The big idea: Use an exponential model y = a·bˣ when the quantity multiplies by the same factor each equal period.
If b > 1, it grows.
If 0 < b < 1, it decays.
| Parameter | Role | Example |
|---|---|---|
| a | Starting value (y when x = 0) | a = 500 → starts at 500 bacteria |
| b > 1 | Growth factor (multiplied each period) | b = 1.2 → 20% increase per period |
| 0 < b < 1 | Decay factor (multiplied each period) | b = 0.8 → 20% decrease per period |
Exponential growth (b > 1)
- Curve rises faster and faster
- Never negative (always above y = 0)
- Example: b = 1.05 means 5% increase per period
Exponential decay (0 < b < 1)
- Curve falls towards y = 0 but never reaches it
- Horizontal asymptote at y = 0
- Example: b = 0.9 means 10% decrease per period
Growth vs decay shapes — y = 2ˣ rises faster and faster; y = 8·(0.5)ˣ falls toward zero but never touches it.
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Growth or decay?: b > 1 → growth (curve rises). 0 < b < 1 → decay (curve falls towards y = 0).
If b = 1 the model is constant — not exponential.
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The big idea: To build y = a·bˣ from a context: identify a (the starting amount), identify b (the multiplier per period), then write the model.
Substitute to make predictions.
Exponential growth model
A colony of 200 bacteria doubles every hour.
Write a model for the number N after t hours, then find N at t = 5.
Step by step
- Write the formula.
- Identify a and b.
- Write the model.
- Substitute t = 5.
Final answer
N = 200 · 2ᵗ. After 5 hours: N = 6400 bacteria.
Exponential decay model
A car worth $24 000 loses 15% of its value each year.
Write a model for value V after t years.
Step by step
- Write the formula.
- A 15% loss means 85% remains each year.
- Write the model.
Final answer
V = 24 000 · (0.85)ᵗ
Decay factor = 1 − rate: If something loses 15% per year, the decay factor is b = 1 − 0.15 = 0.85.
You keep 85% each year.
IB awards a mark for correct identification of b — show this step.
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The big idea: The most common errors: using the percentage rate as b instead of 1 ± rate, and confusing growth factor with decay factor.
Wrong
- b = 0.15 for 15% decay (wrong — use 0.85)
- b = 1.2 for 20% decay (should be 0.80)
- Writing 200 × 2ˣ for decay
- Saying b = 2 means 2% growth
Correct
- 15% decay: b = 1 − 0.15 = 0.85
- 20% decay: b = 1 − 0.20 = 0.80
- Growth: b > 1, decay: 0 < b < 1
- b = 2 means doubling each period (100% increase)
Show the b calculation: Always write b = 1 + rate or b = 1 − rate as a step in your solution.
IB awards a mark for identifying b correctly — even if the final answer has an error.
Avoiding the percentage-rate trap
A bacterial colony of 100 cells doubles every hour, so b = 2.
A student incorrectly writes the SAME model for a 100-cell colony that LOSES 20% per hour.
What is the correct decay model, and what does the student's mistake look like?
Step by step
- Identify a — both situations start with 100 cells.
- Identify b for the decay model. 20% loss means 80% remains each hour.
- Write the decay model.
- The student wrote b = 0.20 (the rate) instead of b = 0.80 (the keep-fraction). That model decays MUCH faster — 100·(0.20)t — and is incorrect.
Final answer
Correct decay model: N = 100·(0.80)ᵗ. Common error: writing b = 0.20 instead of b = 1 − 0.20. Always show b = 1 ± rate as an explicit step.
The big idea: For y = a·bˣ, the horizontal asymptote is y = 0.
As x → +∞ for decay (b < 1), the quantity approaches 0 but never reaches it.
If a constant c is added (y = a·bˣ + c), the asymptote shifts to y = c.
| Model | Horizontal asymptote | What it means |
|---|---|---|
| y = 200 · (0.8)ˣ | y = 0 | Value approaches zero but never reaches it |
| y = 200 · (0.8)ˣ + 50 | y = 50 | Value never drops below 50 |
Long-run behaviour — the curve flattens toward the horizontal asymptote y = 50.
Interactive diagram
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Asymptote in context: If asked 'what does the graph approach as time increases?', state the horizontal asymptote as y = c.
Then explain what this means in context (e.g. 'the temperature approaches 20°C but never goes below it').