Non Linear Functions Pattern - Identify Models

Identifying the Right Model

This pattern gives you a real-world scenario involving growth or decay and asks which equation models it. The answer is almost always an exponential function of the form $f(t) = a \cdot b^t$ (or $a \cdot b^{t/p}$ for non-unit time periods). The key is identifying the initial value $a$, the growth/decay factor $b$, and the time period.

The Core Formula

$f(t) = a \cdot b^{t/p}$

• $a$ = initial value (the amount at $t = 0$)
• $b$ = growth or decay factor per period
• $p$ = length of one period (in the same units as $t$)

Growth factors: "Doubles" → $b = 2$. "Triples" → $b = 3$. "Increases by 50%" → $b = 1.5$.
Decay factors: "Halves" → $b = \dfrac{1}{2}$. "Decreases by 20%" → $b = 0.8$.

Worked Examples

Example 1. At $h = 0$, there are 800 bacteria. The population triples every hour. Which equation models the population $P$ after $h$ hours?

A) $P = 800\left(\dfrac{1}{3}\right)^h$
B) $P = 800^h$
C) $P = 800(3)^h$
D) $P = 3(800)^h$

Initial value: $800$. Triples → factor of $3$. Every hour → period is $1$ hour.
$P = 800 \cdot 3^h$
Gotcha: Option A uses $\dfrac{1}{3}$ (decay instead of growth). Option D swaps the initial value and the base — $3(800)^h$ grows astronomically fast. The initial value goes out front; the growth factor is the base of the exponent.
The answer is C.

Example 2. The number of transistors on a chip was 25,000 at introduction. The number doubles every 3 years. Which equation models $N(t)$ after $t$ years?

A) $N(t) = 25{,}000 \cdot 3^{t/2}$
B) $N(t) = 25{,}000 \cdot \left(\dfrac{1}{2}\right)^{t/3}$
C) $N(t) = 25{,}000 \cdot 2^{3t}$
D) $N(t) = 25{,}000 \cdot 2^{t/3}$

Doubles → $b = 2$. Every $3$ years → $p = 3$.
$N(t) = 25{,}000 \cdot 2^{t/3}$
Gotcha: Option A swaps the base and period ($3^{t/2}$ instead of $2^{t/3}$). Option C uses $2^{3t}$, which would triple the exponent (way too fast). The period goes in the denominator: $t/p$.
The answer is D.

Example 3. A culture starts with 500 bacteria. The population triples every 6 hours. Which equation gives $P(t)$ after $t$ hours?

A) $P(t) = 500 \cdot 3^{t/6}$
B) $P(t) = 500 \cdot \left(\dfrac{1}{3}\right)^{t/6}$
C) $P(t) = 500 \cdot 6^{t/3}$
D) $P(t) = 500 \cdot 3^{6t}$

Triples → $b = 3$. Every $6$ hours → $p = 6$.
$P(t) = 500 \cdot 3^{t/6}$
Check: At $t = 6$, $P = 500 \cdot 3^1 = 1500$ (tripled ✓). At $t = 12$, $P = 500 \cdot 3^2 = 4500$ (tripled again ✓).
The answer is A.

Example 4. A car worth $20,000 depreciates by 15% each year. Which equation models its value $V(t)$ after $t$ years?

"Decreases by 15%" means the car retains 85% of its value each year.
$V(t) = 20{,}000 \cdot (0.85)^t$
Gotcha: "Decreases by 15%" does NOT mean $b = 0.15$. It means $b = 1 - 0.15 = 0.85$. The decay factor is what remains, not what's lost.

Example 5. A sample of a radioactive substance has a half-life of 10 years. If the initial mass is 200 grams, what is the mass after $t$ years?

Half-life means the substance halves every 10 years.
$M(t) = 200 \cdot \left(\dfrac{1}{2}\right)^{t/10}$

What to Do on Test Day

• Initial value = coefficient out front. The number at $t = 0$ goes before the exponential: $a \cdot b^{t/p}$.
• Growth factor = what you multiply by each period. Doubles → $2$. Triples → $3$. Increases by $r$% → $1 + r/100$.
• Decay factor = what remains. Halves → $1/2$. Decreases by $r$% → $1 - r/100$. Not the amount lost.
• Period goes in the denominator of the exponent: $b^{t/p}$. If it doubles every $3$ years, the exponent is $t/3$, not $3t$.
• Don't swap base and period. Doubles every 3 years → $2^{t/3}$, not $3^{t/2}$.
• Quick check: Plug in $t = 0$ to verify you get the initial value ($b^0 = 1$). Plug in $t = p$ to verify you get $a \cdot b$.