Problem 53 - College Algebra with Modeling and Visualization

Question

Decide whether the data are linear or nonlinear. If the data are linear, state the slope \(m\) of the line passing through the data points. $$ \begin{array}{|c|c|c|c|c|c|}\hline\hline x & -4 & 0 & 1 & 2 & 5 \\ \hline y & 5 & 3 & \frac{5}{2} & 2 & \frac{1}{2} \end{array} $$

Step-by-Step Solution

Answer

The data is linear with a slope of \(-\frac{1}{2}\).

Step 1: Understanding Linear vs Nonlinear Data

Data is considered linear if it can be represented by a straight line, which means the change in the y-values over the change in x-values (slope) is constant between all pairs of points.

Step 2: Calculating Slopes Between Points

To determine if the data is linear, calculate the slope between consecutive pairs of points. Slope, \(m\), is calculated as \(m = \frac{y_2 - y_1}{x_2 - x_1}\).- Between \((-4, 5)\) and \((0, 3)\): \[ m = \frac{3 - 5}{0 + 4} = \frac{-2}{4} = -\frac{1}{2} \]- Between \((0, 3)\) and \((1, \frac{5}{2})\): \[ m = \frac{\frac{5}{2} - 3}{1 - 0} = \frac{-\frac{1}{2}}{1} = -\frac{1}{2} \]- Between \((1, \frac{5}{2})\) and \((2, 2)\): \[ m = \frac{2 - \frac{5}{2}}{2 - 1} = \frac{-\frac{1}{2}}{1} = -\frac{1}{2} \]- Between \((2, 2)\) and \((5, \frac{1}{2})\): \[ m = \frac{\frac{1}{2} - 2}{5 - 2} = \frac{-\frac{3}{2}}{3} = -\frac{1}{2} \]

Step 3: Analyzing Slopes for Consistency

Compare the slopes calculated in Step 2. If all the slopes are equal, this indicates that the data points lie on a straight line, meaning the data is linear. Here, all calculated slopes are \(-\frac{1}{2}\), confirming the data is linear.

Step 4: State the Result

The data is linear because the slope between all pairs of consecutive data points is consistent. The slope \(m\) of the line passing through the data points is \(-\frac{1}{2}\).

Key Concepts

Slope Calculation

Understanding how to calculate the slope is fundamental for identifying linear relationships between data points. The slope, often represented by the letter \(m\), indicates how steep a line is. To find the slope between two points, we use the formula:

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Here, \((x_1, y_1)\) and \((x_2, y_2)\) are two points on the line. The numerator (\(y_2 - y_1\)) finds the vertical change or rise between the points, while the denominator (\(x_2 - x_1\)) calculates the horizontal change or run.
This concept is vital because determining the slope helps confirm if data points consistently fit into a linear pattern through the same rate of change. In our exercise, the slope was consistently found to be \(-\frac{1}{2}\) between all points, indicating a linear relationship.

Linear Data Analysis

Linear data analysis revolves around determining if a set of data points can be plotted to form a straight line. A data set is linear if every segment of the line has the same slope. This process typically involves checking the slopes between consecutive data points for consistency.
The exercise demonstrates this by examining data points: \((-4, 5), (0, 3), (1, \frac{5}{2}), (2, 2), (5, \frac{1}{2})\). Calculating the slope between each pair of points ensures that they all have a constant slope of \(-\frac{1}{2}\).

• First, calculate slopes between subsequent pairs.
• Ensure all calculated slopes are identical.

If these conditions are met, the given points fall along the same straight line or are linear. This consistency highlights the beauty of linear data analysis in simplifying complex data into interpretable patterns.

Algebraic Problem Solving

Algebraic problem solving entails using algebra to resolve questions about relationships and patterns. In this exercise, algebra helps determine whether a set of data points is linear and to calculate the slope of the line they form.
The crucial steps are:

• Using algebraic formulas to compute slopes between data points.
• Verifying uniformity in these slopes to establish linearity.
• Interpreting results to clearly understand data behavior.

Such problem solving requires methodically applying algebraic principles to achieve clarity about how variables relate to one another by assessing linearity and consistency. By successfully solving the problem, the students affirm their understanding of these principles, enhancing their competence in handling similar algebraic problems in the future.