Read the gradient of a straight-line graph as a rate with units, read the intercept as a starting value or fixed charge, and use rates from graphs and from two readings to solve problems.
What you will learn
- Work out a gradient from a graph and give it as a rate with units
- Say what the gradient and the intercept mean in context
- Find a rate from two readings, including clock times
- Combine rates to solve problems
The gradient is a rate
On any straight-line graph the gradient is change in the vertical quantity ÷ change in the horizontal quantity.
Its units are "vertical unit per horizontal unit": litres per minute, pounds per mile, centimetres per hour.
- a tank holds 20 litres at 0 minutes and 80 litres at 12 minutes
- change in volume = 80 − 20 = 60 litres
- change in time = 12 minutes
- gradient = 60 ÷ 12 = 5 litres per minute
Use the change in the vertical quantity. Dividing 80 by 12 forgets the 20 litres that were there at the start.
To use a gradient, multiply it by a change across: at 5 litres per minute, 7 more minutes add 7 × 5 = 35 litres. With two lines on one graph, find each gradient; over 10 minutes, a rate of 8 per minute makes (8 − 5) × 10 = 30 more than a rate of 5 per minute.
The rest of this lesson and its 10 practice questions are locked.
In this lesson
- The gradient is a rate
- The intercept is the starting value
- Rates from two readings
- Try it: Gradients on the straight-line simulation
10 practice questions, marked as you go; about 30 minutes in all.
Key terms
Per, Change, Add the rates
In this unit: Ratio, gradients, similarity and growth
More ratio: one quantity as a fraction of another, ratio notation with units, ratios as linear functions, gradients as rates, similar shapes through length, area and volume ratios, and repeated percentage change.