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Pre-Calculus Exponential and logarithmic functions (advanced)

Exponential models (compound, continuous, decay)

20 practice questions 0 video lessons Theory + worked examples

Exponential Functions and Models

Texas Precalculus (TEKS) • Standard P.2(N), P.5(I) • Exponential & Logarithmic Functions

Exponential Functions and Models is the opening topic of Exponential & Logarithmic Functions in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.2(N), P.5(I), which requires students to analyze situations modeled by exponential functions and solve exponential equations.

An exponential function \(f(x)=a\cdot b^x\) grows when \(b>1\) and decays when \(0

Texas Precalculus (TEKS) › Exponential & Logarithmic Functions › Exponential Functions and Models  —  Standard P.2(N), P.5(I)

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Theory

An exponential function has the variable in the exponent:

\[f(x)=a\cdot b^x,\qquad a\neq 0,\ b>0,\ b\neq 1.\]

The base \(b\) sets the behavior: \(b>1\) gives growth, \(0<b<1\) gives decay. Every such graph passes through \((0,a)\), has domain all reals, range \(y>0\) (when \(a>0\)), and a horizontal asymptote \(y=0\).

Constant ratio, not constant difference. Each unit step multiplies the output by \(b\) — that is what makes growth exponential rather than linear.
Exponential growth and decay y equals 2 to the x grows, y equals one half to the x decays; both pass through 0 comma 1 and approach the x-axis. x y y=2ₓ y=(½)ₓ (0,1)
\(y=2^x\) grows and \(y=(\dfrac12)^x\) decays; both pass through \((0,1)\).
Exponential function Exponential function Exponential function f(x) = a · bˣ b > 1: growth 0 < b < 1: decay y-intercept a, asymptote y = 0
The parts of an exponential function.

Exponential function and its growth/decay factor:

\[f(x)=a\cdot b^x;\qquad \text{growth rate } r:\ b=1+r,\quad \text{decay: } b=1-r\]
f of x equals a times b to the x; growth factor is one plus r, decay is one minus r
Percent change to base: \(+3\%\) per period means \(b=1.03\); \(-15\%\) means \(b=0.85\).

How to work with exponential functions

  1. Identify \(a\) and \(b\); check \(b\) for growth vs decay.
  2. Evaluate by substituting into the exponent.
  3. Model: \(a\) is the initial amount, \(b=1\pm r\) from the percent rate.
  4. From data: \(f(0)=a\), then a second point fixes \(b\).
Example 1 — Evaluate
For \(f(x)=3\cdot 2^x\), find \(f(0)\) and \(f(3)\).
Solution

Substitute, using \(2^0=1\).

\(f(0)\)\(=\)\(3\cdot 2^0=3\)
\(f(3)\)\(=\)\(3\cdot 2^3=24\)
f of 0 is 3, f of 3 is 24
Example 2 — Growth or decay
Classify \(f(x)=5(0.85)^x\).
Solution

The base \(0.85\) is between 0 and 1, so the function decays.

\(0<0.85<1\)\(\Rightarrow\)\(\text{decay}\)

It decreases by \(15\%\) each step.

decay of 15 percent each step
Example 3 — Population model
A town of \(8000\) grows \(3\%\) per year. Write a model and find the population after 10 years.
Solution

Growth factor \(1+0.03=1.03\).

\(P(t)\)\(=\)\(8000(1.03)^t\)
\(P(10)\)\(=\)\(8000(1.03)^{10}\approx 10{,}751\)
population after 10 years is about 10751
Example 4 — Find the base from data
An exponential function has \(f(0)=4\) and \(f(1)=12\). Find \(f(x)\).
Solution

\(f(0)=a=4\); then \(f(1)=a\cdot b=12\) gives the base.

\(4b\)\(=\)\(12\)
\(b\)\(=\)\(3\)
\(f(x)\)\(=\)\(4\cdot 3^x\)
the function is 4 times 3 to the x

Common pitfalls

The base is not the exponent. In \(a\cdot b^x\), the variable sits in the exponent, not the base.
Growth vs decay is about \(b\), not \(a\). A base above 1 grows; between 0 and 1 decays.
The asymptote is \(y=0\) (shifted if a constant is added); the graph never reaches it.

Frequently asked questions

What is an exponential function?

A function \(f(x)=a\cdot b^x\) with the variable in the exponent. It grows if \(b>1\) and decays if \(0<b<1\).

How do you tell growth from decay?

Look at the base: greater than 1 is growth, between 0 and 1 is decay.

What is the horizontal asymptote of an exponential function?

\(y=0\) for \(f(x)=a\cdot b^x\); adding a constant \(k\) shifts it to \(y=k\).

How do you find the base from a percent rate?

Add or subtract the rate from 1: \(+3\%\) gives \(b=1.03\), \(-15\%\) gives \(b=0.85\).