1 unit increase in the x variable.”

Use the word increases when the slope is positive and decreases when the slope is negative.

The words in bold are replaced with the appropriate variable names or units of measure for

each variable.

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Making predictions: interpolation and extrapolation

The equation of a linear model may be used to make predictions about the variables. When making predictions from the linear model, these predictions could either be reliable or not reliable.

• When making a prediction within the data range (x-values), the prediction would be reliable. This process is known as interpolation.

• When making a prediction outside the data range (x-values), the prediction would generally not be reliable. This process is known as extrapolation.

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Example 1 – General form of an equation y = a + bx Jessica works for a pizza shop and each day she earns a base pay of $50, plus $3 for every pizza she delivers. a. If n is the number of pizzas Jessica delivers in a day and P is her total pay in dollars, write a formula for P in terms of n.

b. Sketch a graph of the above rule for .200 ≤≤ n c. What does the vertical axis intercept represent? d. How much will Jessica earn for delivering 28 pizzas in a day? e. If Jessica earned $98 today, how many pizzas did she deliver? Solution Initial payment (a) Pay per item (b)

a. P = 50 + 3n {use the correct variables P and n} b. we need to create a graph starting at 0 and finishing at 20 along the x (n) axis.

When n = 0, P = 50 (y-intercept) When n = 20, P = 50 + 3 × 20 = 110 Plot the points (0, 50) and (20, 110)

c. The vertical intercept of 50 would represent Jessica’s pay for coming to work each day. i.e. she receives $50 for each day of work (regardless of whether she delivers any pizzas) d. P = 50 + 3 × 28 = $134 Jessica will be paid $134 for delivering 28 pizzas. e. 98 = 50 + 3n 48 = 3n 16 = n To receive $98 in pay, Jessica needs to deliver 16 pizzas.

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Example 2 – General form of an equation y = a + bx Water is pumped into a water storage tank at a constant rate via a hose. The tank holds 6000 litres of water. The amount of water in the tank at various times after the pumping started is given in the table below.

t (minutes) 15 30 65 90 120

V (litres) 925 1300 2175 2800 3550

a Plot the points on the graph below and use a ruler to draw in a straight line that fits the points.

b. Using a CAS calculator, determine the equation of the line in terms of V and t.

c. Use the equation of the line, to determine the volume of water in the tank when the

pumping started.

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d. At what rate is the water pumped into the tank?

e. Use the equation to determine:

i. how much water is in the tank after 180 minutes.

ii. the time taken to fill the tank.

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Example 3 – Line of best fit The height (cm) and weight (kg) of each student in a Physical Education class was recorded and the results are shown in the scatterplot below.

a. By selecting two points that lie on the line, determine the equation of the line in terms of W and H. Give the coefficients to 2 significant figures. b. Use the equation to predict the weight of a student with height:

i. 150 cm ii. 195 cm

c. Complete the following sentence by filling in the blank: When a student’s height increases by 1 cm, their weight increases by ___________ kg.

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Example 4 The scatter plot below shows the time taken to walk to school and the distance from home to school.

A regression line for this data set is

Time taken to walk = 2.5 + 13.5 × distance to school

a. Draw this line on the scatter plot. b. Interpret the slope. ___________________________________________________________________________

___________________________________________________________________________

___________________________________________________________________________

c. i. Use the regression line to predict the time taken to walk to school for a student that lives 2.1 km from school. Give your answer correct to two significant figures.

_____________________________________________________________________

_____________________________________________________________________

ii. Explain whether this prediction is reliable or not.

_____________________________________________________________________

_____________________________________________________________________