3 eso mathematics

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U Linear functions

3 ESO Mathematics


A8.1 Plotting Linear graphs

Contents

A8 Linear and real-life graphs

A8.2 Gradients and intercepts

A8.3 Analytical expression

A8.5 Distance-time graphs

A8.6 Speed-time graphs

A8.4 Parallel lines


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.

It is very recommendable to add the points of intersection to your table.

xy = 2x + 5

–3 –2 –1 0 1 2 3

For example,

y = 2x + 5

y

x

–1 1 3 5 7 9 11



In the example, y = 2x + 5:

1) If you make x = 0, you obtain y: y = 2 · 0 + 5 y = 5

2) If you make y = 0, you obtain x: 0 = 2x + 5 2x = -5 x = -5 / 2

y = 2x + 5

y

x

You can find the points of intersectionvery easily.



Contents



A8.1 Linear graphs



A8.4 Parallel lines


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.


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,


Calculating gradients

• A8.3 Parallel and perpendicular lines• A8.3 Parallel and perpendicular lines



Contents



A8.1 Linear graphs

A8.4 Parallel lines




Investigating linear graphs


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).


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)


What is the equation of the line?


Match the equations to the graphs


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:


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


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.


A8.4 Parallel lines

Contents



A8.1 Linear graphs





Investigating parallel lines


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.


Matching parallel lines



Contents



A8.1 Linear graphs



A8.4 Parallel lines


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


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


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.


Interpreting 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.



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



Contents

A8.3 Parallel and perpendicular lines


A8.1 Linear graphs



A8.4 Interpreting real-life graphs


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.


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.


Interpreting speed-time graphs

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