graphs

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Finding the gradient of a straight-line graph

It is often useful or necessary to find out what the gradient of a graph is.

For a straight-line graph, pick two points on the graph. The gradient of

the line = (change in y-coordinate)/(change in x-coordinate) .

In this graph, the gradient = (change in y-coordinate)/(change

in x-coordinate) = (8-6)/(10-6) = 2/4 = 1/2

We can of course use this to find the equation of the

line. Since the line crosses the y-axis when y = 2, the equation of

this graph is y = ½x + 2 .

Finding the gradient of a curve

To find the gradient of a curve, you must draw an accurate sketch of the

curve. At the point where you need to know the gradient, draw a tangent

to the curve. A tangent is a straight line which touches the curve at one

point only. You then find the gradient of this tangent.

Example:

Find the gradient of the curve y = x² at the point (3, 9).

Note: this method only gives an approximate

answer. The better your graph is, the closer your answer will be to the

correct answer. If your graph is perfect, you should get an answer of 6 for

the above question.

The Equation of a Straight Line

Equations of straight lines are in the form y = mx + c (m and c are

numbers). m is the gradient of the line and c

is the y-intercept (where the graph crosses the y-axis). NB1: If you are

given the equation of a straight-line and there is a number before the 'y',

divide everything by this number to get y by itself, so that you can see

what m and c are.

NB2: Parallel lines have equal gradients.

The above graph has equation y = (4/3)x - 2 (which is

the same as 3y + 6 = 4x).

Gradient = change in y / change in x = 4 / 3

It cuts the y-axis at -2, and this is the constant in the equation.

Graphs of Quadratic Equations

These are curves and will have a turning point. Remember, quadratic equations are of the form: y

= ax² + bx + c (a, b and c are numbers). If 'a' is positive, the graph will

be 'U' shaped. If 'a' is negative, the graph will be 'n' shaped. The graph

will always cross the y-axis at the point c (so c is the y-intercept

point). Graphs of quadratic functions are sometimes known as parabolas.

Example:

Drawing Other Graphs

Often the easiest way to draw a graph is to construct a table of values.

Example:

Draw y = x² + 3x + 2 for -3 £ x £ 3

x -3 -2

-1 0 1 2

3

x² 9 4

1 0 1

4 9

3x -9 -6 -3

0 3 6

9

2 2 2

2 2 2

2 2

y 2 0

0 2 6

12 20

The table shows that when x = -3, x² = 9, 3x = -9 and 2 = 2. Since y = x² +

3x + 2, we add up the three values in the table to find out what y is when

x = -3, etc.

We then plot the values of x and y on graph paper.

Intersecting Graphs

If we wish to know the coordinates of the point(s) where two graphs

intersect, we solve the equations simultaneously.

This can be done using the graphs.

Simultaneous Equations

You can solve simultaneous

equations by drawing graphs of the two equations you wish to solve.

The x and y values of where the graphs intersect are the solutions to the

equations.

Example:

Solve the simultaneous equations 3y = -2x + 6 and y = 2x -2 by graphical

methods.

From the graph, y = 1 and x = 1.5 (approx.). These are

the answers to the simultaneous equations.

Solving Equations

Any equation can be solved by drawing a graph of the equation in question.

The points where the graph crosses the x-axis are the solutions. So if you

asked to solve x² - 3 = 0 using a graph, draw the graph of y = x² - 3 and

the points where the graph crosses the x-axis are the solutions to the

equation.

If you are asked to draw the graph of y = x² - 3x + 5 and then are asked to

use this graph to solve 3x + 1 - x² = 0 and x² - 3x - 6 = 0, you would

proceed in the following way:

1) Make a table of values for y = x² - 3x + 5 and draw the graph.

2) Make the equations you need to solve like the one you have the graph of.

So for 3x + 1 - x² = 0:

i) multiply both sides by -1 to get x² - 3x -1 = 0

ii) add 6 to both sides: x² - 3x + 5 = 6

Now we know that y = x² - 3x + 5 and x² - 3x + 5 = 6, therefore, y = 6.

Find out what x is when y = 6 and these are the answers (you should get two

answers).

Try solving x² - 3x - 6 = 0 yourself using your graph of y = x² - 3x + 5.

You should get a answers of

Travel Graphs

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Speed, Distance and Time

The following is a basic but important formula which applies when speed is

constant (in other words the speed doesn't change):

Speed = distance

time

Remember, when using any formula, the units must all be consistent. For

example speed could be measured in m/s, distance in metres and time in

seconds.

If speed does change, the average (mean) speed can be calculated:

Average speed = total distance travelled

total time taken

Example:

If a car travels at a speed of 10m/s for 3 minutes, how far will it travel?

Firstly, change the 3 minutes into 180 seconds, so that the units are

consistent. Now rearrange the first equation to get distance = speed ×

time.

Therefore distance travelled = 10 × 180 = 1800m = 1.8km

Units

In calculations, units must be consistent, so if the units in the question

are not all the same (e.g. m/s, m and s or km/h, km and h), change the

units before starting, as above.

The following is an example of how to change the units:

Example:

Change 15km/h into m/s.

15km/h = 15/60 km/min (1)

= 15/3600 km/s = 1/240 km/s (2)

= 1000/240 m/s = 4.167 m/s (3)

In line (1), we divide by 60 because there are 60 minutes in an hour. Often

people have problems working out whether they need to divide or multiply by

a certain number to change the units. If you think about it, in 1 minute,

the object is going to travel less distance than in an hour. So we divide

by 60, not multiply to get a smaller number.

Velocity and Acceleration

Velocity is the speed of a particle and its direction of motion

(therefore velocity is a vector quantity, whereas

speed is a scalar quantity).

When the velocity (speed) of a moving object is increasing we say that the

object is accelerating. If the velocity decreases it is said to be

decelerating. Acceleration is therefore the rate of change of velocity

(change in velocity / time) and is measured in m/s².

Example:

A car starts from rest and within 10 seconds is travelling at 10m/s. What

is its acceleration?

Acceleration = change in velocity = 10 =

1m/s²

time

10

Distance-time graphs:

These have the distance from a certain point on the vertical axis and the

time on the horizontal axis. The velocity can be calculated by finding the

gradient of the graph. If the graph is curved, this can be done by drawing

a chord and finding its gradient (this will give average velocity) or by

finding the gradient of a tangent to the graph (this will give the velocity

at the instant where the tangent is drawn).

Velocity-time graphs/ speed-time graphs:

A velocity-time graph has the velocity or speed of an object on the

vertical axis and time on the horizontal axis. The distance travelled can

be calculated by finding the area under a velocity-time graph. If

the graph is curved, there are a number of ways of estimating the area (see

trapezium rule below). Acceleration is the gradient of a velocity-time

graph and on curves can be calculated using chords or tangents, as above.

On travel graphs, time always goes on the horizontal

axis (because it is the independent variable).

Trapezium Rule

This is a useful method of estimating the area under a graph. You often

need to find the area under a velocity-time graph since this is the

distance travelled.

Area under a curved graph = ½ × d × (first + last + 2(sum of rest))

d is the distance between the values from where you will

take your readings. In the above example, d = 1. Every 1 unit on the

horizontal axis, we draw a line to the graph and across to the y axis.

'first' refers to the first value on the vertical axis, which is about 4

here.

'last' refers to the last value, which is about 5 (green line).]

'sum of rest' refers to the sum of the values on the vertical axis

where the yellow lines meet it.

Therefore area is roughly: ½ × 1 × (4 + 5 + 2(8 + 8.8 + 10.1 + 10.8 + 11.9

+ 12 + 12.7 + 12.9 + 13 + 13.2 + 13.4))

= ½ × (9 + 2(126.8))

= ½ × 262.6

= 131.3 units²

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