© boardworks ltd 2005 1 of 49 a8 linear and real-life graphs 3 eso mathematics

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© Boardworks Ltd 2005 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

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Page 1: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 1 of 49

A8 Linear and real-life graphs

3 ESO Mathematics

Page 2: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 2 of 49

A8.1 Linear graphs

Contents

A8 Linear and real-life graphs

A8.2 Gradients and intercepts

A8.3 Parallel and perpendicular lines

A8.4 Interpreting real-life graphs

A8.5 Distance-time graphs

A8.6 Speed-time graphs

Page 3: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 3 of 49

Coordinate pairs

When we write a coordinate, for example,

Together, the x-coordinate and the y-coordinate are called a coordinate pair.

the first number is called the x-coordinate and the second number is called the y-coordinate.

(3, 5)

x-coordinate

(3, 5)

y-coordinate

(3, 5)

the first number is the x-coordinate and the second number is the y-coordinate.

Page 4: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 4 of 49

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.

Name five other points that will lie on this line.

This line is called x = 2.x = 2

Page 5: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 5 of 49

Graphs parallel to the y-axis

All graphs of the form x = c,

where c is any number, will be parallel to the y-axis and will cut the x-axis at the point (c, 0).

All graphs of the form x = c,

where c is any number, will be parallel to the y-axis and will cut the x-axis at the point (c, 0).

x

y

x = –3x = –10 x = 4 x = 9

Page 6: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 6 of 49

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 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.

This line is called y = 1.

y = 1Name five other points that will lie on this line.

Page 7: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 7 of 49

Graphs parallel to the x-axis

All graphs of the form y = c,

where c is any number, will be parallel to the x-axis and will cut the y-axis at the point (0, c).

All graphs of the form y = c,

where c is any number, will be parallel to the x-axis and will cut the y-axis at the point (0, c).

x

y

y = –2

y = 5

y = –5

y = 3

Page 8: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 8 of 49

Plotting 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:

y = x – 2

We can draw a graph of the function y = x – 2 by plotting points that obey this function.

Page 9: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 9 of 49

Plotting 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 0 1 2 3

–1

(–3, –1)

1 3 5 7 9 11

(–2, 1) (–1, 3) (0, 5) (1, 7) (2, 9) (3, 11)

Page 10: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 10 of 49

Plotting graphs of linear functions

to draw a graph of y = 2x + 5:

1) Complete a table of values:

2) Plot the points on a coordinate grid.

3) Draw a line through the points.

4) Label the line.

5) Check that other points on the line fit the rule.

xy = 2x + 5

–3 –2 –1 0 1 2 3

For example,

y = 2x + 5

y

x

–1 1 3 5 7 9 11

Page 11: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 11 of 49

Plotting graphs of linear functions

Page 12: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 12 of 49

A8.2 Gradients and intercepts

Contents

A8.3 Parallel and perpendicular lines

A8 Linear and real-life graphs

A8.1 Linear graphs

A8.4 Interpreting real-life graphs

A8.5 Distance-time graphs

A8.6 Speed-time graphs

Page 13: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 13 of 49

Gradients of straight-line graphs

The gradient of a line is a measure of how steep the line is.

y

x

a horizontal line

Zero gradient

y

x

a downwards slope

Negative gradient

y

x

an upwards slope

Positive gradient

The gradient of a line can be positive, negative or zero if, moving from left to right, we have

If a line is vertical its gradient cannot by specified.

Page 14: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 14 of 49

Calculating gradients

Page 15: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 15 of 49

Finding the gradient from two given points

If we are given any two points (x1, y1) and (x2, y2) on a line we can calculate the gradient of the line as follows,

the gradient =change in ychange in x

the gradient =y2 – y1

x2 – x1

x

y

x2 – x1

(x1, y1)

(x2, y2)

y2 – y1

Draw a right-angled triangle between the two points on the line as follows,

Page 16: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 16 of 49

Investigating linear graphs

Page 17: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 17 of 49

The general equation of a straight line

The general equation of a straight line can be written as:

y = mx + c

The value of m tells us the gradient of the line.

The value of c tells us where the line crosses the y-axis.

This is called the y-intercept and it has the coordinate (0, c).

For example, the line y = 3x + 4 has a gradient of 3 and crosses the y-axis at the point (0, 4).

Page 18: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 18 of 49

The gradient and the y-intercept

Complete this table:

equation gradient y-intercept

y = 3x + 4

y = – 5

y = 2 – 3x

1

–2

3 (0, 4)

(0, –5)

–3 (0, 2)

y = x

y = –2x – 7

x2

12

(0, 0)

(0, –7)

Page 19: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 19 of 49

Rearranging equations into the form y = mx + c

Sometimes the equation of a straight line graph is not given in the form y = mx + c.

The equation of a straight line is 2y + x = 4.Find the gradient and the y-intercept of the line.

Rearrange the equation by performing the same operations on both sides,

2y + x = 4

y = – x + 212

2y = –x + 4subtract x from both sides:

y =–x + 4

2divide both sides by 2:

Page 20: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 20 of 49

Rearranging equations into the form y = mx + c

Sometimes the equation of a straight line graph is not given in the form y = mx + c.

The equation of a straight line is 2y + x = 4.Find the gradient and the y-intercept of the line.

Once the equation is in the form y = mx + c we can determine the value of the gradient and the y-intercept.

So the gradient of the line is 12

– and the y-intercept is 2.

y = – x + 212

Page 21: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 21 of 49

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, 11 = 3m + 5

6 = 3msubtract 5 from both sides:

2 = mdivide both sides by 3:

m = 2

The equation of the line is therefore y = 2x + 5.

Page 22: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 22 of 49

What is the equation of the line?

Page 23: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 23 of 49

Match the equations to the graphs

Page 24: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 24 of 49

A8.3 Parallel and perpendicular lines

Contents

A8.2 Gradients and intercepts

A8 Linear and real-life graphs

A8.1 Linear graphs

A8.4 Interpreting real-life graphs

A8.5 Distance-time graphs

A8.6 Speed-time graphs

Page 25: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 25 of 49

Investigating parallel lines

Page 26: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 26 of 49

Parallel lines

If two lines have the same gradient they are parallel.If two lines have the same gradient 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 the form y = mx + c.

2y = –6x + 1subtract 6x from both sides:

y =–6x + 1

2divide both sides by 2:

2y + 6x = 1

y = –3x + ½

The gradient m is –3 for both lines and so they are parallel.

Page 27: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 27 of 49

Matching parallel lines

Page 28: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 28 of 49

Investigating perpendicular lines

Page 29: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 29 of 49

Perpendicular lines

If if the gradients of two lines have a product of –1 then they are perpendicular.

If if the gradients of two lines have a product of –1 then they are perpendicular.

In general, if the gradient of a line is m, then the gradient of

the line perpendicular to it is .–1m

Write down the equation of the line that is perpendicular to y = –4x + 3 and passes through the point (0, –5).

The gradient of the line y = –4x + 3 is

The gradient of the line perpendicular to it is therefore 14

The equation of the line with gradient and y-intercept –5 is,

y = x – 5 14

–4.14

.

Page 30: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 30 of 49

Matching perpendicular lines

Page 31: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 31 of 49

A8.4 Interpreting real-life graphs

Contents

A8.3 Parallel and perpendicular lines

A8.2 Gradients and intercepts

A8.1 Linear graphs

A8 Linear and real-life graphs

A8.5 Distance-time graphs

A8.6 Speed-time graphs

Page 32: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 32 of 49

Real-life graphs

The resulting graph shows the rate that one quantity changes with another. For example,

When we use graphs to illustrate real-life situations, instead of plotting y-values against x-values, we plot plot one physical quantity against another physical quantity.

British pounds

Am

eri

can

do

llars

This graph shows the exchange rate from British pounds to American dollars.

It is a straight line graph through the origin and so the equation of the line would be of the form y = mx.

What would the value of m represent?The value of m would be equal to the number of dollars in each pound.

Page 33: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 33 of 49

Real-life graphs

This graph show the value of an investment as it gains interest cumulatively over time.

time

inve

stm

en

t va

lue

The graph increases by increasing amounts.

Each time interest is added it is calculated on an ever greater amount.

This makes a small difference at first but as time goes on it makes a much greater difference.

This is an example of an exponential increase.

Page 34: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 34 of 49

Real-life graphs

This graph show the mass of a newborn baby over the first month from birth.

The baby’s mass decreases slightly during the first week.

Its mass then increases in decreasing amounts over the rest of the month.

time

ma

ss

Page 35: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 35 of 49

Rates of change

Page 36: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 36 of 49

Filling flasks

Page 37: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 37 of 49

A8.5 Distance-time graphs

Contents

A8.3 Parallel and perpendicular lines

A8.2 Gradients and intercepts

A8.1 Linear graphs

A8 Linear and real-life graphs

A8.6 Speed-time graphs

A8.4 Interpreting real-life graphs

Page 38: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 38 of 49

Formulae relating distance, time and speed

It is important to remember how distance, time and speed are related.

Using a formula triangle can help,

distance = speed × timedistance = speed × time

DISTANCE

SPEED TIME

time =distance

speed

speed =distance

time

Page 39: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 39 of 49

Distance-time graphs

In a distance-time graph the horizontal axis shows time and the vertical axis shows distance.

For example, John takes his car to visit a friend. There are three parts to the journey:

John drives at constant speed for 30 minutes until he reaches his friend’s house 20 miles away.

He stays at his friend’s house for 45 minutes.

He then drives home at a constant speed and arrives 45 minutes later.

0

time (mins)

dist

ance

(m

iles)

15 30 45 60 75 90 105 120

5

10

15

20

0

Page 40: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 40 of 49

Finding speed from distance-time graphs

How do we calculate speed?

Speed is calculated by dividing distance by time.

The steeper the line, the faster the object is moving.

time

dis

tan

ce

In a distance-time graph this is given by the gradient of the graph.

change in distance

change in time

gradient =change in distance

change in time

= speed

A zero gradient means that the object is not moving.

Page 41: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 41 of 49

Interpreting distance-time graphs

Page 42: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 42 of 49

Distance-time graphs

When a distance-time graph is linear the objects involved are moving at a constant speed.

Most real-life objects do not always move at constant speed, however. It is more likely that they will speed up and slow down during the journey.

Increase in speed over time is called acceleration.

acceleration =change in speed

time

It is measured in metres per second per second or m/s2.

When speed decreases over time is often is called deceleration.

Page 43: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 43 of 49

Distance-time graphs

Distance-time graphs that show acceleration or deceleration are curved. For example,

This distance-time graph shows an object decelerating from constant speed before coming to rest.

time

dis

tan

ce

This distance-time graph shows an object accelerating from rest before continuing at a constant speed.

time

dis

tan

ce

Page 44: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 44 of 49

A8.6 Speed-time graphs

Contents

A8.3 Parallel and perpendicular lines

A8.2 Gradients and intercepts

A8.1 Linear graphs

A8 Linear and real-life graphs

A8.5 Distance-time graphs

A8.4 Interpreting real-life graphs

Page 45: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 45 of 49

Speed-time graphs

Travel graphs can also be used to show change in speed over time.

For example, this graph shows a car accelerating steadily from rest to a speed of 20 m/s.

0

time (s)

spee

d (m

/s)

5 10 15 20 25 30 35 40

5

10

15

20

0

It then continues at a constant speed for 15 seconds.

The brakes are then applied and it decelerates steadily to a stop.

The car is moving in the same direction throughout.

Page 46: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 46 of 49

Finding acceleration from speed-time graphs

Acceleration is calculated by dividing speed by time.

The steeper the line, the greater the acceleration.

time

spe

ed

In a speed-time graph this is given by the gradient of the graph.

gradient =change in speedchange in time

= acceleration

A zero gradient means that the object is moving at a constant speed.

change in speed

change in time

A negative gradient means that the object is decelerating.

Page 47: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 47 of 49

Finding distance from speed-time graphs

The following speed-time graph shows a car driving at a constant speed of 30 m/s for 2 minutes.

What is the area under the graph?

0

time (s)

spee

d (m

/s)

15 30 45 60 75 90 105 120

5

10

15

20

0

The area under the graph is rectangular and so we can find its area by multiplying its length by its height.

Area under graph = 20 × 120 = 240

What does this amount correspond to?

Page 48: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 48 of 49

The area under a speed-time graph

This area under a speed-time graph corresponds to the distance travelled.This area under a speed-time graph

corresponds to the distance travelled.

0

time (s)

spee

d (m

/s)

5 10 15 20 25 30 35 40

5

10

15

20

0

For example, to find the distance travelled for the journey shown in this graph we find the area under it.

The shape under the graph is a trapezium so,

Area = ½(15 + 40) × 20= ½ × 55 × 20= 550

So, distance travelled= 550 m

15

Page 49: © Boardworks Ltd 2005 1 of 49 A8 Linear and real-life graphs 3 ESO Mathematics

© Boardworks Ltd 2005 49 of 49

Interpreting speed-time graphs