Problem 5 - College Algebra with Modeling and Visualization

Question

Make a scatterplot of the data. Then find an exponential, logarithmic, or logistic function \(f\) that best models the data. $$\begin{array}{ccccc}x & 1 & 2 & 3 & 4 \\\\\hline y & 2.04 & 3.47 & 5.90 & 10.02\end{array}$$

Step-by-Step Solution

Answer

The exponential function that best models the data is \(y \approx 1.92 \cdot 1.73^x\).

Step 1: Organize the Data

First, list the given data points as ordered pairs: \((1, 2.04), (2, 3.47), (3, 5.90), (4, 10.02)\).

Step 2: Create a Scatterplot

Plot each data point on a graph with the x-axis representing the \(x\) values and the y-axis representing the \(y\) values. This provides a visual representation of the data's trend.

Step 3: Analyze the Trend

Examine the scatterplot to assess the nature of the relationship between \(x\) and \(y\). The data looks to be following a curve, indicating a possible exponential relationship.

Step 4: Choose a Model

Since the data suggests exponential growth, select an exponential model of the form \(y = ab^x\).

Step 5: Transform the Data

Take the natural logarithm of the \(y\) values to linearize the data: \[\ln(y) = \ln(ab^x) = \ln(a) + x \ln(b)\]This allows us to apply linear regression techniques.

Step 6: Perform Linear Regression

Calculate \(\ln(y)\) for each \(y\) value: \((1, \ln(2.04)), (2, \ln(3.47)), (3, \ln(5.90)), (4, \ln(10.02))\).Perform linear regression on the transformed data to determine \(\ln(a)\) and \(\ln(b)\).

Step 7: Compute the Exponential Function

After finding \(\ln(a)\) and \(\ln(b)\) from the linear regression, compute \(a = e^{\ln(a)}\) and \(b = e^{\ln(b)}\). These values form the exponential function \(y = ab^x\).

Step 8: Verify the Model

Substitute the \(x\) values back into the exponential model and check if the predicted \(y\) values closely match the original data. If they do, the model accurately fits the data.

Key Concepts

Scatterplot Creation

Creating a scatterplot is a fundamental tool in analyzing relationships between two variables. It involves plotting data points on a graph, showing one variable on the x-axis and another on the y-axis. In our example, the given data points are organized as

• \((1, 2.04)\)
• \((2, 3.47)\)
• \((3, 5.90)\)
• \((4, 10.02)\)

To make a scatterplot, place each ordered pair onto a Cartesian plane. This offers a visual representation of whether there's a recognizable pattern or trend in the data. When executed properly, scatterplots can reveal nonlinear associations that aren't easily identified using other methods, providing a foundation for deeper statistical analysis.
This is crucial when we are in pursuit of finding an appropriate model to describe the data's behavior.

Data Trend Analysis

Understanding the trend in the data involves examining the shape and direction of the plotted points on a scatterplot. This insight is crucial before applying any mathematical models for prediction. If data points form a distinct upward or downward curve, it signals a possible exponential relationship. Analyzing a scatterplot requires a few essential steps to determine the trend:

• Identify if the data points are moving along a straight line or a curve.
• Observe if the spacing between points is increasing or decreasing.
• Consider if there is a consistent rate of change.

In our case, the points appear to be following a curve which suggests an exponential growth model might suit this data best. Such an observation can prelude the selection of a mathematical model like exponential, logarithmic, or logistic functions to best fit the underlying trend.

Linear Regression

Once a potential mathematical model is identified through trend analysis, linear regression is used to determine the best-fitting line that approximates the data. For exponential models, the data is transformed to fit into a linear context. This involves taking the natural logarithm of the y-values, converting our data into a form suitable for linear analysis.
The linear regression on transformed data involves:

• Calculating \(\ln(y)\) for each y-value.
• Running linear regression on the points \((x, \ln(y))\).
• Estimating model parameters \(\ln(a)\) and \(\ln(b)\) necessary to devise the exponential model.

By reverse-engineering the parameters using exponential functions \(a = e^{\ln(a)}\) and \(b = e^{\ln(b)}\), we derive the exponential equation that best approximates the original data. This intersection of nonlinear and linear techniques is what makes linear regression a versatile tool for tackling a variety of data modeling challenges.