Exponential & Logarithmic Functions

Exponential Growth & Decay

Graphing and interpreting exponential growth and decay functions.

Advanced20 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Growing by Addition vs. Growing by Multiplication

One account earns a flat $50 every month, no matter the balance. A different account earns 5% of its current balance every month. Before reading on, predict: after many years, which account do you think ends up larger — and why might the gap between them keep growing rather than staying fixed?

Definition — Exponential Function

A function of the form y = a · bˣ, where a is the starting value and b is the growth factor. If b > 1, the function models exponential growth; if 0 < b < 1, it models exponential decay.

The flat $50 account grows linearly — additively, the same fixed amount every time. The 5% account grows exponentially — multiplicatively, and crucially, each new gain is calculated on an already-larger balance than before. That compounding effect is why exponential growth eventually overtakes any linear (or even any polynomial) growth, no matter how large the linear rate is at first: multiplicative growth builds on itself, while additive growth just keeps adding the same fixed amount forever.

Worked Example — Modeling Exponential Growth

A population of 200 bacteria doubles every hour. Write a function for the population after t hours, and find the population after 5 hours. The growth factor is 2 (doubling), so P(t) = 200 · 2ᵗ. After 5 hours: P(5) = 200 · 2⁵ = 200 · 32 = 6,400.

Worked Example — Modeling Exponential Decay

A radioactive sample of 80 grams loses 15% of its mass every year. Write a function for the mass after t years, and find the mass after 4 years. Losing 15% means 85% remains each year, so the decay factor is 0.85: M(t) = 80 · (0.85)ᵗ. After 4 years: M(4) = 80 · (0.85)⁴ ≈ 80 · 0.522 ≈ 41.8 grams.

y = a \cdot b^{x} \qquad b>1 \text{: growth} \qquad 0<b<1 \text{: decay}

Function Explorer

f(x) =

sin(x)

Transform: g(x) = a·f(b(x − h)) + k

a — vertical stretch/reflect1

b — horizontal stretch/reflect1

h — horizontal shift0

k — vertical shift0

Composition

Compose with another function

Trace mode

Show inverse view

Analysis (of the transformed function, in view)

y-intercept: (0, 0)
x-intercepts: (-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Extrema: local min at (-7.85, -1); local max at (-4.71, 1); local min at (-1.57, -1); local max at (1.57, 1); local min at (4.71, -1); local max at (7.85, 1)
Inflection points: (-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [-1, 1]

Tip

For a percent increase, the growth factor is 1 plus the rate (as a decimal); for a percent decrease, it's 1 minus the rate — the same multiplier idea from Grade 7's percent-change lessons, now built directly into a full function of time.

Common Mistakes

Using the percent rate itself as the growth factor, such as using 0.15 instead of 0.85 for a 15% decay.
A decay rate of 15% means 85% remains — the growth factor is 1 minus the decay rate (or 1 plus the growth rate for growth), not the rate itself.
Treating exponential growth as if it were the same as linear growth over long time periods, expecting a roughly constant amount of change per time step.
Exponential growth compounds — the amount of change itself grows over time, unlike linear growth's constant rate — which is why exponential functions eventually outpace linear ones by an ever-widening margin.

Key Takeaways

Exponential functions grow or decay by a constant multiplying factor, not a constant additive amount.
A growth factor greater than 1 models growth; between 0 and 1 models decay.
Exponential growth compounds on itself, which is why it eventually outpaces any linear growth rate.

Summary

Exponential functions model quantities that grow or shrink multiplicatively. The next lesson asks the natural reverse question: given an exponential relationship, how do you solve for the exponent itself?

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