3.6 Exponential and Quadratic Models in Context: Growth, Decay, Compound Interest, and Depreciation
Key Takeaways
Linear models add the same amount each period; exponential models multiply by the same factor each period.
In y = a · bᵗ, a is the starting value and b is the growth factor: b = 1 + r for growth and b = 1 − r for decay at rate r.
Half-life and doubling models use y = a · (1/2)^(t/h) and y = a · 2^(t/d), where h and d are the half-life and doubling time.
Exponential equations with a common base are solved by setting the exponents equal: 9ˣ = 27^(x−1) becomes 2x = 3x − 3, so x = 3.
For quadratic models such as h(t) = −16t² + vt + s, the vertex gives the maximum and the positive root gives the landing time.
Exponential and Quadratic Models in Context: Growth, Decay, Compound Interest, and Depreciation
Quick Answer: The College Board Mathematics Test Specifications list "solve quadratic and exponential relationship problems in context (e.g., exponential decay/growth, compound interest, and depreciation)" as an Algebraic Reasoning subcategory on both the CRC and the Diagnostic. An exponential model multiplies by a constant growth factor each period: y = a · bᵗ. A quadratic model has a squared term and a turning point (vertex) that often answers "maximum" or "minimum" questions.
1. Linear vs. Exponential Change
| Feature | Linear model | Exponential model |
|---|---|---|
| Pattern each period | Adds or subtracts the same amount | Multiplies by the same factor |
| Equation form | y = mx + b | y = a · bˣ |
| Table clue | Constant differences in y | Constant ratios in y |
| Context words | "increases by $40 each month" | "increases by 4% each month," "doubles," "loses half" |
Table test:
x: 0 1 2 3
y: 200 240 288 345.6
Differences: 40, 48, 57.6 (not constant, so not linear)
Ratios: 240/200 = 1.2, 288/240 = 1.2, 345.6/288 = 1.2 (constant)
Model: y = 200(1.2)ˣ, a 20% increase per step
2. The General Exponential Model y = a · bᵗ
- a is the initial value (the output when t = 0).
- b is the growth or decay factor per time unit.
- Growth at rate r: b = 1 + r. A 5% annual growth rate gives b = 1.05.
- Decay at rate r: b = 1 − r. An 8% annual decline gives b = 0.92.
- If 0 < b < 1 the model decays. If b > 1 it grows.
Interpreting a Model
For P(t) = 1,200(1.035)ᵗ:
- 1,200 is the starting amount.
- 1.035 means the quantity grows 3.5% per period. It does not mean "grows by 1.035 per year," and it does not mean "grows 103.5%."
For V(t) = 350(0.92)ᵗ:
- 350 is the starting value.
- 0.92 means the value keeps 92% of itself each period, which is an 8% decrease per period.
Worked Growth Problem
A town of 2,000 people grows 5% per year. Estimate the population after 3 years.
P(3) = 2,000(1.05)³
1.05² = 1.1025
1.05³ = 1.157625
P(3) = 2,000 × 1.157625 = 2,315.25, or about 2,315 people
Note that 5% growth for 3 years is more than a 15% total increase (2,300), because each year's growth is calculated on a larger base.
3. Depreciation: Linear vs. Exponential
Depreciation problems may use either model, so read carefully:
| Wording | Model | Example: $24,000 vehicle over 2 years |
|---|---|---|
| "loses $3,000 of value each year" (straight-line) | Linear: V = 24,000 − 3,000t | 24,000 − 6,000 = $18,000 |
| "loses 15% of its value each year" | Exponential: V = 24,000(0.85)ᵗ | 24,000 × 0.7225 = $17,340 |
A common distractor subtracts 15% twice from the original value (24,000 × 0.70 = $16,800). That is wrong, because the second year's 15% is taken from the reduced value.
4. Compound Interest as an Exponential Model
The compound interest formula from Section 2.3, A = P(1 + r/n)^(nt), is an exponential model with factor (1 + r/n) applied nt times.
$2,000 is invested at 6% annual interest compounded quarterly for 2 years.
r/n = 0.06/4 = 0.015 per quarter
nt = 4 × 2 = 8 quarters
A = 2,000(1.015)⁸
1.015² = 1.030225; 1.015⁴ ≈ 1.061364; 1.015⁸ ≈ 1.126493
A ≈ 2,000 × 1.126493 ≈ $2,252.99
5. Doubling and Half-Life Models
When a quantity doubles every d time units, or halves every h time units:
Doubling: y = a · 2^(t/d)
Half-life: y = a · (1/2)^(t/h)
Doubling example: A culture of 500 bacteria doubles every 20 minutes. After 2 hours (120 minutes) there have been 120 ÷ 20 = 6 doublings, so y = 500 × 2⁶ = 500 × 64 = 32,000.
Half-life example: A medication dose of 80 mg has a half-life of 6 hours. After 18 hours (3 half-lives), y = 80 × (1/2)³ = 80 ÷ 8 = 10 mg.
The official sample questions include this style: a population that "doubles every nine years" starting at 100 individuals is modeled by P = 100 · 2^(t/9).
6. Solving Exponential Equations with a Common Base
If bᵐ = bⁿ (with b > 0 and b ≠ 1), then m = n. Rewrite both sides with the same base:
3^(x + 1) = 81 → 3^(x + 1) = 3⁴ → x + 1 = 4 → x = 3
25ˣ = 125 → 5^(2x) = 5³ → 2x = 3 → x = 3/2
9ˣ = 27^(x − 1) → 3^(2x) = 3^(3x − 3) → 2x = 3x − 3 → x = 3
2^(3x) = 1/16 → 2^(3x) = 2⁻⁴ → 3x = −4 → x = −4/3
Rewriting Exponential Expressions
Exponent laws let you rewrite a model to reveal a different rate:
- 500(1.04)^(2t) = 500[(1.04)²]ᵗ = 500(1.0816)ᵗ. A 4% rate applied twice per unit of t equals about an 8.16% rate per unit of t.
- 1.21^(t/2) = (1.1²)^(t/2) = 1.1ᵗ.
7. Quadratic Models in Context
Quadratic models appear when a quantity rises and then falls, or when area or revenue depends on a product of two linear factors.
Projectile Height
A ball is thrown upward from a 160-foot platform: h(t) = −16t² + 48t + 160
Maximum height: t = −b/(2a) = −48/(2 × −16) = 1.5 seconds
h(1.5) = −16(2.25) + 48(1.5) + 160 = −36 + 72 + 160 = 196 feet
Landing time: −16t² + 48t + 160 = 0 → divide by −16 → t² − 3t − 10 = 0
(t − 5)(t + 2) = 0 → t = 5 seconds (reject t = −2, since time cannot be negative)
Revenue and Area
A club sells t-shirts. At price p dollars, it sells (200 − 4p) shirts.
Revenue R(p) = p(200 − 4p) = −4p² + 200p
Maximum at p = −200/(2 × −4) = 25 dollars
R(25) = 25 × (200 − 100) = 25 × 100 = $2,500
8. TSIA2 Traps for Exponential and Quadratic Models
- Using the rate instead of the factor. A 6% decline is multiplied by 0.94, not by 0.06.
- Treating repeated percent change as linear. Twenty percent off for three years is not 60% off.
- Mismatched time units. If a half-life is in hours and t is in days, convert first.
- Keeping a negative root. In context, time, length, and price cannot be negative.
- Confusing the vertex's x-value with the maximum itself. The x-value tells when or at what price. Plug it back in to get the maximum value.
Lab equipment is purchased for $30,000 and depreciates by 20% of its value each year. What is the value of the equipment after 3 years?
$12,000
$15,360
$18,000
$19,200
A hospital pharmacy stores 96 mg of a radioactive tracer that has a half-life of 4 hours. How many milligrams remain after 12 hours?
48 mg
24 mg
12 mg
8 mg
The population of a town is modeled by P(t) = 1,200(1.035)ᵗ, where t is the number of years after 2020. What does the number 1.035 tell you?
The population increases by 35 people each year
The population in 2020 was 1,035 people
The population increases by 103.5% each year
The population grows by 3.5% each year
What value of x satisfies the equation 9ˣ = 27^(x − 1)?
3
1
−3
3/5
A ball is thrown upward from the top of a 160-foot platform, and its height in feet after t seconds is h(t) = −16t² + 48t + 160. What is the maximum height of the ball?
160 feet
208 feet
232 feet
196 feet
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