Direct and inverse proportion
When two quantities are in direct proportion, as one increases the other does too.
We can display this relationship in a graph. Two quantities that are in direct proportion will always produce a straight-line graph that passes through the origin.
If the constant of proportionality is positive, the graph will have a positive gradient. If the constant is negative, the graph will have a negative gradient.
Example 1
A babysitter’s earnings are directly proportional to the number of hours worked.
If they are paid £9 for each hour of work we can write this as a formula:
\(\text {Earnings = £9} \times \text {hours worked}\)
To plot this we need three points, we can use a table of values to help us:
When hours = 0, earnings = £9 × 0 = £0
When hours = 1, earnings = £9 × 1 = £9
When hours = 2, earnings = £9 × 2 = £18
Giving the points (\({0} \text{,} {~0}\)), (\({1} \text{,} {~9}\)) and (\({2} \text{,} {~18}\)).
Once you’ve plotted these points, draw a line through all three extending it as far as you can.
Example 2
The number of packets of crisps and the total cost are shown in the table.
When we graph these points we see they can be connected together in a straight line:
For every 1 on the \(\text {x}\) axis, the \(\text {y}\) axis increases by 30. This means the gradient of the graph is 30 and the graph can be written as:
\(\text {y = 30 x}\)
Finding relationships from a graph
\(\text {y}\) is directly proportional to \(\text {x}\). From this graph, write an equation to show the relationship between \(\text {x}\) and \(\text {y}\).
When \(\text {x}\) = 2, \(\text {y}\) = 30
\({30} \text{~÷} {~2} = {15}\)
\(\text {y = 15~\times~x}\)
\(\text {y = 15 x}\)
When two quantities are in inverse proportion, as one increases the other decreases.
When we graph this relationship we get a curved graph.
Example
\(\text {y}\) is inversely proportional to \(\text {x}\) and when \(\text {x}\) = 2, \(\text {y}\) = 10
Draw a graph by completing the table of values:
1. As \(\text {y}\) is inversely proportional to \(\text {x}\) we can write:
\(\text {y = k / x}\)
2. Substitute in the known values and rearranging finds the value of \(\text {k}\):
\(\text {10 = k / 2}\)
\(\text {10 \times~2 = k}\)
\(\text {k = 20}\)
3. We can now complete the table of values:
When \(\text {x = 1}\), \(\text {y = 20 / 1} = 20\)
When \(\text {x = 4}\), \(\text {y = 20 / 4} = 5\)
When \(\text {x = 5}\), \(\text {y = 20 / 5} = 4\)
4. We can now plot these points and join them together in a curve.