Competing Function Model Validation

AP Precalculus· AP Precalculus CED — Exponential and Logarithmic Functions· 14 min read

1. What is Competing Function Model Validation?★★☆☆☆⏱ 2 min

Competing function model validation is the process of testing two or more candidate function models (most commonly linear, exponential, and power) against real-world bivariate data to select the model that best describes the underlying relationship. This topic accounts for 2-3% of the total AP Precalculus exam score, and appears in both multiple-choice and free-response sections.

On the exam, you will typically be given a scatterplot, data table, or pre-fit candidate models, then asked to justify which model is most appropriate using quantitative or graphical evidence. Unlike fitting a single model, validation focuses on comparing competing options, a critical skill for applied data analysis that is heavily weighted for justification points on FRQs.

2. Graphical Residual Analysis★★☆☆☆⏱ 3 min

📘 Definition

Residual

$e_{i}$

The difference between the observed response value $y_{i}$ and the model's predicted response value $i l d e y_{i}$ for a given data point $(x_{i}, y_{i})$ : $e_{i} = y_{i} - \overset{y}{^}_{i}$

Example:

For model $\overset{y}{^} = 2 x + 1$ and point $(3, 8)$ , $e_{i} = 8 - 7 = 1$

Residual analysis is the most intuitive and widely tested method for comparing model fit on the AP exam. If a model fits well, residuals will be randomly scattered around the horizontal axis $e = 0$ , with no clear systematic pattern (like a curve, trend, or funnel shape). If residuals show a clear pattern, the model is missing the underlying trend, so another competing model is a better choice.

📐 Worked Example

An ecologist measures the population of deer in a protected forest over 12 years. Residual plots for two candidate models: Model 1 (linear growth) has residuals that start positive, turn negative in the middle, then positive again, forming a clear upward-opening parabolic pattern. Model 2 (exponential growth) has residuals randomly scattered between $- 18$ and $15$ deer, centered evenly around $e = 0$ . Which model is a better fit? Justify your choice.

1
Recall that a well-fitting model produces residuals with no systematic pattern, randomly distributed around the zero line.
2
Evaluate Model 1: The clear parabolic pattern means the linear model fails to capture the non-linear trend in deer population growth, so it is a poor fit.
3
Evaluate Model 2: Residuals have no systematic pattern and are evenly scattered around zero, which indicates the model correctly captures the underlying trend.
4
Conclude that Model 2 (exponential growth) is the better fitting model.

Exam tip:

On AP FRQ, you must explicitly reference the presence/absence of a pattern in residuals to earn the justification point; just saying "the residuals are better" will not get you full credit.

3. Coefficient of Determination ($R^2$) for Model Comparison★★★☆☆⏱ 3 min

The coefficient of determination, written $R^{2}$ , is a quantitative measure of the proportion of variation in the response variable $y$ that is explained by the explanatory variable $x$ in the fitted model. $R^{2}$ ranges from $0$ to $1$ (or 0% to 100%). When comparing two competing models fit to the same data, the model with the higher $R^{2}$ explains more variation in $y$ , so it is generally the better fitting model.

📐 Worked Example

A small business owner compares two models for annual revenue over 10 years: a linear model $R (t) = 125000 + 18000 t$ and an exponential model $R (t) = 120000 (1.11)^{t}$ . Both models are fit to the original untransformed annual revenue data. The linear model has $R^{2} = 0.84$ , and the exponential model has $R^{2} = 0.95$ . Which model is a better fit? Justify.

1
Confirm that both models use the same response variable (original untransformed annual revenue $R$ ), so direct comparison of $R^{2}$ is valid.
2
Recall that for competing models with the same response variable, a higher $R^{2}$ indicates more variation explained and a better overall fit.
3
Compare values: $0.95 > 0.84$ , so the exponential model explains 11% more variation in annual revenue than the linear model.
4
Conclude the exponential model is the better fit for this revenue data.

Exam tip:

Always check that the response variable is identical for both models before comparing $R^{2}$ ; transformed models have $R^{2}$ values that only compare to other models of the same transformed response.

4. Log-Transformation for Linearization of Non-Linear Models★★★★☆⏱ 4 min

When comparing two non-linear models (exponential vs power), we use log-transformation to linearize both models, then compare the fit of the linearized versions to select the best original model. This method is only valid if all $x$ and $y$ values are positive, since the logarithm is only defined for positive inputs.

An exponential model has the form $y = a b^{x}$ . Taking the natural log of both sides gives the linear form:

y = ab^x \implies \ln(y) = \ln(a) + x\ln(b)

A power model has the form $y = a x^{b}$ . Taking the natural log of both sides gives its linear form:

y = ax^b \implies \ln(y) = \ln(a) + b\ln(x)

To compare which non-linear model fits better, we check the $R^{2}$ of the linearized regression: the model whose linearized form has a higher $R^{2}$ (and random residuals after regression) is the better original non-linear model.

📐 Worked Example

We have 9 data points $(x, y)$ with $x > 0$ and $y > 0$ , and we want to choose between an exponential model $y = a b^{x}$ and a power model $y = a x^{b}$ . After correct transformation and regression, the exponential model has $R^{2} = 0.88$ , and the power model has $R^{2} = 0.97$ . Both linearized models have randomly scattered residuals. Which original model is better? Justify.

1
Confirm all $x$ and $y$ are positive, so log-transformation is valid for both models.
2
Confirm each model was correctly linearized: exponential is regressed as $ln (y)$ vs $x$ , power is regressed as $ln (y)$ vs $ln (x)$ .
3
Compare the $R^{2}$ of the correctly linearized models: $0.97 > 0.88$ , so the power model's linearization has better fit.
4
A better fit for the linearized transformation corresponds to a better fit for the original non-linear model, so the original power model is preferred.

Exam tip:

Always remember that exponential models linearize against $x$ , while power models linearize against $ln (x)$ ; mixing up the predictor variable will give an incorrect $R^{2}$ and wrong conclusion.

5. AP-Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with this multiple-choice question:

A researcher compares three models for the height of a growing tomato plant over time, all fit to the original height data (in cm). The results are below:
- Linear Model: $R^{2} = 0.82$ , residuals form a clear upward curved trend
- Exponential Model: $R^{2} = 0.91$ , residuals are randomly scattered around 0
- Quadratic Model: $R^{2} = 0.92$ , residuals are randomly scattered around 0 Based on this information, which model is the most appropriate?
- Linear model, because it has the lowest $R^{2}$
- Quadratic model, because it has the highest $R^{2}$ and random residuals
- Exponential model, because it has a higher $R^{2}$ than the linear model
- Quadratic model, because growth is always curved
Reveal answer
1 —
Correct. The linear model is eliminated by its clear curved residual pattern. Both remaining models are fit to the same response variable and have random residuals, so we select the one with higher $R^{2}$ , which is the quadratic model.

6. Common Pitfalls

Wrong move:

Comparing $R^{2}$ of a linear model fit to original $y$ directly to the $R^{2}$ of a linearized exponential model fit to $ (y)$

Why:

$R^{2}$ measures variation in the response variable, so it is only comparable when the response variable is identical across models

Correct move:

If you need to compare on the original scale, calculate $R^{2}$ for the exponential model using predicted $i l d e y$ (not $ln (\overset{y}{^})$ ) on the original scale, then compare

Wrong move:

Claiming a model is bad just because one residual is much larger than the rest

Why:

A single outlier is not the same as a systematic pattern across all data points

Correct move:

Judge model fit based on the overall pattern of all residuals, not just one extreme outlier

Wrong move:

Linearizing a power model by regressing $ (y) $o n$ x $in s t e a d o f$ (x)$

Why:

Students often mix up the linearization formulas for exponential and power models

Correct move:

Memorize: exponential uses predictor $x$ , power uses predictor $ln (x)$ , match to the original model form

Wrong move:

Concluding a model is better just because it has a higher $R^{2}$ , even when it has a clear systematic residual pattern

Why:

Students over-rely on $R^{2}$ and ignore critical graphical evidence of poor fit

Correct move:

Always check residual patterns first; a model with slightly lower $R^{2}$ but random residuals is better than a higher $R^{2}$ model with a clear systematic pattern

Wrong move:

Attempting to take the logarithm of zero or a negative $y$ -value when linearizing

Why:

Students do not check the domain of the data before applying transformation

Correct move:

Confirm all $x$ and $y$ values are positive before using log-transformation; if non-positive values exist, use residual analysis on original data for comparison instead

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Residual Calculation	$e_{i} = y_{i} - \overset{y}{^}_{i}$	$y_{i}$ = observed $y$ , $ hat{y}_i $= p r e d i c t e d$ y$ from the model
Good Residual Pattern	No systematic pattern, random around $e = 0$	Indicates a well-fitting model
Bad Residual Pattern	Curve, linear trend, or funnel shape	Indicates a poorly fitting model
$R^{2}$ Comparison Rule	Higher $R^{2}$ = better fit	Only valid when both models have the same response variable
Exponential Linearization	$ln (y) = ln (a) + (ln b) x$	Response = $ln (y)$ , Predictor = $x$ , only valid for $y > 0$
Power Model Linearization	$ln (y) = ln (a) + b ln (x)$	Response = $ln (y)$ , Predictor = $ln (x)$ , only valid for $x > 0, y > 0$
Model Selection Priority	Check residual pattern first 2. Compare $R^{2}$ if multiple models have good residuals	Residual pattern always takes priority over $R^{2}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · MCQ
Compare residual plots for two models
2023 · FRQ
Linearize and compare exponential/power models

Going deeper

unit overviewAP Precalculus Unit 2 OverviewFull unit content outline

What's Next

Competing function model validation is the capstone modeling skill for Unit 2: Exponential and Logarithmic Functions, and it prepares you for all future modeling-focused content on the AP Precalculus exam. The core skills of residual analysis, goodness-of-fit comparison, and model justification you learn here apply to every type of model you will encounter for the rest of the course and on the final exam. Immediately after mastering this topic, you will move into Unit 3, which covers polynomial and rational functions, where you will apply the same core model validation skills to compare polynomial models of different degrees to bivariate data. Without mastering these foundational skills here, justifying model selection for non-linear polynomial models will be much more difficult, as the same principles apply directly.