Unit 11 · Lesson 1 of 5

Exponential functions, growth, decay, and interest

About 25 minutes · 5 steps · one idea: the variable is in the exponent

You'll be able to: evaluate \(a \cdot b^x\), tell growth from decay, read the y-intercept and asymptote, and use the interest formulas.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first, then check.

Find the inverse of \(f(x) = 3x - 6\). Type \(f^{-1}(x)\).

Write \(y = 3x - 6\), then swap \(x\) and \(y\): \(x = 3y - 6\).

Solve for \(y\): \(x + 6 = 3y\), so \(y = \dfrac{x+6}{3}\).

Answer: \(f^{-1}(x) = \dfrac{x+6}{3}\).

\(2^{-3}\)

A negative exponent means "one over": \(2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}\).

Step 2 of 5The idea

An exponential function has the variable in the exponent, like \(f(x) = 3 \cdot 2^x\). Each time \(x\) goes up by 1, the output gets multiplied by the same number.

The form. \(f(x) = a \cdot b^x\), with \(b \gt 0\) and \(b \ne 1\).

\(a\) is the starting value. \(f(0) = a \cdot b^0 = a\), so the graph crosses the y-axis at \((0, a)\).

\(b\) is the growth or decay factor. \(b \gt 1\): growth (rises left to right). \(0 \lt b \lt 1\): decay (falls left to right).

Interest. Compounded \(n\) times a year: \(A = P\left(1 + \dfrac{r}{n}\right)^{nt}\). Compounded continuously: \(A = Pe^{rt}\).

The graph of \(a \cdot b^x\) gets closer and closer to the x-axis on one side but never touches it. That line, \(y = 0\), is called the horizontal asymptote. If a number is added, like \(2^x + 3\), the graph shifts up 3 and the asymptote becomes \(y = 3\) (shifts were in Lesson 0052).

Growth \(y = 2^x\) and decay \(y = \left(\tfrac12\right)^x\). Both cross at \((0, 1)\) and hug the line \(y = 0\) on one side.

In the interest formulas, \(P\) is the starting money, \(r\) is the yearly rate as a decimal (5% = 0.05), \(t\) is the number of years, and \(e \approx 2.718\) is the natural base. Problems with interest say "calculator OK".

Why this shows up on the placement test: College Algebra starts its exponential unit assuming you can evaluate \(a \cdot b^x\) and plug into an interest formula.

Step 3 of 5Worked examples

Example 1: \(f(x) = 3 \cdot 2^x\). Find \(f(0)\), \(f(2)\), and \(f(-1)\).

  1. \(f(0) = 3 \cdot 2^0 = 3 \cdot 1 = 3\). So the y-intercept is \((0, 3)\).
  2. \(f(2) = 3 \cdot 2^2 = 3 \cdot 4 = 12\). Power first, then multiply.
  3. \(f(-1) = 3 \cdot 2^{-1} = 3 \cdot \dfrac{1}{2} = \dfrac{3}{2}\). A negative exponent flips to "one over".
  4. The base is \(2 \gt 1\), so this is growth, and the asymptote is \(y = 0\).

Example 2 (calculator OK): $2000 at 5% for 3 years

  1. Compounded monthly: \(P = 2000\), \(r = 0.05\), \(n = 12\), \(t = 3\). "Monthly" means 12 times a year.
  2. \(A = 2000\left(1 + \dfrac{0.05}{12}\right)^{12 \cdot 3} = 2000\left(1 + \dfrac{0.05}{12}\right)^{36}\).
  3. Calculator: \(A \approx 2322.94\), so about $2322.94. Rounded to the nearest cent at the very end.
  4. Compounded continuously: \(A = 2000e^{0.05 \cdot 3} = 2000e^{0.15} \approx 2323.67\), so about $2323.67, a little more.

Step 4 of 5Practice

Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Use a calculator only where the problem says so.

\(f(x) = 5 \cdot 3^x\). Find \(f(2)\).

Power first: \(3^2 = 9\). Then multiply: \(5 \cdot 9 = 45\). Answer: \(45\).

\(g(x) = 16\left(\dfrac{1}{2}\right)^x\). Find \(g(3)\).

\(\left(\dfrac12\right)^3 = \dfrac{1}{8}\). Then \(16 \cdot \dfrac18 = 2\). Answer: \(2\).

\(f(x) = 4 \cdot 2^x\). Find \(f(-2)\).

\(2^{-2} = \dfrac{1}{2^2} = \dfrac14\). Then \(4 \cdot \dfrac14 = 1\). Answer: \(1\).

Where does the graph of \(f(x) = 7(0.8)^x\) cross the y-axis? Type the point.

The y-axis is where \(x = 0\): \(f(0) = 7(0.8)^0 = 7 \cdot 1 = 7\). Answer: \((0, 7)\).

Which function shows exponential decay?

  1. \(f(x) = 3(1.2)^x\)
  2. \(f(x) = 0.5(2)^x\)
  3. \(f(x) = 2(1.25)^x\)
  4. \(f(x) = 4(0.75)^x\)

Look only at the base \(b\), the number being raised to the power \(x\).

(a) \(b = 1.2 \gt 1\): growth. (b) \(b = 2 \gt 1\): growth. The 0.5 in front is only the starting value. (c) \(b = 1.25 \gt 1\): growth.

(d) \(b = 0.75\), which is between 0 and 1: decay. This one.

What is the horizontal asymptote of \(f(x) = 3 \cdot 2^x - 4\)?

\(3 \cdot 2^x\) is always positive, and it gets closer and closer to 0 as \(x\) goes far to the left.

So \(f(x)\) gets closer and closer to \(0 - 4 = -4\). The \(-4\) shifts the whole graph down 4.

Answer: \(y = -4\).

Calculator OK. You put $1500 in an account that pays 4% interest compounded quarterly (4 times a year). How much is in the account after 5 years? Round to the nearest cent.

\(P = 1500\), \(r = 0.04\), \(n = 4\), \(t = 5\), so \(nt = 20\).

\(A = 1500\left(1 + \dfrac{0.04}{4}\right)^{20} = 1500(1.01)^{20}\).

Calculator: \(A \approx 1830.285...\). Answer: about $1830.29.

Calculator OK. $900 is invested at 5% compounded continuously. How much is it worth after 8 years? Round to the nearest cent.

"Continuously" means \(A = Pe^{rt}\), with \(P = 900\), \(r = 0.05\), \(t = 8\).

\(rt = 0.05 \cdot 8 = 0.4\), so \(A = 900e^{0.4}\).

Calculator: \(A \approx 1342.642...\). Answer: about $1342.64.

Step 5 of 5Wrap-up

You can now read \(a \cdot b^x\): \(a\) is where it starts, \(b\) tells growth or decay, the asymptote is \(y = 0\) unless the graph is shifted, and the interest formulas are plug-in and calculate.

Read or watch next: OpenStax Intermediate Algebra 2e, 10.2 Evaluate and Graph Exponential Functions (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Exponential growth & decay (short videos). Quick rules: Exponentials and logs cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why does compounding more often give a little more money?" It knows this course and can explain things another way.