Problem 50 - 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. $$ \begin{array}{|c|c|c|c|c|c|}\hline\hline x & -4 & -2 & 0 & 2 & 4 \\ \hline y & 1 & -\frac{1}{2} & -2 & -\frac{7}{2} & -5 \end{array} $$

Step-by-Step Solution

Answer

The data are linear with a slope \( m = -\frac{3}{4} \).

Step 1: Identify Data Points

We start by identifying the data points from the given table. The data points are \((-4, 1)\), \((-2, -\frac{1}{2})\), \((0, -2)\), \((2, -\frac{7}{2})\), \((4, -5)\).

Step 2: Calculate Differences in Y

Calculate the difference \( \Delta y \) between successive \( y \)-values: \(-\frac{1}{2} - 1 = -\frac{3}{2}\), \(-2 - (-\frac{1}{2}) = -\frac{3}{2}\), \(-\frac{7}{2} - (-2) = -\frac{3}{2}\), \(-5 - (-\frac{7}{2}) = -\frac{3}{2}\).

Step 3: Calculate Differences in X

Calculate the difference \( \Delta x \) between successive \( x \)-values, which all have a constant value of \(2\): \(-2 - (-4) = 2\), \(0 - (-2) = 2\), \(2 - 0 = 2\), \(4 - 2 = 2\).

Step 4: Check for Constant Slope

Calculate the slope \( m \) for each pair of successive points using the formula \( m = \frac{\Delta y}{\Delta x} \): \(\frac{-\frac{3}{2}}{2} = -\frac{3}{4}\).This calculation applies to each pair of points; hence, the slope is constant.

Step 5: Conclude If Data Are Linear or Nonlinear

Since the slope \( m = -\frac{3}{4} \) is constant for all intervals, the data points form a linear pattern.

Key Concepts

Slope Calculation

To determine if a set of data points is linear, one of the first steps is to calculate the slope between successive points. The slope, often symbolized as \(m\), tells us how much the \(y\)-value changes for each unit change in \(x\). In essence, it represents the steepness or incline of the line formed by the data points.

To perform this calculation, we use the slope formula:

• \(m = \frac{\Delta y}{\Delta x} \)

This formula calculates the ratio of the change in \(y\) (\(\Delta y\)) to the change in \(x\) (\(\Delta x\)). If the data is linear, this slope will remain the same for any two points on the line. In our case, we find consistent slopes, such as calculating for pairs:

• \(\frac{-\frac{3}{2}}{2} = -\frac{3}{4}\)

When \(m\) is constant, it strongly suggests linearity.

Constant Slope

Having a constant slope is a crucial characteristic of linear data. This means that regardless of which two points you choose to calculate the slope from, the value of \(m\) should remain the same. It's this uniformity that defines a straight line in a graph.

Given the dataset from our exercise, we've determined that each set of successive points results in a slope of \(-\frac{3}{4}\). Calculating the differences in the \(y\)-values and \(x\)-values separately for each pair yields the same slope, confirming the characteristic of linear data. The steps are straightforward:

• Identify changes in \(x\) (\(\Delta x\) is constant at \(2\))
• Identify changes in \(y\) (each \(\Delta y = -\frac{3}{2}\))
• Use the formula \(m = \frac{\Delta y}{\Delta x}\) for each pair

Upon seeing a consistent slope across all intervals, we can confidently state the data forms a linear relationship.

Data Patterns

Recognizing patterns in data is an essential skill, particularly in identifying linear versus nonlinear patterns. Linear data can be visualized and understood easily due to its uniformity—a straight line displays this consistency clearly.

Here, our data points demonstrate such a pattern. By checking if each segment joins seamlessly along the line with a constant slope, we ensure that what we see visually aligns with our mathematical findings. If the slope results were mixed or inconsistent, the pattern would imply nonlinearity, such as curves or other irregularities.

In practice, identifying linear data allows us to predict other values along the line. With a known slope and one point, you could determine every other point through extrapolation—a powerful tool in data analysis and predictions. In summary, the pattern found in our exercise confirms that the data is linear, which is important for effective data description and analysis.