Unit 2 · Topic 2.3

2.3 Exponential Functions

An exponential function f(x) = a · bˣ multiplies by the same factor b for each 1-unit step in x. It always rises or always falls, it always bends the same way, and at one end it levels off toward a horizontal asymptote.

Key terms

exponential growth
exponential decay
base
initial value
horizontal asymptote

The general form

An exponential function in general form is f(x) = a · bˣ, with a ≠ 0, b > 0 and b ≠ 1. The initial value is a = f(0), and b is the base.

When a > 0 and b > 1, the function shows exponential growth. When a > 0 and 0 < b < 1, it shows exponential decay.

For whole-number inputs, x counts how many factors of b multiply the starting value: f(3) = a · b · b · b. But the domain is all real numbers, so f(2.5) makes sense too.

Shape: monotonic and one concavity

Every input step of the same length multiplies the output by the same factor. That steady multiplying means an exponential function moves in one direction the whole way (always up or always down) and bends one way the whole way (always concave up or always concave down).

So an exponential function has no turning points and no points of inflection. It has a maximum or minimum only on a closed interval, at an endpoint.

Signs	Direction	Concavity	Example
a > 0, b > 1	Increasing	Concave up	3 · 2ˣ
a > 0, 0 < b < 1	Decreasing	Concave up	3 · (0.5)ˣ
a < 0, b > 1	Decreasing	Concave down	−3 · 2ˣ
a < 0, 0 < b < 1	Increasing	Concave down	−3 · (0.5)ˣ

End behavior and the asymptote

At one end, the outputs head toward 0, so the graph has a horizontal asymptote at y = 0. At the other end, the outputs grow or fall without bound.

For 2ˣ: lim (x→−∞) 2ˣ = 0 and lim (x→∞) 2ˣ = ∞. For (0.5)ˣ it's the other way round: lim (x→∞) (0.5)ˣ = 0 and lim (x→−∞) (0.5)ˣ = ∞. With a < 0, the unbounded end goes to −∞ instead.

Shifted exponentials

Adding a constant, f(x) = a · bˣ + k, moves the asymptote to y = k. The outputs are no longer proportional, but the outputs minus k are.

This gives a test for data: if the ratios of consecutive outputs aren't constant, try subtracting a constant first. If the adjusted values change by a constant ratio, an additive transformation of an exponential function models the data.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Describing an exponential decay function
Describe f(x) = 5(0.8)ˣ: its initial value, direction, concavity, asymptote and end behavior.
Show the solutionHide the solution
1. Step 1: a = 5 > 0 and b = 0.8, between 0 and 1, so this is exponential decay. f(0) = 5.
2. Step 2: Each 1-unit step multiplies the output by 0.8, a 20% drop. The function is always decreasing.
3. Step 3: The drops get smaller as the values shrink (5 to 4 is a drop of 1; 4 to 3.2 is a drop of 0.8), so the rate of change is increasing toward 0: concave up.
4. Step 4: As x → ∞, the outputs approach 0. As x → −∞, they grow without bound.
Answer: Initial value 5; decreasing; concave up; horizontal asymptote y = 0; lim (x→∞) f(x) = 0 and lim (x→−∞) f(x) = ∞.
Example 2
A reflected and shifted exponential
Describe g(x) = −2 · 3ˣ + 4.
Show the solutionHide the solution
1. Step 1: Start with 3ˣ: increasing, concave up, asymptote y = 0 on the left.
2. Step 2: Multiplying by −2 reflects it over the x-axis and stretches it: −2 · 3ˣ is decreasing and concave down.
3. Step 3: Adding 4 shifts it up, so the asymptote becomes y = 4.
4. Step 4: As x → −∞, 3ˣ → 0, so g(x) → 4. As x → ∞, −2 · 3ˣ → −∞.
Answer: Decreasing and concave down; horizontal asymptote y = 4; lim (x→−∞) g(x) = 4 and lim (x→∞) g(x) = −∞; the range is y < 4.
Example 3
Trap: data that needs a shift
A function has f(0) = 7, f(1) = 9, f(2) = 13, f(3) = 21. Is it exponential? Find a model.
Show the solutionHide the solution
1. Step 1: Ratios: 9/7 ≈ 1.29, 13/9 ≈ 1.44, 21/13 ≈ 1.62. Not constant, so f is not of the form a · bˣ. Don't stop here.
2. Step 2: Differences: 2, 4, 8. They double each step, which is the signature of c · 2ˣ plus a constant.
3. Step 3: If f(x) = c · 2ˣ + k, the difference from x to x + 1 is c · 2ˣ. At x = 0 that is c = 2. Then k = f(0) − c = 7 − 2 = 5.
4. Step 4: Check: subtracting 5 gives 2, 4, 8, 16, which have a constant ratio 2. And f(3) = 2 · 8 + 5 = 21.
Answer: f(x) = 2 · 2ˣ + 5, an exponential function shifted up 5 (asymptote y = 5).

Common mistakes

Assuming every exponential function is increasing. Decay (0 < b < 1) and negative a both produce decreasing functions.
Saying an exponential graph eventually reaches its asymptote. It gets as close as you like but never equals 0 (or k, after a shift).
Giving up on an exponential model when ratios aren't constant. Check whether subtracting a constant makes them constant.

On the exam

Expect multiple-choice questions on growth vs decay, concavity, asymptotes and limits of exponential functions, often with a transformation included.
For a free-response description, state direction and concavity and give limit statements for the end behavior.

Connected topics

Videos

AP Precalculus – 2.3 Exponential Functions
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Graphing exponential functions | Exponential and logarithmic functions | Algebra II | Khan Academy
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AP Precalculus Notes (Topic 2.3) Exponential Functions
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Quick Review Topic 2.3 Exponential Functions - AP Precalculus
Michael Porinchak - AP Statistics & AP PrecalculusWatch on YouTube (opens in a new tab)
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Professor Dave ExplainsWatch on YouTube (opens in a new tab)