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Lesson 7 of 16
  • Year 11
  • Higher

Interpreting and drawing more real-life graphs

I can interpret and draw more real-life graphs.

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View unit
Lesson 7 of 16
New
New
  • Year 11
  • Higher

Interpreting and drawing more real-life graphs

I can interpret and draw more real-life graphs.

Download all resources
Share activities with pupils

These resources will be removed by end of Summer Term 2025.

Switch to our new teaching resources now - designed by teachers and leading subject experts, and tested in classrooms.

These resources were created for remote use during the pandemic and are not designed for classroom teaching.

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Lesson details

Key learning points

  1. Not all real-life graphs can be drawn using sections of different linear graphs.
  2. The graphs are models, which means that we want them to be useful.

Keywords

  • Simple interest - Interest is money added to savings or loans. Simple interest is always calculated on the original amount.

  • Compound interest - Compound interest is calculated on the original amount and the interest accumulated over the previous period.

  • Exponential - The general form for an exponential equation is $$y = ab^x$$ where $$a$$ is the coefficient, $$b$$ is the base and $$x$$ is the exponent.

  • Speed - Speed is the rate at which something is moving. It is measured as the distance travelled per unit of time.

  • Acceleration - Acceleration is the rate of change of speed with respect to time.

Common misconception

Pupils may think the gradient of the line to model simple interest is always the interest rate.

This works when investing £100 but not with other start values. The gradient is the amount of interest added each time which depends on the amount invested and the interest rate. There are examples to show this in the lesson.


To help you plan your year 11 maths lesson on: Interpreting and drawing more real-life graphs, download all teaching resources for free and adapt to suit your pupils' needs...

Graphing software may be helpful in learning cycle 1 to help students explore the general form of exponential curves as well as the growth rate of different interest rates and different start values. There are lots of ideas to explore in this lesson and you may feel this is worth additional time.
Teacher tip

Licence

This content is © Oak National Academy Limited (2025), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Lesson video

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Prior knowledge starter quiz

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6 Questions

Q1.
Interest which is calculated on the original amount and the interest accumulated over the previous period is called __________.

Correct answer: compound interest
simple interest

Q2.
£200 is invested into an account with 3% simple interest each year. How much will be in the account after 5 years?

Correct Answer: £230, 230

Q3.
£200 is invested into an account with 3% compound interest each year. Which of these calculations will give the amount in the account after 5 years?

Correct answer: $$200 \times 1.03^5$$
$$200 \times 1.15$$
$$200 \times 1.3^5$$

Q4.
£1000 is invested into an account with 2% compound interest paid monthly. How much would be in the account after 1 year? Round to the nearest pence.

Correct Answer: £1268.24, 1268.24

Q5.
Select the coordinates that lie on the curve $$y=3^x$$

$$\left({1\over 3}, 1\right)$$
Correct answer: (1,3)
(2,6)
Correct answer: (3, 27)
(4,64)

Q6.
The general form for an exponential equation is:

$$x = a$$
$$y=mx+c$$
Correct answer: $$y=ab^x$$
$$y=ax^2+bx + c$$

Assessment exit quiz

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6 Questions

Q1.
Which would be the equation of the line if graphing £300 invested at a simple interest rate of 2% per month?

$$y=2x+300$$
Correct answer: $$y=6x+300$$
$$y=300x + 2$$
$$ y= 2 \times 300^x$$
$$ y = 300 \times 1.20^x$$

Q2.
Which would be the equation of the exponential curve if graphing £500 invested at a simple interest rate of 3% per month?

$$y = 500 \times 0.03^x$$
$$y = 500 \times x^3$$
$$y = 500 \times 3^x$$
Correct answer: $$y = 500 \times 1.03^x$$
$$y = 500 \times 1.3^x$$

Q3.
Which of these equations could model the value of a car depreciating over time?

$$y = 20 000 \times -1.15^x$$
$$y = 20 000 \times -0.85^x$$
Correct answer: $$y = 20 000 \times 0.85^x$$
$$y = 20 000 \times 1.00015^x$$
$$y = 20 000 \times 1.15^x$$

Q4.
Which of these journey segments show deceleration on a displacement-time graph?

An image in a quiz
Correct Answer: An image in a quiz
An image in a quiz
An image in a quiz
An image in a quiz

Q5.
Match the sections of the graph labelled a)-f) with the correct part of the journey.

An image in a quiz
Correct Answer:a),Increasing in speed moving away from the fixed start point.

Increasing in speed moving away from the fixed start point.

Correct Answer:b),Decelerating moving away from the fixed start point.

Decelerating moving away from the fixed start point.

Correct Answer:c),Object is stationary.

Object is stationary.

Correct Answer:d),Increasing in speed moving towards the fixed start point.

Increasing in speed moving towards the fixed start point.

Correct Answer:e),Travelling at a constant speed.

Travelling at a constant speed.

Correct Answer:f),Decelerating moving back towards the fixed start point.

Decelerating moving back towards the fixed start point.

Q6.
Here is a displacement-time graph for a journey. Which of these could be a speed-time graph of the same journey?

An image in a quiz
An image in a quiz
Correct Answer: An image in a quiz
An image in a quiz
An image in a quiz