Questions about graphs are certain to appear on the exam paper. This revision bite will take you through some of the basic skills you will need to tackle them.
Coordinates are the numbers used to tell us how to get to a certain point on a grid - eg, a graph or map. The grid will have an x axis which runs horizontally, and a y axis that runs vertically.
Coordinates are written in pairs. The coordinate of P on the graph above is (2, 1). The first number in the pair is known as the x coordinate (2, 1).
The x coordinate tells us how many units to go across (to the left if it is a negative number, or the right if it is positive).
The y coordinate tells us how many units to go up if the number is positive, or down if the number is negative.
Look at these coordinates and work out which letter represents them on the graph. Write the letters down and they will form a word.
(5, 0) (-3, 4) (5, -2) (-4, -1)
The answer is 'EASY'.
Here is a breakdown of the answer:
Now spell the word (0, 5) (4, -3) (2, -5) (5, 0), and check your answer at the bottom of the page.
The answer is BITE.
Straight-line graphs are easy to spot.
The equation of a straight-line graph can contain;
Although some equations for straight lines might have only two of those, like an x term and a number, eg x=5.
These are all equations of straight lines:
In a typical question, you will be asked to fill in a table of values, plot them and join them up to make a straight-line graph.
Complete the table of values for y = x + 3, and then draw the graph.
|y = x + 3||2||4||5|
Choose suitable possible values of x. Six values are usually enough.
With the equation y = x + 3, we have to add 3 to each x value to get the the value of y.
Once the values of y have been worked out, the table looks like this:
|y= x +3||0||1||2||3||4||5||6|
The graph is plotted on the next page.
On the previous page we worked out the values of y = x +3. These points can be plotted on on a graph. Each pair of values in the table is an (x, y) coordinate - eg (-3, 0) (-2, 1) (-1, 2) (0, 3) etc.
|y= x +3||0||1||2||3||4||5||6|
Take a look at the graph y = x + 3 and see how the values are plotted.
If the question does not ask you to complete a table of values first, you can still create one by making up your own values for x. You should work out a minimum of 3 points for a straight-line graph, in case one of them is wrong.
Now practise drawing some more graphs in the activity below.
Sometimes you are given a graph of a straight line and you need to find its gradient.
To find the gradient of a straight line:
The following graph shows the exchange rate for euros € and United States dollars $ in March 2011.
There are two marked points at (0, 0) and (70, 98).
By working out the gradient of the graph, we can find the exchange rate from euros to United States dollars.
The vertical distance between (0, 0) and (70, 98) is 98.
The horizontal distance between (0, 0) and (70, 98) is 70.
98 ÷ 70 = 1.4
This means that 1 euro is equal to 1.4 United States dollars.
Drawing a curved graph is similar to drawing a straight-line graph and you also have to substitute numbers into the equation.
With a curved line graph the formula will include x2, or some other power of x.
For a curved graph, you need as many points as possible to make it accurate.
Complete the table for y = x2 + 2.
|y = x2 + 2|
We know that y = x2 + 2, so we need to square (multiply by itself) each x value and add 2.
This can be used to complete a table
|y = x2 + 2||2||3||6||11||18||27||38||51|
We could plot these points on a grid. Each pair of values is an (x, y) coordinate - eg, (0, 2), (1, 3), (2, 6) etc.
Complete the table for y = 2x2 - 10. Then draw the graph for the equation.
Use your graph to find the value of x when y = 15
|y = 2x2 - 10.||-8||8||40||62|
Here's how the completed table looks:
|y = 2x2 - 10.||-10||-8||-2||8||22||40||62|
Draw the graph of y = 2x2 - 10.
The value of x when y = 15 should be about 3.5.
Now practise drawing some more curved graphs in the activity below.
Graphs are a good way to show formulae in visual form.
For example, a taxi firm uses the following graph to work out the cost of a journey.
We can find out lots of different information from this graph.
The labels on each axis are very important. The horizontal line - or x axis - shows the number of miles. The vertical line - or y axis - tells us the cost.
There are two marked points at (0, 2) and (10, 7).
Read up from 6 miles on the x axis to the straight line. When you reach the line move across horizontally until you reach the y axis. The cost is £5. This means a journey of 6 miles will cost £5.
Now try these questions.
Jo pays £7 for a taxi journey. What distance did she travel?
This time we're told the cost.
Reading from £7, we see the journey was 10 miles.
The distance was 10 miles.
The graph ranges up to a distance of only 11 miles.
If we want to find out the cost of a journey longer than this, we first need to find more information from the graph.
The graph starts at 0 miles.
The point (0, 2) tells us that 0 miles costs £2.
£2 is the fixed charge.
We can work out the gradient by comparing the distance between two marked points on the line.
The vertical distance between (0, 2) and (10, 7) is 5.
The horizontal distance between (0, 2) and (10, 7) is 10.
Using the formula vertical length ÷ horizontal length, we find:
5 ÷ 10 = 0.5
The gradient of the line is 0.5.
This tells us that for each mile the cost is £0.50.
From this information, we can find that a journey of 18 miles would cost:
We can use this information to find a formula for the graph.
The formula for the cost, C, for a distance of n miles, is:
C = 2 + 0.5n
Using the formula we can find the cost for a mileage greater than the range of the graph.
Now practise finding information from graphs that model real situations.
John was on a cycling holiday. He drew this graph to show how far he cycled on one day.
How far did John cycle that day?
The end point on the line on the graph is at 50 km, so John cycled 50 km that day.
John stopped to mend a puncture. What time was that, and how long did it take?
Part of the line on the graph is horizontal. This tells us that John did not cycle any distance during this time. The horizontal line starts at 11:30 and ends at 12:00. It therefore took John half an hour to mend the puncture.
When did John travel at his fastest speed? What was that speed?
The fastest speed is the line with the biggest gradient.
The line from 10:00 to 11:30 has a greater gradient than the line from 12:00 to 14:30. This means that John‘s speed was greatest from 10:00 to 11:30.
The speed is the gradient of the line.
John travelled 30 km from 10:00 to 11:30. It took him 1.5 hours to cycle this distance.
Speed = distance ÷ time
30 ÷ 1.5 = 20
We have found the speed using the same method, vertical ÷ horizontal.
We work out the units of the answer (km/h) by referring to the axes on the graph, in this case km and hours.
Therefore John's fastest speed was 20 km/h.