8.3 Exponential Growth & Decay Models, & Comparing Function Families

Key Takeaways

  • An exponential function has the algebraic form f(x)=a⋅bxf(x) = a \cdot b^x, where aa represents the non-zero initial amount (yy-intercept when x=0x = 0) and bb represents the constant multiplicative base factor (b>0,b≠1b > 0, b \neq 1).

  • In exponential growth models (b>1b > 1), the growth factor is b=1+rb = 1 + r, whereas in exponential decay models (0<b<10 < b < 1), the decay factor is b=1−rb = 1 - r, where rr is the fractional growth/decay rate per time interval.

  • Key real-world exponential formulas include compound interest A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}, asset depreciation V(t)=P(1−r)tV(t) = P(1 - r)^t, and radioactive half-life decay N(t)=N0(12)t/t1/2N(t) = N_0 \left(\frac{1}{2}\right)^{t / t_{1/2}}.

  • Function families are classified from data tables by inspecting rates of change: linear functions have constant first differences (Δy\Delta y), quadratic functions have constant second differences (Δ2y\Delta^2 y), and exponential functions have constant multiplicative ratios (ynyn−1\frac{y_{n}}{y_{n-1}}).

Last updated: August 2026

8.3 Exponential Growth & Decay Models, & Comparing Function Families

While linear functions model situations with a constant additive change, many real-world phenomena—including compound interest, biological population growth, epidemic spread, equipment depreciation, and radioactive decay—change by a constant multiplicative factor over equal intervals of time. These relationships are modeled by exponential functions.


1. Anatomy of an Exponential Function

The standard algebraic form of an exponential function is:

f(x)=a⋅bxf(x) = a \cdot b^x
Anatomy of f(x) = a · b^x:
├── a: Initial Value / Starting Amount (y-intercept at (0, a) since b^0 = 1, a ≠ 0)
├── b: Base Factor / Multiplier (b > 0 and b ≠ 1)
└── x: Independent Variable (Exponent representing time intervals or cycles)

Key Structural Features

  • Domain: All real numbers (−∞,∞)(-\infty, \infty).
  • Range (when a>0a > 0): All positive real numbers (0,∞)(0, \infty). The function values approach but never reach zero.
  • Horizontal Asymptote: The line y=0y = 0 (xx-axis) acts as a horizontal asymptote.

2. Exponential Growth vs. Exponential Decay

The behavior of an exponential function is dictated entirely by its base factor bb:

Exponential Growth vs Decay:

  EXPONENTIAL GROWTH (b > 1)                 EXPONENTIAL DECAY (0 < b < 1)
             |          /                               |  \
             |        /                                 |   \
             |      /                                   |    \ 
             |    /                                     |     \___
     --------+---/---------- x                  --------+---------────── x
             |  / (Asymptote y=0)                       |         (Asymptote y=0)
FeatureExponential GrowthExponential Decay
Base Factor Conditionb>1b > 10<b<10 < b < 1
Formulation with Rate rrb=1+rb = 1 + r (r=growth rate as decimalr = \text{growth rate as decimal})b=1−rb = 1 - r (r=decay rate as decimalr = \text{decay rate as decimal})
Curve DirectionRises steeply from left to rightFalls rapidly, then levels off toward y=0y = 0
Real-World ExamplesCompound interest, population growthMachinery depreciation, drug clearance, half-life

Caution

The Percentage Rate vs. Base Factor Trap: If an investment grows by 7%7\% annually, the base factor is b=1+0.07=1.07b = 1 + 0.07 = 1.07 (NOT 0.070.07). If a vehicle depreciates by 15%15\% annually, the base factor is b=1−0.15=0.85b = 1 - 0.15 = 0.85 (NOT 0.150.15 or 0.850.85 subtracted from xx).


3. Applied Exponential Growth & Decay Models

A. The Compound Interest Formula

When interest is compounded nn times per year, interest earned in earlier periods begins earning interest itself:

A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}
  • A=Total accumulated balance (principal + interest)A = \text{Total accumulated balance (principal + interest)}
  • P=Principal (initial investment amount)P = \text{Principal (initial investment amount)}
  • r=Annual interest rate expressed as a decimalr = \text{Annual interest rate expressed as a decimal} (6%=0.066\% = 0.06)
  • n=Number of compounding periods per yearn = \text{Number of compounding periods per year}
    • Annually: n=1n = 1
    • Semi-annually: n=2n = 2
    • Quarterly: n=4n = 4
    • Monthly: n=12n = 12
    • Daily: n=365n = 365
  • t=Time in yearst = \text{Time in years}

Worked Example 1: A commercial reserve account deposits $8,000\text{\textdollar}8,000 at an annual interest rate of 6%6\% compounded quarterly. Find the balance after 55 years.

  1. Identify parameters: P=8000P = 8000, r=0.06r = 0.06, n=4n = 4, t=5t = 5.
  2. Calculate periodic rate: rn=0.064=0.015\frac{r}{n} = \frac{0.06}{4} = 0.015.
  3. Calculate total periods: nt=4×5=20nt = 4 \times 5 = 20.
  4. Apply formula: A=8000(1+0.015)20=8000(1.015)20A = 8000(1 + 0.015)^{20} = 8000(1.015)^{20} A≈8000(1.346855)=$10,774.84A \approx 8000(1.346855) = \$10,774.84

B. Equipment & Asset Depreciation

Physical machinery, commercial vehicles, and electronics lose value at an annual percentage rate:

V(t)=P(1−r)tV(t) = P(1 - r)^t

Worked Example 2: A CNC manufacturing mill is purchased for $45,000\text{\textdollar}45,000 and depreciates at 12%12\% per year. Find its value after 33 years.

V(t)=45000(1−0.12)t=45000(0.88)tV(3)=45000(0.88)3=45000(0.681472)=$30,666.24\begin{aligned} V(t) &= 45000(1 - 0.12)^t = 45000(0.88)^t \\ V(3) &= 45000(0.88)^3 = 45000(0.681472) = \$30,666.24 \end{aligned}

C. Radioactive Decay & Half-Life

The half-life (t1/2t_{1/2}) is the time required for half of a radioactive substance to decay:

N(t)=N0(12)tt1/2N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}
  • If a 200 mg200\text{ mg} sample has a half-life of 6 hours6\text{ hours}, the amount remaining after 18 hours18\text{ hours} (33 half-life cycles) is: N(18)=200(12)186=200(12)3=200(18)=25 mgN(18) = 200\left(\frac{1}{2}\right)^{\frac{18}{6}} = 200\left(\frac{1}{2}\right)^3 = 200\left(\frac{1}{8}\right) = 25\text{ mg}

4. Comparing Function Families: Linear, Quadratic, & Exponential

On TABE Mathematics Level A, recognizing and comparing different function families from data tables and graphs is a heavily tested competency:

Tabular Difference Method for Function Identification (Equal x-intervals):
├── 1. First Differences Constant (Δy = k) ─────────> LINEAR (y = mx + b)
├── 2. Second Differences Constant (Δ²y = k) ───────> QUADRATIC (y = ax² + bx + c)
└── 3. Ratios of Consecutive Outputs Constant ──────> EXPONENTIAL (y = a · b^x)
Function FamilyStandard EquationGraphical ShapeTabular Identification RuleLong-Term Rate of Growth
Linearf(x)=mx+bf(x) = mx + bStraight lineConstant 1st differences (Δy=c\Delta y = c)Constant additive rate
Quadraticf(x)=ax2+bx+cf(x) = ax^2 + bx + cU-shaped parabolaConstant 2nd differences (Δ2y=c\Delta^2 y = c)Accelerated polynomial growth
Exponentialf(x)=a⋅bxf(x) = a \cdot b^xJ-shaped exponential curveConstant ratios (yk+1yk=b\frac{y_{k+1}}{y_k} = b)Dominates all polynomial growth

Step-by-Step Tabular Analysis Example

Analyze the data table below where xx increases by uniform steps of 11:

xxyy1st Differences (Δy\Delta y)2nd Differences (Δ2y\Delta^2 y)Output Ratio (yk+1yk\frac{y_{k+1}}{y_k})
0055———
11111111−5=611 - 5 = \mathbf{6}—11/5=2.2011 / 5 = 2.20
22212121−11=1021 - 11 = \mathbf{10}10−6=410 - 6 = \mathbf{4}21/11≈1.9121 / 11 \approx 1.91
33353535−21=1435 - 21 = \mathbf{14}14−10=414 - 10 = \mathbf{4}35/21≈1.6735 / 21 \approx 1.67
44535353−35=1853 - 35 = \mathbf{18}18−14=418 - 14 = \mathbf{4}53/35≈1.5153 / 35 \approx 1.51
  • First differences: 6,10,14,186, 10, 14, 18 (Not constant   ⟹  \implies Not linear).
  • Second differences: 4,4,44, 4, 4 (Constant +4  ⟹  +4 \implies Quadratic Function).
  • Ratios: 2.20,1.91,1.67,1.512.20, 1.91, 1.67, 1.51 (Not constant   ⟹  \implies Not exponential).
  • Conclusion: The data are governed by a quadratic function (specifically y=2x2+4x+5y = 2x^2 + 4x + 5).

5. The Principle of Exponential Dominance

A fundamental mathematical property tested on the TABE is that an exponential growth function with base b>1b > 1 will eventually exceed any linear or polynomial function, regardless of how large the linear slope or polynomial coefficient may be.

xxLinear: L(x)=500xL(x) = 500xQuadratic: Q(x)=5x2Q(x) = 5x^2Exponential: E(x)=2xE(x) = 2^x
115005005522
552,5002,5001251253232
10105,0005,0005005001,0241,024
15157,5007,5001,1251,12532,76832,768 (Exceeds Linear!)
202010,00010,0002,0002,0001,048,5761,048,576 (Exceeds Quadratic by 500x!)
Loading diagram...
Function Family Identification & Exponential Modeling Flowchart
Test Your Knowledge

A small business deposits $8,000 into a commercial reserve account earning an annual interest rate of 6% compounded quarterly. What will be the account balance after 5 years? (Round to the nearest cent.)

A

$10,400.00

B

$10,774.84

C

$10,707.04

D

$11,358.24

Test Your Knowledge

A data table records the following values: x: 0, 1, 2, 3, 4 y: 5, 11, 21, 35, 53 Which function family best models these data, and why?

A

Linear function, because the first differences are constantly 6

B

Exponential function, because the ratio between consecutive y-values is constantly 2.2

C

Linear function, because the second differences are zero

D

Quadratic function, because the second differences are constantly 4

Test Your Knowledge

A manufacturing company purchases a CNC milling machine for $45,000. The machine depreciates at a continuous annual rate of 12% per year. Which exponential equation models the machine's value V(t) after t years, and what is its estimated value after 3 years?

A

V(t) = 45,000(0.88)^t; after 3 years, value is $30,666.24

B

V(t) = 45,000(0.12)^t; after 3 years, value is $77.76

C

V(t) = 45,000(1.12)^t; after 3 years, value is $63,221.76

D

V(t) = 45,000 - 5,400t; after 3 years, value is $28,800.00

Sections you finish are checked off in the contents.