6.5Exponential Growth & Decay
Model real-world growth and decay situations using exponential equations, tables, and graphs. Apply compound interest, half-life, and percent change to solve problems.
Why This Matters
Exponential growth and decay describe how investments compound, how diseases spread, and how carbon-14 dating works in archaeology. This is one of the most directly applicable math topics you'll ever study.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
How can exponential functions be used to model and predict real-world growth and decay?
Lesson Overview
In Chapter 4 you learned the general form f(x) = ab^x. In this chapter you connect that form to real-world situations. When a quantity grows by a fixed percentage each period, the base b = 1 + r is called the growth factor. When it shrinks, b = 1 − r is the decay factor. You will write, evaluate, and graph both types of models, then apply them to compound interest, half-life, population growth, and more.
The key insight: exponential change is multiplicative — each output is multiplied by the same factor. Linear change is additive — each output increases by the same amount. Recognizing this difference lets you choose the right model for any real-world data set.
Growth: f(x) = 1·2ˣ (b = 2 > 1)
Exponential growth curve y = 2^x
Decay: f(x) = 8·(0.5)ˣ (b = 0.5 < 1)
Exponential decay curve
Exponential Growth
f(x) = a(1 + r)^x
- • Base b = 1 + r > 1
- • Graph rises left to right
- • y-intercept = a
- • Asymptote: y = 0
Exponential Decay
f(x) = a(1 − r)^x
- • Base b = 1 − r, 0 < b < 1
- • Graph falls left to right
- • y-intercept = a
- • Asymptote: y = 0
Growth Table → Equation: f(x) = 2(3)^x
Table of values for f(x) = 2·3^x showing constant ratio of 3
Exponential Growth Formula
f(x) = a(1 + r)^x
- a = initial value (y-intercept)
- r = growth rate (as a decimal)
- 1 + r = growth factor (b > 1)
- x = number of time periods
Exponential Decay Formula
f(x) = a(1 − r)^x
- a = initial value (y-intercept)
- r = decay rate (as a decimal)
- 1 − r = decay factor (0 < b < 1)
- x = number of time periods
Growth vs. Decay — Side by Side
Growth: b = 2 > 1 ↑
Both on same axes
Decay: b = 0.5, 0 < b < 1 ↓
Worked Examples
Identify whether each function represents growth or decay, and state the growth/decay factor and rate. (a) f(x) = 3(1.25)^x (b) g(x) = 500(0.90)^x
(a) f(x) = 3(1.25)^x: base b = 1.25 > 1 → GROWTH
Growth factor = 1.25; growth rate r = 1.25 − 1 = 0.25 = 25%
(b) g(x) = 500(0.90)^x: base b = 0.90, and 0 < 0.90 < 1 → DECAY
Decay factor = 0.90; decay rate r = 1 − 0.90 = 0.10 = 10%
f(x) = 3(1.25)^x — Growth
g(x) = 500(0.90)^x — Decay
A town has a population of 4,000 and grows at 5% per year. Write an exponential growth model and find the population after 10 years.
Identify: a = 4,000 (initial), r = 5% = 0.05
Growth factor: b = 1 + r = 1 + 0.05 = 1.05
Model: P(t) = 4000(1.05)^t
After 10 years: P(10) = 4000(1.05)^10
P(10) = 4000 × 1.6289 ≈ 6,516
A car worth $24,000 depreciates at 15% per year. Write a decay model and find its value after 6 years.
Identify: a = 24,000 (initial), r = 15% = 0.15
Decay factor: b = 1 − r = 1 − 0.15 = 0.85
Model: V(t) = 24000(0.85)^t
After 6 years: V(6) = 24000(0.85)^6
V(6) = 24000 × 0.3771 ≈ $9,051
Build a table of values for f(x) = 2(3)^x for x = 0, 1, 2, 3, 4. Then describe the pattern.
x = 0: f(0) = 2(3)^0 = 2(1) = 2
x = 1: f(1) = 2(3)^1 = 2(3) = 6
x = 2: f(2) = 2(3)^2 = 2(9) = 18
x = 3: f(3) = 2(3)^3 = 2(27) = 54
x = 4: f(4) = 2(3)^4 = 2(81) = 162
Pattern: each output is multiplied by 3 (the growth factor).
| x | f(x) = 2(3)^x | Ratio |
|---|---|---|
| 0 | 2 | — |
| 1 | 6 | ×3 |
| 2 | 18 | ×3 |
| 3 | 54 | ×3 |
| 4 | 162 | ×3 |
$2,500 is invested at 6% annual interest, compounded monthly. Find the balance after 5 years.
Formula: A = P(1 + r/n)^(nt)
P = 2500, r = 0.06, n = 12 (monthly), t = 5
A = 2500(1 + 0.06/12)^(12×5)
A = 2500(1.005)^60
A = 2500 × 1.3489 ≈ $3,372.33
Guided Practice
Guided Practice Video: Exponential Growth and Decay
Watch the guided practice walkthrough for exponential growth and decay, then complete the problems below.
Video by Sang Real Math
Watch on YouTube ↗A savings account has $800 and earns 4% annual interest. Write the growth model and find the balance after 7 years.
Hint: Use f(t) = 800(1.04)^t. Compute (1.04)^7 ≈ 1.3159.
A radioactive element decays at 20% per year. Starting with 500 g, write the decay model and find the amount after 5 years.
Hint: Decay factor = 1 − 0.20 = 0.80. Model: A(t) = 500(0.80)^t.
Determine if f(x) = 6(1.5)^x is growth or decay. Find f(4).
Hint: b = 1.5 > 1 → growth. Compute 6 × (1.5)^4 = 6 × 5.0625.
A car costs $30,000 and depreciates 12% per year. Find its value after 8 years.
Hint: V(t) = 30000(0.88)^t. Compute (0.88)^8 ≈ 0.3596.
$5,000 is invested at 8% compounded quarterly for 3 years. Find the final amount.
Hint: A = 5000(1 + 0.08/4)^(4×3) = 5000(1.02)^12.
Key Vocabulary
Exponential Growth
A function f(x) = ab^x where b > 1; the output increases as x increases.
Exponential Decay
A function f(x) = ab^x where 0 < b < 1; the output decreases as x increases.
Growth Factor
The base b = 1 + r in a growth model; the multiplier applied each period.
Decay Factor
The base b = 1 − r in a decay model; a number between 0 and 1.
Percent Rate of Change (r)
The percent increase or decrease per period, written as a decimal.
Compound Interest
Interest calculated on both principal and previously earned interest: A = P(1 + r/n)^(nt).
Half-Life
The time it takes for a quantity to decrease to half its original amount.
Asymptote
A line the graph approaches but never touches; exponential functions have a horizontal asymptote at y = 0.
Interactive Practice — 5 Questions
Which function represents exponential DECAY?
A population of 2,000 grows at 4% per year. Which equation models this?
What is the growth factor in f(x) = 500(1.15)^x?
$1,000 invested at 5% compounded annually for 3 years gives:
A table shows consecutive ratios of 0.75. This is:
Independent Practice
Independent Practice
Identify growth or decay and state the rate: (a) f(x) = 7(1.3)^x (b) h(x) = 200(0.65)^x. Explain how you know from the base b.
A town has a population of 5,000 and grows at 2.5% per year. Write the exponential model and find the population after 20 years.
A car costs $30,000 and depreciates 12% per year. Write the decay model and find its value after 8 years.
$3,000 is invested at 5% compounded annually. Write the compound interest model and find the balance after 10 years.
A radioactive substance has a half-life of 8 years. Starting with 640 g, write the half-life model and find the amount remaining after 32 years.
Common Mistakes
Using the growth rate r as the base instead of (1 + r) — e.g., writing A(t) = 500(0.04)^t for 4% growth.
The base is (1 + r) for growth: A(t) = 500(1.04)^t. The rate r is added to 1 to get the growth factor.
Using (1 + r) for decay instead of (1 − r) — e.g., writing A(t) = 200(1.15)^t for 15% decay.
For decay, subtract the rate: A(t) = 200(1 − 0.15)^t = 200(0.85)^t.
Confusing the decay factor with the decay rate — e.g., saying the decay rate is 0.85 when it's actually 15%.
The decay factor is 0.85. The decay rate is 1 − 0.85 = 0.15 = 15%. They are different values.
Forgetting to identify the initial value a before writing the model.
The initial value a is the starting amount at t = 0. Always identify it first: A(t) = a · b^t.
Math Tips
Always convert percent to decimal first: 8% growth → b = 1 + 0.08 = 1.08, not 1.8.
Growth factor b = 1 + r (add rate). Decay factor b = 1 − r (subtract rate).
The y-intercept is always (0, a) — the initial value before any growth or decay.
For compound interest A = P(1 + r/n)^(nt): divide the rate by n and multiply time by n.
Half-life model: N(t) = N₀ · (0.5)^(t/h) where h is the half-life period.