How do I know whether data are linear or exponential?

With equally spaced inputs, check the outputs. About constant differences point to linear, and about constant ratios point to exponential.

What does a residual plot tell me?

Whether the model has the right shape. Small residuals scattered above and below zero with no pattern mean the model fits, and a curved pattern means it misses the shape of the data.

What does a linear pattern on a semi-log plot mean?

The data are exponential. A line for ln y with slope m and intercept k means y = a times b to the x, where a is e to the k and b is e to the m.

How do I type an exponential regression in Desmos?

Put the data in a table, then type y_1 ~ ab^(x_1) on a new line, with the tilde in place of an equals sign. Desmos prints the best values of a and b.

Regression and choosing a model for AP Precalculus

Updated November 10, 2026

AP Precalculus asks you to build models from data, not just use them. You look at a table, decide which kind of function fits, run a regression in Desmos, and then judge the fit with residuals. The second free-response question, Modeling a Non-Periodic Context, is built around exactly this, and it's on a calculator part of the exam.

This guide covers how to choose a model, the Desmos syntax for each regression, how to read a residual plot, and what a semi-log plot tells you. Every example uses coefficients from a regression that was actually run on the data shown.

Choosing a model from a table

When the inputs are equally spaced, look at how the outputs change from one row to the next. The pattern points to a model type.

Linear: the differences between outputs are about constant. The output changes by the same amount each step.
Quadratic: the differences change by about the same amount each step, so the second differences are about constant.
Cubic: the third differences are about constant.
Exponential: the ratios of outputs are about constant. The output is multiplied by the same factor each step.
Logarithmic: multiplying the input by a constant factor adds about the same amount to the output. Look at inputs like $1, 2, 4, 8$ .
Sinusoidal: the outputs rise and fall in a repeating cycle.

Real data won't match a pattern exactly, so check two more things. Shape: does the scatter plot curve, level off, or turn around? A quadratic turns around once, and an exponential never does. Context: a drug in the bloodstream decays toward $0$ , so a linear model that goes negative can't be right for long. Population growth by a percent each year suggests exponential.

Running a regression in Desmos

1Click the + in the expression list and choose table. Desmos names the columns $x_1$ and $y_1$ .
2Type the inputs in the $x_1$ column and the outputs in the $y_1$ column. The points appear on the graph.
3On a new line, type the model with a tilde, ~, in place of the equals sign. Desmos finds the best values of the letters and prints them below the line.

Linear: $y_1 \sim m x_1 + b$
Quadratic: $y_1 \sim a x_1^2 + b x_1 + c$
Cubic: $y_1 \sim a x_1^3 + b x_1^2 + c x_1 + d$
Exponential: $y_1 \sim a b^{x_1}$
Logarithmic: $y_1 \sim a + b \ln(x_1)$
Sinusoidal: $y_1 \sim a \sin(b x_1 + c) + d$

After the regression runs, the letters are defined everywhere in that graph. Type $f(x) = a b^{x}$ on a new line and you can evaluate $f(8)$ , or graph $y = 6$ and click where it crosses $f$ . Desmos uses the full unrounded values, so your predictions don't pick up rounding error.

Worked example: an exponential model

A car losing value

The value

V

of a used car, in thousands of dollars,

t

years after it was bought is shown.

t

0, 1, 2, 3, 4, 5

V

24.0, 20.5, 17.6, 15.1, 12.9, 11.0

. (a) Explain why an exponential model is appropriate. (b) Find an exponential regression

V = a b^{t}

. (c) Use the model to predict the value at

t = 8

, and say what

b

means in context.

(a) The ratios of consecutive values are $\frac{20.5}{24.0} \approx 0.854$ , then $0.859$ , $0.858$ , $0.854$ , and $0.853$ . Over equal one-year steps, the value is multiplied by about the same factor, which is the pattern of an exponential model.

(b) Enter the table and type $y_1 \sim a b^{x_1}$ . Desmos gives $a \approx 23.990$ and $b \approx 0.856$ , so $V = 23.990(0.856)^{t}$ .

y = 23.990(0.856)^{x}

The data points and the exponential regression. The curve passes very close to every point.

(c) Type $f(x) = a b^{x}$ and then $f(8)$ . Desmos shows $6.927$ , so the model predicts a value of about $6.927$ thousand dollars. If you use the rounded values instead, $23.990(0.856)^{8} \approx 6.915$ , which is slightly off. The base $b \approx 0.856$ means the car keeps about $85.6\%$ of its value each year, so it loses about $14.4\%$ per year.

Answer: (a) Nearly constant ratios, about $0.85$ to $0.86$ . (b) $V = 23.990(0.856)^{t}$ . (c) About $6.927$ thousand dollars; the car loses about $14.4\%$ of its value each year.

Reading residuals

A residual is the actual value minus the predicted value: residual $= y - \hat{y}$ . A positive residual means the model predicted too low for that point. A negative one means it predicted too high.

Under a regression line, Desmos lists the residuals as a new variable, $e_1$ , with an option to plot them. Plotting $e_1$ against $x_1$ gives a residual plot. A good model leaves small residuals scattered above and below $0$ with no pattern. A curved pattern, such as positive at both ends and negative in the middle, means the model is missing the shape of the data.

Linear or quadratic for stopping distance

A car's stopping distance

d

, in feet, at speed

s

, in miles per hour, is measured.

s

20, 30, 40, 50, 60, 70

d

40, 73, 117, 169, 231, 305

. Use residuals to decide whether a linear or a quadratic model is more appropriate, then use that model to predict the stopping distance at

45

miles per hour.

The first differences are $33, 44, 52, 62, 74$ , so the data aren't linear. The second differences are $11, 8, 10, 12$ , close to constant, which already hints at a quadratic.

A linear regression, $y_1 \sim m x_1 + b$ , gives $d = 5.289s - 82.152$ . Its residuals are about $16.4, -3.5, -12.4, -13.3, -4.2, 17.0$ : positive at both ends and negative in the middle.

y = 0

Residuals of the linear model. The U shape means a line misses the curve in the data.

A quadratic regression, $y_1 \sim a x_1^2 + b x_1 + c$ , gives $a \approx 0.049$ , $b \approx 0.837$ , and $c \approx 3.586$ . Its residuals are $-0.1, -0.2, 0.8, -0.1, -0.9, 0.5$ : all under $1$ foot, with no pattern.

y = 0

Residuals of the quadratic model, on the same scale.

The quadratic model is more appropriate. Define $f(x) = a x^2 + b x + c$ and evaluate $f(45)$ : about $141.406$ feet. With the rounded coefficients you'd get about $140.476$ , a full foot off, which is why you let Desmos keep the unrounded values.

Answer: Quadratic, because its residuals are small and patternless while the linear residuals form a U shape; about $141.406$ feet at $45$ mph.

Semi-log plots

A semi-log plot puts $\ln y$ (or $\log y$ ) on the vertical axis and keeps $x$ as it is. If the points on a semi-log plot form a line, the data are exponential. Here's why: if $y = a b^{x}$ , then $\ln y = \ln a + x \ln b$ , which is linear in $x$ with slope $\ln b$ and intercept $\ln a$ .

To go back from a line $\ln y = mx + k$ to an exponential model, undo the log: $y = e^{mx + k} = e^{k}(e^{m})^{x}$ . So $a = e^{k}$ and $b = e^{m}$ . For a base-10 log, use $10$ in place of $e$ . In Desmos, make the new column by typing $L = \ln(y_1)$ , then fit $L \sim m x_1 + k$ .

Views of a new video

A video's views

v

on day

d

after posting were

150, 260, 470, 820, 1450, 2560

for

d = 1

6

. (a) Fit a line to the points

(d, \ln v)

. (b) Turn it into an exponential model for

v

. (c) Predict the views on day

8

(a) The logs are about $5.011, 5.561, 6.153, 6.709, 7.279, 7.848$ . They go up by about $0.57$ each day, so the semi-log plot is close to a line. The regression $L \sim m x_1 + k$ gives $m \approx 0.569$ and $k \approx 4.437$ .

y = 0.569x + 4.437

On a semi-log plot the points fall on a line, so the views grow exponentially.

(b) $v = e^{4.437}(e^{0.569})^{d}$ . With the unrounded values, $a = e^{k} \approx 84.515$ and $b = e^{m} \approx 1.766$ , so $v = 84.515(1.766)^{d}$ . The views grow by about $76.6\%$ per day.

(c) $v = e^{8m + k} \approx 7983$ views. That's extrapolation past the last measured day, and growth this fast can't last, so treat it as a rough estimate.

Answer: (a) $\ln v = 0.569d + 4.437$ . (b) $v = 84.515(1.766)^{d}$ . (c) About $7983$ views.

Worked example: a logarithmic model

A new worker speeding up

A new warehouse worker packs

y

boxes per hour after

x

days on the job.

x

1, 2, 4, 8, 16

y

20, 26, 31, 37, 42

. Explain why a logarithmic model fits, find a logarithmic regression model for

y

, and predict the rate after

30

days.

Each input is double the one before, and each doubling adds about the same amount: $6, 5, 6, 5$ . Equal factors in the input giving equal changes in the output is the logarithmic pattern. A linear model would need equal changes for equal steps in $x$ , and these steps get longer.

Type $y_1 \sim a + b\ln(x_1)$ . Desmos gives $a \approx 20.200$ and $b \approx 7.935$ .

y = 20.200 + 7.935\ln(x)

The logarithmic curve rises quickly at first, then slows down.

$20.200 + 7.935\ln(30) \approx 47.188$ boxes per hour. The model keeps rising without bound, which can't be true forever, so it's best used near the measured days.

Answer: $y = 20.200 + 7.935\ln(x)$ ; about $47.188$ boxes per hour after $30$ days.

A quick check

From a semi-log line to a model

A data set of positive values lies very close to the line

\ln y = 0.2x + 1.5

on a semi-log plot. Which model fits the data?

A. $y = 4.482(1.221)^{x}$
B. $y = 1.5(0.2)^{x}$
C. $y = 0.2x + 1.5$
D. $y = 1.5(1.221)^{x}$

Undo the log: $y = e^{1.5}(e^{0.2})^{x}$ , and $e^{1.5} \approx 4.482$ and $e^{0.2} \approx 1.221$ . Choice C is the line for $\ln y$ , not for $y$ . Choice D forgets to exponentiate the intercept.

Answer: A

The algebra behind these models, like turning $e^{0.2x}$ into $(1.221)^{x}$ , is in the exponential and logarithmic equations guide. For everything else Desmos does on test day, see Desmos on the AP Precalculus exam.

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Questions

How do I know whether data are linear or exponential?: With equally spaced inputs, check the outputs. About constant differences point to linear, and about constant ratios point to exponential.
What does a residual plot tell me?: Whether the model has the right shape. Small residuals scattered above and below zero with no pattern mean the model fits, and a curved pattern means it misses the shape of the data.
What does a linear pattern on a semi-log plot mean?: The data are exponential. A line for ln y with slope m and intercept k means y = a times b to the x, where a is e to the k and b is e to the m.
How do I type an exponential regression in Desmos?: Put the data in a table, then type y_1 ~ ab^(x_1) on a new line, with the tilde in place of an equals sign. Desmos prints the best values of a and b.