AP Precalculus: Exponential Functions

Linear, quadratic, and exponential models can all describe data that changes at a slowly changing rate, so you must validate the model. A scatterplot shows the overall shape. The coefficient of determination r² measures how closely predictions follow the data: values near 1 mean a close fit. A residual is observed value − predicted value. A positive residual means the model underestimates that point, and a negative one means it overestimates. A residual plot graphs the residuals against x. If the points are randomly scattered around the x-axis the model is appropriate, but a clear pattern (such as a U shape) means a better model exists. Finally, predictions far outside the data (extrapolation) are less reliable, and the model’s error is usually largest there.

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1. Exponential Functions and Their Graphs

An exponential function has the variable in the exponent. Compare y = 2x (a line), y = x² (a parabola) and y = 2^x: the first two raise the variable to a power, but in the third the variable is the power. The general form is f(x) = a·b^x, where a ≠ 0 is the initial value and b > 0, b ≠ 1 is the base (the growth or decay factor). Key properties when a > 0:

Domain: all real numbers. Range: y > 0, that is (0, ∞).
y-intercept: (0, a), since b⁰ = 1. There is no x-intercept.
Horizontal asymptote: y = 0.
If b > 1 the function always increases (exponential growth). If 0 < b < 1 it always decreases (exponential decay).
It is always concave up, has no extrema on its whole domain, and has no inflection points.

If a < 0, the graph is flipped over the x-axis: the range becomes y < 0 and the graph is always concave down.

Worked Example: For f(x) = 3·2^x, identify a and b, the y-intercept, the range, and f(−1), f(1), f(2).

Here a = 3 and b = 2. Since b > 1 and a > 0, the function grows. The y-intercept is (0, 3) and the range is y > 0.

f(−1) = 3·2⁻¹ = 1.5, f(1) = 3·2 = 6, and f(2) = 3·4 = 12. Each step right doubles the output.

a = 3, b = 2; y-intercept (0, 3); range (0, ∞)

y = 3·2^x (growth, b > 1) and y = 3·(1/2)^x (decay, 0 < b < 1). Both pass through (0, 3) and flatten toward the horizontal asymptote y = 0, the dashed line.

1. Which statement describes y = 5(0.8)^x?

Explanation: The initial value is a = 5, so the y-intercept is (0, 5). The base 0.8 is between 0 and 1, so the outputs shrink by 20% for each unit step right. That is decay.

2. What is the range of y = −4·3^x?

Explanation: 3^x is always positive, so multiplying by −4 makes every output negative. The graph never reaches 0 (the asymptote), so the range is y < 0.

3. Which is true of every function y = a·b^x with a > 0 and b > 0, b ≠ 1?

Explanation: The graph always approaches the x-axis on one side without touching it. The y-intercept is (0, a), which equals (0, 1) only when a = 1, and the function decreases when 0 < b < 1.

4. If f(x) = 2(1/2)^x, what is f(−3)?

Explanation: f(−3) = 2(1/2)⁻³ = 2·2³ = 2·8 = 16. A negative exponent flips the fraction.

2. End Behavior and Limits

The end behavior of f(x) = a·b^x describes what happens as x grows without bound in either direction. Every exponential function has exactly one end that levels off at the horizontal asymptote and one end that runs off to ∞ or −∞, so every end-behavior limit is one of three values: ∞, −∞, or 0. To decide, ask two questions: Is the output growing or shrinking as x moves in that direction? and What is the sign of a? For b > 1, 3^x → ∞ as x → ∞ and 3^x → 0 as x → −∞. For 0 < b < 1 it is the other way around. A negative a flips the sign of the unbounded end.

Worked Example: Evaluate the four end-behavior limits of p(x) = 4(1/2)^x and q(x) = −2·3^x.

p: the base is 1/2, so as x → ∞ the factors keep halving, giving lim p(x) = 0. As x → −∞, (1/2)^x = 2^−x becomes huge, so lim p(x) = ∞.

q: as x → ∞, 3^x → ∞ and the −2 makes it negative, so lim q(x) = −∞. As x → −∞, 3^x → 0, so lim q(x) = 0.

p: 0 at +∞, ∞ at −∞. q: −∞ at +∞, 0 at −∞.

p(x) = 4(1/2)^x approaches 0 as x → ∞ and grows without bound as x → −∞. q(x) = −2·3^x approaches 0 as x → −∞ and falls without bound as x → ∞.

1. What is lim_x→−∞ 3·2^x?

Explanation: As x becomes very negative, 2^x = 1/2^|x| gets arbitrarily close to 0, and 3 times a number near 0 is near 0.

2. What is lim_x→∞ (−5)(1.2)^x?

Explanation: The base 1.2 is greater than 1, so (1.2)^x grows without bound. Multiplying by −5 makes the outputs grow without bound in the negative direction.

3. What is lim_x→−∞ 6(1/4)^x?

Explanation: (1/4)^x = 4^−x. As x → −∞, −x → ∞, so 4^−x → ∞. Multiplying by 6 keeps it at ∞.

4. What is lim_x→∞ 7(0.9)^x?

Explanation: A base between 0 and 1 makes the outputs shrink toward the horizontal asymptote y = 0 as x increases, so the limit is 0.

3. Laws of Exponents with Variable Exponents

The familiar exponent laws still work when the exponent contains a variable, and they let you rewrite exponential expressions in a convenient form.

Product: b^m·bⁿ = b^m+n
Quotient: b^m/bⁿ = b^m−n
Power of a power: (b^m)ⁿ = b^mn
Negative exponent: b⁻ⁿ = 1/bⁿ
Rational exponent: b^m/n = (ⁿ√b)^m
Zero exponent: b⁰ = 1 for b ≠ 0

A good habit is to write every term with the same base, then combine exponents. Remember that the laws only combine exponents when the bases match: 2^x·3^x does not equal 5^x.

Worked Example: Simplify (4^x+1·2^3x)/8^x and write it in the form a·b^x.

Rewrite with base 2: 4^x+1 = (2²)^x+1 = 2^2x+2, and 8^x = (2³)^x = 2^3x.

Numerator: 2^2x+2·2^3x = 2^5x+2. Divide by 2^3x: 2^5x+2−3x = 2^2x+2.

Finish: 2^2x+2 = 2²·(2²)^x = 4·4^x.

The expression equals 4·4^x

1. What is 3^x+2/3^x−1?

Explanation: Subtract the exponents: (x + 2) − (x − 1) = 3, so the quotient is 3³ = 27. The variable cancels completely.

2. Which expression equals 9^x?

Explanation: Since 9 = 3², 9^x = (3²)^x = 3^2x.

3. Which expression equals the cube root of 8^x?

Explanation: The cube root is the 1/3 power: (8^x)^1/3 = 8^x/3 = (2³)^x/3 = 2^x.

4. Which expression equals 5^x−2?

Explanation: 5^x−2 = 5^x·5⁻² = 5^x/25. Choice C wrongly pulls the −2 outside as a subtraction.

4. Transformations of Exponential Graphs

Exponent laws turn some transformations into others. Product property: b^x+k = b^k·b^x, so every horizontal shift of an exponential function is also a vertical dilation by the factor a = b^k. Power property: b^cx = (b^c)^x, so every horizontal dilation is the same as a change of base. And since b^−x = (1/b)^x, replacing x with −x is a reflection over the y-axis that turns growth into decay. The usual rules also apply: f(x) + d shifts up, −f(x) reflects over the x-axis, and the horizontal asymptote moves to y = d.

Worked Example: Describe g(x) = 3^x+2 as a transformation of f(x) = 3^x in two ways.

As a shift: g(x) = f(x + 2), a horizontal shift 2 units left.

As a dilation: 3^x+2 = 3²·3^x = 9·3^x, a vertical stretch by a factor of 9. The graph is identical either way: g(0) = 9, which is f(2), because the shifted graph reaches at x = 0 the height f reaches at x = 2.

Change of base: h(x) = 81^x = (3⁴)^x = 3^4x, so h is f compressed horizontally by a factor of 1/4. And (1/3)^x = 3^−x is f reflected over the y-axis.

3^x+2 = 9·3^x

y = 3^x+2 is y = 3^x shifted 2 units left. It is also y = 9·3^x, a vertical stretch by a factor of 9.

1. Which expression is equivalent to 2^x−3?

Explanation: 2^x−3 = 2^x·2⁻³ = 2^x/8. A shift 3 units right matches a vertical shrink by a factor of 1/8.

2. Compared with y = 3^x, the graph of y = 27^x is:

Explanation: 27 = 3³, so 27^x = 3^3x. Multiplying the input by 3 compresses the graph horizontally by a factor of 1/3.

3. How does the graph of y = (1/4)^x compare with y = 4^x?

Explanation: (1/4)^x = 4^−x = f(−x), so it is the reflection of 4^x over the y-axis. A reflection over the x-axis would give −4^x.

4. Which expression equals 4^x+1?

Explanation: 4^x+1 = 4^x·4¹ = 4·4^x. Adding 1 inside the exponent multiplies the whole function by 4.

5. Linear or Exponential? Reading Tables

For a table with equally spaced inputs, compare two things. A constant difference in outputs (a constant average rate of change) means the data are linear, and the difference is the slope m. A constant ratio of consecutive outputs means the data are exponential, and that ratio is the base b. The value at x = 0 gives the y-intercept b in y = mx + b, or the initial value a in y = a·b^x. If neither the differences nor the ratios are constant, the model is neither. Always make sure the x-values are equally spaced before comparing.

Worked Example: The tables below have x = −1, 0, 1, 2, 3. Classify each and write a model.

f: 11, 8, 5, 2, −1. The differences are −3 each time, so f is linear. The value at x = 0 is 8, so f(x) = −3x + 8.

g: 3, 6, 12, 24, 48. The differences are 3, 6, 12, 24 (not constant), but the ratios are all 2. So g is exponential with b = 2 and a = g(0) = 6: g(x) = 6·2^x.

h: 2, 3, 5, 8, 12. The differences are 1, 2, 3, 4 and the ratios are 1.5, 1.67, 1.6, 1.5, so h is neither linear nor exponential.

f(x) = −3x + 8; g(x) = 6·2^x; h is neither

1. The table gives y = 5, 10, 20, 40 for x = 0, 1, 2, 3. Which function models it?

Explanation: The ratios are 10/5 = 2, 20/10 = 2, 40/20 = 2, so it is exponential with b = 2. The value at x = 0 is 5, so y = 5·2^x.

2. The table gives y = 81, 27, 9, 3 for x = 0, 1, 2, 3. Which function models it?

Explanation: Each output is one third of the previous one (27/81 = 1/3, 9/27 = 1/3, 3/9 = 1/3), so b = 1/3 and a = 81. The differences are −54, −18, −6, which are not constant, so it is not linear.

3. The table gives y = 4, 7, 10, 13 for x = 0, 1, 2, 3. Which describes it?

Explanation: The differences are all 3 (7 − 4, 10 − 7, 13 − 10), so it is linear with slope 3 and intercept 4. The ratios 1.75, 1.43, 1.3 are not constant.

4. The table gives y = 2, 4, 7, 11 for x = 0, 1, 2, 3. Which describes it?

Explanation: The differences are 2, 3, 4 (not constant) and the ratios are 2, 1.75, 1.57 (not constant), so the data are neither linear nor exponential.

6. Constructing an Exponential Model from Two Points

An exponential model y = a·b^x has two unknowns, so two points are enough to find both. Substitute each point to get two equations. If the points are (x₁, y₁) and (x₂, y₂), divide the equations: the a cancels, leaving b^x₂−x₁ = y₂/y₁. Take the matching root to get b (keep the positive one since b > 0), then substitute back to find a. If one of the points is the y-intercept (0, a), then a is already known and only b needs solving.

Worked Example: Find the exponential function through (1, 6) and (3, 54).

Write 6 = a·b and 54 = a·b³. Divide the second by the first: 54/6 = b³/b, so b² = 9 and b = 3 (b must be positive).

Substitute into 6 = a·3 to get a = 2.

Check with the second point: 2·3³ = 54. ✓

f(x) = 2·3^x

f(x) = 2·3^x passes through the two given points (1, 6) and (3, 54). The y-intercept (0, 2) is also marked.

1. An exponential function passes through (0, 5) and (3, 40). Which equation is it?

Explanation: The point (0, 5) gives a = 5. Then 40 = 5·b³ gives b³ = 8, so b = 2.

2. An exponential function passes through (1, 12) and (3, 3). Which equation is it?

Explanation: Dividing: 3/12 = b², so b² = 1/4 and b = 1/2. Then 12 = a(1/2) gives a = 24. Check: 24(1/2)³ = 3. ✓

3. If f is exponential with f(2) = 18 and f(5) = 486, what is f(4)?

Explanation: Dividing: 486/18 = 27 = b³, so b = 3. Then 18 = a·9 gives a = 2, so f(x) = 2·3^x and f(4) = 2·81 = 162.

4. An exponential function passes through (0, 10) and (2, 22.5). What is its base?

Explanation: With a = 10, 22.5 = 10·b², so b² = 2.25 and b = 1.5. The value 2.25 is b², not b.

7. Modeling Growth and Decay

When a quantity changes by a fixed percent each period, it is exponential: f(t) = a(1 + r)^t, where a is the starting amount, r is the rate written as a decimal, and 1 + r is the growth factor. For growth r > 0 (a 6% increase gives factor 1.06). For decay r < 0 (a 12% decrease gives factor 1 − 0.12 = 0.88, not 0.12). Two special cases are common: a quantity that doubles every period has factor 2, and a half-life H means the amount is multiplied by 1/2 every H units of time, so f(t) = a(1/2)^t/H. To find when a model reaches a target, graph the model and the line y = target and find the intersection with technology.

Worked Example: $2,000 is deposited at 6% interest compounded annually. Write the balance model, find the balance after 5 years, and estimate when it reaches $4,000.

The factor is 1 + 0.06 = 1.06, so B(t) = 2000(1.06)^t.

B(5) = 2000(1.06)⁵ ≈ 2000(1.3382) = $2,676.45.

For $4,000, solve 2000(1.06)^t = 4000, that is (1.06)^t = 2. Graphing y = B(t) and y = 4000 and finding the intersection gives t ≈ 11.9.

B(5) ≈ $2,676.45; doubles in about 11.9 years

B(t) = 2000(1.06)^t. The balance reaches $4,000 (the dashed line) at about t = 11.9 years, where the graph and the line intersect.

1. A $24,000 car loses 12% of its value each year. What is its value after 3 years, to the nearest cent?

Explanation: The decay factor is 1 − 0.12 = 0.88, so V(3) = 24000(0.88)³ = 24000(0.681472) = $16,355.33. Choice B subtracts 12% of the value three times from the original and does not compound.

2. A culture starts with 50 bacteria and triples every hour. How many are there after 4 hours?

Explanation: P(t) = 50·3^t, so P(4) = 50·81 = 4,050. Tripling is a factor of 3, not 1 + 3.

3. A sample has a half-life of 8 days and starts at 40 grams. How much remains after 24 days?

Explanation: 24 days is 3 half-lives, so the amount is 40(1/2)³ = 40/8 = 5 grams.

4. Which function models a quantity that decreases by 7% each period from a starting amount of a?

Explanation: A 7% decrease keeps 93% of the previous amount, so the factor is 1 − 0.07 = 0.93.

8. Continuous Growth and Exponential Regression

When growth or decay happens continuously, the model is f(t) = a·e^kt, where a is the initial amount, e ≈ 2.718 is the natural base, and k is the continuous rate: k > 0 means growth and k < 0 means decay. This is the same shape as a·b^t with b = e^k. When you have a data set and no formula, exponential regression on a calculator (ExpReg) returns the best-fit a and b for y = a·b^x. Enter the data in lists, run the regression, and then substitute into the model to predict. Keep full precision in a and b until the final answer.

Worked Example: $5,000 is invested at 4% interest compounded continuously. What is the balance after 10 years?

Here a = 5000, k = 0.04, and t = 10, so A(10) = 5000e^(0.04)(10) = 5000e^0.4.

Since e^0.4 ≈ 1.49182, A(10) ≈ $7,459.12.

Regression example: suppose ExpReg returns y = 3.1(1.9)^x for a data set. A prediction at x = 6 is 3.1(1.9)⁶ ≈ 3.1(47.046) ≈ 145.8.

A(10) ≈ $7,459.12

A(t) = 5000e^0.04t. The green point marks t = 10, where A ≈ $7,459.12.

1. A model is f(t) = 800e^0.03t. What is f(5), to the nearest hundredth?

Explanation: f(5) = 800e^0.15 = 800(1.161834) = 929.47. The exponent is kt = 0.03 × 5 = 0.15.

2. Which function models continuous decay?

Explanation: Decay needs a negative k. With k = −0.2, f(t) = 50e^−0.2t decreases toward 0 from 50. Choice C is negative-valued growth, not decay.

3. $1,500 is invested at 3.5% compounded continuously. What is the balance after 8 years, to the nearest cent?

Explanation: A(8) = 1500e^(0.035)(8) = 1500e^0.28 = 1500(1.32313) = $1,984.69.

4. Technology gives the regression model y = 2.4(1.5)^x. What is the predicted y at x = 5, to the nearest tenth?

Explanation: y = 2.4(1.5)⁵ = 2.4(7.59375) = 18.225, which rounds to 18.2.

9. Choosing and Validating Models: r² and Residuals

Worked Example: A model is f(x) = 5·2^x, and the observed values at x = 0, 1, 2, 3 are 5, 11, 22, 46. Find and interpret the residuals.

Predicted values: 5, 10, 20, 40.

Residuals (observed − predicted): 0, 1, 2, 6.

The residuals are all non-negative and growing, so the model underestimates and the gap increases. That is a pattern, not random scatter, so a base larger than 2 would fit better. The two residual plots below show what to look for.

Residuals 0, 1, 2, 6: the model underestimates

Random scatter: the model is appropriate.

U-shaped pattern: the model is not appropriate and a different (nonlinear) model is needed.

1. An observed value is 38 and the model predicts 41.5. What is the residual?

Explanation: Residual = observed − predicted = 38 − 41.5 = −3.5. The negative sign means the model overestimates this point.

2. A residual plot for a linear model shows a clear U shape. What does that indicate?

Explanation: A pattern in the residuals means the linear model is missing a systematic trend. Only random scatter around the horizontal axis supports the model.

3. An exponential model fits data from years 0 to 40 with r² = 0.998. It is used to predict year 90. Which statement is best?

Explanation: A high r² only describes the data range. Extrapolating far beyond it assumes the pattern continues, which may not hold.

4. A point has a positive residual. What does that mean?

Explanation: Residual = observed − predicted, so a positive residual means observed > predicted: the model underestimates that point.

Common Mistakes to Avoid

Mixing up x² and 2^x. In x² the variable is the base, and in 2^x it is the exponent. Their graphs and behavior are completely different.
Saying every exponential has y-intercept (0, 1). The y-intercept is (0, a). It equals (0, 1) only when a = 1.
Using the rate instead of the factor. A 12% decrease means the factor is 0.88 (1 − 0.12), and a 6% increase means 1.06, never 0.12 or 0.06.
Combining exponents of different bases. b^m·bⁿ = b^m+n only when the bases match, so 2^x·3^x ≠ 5^x. Rewrite to a common base first.
Going the wrong way on horizontal shifts. 2^x+3 shifts the graph 3 units left, which equals a vertical stretch by 8.
Comparing outputs of unequally spaced inputs. Differences and ratios only identify linear or exponential data when the x-values are equally spaced.
Trusting r² alone. A high r² can hide a patterned residual plot, and predictions far outside the data may be unreliable.

Frequently Asked Questions

What is the difference between exponential growth and linear growth?

Linear growth adds the same amount each step, while exponential growth multiplies by the same factor each step. That is why exponential data show constant ratios and linear data show constant differences.

Why must the base b be positive and not equal to 1?

If b = 1 the function is the constant a, which is linear. If b is negative, b^x is not defined for many real x, such as x = 1/2, so it would not be a function on all real numbers.

What is e, and when do I use it?

e ≈ 2.718 is a special base that appears in continuous growth and decay. Use f(t) = a·e^kt when the problem says continuously compounded or continuous growth, and a(1 + r)^t when the change happens once per period.

Where can I practice more problems like these?

The Exponential and Logarithmic Functions quiz in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.