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Exponential growth and decay Why does multiplying by the same number each year make a curve, not a line?

Foundation and Higher · AQA · 45 minutes

Students race a line that adds against a curve that doubles, then find why multiplying makes a curve, write compound interest and depreciation as powers, and read a and b from the graph of y = a × bˣ. Slides to present, a vote that shows who's stuck, practice and homework.

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Two plans both start with £1. The line adds £k a day; the curve doubles every day. Move k: can the line stay ahead for ever?

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They cross at (0, 1) and (2.66, 6.32)

The idea in one minute

Adding the same amount each step makes a straight line: y = 1000 + 50x. Multiplying by the same number each step makes an exponential curve: y = 1000 × 1.05ˣ. In y = a × bˣ, a is the start, the value at x = 0, where the curve crosses the y-axis, and b is the multiplier: more than 1 grows, between 0 and 1 decays. Compound interest and depreciation are exactly this: x years means multiplying by b x times, so bˣ, never b × x. A decay curve gets closer and closer to 0 but never reaches it, because a part of something is never nothing.

The lesson goes picture first. Students race a line that adds against a curve that doubles before anyone explains, then compare a table that adds with one that multiplies. Only then do they write multipliers, work out compound interest and depreciation with a power, and read a and b from a graph. Multipliers, compound interest and depreciation are for Foundation and Higher; the graphs of y = a × bˣ and finding b from two points are Higher.

By the end, students can

  • tell growth that adds from growth that multiplies, in a table or a graph
  • write the multiplier for a percentage rise or fall, such as 1.05 or 0.85
  • work out compound interest and depreciation with a power
  • read a and b from the graph of y = a × bˣ, and say why a decay curve never reaches 0

Key words

  • exponential
  • multiplier
  • compound interest
  • depreciation
  • growth
  • decay
  • asymptote

Before this lesson

Students should find a percentage of an amount, know that 2³ = 2 × 2 × 2, and know straight-line graphs, which add the same each step.

Taught in Straight-line graphs.

The lesson, step by step

45 minutes in total. The stretch is optional.

  1. 1 · Think5 min

    Add, or double?

    Both plans start with £1.

    Plan A: add £2 a day.

    Plan B: double every day.

    Which would you take for a week?

    Choose one. Say why.

    Write here

    5 min 1 · Hook · students think · the slides say “Think”

    Add, or double?

    Students choose between a plan that adds and one that doubles before anyone explains.

    Say

    “Two plans both start with £1. Plan A adds £2 a day. Plan B doubles every day. Which would you take for a week? Write your choice, and why.”

    Ask
    • After 2 days, which plan is ahead? After 3?
    • Is there a k so big that the line stays ahead for ever?
    Watch for

    Choosing Plan A because it starts faster: keep it. The next step is about why doubling wins in the end.

    For the teacher

    Give a minute to decide, then vote. Plan A leads at first: 3 against 2 after one day, 5 against 4 after two.

    Don't explain yet. Move k: the line starts steeper each time, but the curve always overtakes it. The question is why.

    Take a vote with Ask the class: Plan A or Plan B for a week? Then move k up to 8 and watch where the curve overtakes the line each time.

    Shorter: Vote only; move k in the next step (−2 min).

  2. 2 · Watch12 min

    Add, or multiply?

    Plan A: 1, 3, 5, 7, 9

    +2 each day: a straight line

    Plan B: 1, 2, 4, 8, 16

    ×2 each day: a curve

    Doubling adds more each time.

    Same difference: line. Same ratio: curve.

    2 · Watch

    y = a × bˣ

    At x = 0, b⁰ = 1, so y = a.

    a: the start, on the y-axis

    b: the multiplier each step

    b = 2 grows; b = 0.5 decays.

    Decay never reaches 0.

    Predict, then move a.

    2 · Watch

    Compound interest: a power

    £500 at 4% a year, for 3 years

    Each year keeps 100% and adds 4%: × 1.04

    500 × 1.04 × 1.04 × 1.04 = 500 × 1.04³

    = 562.432, so £562.43

    Not 500 × 1.04 × 3. Not 500 + 3 × 20.

    x years: multiply by the multiplier x times.

    2 · Watch

    Depreciation: what is kept

    £12 000, losing 15% a year

    It keeps 85%: × 0.85

    Not × 0.15: that is the loss.

    2 years: 12 000 × 0.85² = £8670

    It never reaches £0.

    Down: a multiplier below 1.

    2 · Watch

    Find a and b (Higher)

    Through (0, 2) and (2, 8)

    At x = 0, y = a: so a = 2

    8 = 2 × b², so b² = 4

    b = 2 (b is positive)

    y = 2 × 2ˣ

    a from the y-axis, then b.

    12 min 2 · Explain · you explain · the slides say “Watch”

    Adding, multiplying, y = a × bˣ

    A table that adds against one that multiplies; a and b; compound interest and depreciation as powers; a and b from a graph.

    Say

    “Adding the same each step makes a line. Multiplying by the same each step makes a curve.”

    Ask
    • 1, 2, 4, 8: what do you do to get the next number? And for 1, 3, 5, 7?
    • y = a × bˣ: what is y when x = 0?
    • A car loses 15% a year. What part of its value does it keep?
    Watch for

    Writing 500 × 1.04 × 3 for three years: each year multiplies again, so it is 1.04³.

    Compound interest: a power

    £500 at 4% a year, for 3 years

    Each year keeps 100% and adds 4%: × 1.04

    500 × 1.04³ = 562.432, so £562.43

    Depreciation: what is kept

    A car worth £12 000 loses 15% a year

    It keeps 85%: × 0.85

    After 2 years: 12 000 × 0.85² = £8670

    a and b from a graph (Higher)

    y = a × bˣ goes through (0, 2) and (2, 8)

    At x = 0, y = a, so a = 2

    8 = 2 × b², so b² = 4 and b = 2 (b is positive)

    For the teacher

    First slide: Plan A goes 1, 3, 5, 7: the same difference each day, a straight line. Plan B goes 1, 2, 4, 8: the same ratio each day, a curve that bends upwards ever more steeply. Adding 2 can't keep up with doubling, because doubling adds more each time. That answers the hook.

    Second slide: y = a × bˣ. At x = 0, b⁰ = 1, so y = a: a is the start, on the y-axis. b is the multiplier. Move a; then compare ×2 with ×0.5, mirror images of each other. A decay curve halves for ever and never reaches 0: the x-axis is an asymptote.

    Third and fourth slides: compound interest is y = a × bˣ in money. 4% up is × 1.04; 15% down keeps 85%, × 0.85. Three years is three multiplications, a power. Fifth slide (Higher): read a from the y-axis, then put the second point in and solve for b, which is positive.

    Vote before each slide (Ask the class): which table makes a line? Where will y = 3 × 2ˣ cross the y-axis?

    Shorter: Skip the slide that finds a and b from a graph (Higher) (−4 min).

  3. 3 · Your turn8 min

    Worked, then yours

    I do: £2000 at 5% a year for 2 years

    × 1.05 each year: 2000 × 1.05² = £2205

    You do: £4000 at 2% a year for 3 years

    The multiplier first, then the power.

    Write here

    3 · Your turn

    Two more pairs

    I do: a £9000 van loses 20% a year: 9000 × 0.8³ = £4608

    You do: a £15 000 boat loses 10% a year. After 2 years?

    I do: y = a × bˣ through (0, 5) and (1, 15)

    a = 5; 15 = 5 × b, so b = 3

    You do: y = a × bˣ through (0, 4) and (2, 36)

    Losing 20%? It keeps 80%.

    Write here

    Its answers' slide comes with your copy.

    8 min 3 · Your turn · students write

    Worked, then yours

    Each worked example beside a twin that students solve themselves.

    Say

    “Watch me do one. Then you do its twin, which is almost the same.”

    Ask
    • Up or down? So is the multiplier more or less than 1?
    • How many times do you multiply?
    Watch for

    Using the part lost (0.2) as the multiplier, not the part kept (0.8).

    Compound interest

    £2000 at 5% a year for 2 years

    × 1.05 each year: 2000 × 1.05² = £2205

    Then theirs

    £4000 at 2% a year for 3 years

    Depreciation

    A £9000 van loses 20% a year

    It keeps 80%: 9000 × 0.8³ = £4608 after 3 years

    Then theirs

    A £15 000 boat loses 10% a year: its value after 2 years

    Find a and b (Higher)

    y = a × bˣ through (0, 5) and (1, 15)

    a = 5; 15 = 5 × b, so b = 3: y = 5 × 3ˣ

    Then theirs

    y = a × bˣ through (0, 4) and (2, 36)

    For the teacher

    Work the first example aloud, writing each line. Then students solve its twin while you watch who writes a power and who multiplies by the number of years.

    Do the same for each pair. The second pair catches students who use the part lost; the third, students who halve 36 ÷ 4 = 9 instead of taking its square root.

    Open the answers slide once most have finished their twins.

    Shorter: Do the first two pairs only (−3 min).

  4. 4 · Vote and practise15 min

    Vote first

    £3000 at 4% compound interest a year.

    Which gives its value after 5 years?

    A: 3000 × 1.04⁵

    B: 3000 × 1.04 × 5

    C: 3000 + 5 × 120

    D: 3000 × 0.04⁵

    Choose on your own, then say why.

    4 · Vote and practise

    Practise: the multiplier first

    1. The multipliers for 6% up and for 6% down

    2. £800 at 3% compound interest: after 2 years?

    3. A £600 phone loses 25% a year: after 2 years?

    4. Exponential growth: 5, 10, 20, 40 or 5, 10, 15, 20?

    Stuck on 2? 3% up: times what? How often?

    Write here

    4 · Vote and practise

    Reason it out

    5. Where does y = 6 × 0.5ˣ cross the y-axis?

    6. Does y = 6 × 0.5ˣ ever reach 0?

    7. 20 000 people, growing 10% a year: after 3 years?

    8. Higher: y = a × bˣ through (0, 3) and (2, 48)

    Stretch: £1000 at 8%. When is it first over £2000?

    15 min 4 · Who's stuck? · students vote · the slides say “Vote and practise”

    Vote, then practise

    One question shows who's stuck; then practice in levels, and a stretch.

    Say

    “Choose an answer on your own. Nobody talks until everyone has chosen.”

    Ask
    • What is the multiplier for one year?
    • How many times do you multiply by it?
    Watch for

    Answers not checked against the story: ask what the amount is after year 1, and what the second year's interest is worked out on.

    Write the multiplier first, then the working. Give money to the nearest penny. The stretch is optional.

    The question that shows who's stuck

    1. £3000 is invested at 4% compound interest a year. Which works out its value after 5 years?
      1. 3000 × 1.04⁵
      2. 3000 × 1.04 × 5
      3. 3000 + 5 × 120
      4. 3000 × 0.04⁵
    In your Classroom, as they answer (an example)
    • A11
    • B6
    • C6
    • D2

    Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.

    1. The multiplier for a 6% increase is ___, and for a 6% decrease it is ___
    2. £800 is invested at 3% compound interest a year. What is it worth after 2 years? Type the number of pounds to the penny, without the £ sign.
    3. A phone bought for £600 loses 25% of its value each year. What is it worth after 2 years? Type the number of pounds, without the £ sign.
    4. Each table gives y for x = 0, 1, 2, 3. Which shows exponential growth?
      1. 5, 10, 20, 40
      2. 5, 10, 15, 20
      3. 40, 20, 10, 5
    5. Where does the graph of y = 6 × 0.5ˣ cross the y-axis?
      1. (0, 6)
      2. (0, 0.5)
      3. (0, 3)
      4. (0, 0)
    6. Does the graph of y = 6 × 0.5ˣ ever reach 0?
      1. No: half of something is never nothing
      2. Yes, at x = 12
      3. Yes, at x = 0
    7. A town has 20 000 people, and its population grows by 10% a year. After 3 years it has ___ people
    8. Higher: y = a × bˣ goes through (0, 3) and (2, 48), with b positive. a = ___ and b = ___
    9. Stretch: £1000 is invested at 8% compound interest a year. After how many whole years is it first worth more than £2000?

    Support · For question 2: 3% up keeps all of it and adds 3%. What do you multiply by for one year? Two years: how many times do you multiply by it?

    For the teacher

    Everyone chooses before anyone explains. Ask them to work out year 1 and year 2 by hand: is year 2's interest on £3000 or on £3120? Then ask how five multiplications by the same number are written.

    Questions 1–2 build fluency; 3–4 change the look; 5–6 ask for reasons; 7–8 are exam-style, 8 for Higher. The stretch asks when an investment first doubles.

    Students show their working on paper or the board; the quiz keeps their final answers.

    Whatever the vote, before they practise: work out year 1 and year 2 aloud with the class, and ask which option gives the same numbers.

    Shorter: After the vote, questions 1–4 only (−6 min).

    Longer: The stretch, then: how long does any amount take to double at 8%? (+10 min).

  5. 5 · Show what you know5 min

    Write it, work it, say why

    A bike costs £400 and loses 20% a year.

    Its value after x years as y = a × bˣ?

    Its value after 2 years?

    Will it ever be worth £0? Why?

    The multiplier first.

    Write here

    5 min 5 · Exit · students show · the slides say “Show what you know”

    Write it, work it, say why

    One depreciation question shows who writes the multiplier and the power, and why it never reaches 0.

    Say

    “A bike costs £400 and loses 20% of its value each year. Write its value after x years as y = a × bˣ, and work out its value after 2 years.”

    Watch for

    A multiplier of 0.2, or a value that reaches £0.

    The equation, the working, and one sentence.

    1. A bike costs £400 and loses 20% of its value each year. Write its value after x years as y = a × bˣ, and work out its value after 2 years. Will it ever be worth £0? Say why.
    For the teacher

    A bike costs £400 and loses 20% of its value each year. Write its value after x years as y = a × bˣ, and work out its value after 2 years.

    Then say whether it will ever be worth £0, and why.

    The answers show who still uses the part lost, or multiplies by the number of years: start the next lesson with one of each. Set the homework.

    Longer: Sketch y = 400 × 0.8ˣ and mark where it crosses the y-axis (+5 min).

One to one: ask the student to say the multiplier aloud before any working ("keeps 85%, so times 0.85"), and to check a compound answer against the simple-interest one: compound should be a little more.

Homework: multipliers, powers and graphs

Write the multiplier before any working. Give money to the nearest penny. Question 6 is for Higher.

  1. Write the multiplier for a 7% increase, for a 30% decrease and for a 2.5% increase.
  2. £5000 is invested at 3% compound interest a year. Work out its value after 4 years.
  3. A car bought for £16 000 loses 12% of its value each year. Work out its value after 3 years.
  4. Complete a table for y = 2 × 3ˣ for x = 0, 1, 2, 3, then draw the graph.
  5. £1000 is saved for 3 years at 5% a year. Work out its value with simple interest and with compound interest. Which is more, and why?
  6. The graph of y = a × bˣ goes through (0, 80) and (2, 20), with b positive. Find a and b, then the value of y when x = 3.

Print the worksheet with working space and homework. Teacher answers come with your Library copy.

Teaching exponential growth and decay: common questions

How do you explain the difference between linear and exponential growth?

Put two tables side by side: one that adds the same amount each step (1, 3, 5, 7) and one that multiplies by the same number (1, 2, 4, 8). The first has the same difference each time and makes a straight line; the second has the same ratio and makes a curve. Multiplying wins in the end because each step adds more than the one before.

Why is the multiplier for a 15% decrease 0.85 and not 0.15?

Because the multiplier gives what is left, not what is lost. A 15% fall keeps 100% − 15% = 85% of the value, so you multiply by 0.85. Multiplying by 0.15 gives the amount lost in the first year, not the new value.

Why does an exponential decay graph never reach zero?

Each step keeps a part of what was there: half, or 85%, of something is never nothing. The values get closer and closer to 0, so the x-axis is an asymptote, but no step ever takes the value to 0 itself.

Is this exponential graphs lesson Foundation or Higher?

Both. Multipliers, compound interest and depreciation are for Foundation and Higher. Recognising and sketching y = a × bˣ, and finding a and b from points on its graph, are Higher; the lesson marks those parts, and can leave them out at its 30-minute length.

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Curriculum reference: AQA GCSE Mathematics 8300: R16 (growth and decay, compound interest and repeated percentage change) and A12 (exponential graphs, Higher).

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