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Exponential Functions

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Exponential functions describe anything that doubles, halves, or multiplies by a constant — population growth, compound interest, radioactive decay. The pattern: the variable lives in the exponent.

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Equivalent Expressions→Linear Functions→

Exponential Functions

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An exponential function has the form $f(x) = a \cdot b^x$ where:

$a$ = the starting value (the value when $x = 0$ ).
$b$ = the growth factor — what you multiply by each time $x$ increases by 1.

$b > 1$ : growth (function increases). $f(x) = 100 \cdot 2^x$ doubles every step.

$0 < b < 1$ : decay (function decreases). $f(x) = 100 \cdot (0.5)^x$ halves every step.

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Growth vs. decay — what each factor does

Scenario	$b$	Behavior
Doubles each year	2.0	Strong growth
Grows 10% per year	1.10	Moderate growth
Constant	1.0	No change
Decays 10% per year	0.90	Moderate decay
Half each year	0.5	Strong decay

The crucial difference from linear functions: linear adds a constant; exponential multiplies by a constant.

Linear: $y = mx + b$ — $y$ increases by $m$ for every $+1$ in $x$ .
Exponential: $y = a \cdot b^x$ — $y$ multiplies by $b$ for every $+1$ in $x$ .

Exponentials grow much faster than linear functions over time, even if they start slower.

Percent growth/decay translation:

Growing by 5% per year: $b = 1 + 0.05 = 1.05$ . So $f(t) = a \cdot (1.05)^t$ .
Decaying by 8% per year: $b = 1 - 0.08 = 0.92$ . So $f(t) = a \cdot (0.92)^t$ .
Doubling every $k$ years: $f(t) = a \cdot 2^{t/k}$ .
Half-life of $h$ years: $f(t) = a \cdot (0.5)^{t/h}$ .

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Word problem → exponential function

Description	Function	Why
Starts at 200, grows 5% per year	$200 \cdot (1.05)^t$	Factor = $1 + 0.05$
Starts at 200, decays 5% per year	$200 \cdot (0.95)^t$	Factor = $1 - 0.05$
Starts at 80, doubles every 3 years	$80 \cdot 2^{t/3}$	$/3$ in exponent for 3-year period
Half-life of 8 years, starts at 80	$80 \cdot (0.5)^{t/8}$	$/8$ in exponent for 8-year half-life
\$1000 invested at 6% annual interest	$1000 \cdot (1.06)^t$	Annually-compounded interest

SAT word-problem template:

"A bacteria population starts at 200 and doubles every 3 hours."

→ $P(t) = 200 \cdot 2^{t/3}$ .

The $/3$ in the exponent says it takes 3 hours for the doubling to happen once.

Compound interest — a special case of exponential growth:

A = P\left(1 + \frac{r}{n}\right)^{nt}

where $P$ is principal, $r$ is annual interest rate (as decimal), $n$ is times compounded per year, $t$ is years. For annual compounding, simplify to $A = P(1 + r)^t$ .

Reading exponential graphs:

$y$ -intercept $= a$ (the starting value).
If the curve rises sharply, $b > 1$ and $b$ is large.
If the curve approaches but never touches the $x$ -axis from above, $0 < b < 1$ .

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Quick check

Identify $a$ (starting value) and $b$ (growth factor). Convert percent rates to factors: $1 + r$ for growth, $1 - r$ for decay.

What is the vertex of f(x) = x² - 4x + 3?

(4, 3)

(2, -1)

(-2, -1)

(2, 3)

Worked examples

Example 1

A town's population was 12,000 in 2020 and is decreasing by 4% per year. Which function models the population $t$ years after 2020?

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Example 2

A radioactive isotope has a half-life of 8 years. If the initial amount is 80 grams, how many grams remain after 24 years?

Common pitfalls

Confusing exponential with linear

Decreases by 4% per year is exponential (multiplicative). Decreases by 4 per year is linear (additive). The word percent almost always signals exponential.

Using the wrong growth factor

Growing by 5% → factor is $1.05$ , not $0.05$ . Decaying by 5% → factor is $0.95$ , not $-0.05$ . Always: $b = 1 \pm$ (rate as decimal).

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Forgetting to divide by the period in the exponent

Doubles every 3 years → $f(t) = a \cdot 2^{t/3}$ , NOT $a \cdot 2^t$ or $a \cdot 2 \cdot 3$ . The $/3$ scales time so the doubling happens once per 3 years.

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Plugging in $b$ when you should plug in $b - 1$

If the function is $a \cdot (1.07)^t$ , the growth rate is 7% per year (not 107%). The factor is $1.07$ . The rate is $0.07$ . Don't conflate them.

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Key takeaways

Exponential function: $f(x) = a \cdot b^x$ . $a$ = starting value. $b$ = growth factor.
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$b > 1$ : growth. $0 < b < 1$ : decay.
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Percent translation: growing by $r$ → $b = 1 + r$ . Decaying by $r$ → $b = 1 - r$ .
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Doubling every $k$ years: $a \cdot 2^{t/k}$ . Half-life $h$ : $a \cdot (0.5)^{t/h}$ .
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Linear adds; exponential multiplies. Look for percent or each year to know which model fits.

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Where this lesson fits

Before

Equivalent Expressions→Linear Functions→

Exponential Functions

You are here

→Nonlinear Equations in 1 Variable and Systems of Equations in 2 Variables