Recognizing Percent Growth and Decay - Algebra 2

Section 1

Historical Context & Motivation

Long before algebra students graphed exponential curves, merchants, bankers, and natural philosophers observed a fundamental pattern: quantities that change by a fixed proportion at each step behave very differently from those that change by a fixed amount. A population of rabbits that doubles every season grows far faster than one that adds a fixed number of offspring each year. A loan accumulating interest "on the interest" spirals in ways that simple addition cannot capture. Understanding this distinction — between additive (linear) change and multiplicative (exponential) change — is one of the most powerful ideas in all of mathematics.

~2000 BCE

Babylonian clay tablets record compound interest calculations, tracking how debts grow when interest is charged on previously accumulated interest — one of the earliest implicit uses of exponential growth.

1614 CE

John Napier publishes Mirifici Logarithmorum Canonis Descriptio, introducing logarithms as a computational tool. By creating the inverse of exponentiation, Napier gave mathematicians a way to "undo" exponential growth and solve for unknown time periods or rates.

1683 CE

Jacob Bernoulli investigates continuous compounding and discovers that the limit of (1 + 1/n)ⁿ as n → ∞ converges to a constant — later named e ≈ 2.71828 — establishing the natural base of exponential functions.

1798 CE

Thomas Malthus publishes An Essay on the Principle of Population, arguing that populations grow exponentially while food supply grows linearly, bringing percent growth into social science and policy debates.

Present Day

Exponential models underpin fields from epidemiology (virus spread) and pharmacology (drug half-lives) to finance (compound interest, depreciation) and computer science (Moore's Law). Recognizing whether a scenario involves percent growth or decay is a foundational algebraic skill.

The central question this lesson addresses is deceptively simple: given a function, a table, an equation, or a real-world scenario, how do you recognize whether the quantity is growing or decaying by a fixed percentage? Answering that question lets you choose the right model, make accurate predictions, and compare different exponential processes on equal footing.

Section 2

Core Principles & Definitions

Before we can recognize percent growth and decay, we need precise definitions of the building blocks. Every exponential function expressing percent change shares a common anatomy, and learning to read that anatomy is the key to classification.

Exponential Growth

A quantity exhibits exponential growth when it increases by a fixed percentage during each equal time interval. The growth factor b is greater than 1, meaning the output is multiplied by more than 1 each step.

Exponential Decay

A quantity exhibits exponential decay when it decreases by a fixed percentage during each equal time interval. The decay factor b is between 0 and 1 (exclusive), so the output shrinks with each step.

Growth/Decay Rate (r)

The rate is the percent change per period, expressed as a decimal. For growth, b = 1 + r. For decay, b = 1 − r. A 7% growth rate means r = 0.07 and b = 1.07.

Initial Value (a)

The initial value a is the quantity at time t = 0. It is the y-intercept of the exponential function and sets the scale for all future outputs. It must be positive for the standard model to apply.

✦

Key Takeaway

Think of the growth factor b as a "multiplier dial." If you set the dial above 1, the quantity swells over time — like inflating a balloon a little more with each breath. If you set it below 1, the quantity shrinks — like a puddle evaporating, losing the same fraction of its water every hour. The crucial insight is that the percentage stays constant, but the amount of change varies because it is always a fraction of the current value, not a fixed number.

Section 3

Visual Explanation

The graph below is the visual centerpiece of this lesson. It shows two exponential functions on the same coordinate plane: one representing 8% growth (green curve) and one representing 8% decay (red curve). Both start at the same initial value of 100. Notice how the growth curve accelerates upward while the decay curve flattens toward zero — yet neither ever reaches infinity or zero in finite time. The dashed blue line shows a linear function growing at 8 units per period for comparison.

Several observations emerge from this graph. First, the growth curve bends upward with increasing steepness — it is concave up. The decay curve also bends upward (concave up), but it is falling toward the horizontal axis as an asymptote — it approaches zero but never reaches it. Second, the linear function grows steadily, adding 8 units per period, and it actually starts out ahead of the growth curve for the first few periods. But eventually the exponential overtakes it, because each 8% step is applied to an ever-larger base. This is the hallmark of multiplicative versus additive change.

Section 4

Mathematical Framework

Every percent-growth or percent-decay scenario can be captured by a single general form. Mastering this equation and its parts lets you move fluently between word problems, tables, graphs, and algebraic expressions.

General Exponential Model

y = a · bᵗ

a = initial value (y-intercept), b = growth/decay factor, t = time (number of periods)

The factor b encodes the percent change. When a quantity changes by r percent per period (expressed as a decimal), the relationship between b and r takes one of two forms depending on whether the quantity is increasing or decreasing.

Percent Growth (r > 0)

b = 1 + r → y = a(1 + r)ᵗ

If r = 0.05 (a 5% increase), then b = 1.05. Since b > 1, the function rises.

Percent Decay (r > 0)

b = 1 − r → y = a(1 − r)ᵗ

If r = 0.05 (a 5% decrease), then b = 0.95. Since 0 < b < 1, the function falls.

Notice that in both cases the rate r is a positive number. The direction of change — growth or decay — is determined entirely by whether you add or subtract r from 1. This is a common source of confusion: a "12% decay rate" means r = 0.12 and b = 0.88, not b = −0.12.

Extracting the Rate from the Factor

r = |b − 1|

If b = 1.15, then r = 0.15 (15% growth). If b = 0.72, then r = 0.28 (28% decay).

This extraction formula is especially useful when you are handed an equation like y = 500(0.83)ᵗ and asked to interpret it. The factor 0.83 is less than 1, so this is decay. The rate is |0.83 − 1| = 0.17, meaning the quantity decreases by 17% per period.

✦

Key Takeaway

The single number b tells you everything: if b > 1 it is growth, if 0 < b < 1 it is decay, and the distance from 1 gives you the percent rate. Learning to read b at a glance is like learning to read a thermometer — the scale is the same every time, and the number tells the whole story.

Section 5

Detailed Breakdown & Classification

In practice, you will encounter percent-change scenarios in many disguises: word problems, data tables, equations, and graphs. The diagram below provides a decision flowchart for classifying any exponential function you meet.

Let's also organize the key signatures of growth versus decay in a reference table. When you see a new function or dataset, quickly checking these features will tell you which type you are dealing with.

Feature	Percent Growth	Percent Decay
Factor b	b > 1	0 < b < 1
Rate formula	r = b − 1	r = 1 − b
Graph shape	Rising, concave up	Falling, concave up
Table pattern	Successive ratios > 1	Successive ratios < 1
End behavior (t → ∞)	y → ∞	y → 0⁺ (horizontal asymptote)
Real-world examples	Compound interest, population growth, inflation	Radioactive decay, depreciation, cooling

The Growth Factor Spectrum

b → 0 (rapid decay) b → ∞ (rapid growth)

b = 1 (50%)

Section 6

Worked Example

A new car is purchased for $28,000. Its value depreciates by 14% each year. Write an exponential model for the car's value, determine whether it represents growth or decay, state the rate, and find the car's value after 6 years.

Car Depreciation

1
Step 1 — Identify the Initial Value

The car is purchased for $28,000, so the initial value is a = 28,000. This is the value at time t = 0 (the moment of purchase).
2
Step 2 — Determine the Rate and Factor

The car depreciates (loses value), so this is a decay problem. The rate of depreciation is 14%, which means r = 0.14. The decay factor is:

b = 1 − r = 1 − 0.14 = 0.86 Since 0 < 0.86 < 1, this confirms we have exponential decay.
3
Step 3 — Write the Exponential Model

Substituting into y = a · bᵗ:

V(t) = 28000 × (0.86)ᵗ Here V(t) represents the car's value in dollars after t years.
4
Step 4 — Calculate the Value After 6 Years

Substitute t = 6: V(6) = 28000 × (0.86)⁶. First, compute (0.86)⁶. We can build this up:

0.86² = 0.7396 → 0.86³ ≈ 0.6361 → 0.86⁶ = (0.86³)² ≈ 0.6361² ≈ 0.4046 Now multiply: V(6) = 28000 × 0.4046 ≈ $11,329
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Step 5 — Interpret the Result

After 6 years, the car is worth approximately $11,329 — it has lost about $16,671, or roughly 60% of its original value. Each year the car loses 14% of its current value (not 14% of the original price), which is why the dollar amount lost decreases each year even though the percentage remains constant. In year 1 the car loses 28,000 × 0.14 = $3,920, but by year 6 it loses only about 13,174 × 0.14 ≈ $1,844.

Section 7

Strengths, Limitations & Comparisons

The exponential model y = a · bᵗ is elegant and widely applicable, but it has important boundaries. Comparing it with the linear model highlights both its power and its limitations.

Criterion	Linear Model (y = mx + c)	Exponential Model (y = a·bᵗ)
Type of change	Constant amount per period	Constant percentage per period
Graph shape	Straight line	Curve (concave up)
Table test	Constant first differences	Constant ratios of successive terms
Long-term behavior	Grows/shrinks without bound at a steady pace	Growth accelerates; decay approaches zero asymptotically
Realistic for	Short-term, controlled processes	Populations, finance, radioactive decay — but may need logistic cap for growth
Key parameter	Slope m	Base b (growth/decay factor)

A critical limitation of the exponential growth model is that it assumes unlimited resources. Real populations, for instance, eventually encounter carrying-capacity limits and transition to logistic growth. Similarly, exponential decay models a quantity that approaches zero but mathematically never reaches it, which is physically accurate for processes like radioactive half-life but less so for, say, the number of cookies on a plate.

✦

Key Takeaway

The fastest way to distinguish a linear scenario from an exponential one is to check the table test: compute the differences between consecutive outputs for linear, or the ratios of consecutive outputs for exponential. If the differences are constant, the relationship is linear. If the ratios are constant, it is exponential. A real-world scenario that says "increases by 50 per year" is linear; one that says "increases by 5% per year" is exponential.

Section 8

Connection to Advanced Theory

The percent-change model y = a(1 ± r)ᵗ is actually a special case of the broader continuous exponential model that uses Euler's number e ≈ 2.71828 as the base. In more advanced courses (Precalculus and Calculus), you will see that any exponential function can be rewritten in terms of e:

Continuous Form

y = a · eᵏᵗ

Where k = ln(b). If k > 0 → growth. If k < 0 → decay.

The connection is straightforward: since b = e^k, we can always convert between the two forms. The continuous model is preferred in calculus because its derivative is proportional to itself — the rate of change at any instant is proportional to the current value, which is the mathematical definition of exponential behavior.

Feature	Discrete: y = a·bᵗ	Continuous: y = a·eᵏᵗ
When to use	Percent change per fixed period	Continuous compounding or calculus
Growth indicator	b > 1	k > 0
Decay indicator	0 < b < 1	k < 0
Conversion	b = eᵏ	k = ln(b)
Half-life / doubling time	t = ln(2) / \|ln(b)\|	t = ln(2) / \|k\|

Another important extension is the logistic model, y = L / (1 + Ce^−kt), which starts out looking exponential but levels off at a carrying capacity L. If you continue in mathematical biology or economics, you will see this model emerge naturally when resources are limited. For now, recognizing that the simple percent-change model is the first building block toward these richer models provides valuable perspective.

Section 9

Practice Problems

Question 1: Problem 1 — Conceptual

A function is defined as f(t) = 250(0.88)^t. Without performing any calculations, determine whether this function represents exponential growth or exponential decay, and explain how you know.

Show answer

This function represents exponential decay. The key indicator is the base (growth/decay factor) b = 0.88, which falls between 0 and 1. Whenever 0 < b < 1, multiplying by b repeatedly makes the output smaller and smaller over time. Specifically, the decay rate is r = 1 − 0.88 = 0.12, meaning the quantity decreases by 12% per period. The initial value is 250, and as t increases, f(t) approaches zero but never reaches it — zero serves as a horizontal asymptote.

Question 2: Problem 2 — Basic Identification

A town had a population of 12,400 in 2020 and grows by 3.2% per year. Write an exponential model for the population P(t) where t is the number of years after 2020. State the growth factor.

Show answer

The initial value is a = 12,400 and the growth rate is r = 0.032 (3.2% expressed as a decimal). Since the population is growing, the growth factor is b = 1 + 0.032 = 1.032. The exponential model is: P(t) = 12400 × (1.032)^t. Each year, the population is multiplied by 1.032, meaning it retains 100% of the previous year's population and adds an additional 3.2%. After 1 year: P(1) ≈ 12,797. After 10 years: P(10) ≈ 12,400 × 1.032¹⁰ ≈ 12,400 × 1.3711 ≈ 17,002.

Question 3: Problem 3 — Intermediate

The following table shows the value of an investment over time. Determine whether the data is best modeled by a linear or exponential function. If exponential, state whether it represents growth or decay and find the rate.

Year: 0, 1, 2, 3, 4
Value ($): 5,000 | 5,300 | 5,618 | 5,955 | 6,312

Show answer

First, check the differences: 5300 − 5000 = 300; 5618 − 5300 = 318; 5955 − 5618 = 337; 6312 − 5955 = 357. These are not constant, so the model is not linear. Now check the ratios: 5300/5000 = 1.06; 5618/5300 ≈ 1.06; 5955/5618 ≈ 1.06; 6312/5955 ≈ 1.06. The ratios are constant at approximately 1.06. Since the ratios are constant and greater than 1, this is exponential growth. The growth factor is b = 1.06, so the growth rate is r = 1.06 − 1 = 0.06, or 6% per year. The model is V(t) = 5000(1.06)^t.

Question 4: Problem 4 — Applied / Multi-Step

A pharmaceutical drug has a half-life of 4 hours in the bloodstream. A patient takes a 400 mg dose. (a) Find the hourly decay rate. (b) Write the exponential model. (c) How much of the drug remains after 10 hours?

Show answer

(a) A half-life of 4 hours means after 4 hours, exactly half remains. We need b⁴ = 0.5. Solving: b = 0.5^1/4 ≈ 0.8409. The hourly decay rate is r = 1 − 0.8409 ≈ 0.1591, or about 15.9% per hour. (b) With a = 400 mg and b ≈ 0.8409: D(t) = 400 × (0.8409)ᵗ, where t is in hours. (c) At t = 10: D(10) = 400 × (0.8409)¹⁰. Since 0.8409⁴ = 0.5 and 0.8409⁸ = 0.25, we get 0.8409¹⁰ = 0.8409⁸ × 0.8409² ≈ 0.25 × 0.7071 ≈ 0.1768. So D(10) ≈ 400 × 0.1768 ≈ 70.7 mg. After 10 hours, roughly 70.7 mg of the drug remains in the bloodstream.

Question 5: Problem 5 — Critical Thinking / Synthesis

Two savings accounts are opened on the same day. Account A starts with $10,000 and grows at 5% per year. Account B starts with $8,000 and grows at 8% per year. Which account will be worth more after 15 years? After how many full years does Account B first surpass Account A? Explain why the account with the lower initial value eventually wins.

Show answer

Models: A(t) = 10000(1.05)^t and B(t) = 8000(1.08)^t. After 15 years: A(15) = 10000 × 1.05¹⁵ ≈ 10000 × 2.0789 ≈ $20,789. B(15) = 8000 × 1.08¹⁵ ≈ 8000 × 3.1722 ≈ $25,378. Account B is worth more. Finding the crossover: Set 10000(1.05)^t = 8000(1.08)^t. Divide both sides by 8000(1.05)^t: 1.25 = (1.08/1.05)^t = (1.02857)^t. Take ln of both sides: ln(1.25) = t × ln(1.02857), so t = ln(1.25)/ln(1.02857) ≈ 0.2231/0.02817 ≈ 7.92. Account B first surpasses Account A after 8 full years. Why: Even though Account B starts $2,000 behind, its higher growth rate means it multiplies by a larger factor each year. Over time, the compounding effect on the larger percentage overwhelms the initial deficit. This illustrates a fundamental property of exponential functions: in the long run, the rate matters more than the starting value, because the rate controls the exponent while the starting value is just a multiplicative constant.

Lesson Summary

Putting It All Together

Summary

Recognizing percent growth and decay begins with understanding that any function of the form y = a · bᵗ describes a quantity changing by a constant percentage over equal time intervals. The growth/decay factor b is the single most important number to examine: if b > 1, the function models exponential growth at a rate of r = b − 1; if 0 < b < 1, it models exponential decay at a rate of r = 1 − b. The initial value a sets the starting point, and the exponent t counts the number of periods elapsed.

To classify a scenario, look for the constant-ratio signature in tables (divide each output by the previous one) or the characteristic concave-up curve on graphs. In word problems, phrases like "increases by ___%" or "loses ___% of its value" signal exponential behavior, as opposed to linear phrases like "increases by ___ units." These percent-change models connect forward to continuous exponential functions using base e and to logistic models that incorporate carrying capacities — but the algebraic foundation you have built here, centered on reading and interpreting the factor b, remains the essential skill at every level.