Problem 56 - College Algebra with Modeling and Visualization

Question

Analyzing Real Data For the given data set complete the following. (a) Make a line graph of the data. Let this graph represent a function \(f\) (b) Decide whether \(f\) is linear or nonlinear. Interest income after 1 year on an investment earning 7% per year. $$ \begin{array}{|r|r|r|r|r|}\hline \hline \text { Investment } & \$ 500 & \$ 1000 & \$ 2000 & \$ 3500 \\ \hline \text { Interest } & \$ 35 & \$ 70 & \$ 140 & \$ 245 \end{array} $$

Step-by-Step Solution

Answer

The function \(f\) is linear as the interest increases at a constant rate with investment.

Step 1: Organize the Data

First, let's organize the given data into ordered pairs, where the investment amount is the independent variable (x) and the interest is the dependent variable (y). The pairs are: \((500, 35)\), \((1000, 70)\), \((2000, 140)\), \((3500, 245)\).

Step 2: Plot the Data Points

Plot the ordered pairs on a graph with the x-axis representing the investment amount and the y-axis representing the interest. This visually helps in determining the trend of the function \(f\).

Step 3: Draw the Line

Once the data points are plotted, draw a line through them. Observe the pattern to determine if it forms a straight line.

Step 4: Analyze the Linearity

For a function \(f\) to be linear, the rate of change between any two points should be constant. Calculate the slope between each pair of successive points: \\[\text{slope}_{1,2} = \frac{70 - 35}{1000 - 500} = 0.07\] \\[\text{slope}_{2,3} = \frac{140 - 70}{2000 - 1000} = 0.07\] \\[\text{slope}_{3,4} = \frac{245 - 140}{3500 - 2000} = 0.07\] \As the slope is consistent across the data points, \(f\) is linear.

Key Concepts

Interest Income

Understanding interest income is crucial when analyzing investment returns. Interest income is the amount earned from lending or investing money at a specified interest rate per year. In this case, the data reflects the interest generated after a year on various investments at a 7% interest rate. When talking about interest income:

• The larger the principal amount invested, the higher the interest earned.
• The interest is directly proportional to both the principal and the interest rate.
• Compound interest, which involves reinvesting interest, can generate more income over time compared to simple interest.

For an investment, maintaining a steady interest rate means predictable and reliable income over the term of the investment. This consistent income stream helps investors plan their financial futures more effectively.

Data Analysis

Data analysis involves observing and interpreting data to draw meaningful conclusions. In this scenario, the data points show the relationship between investment amounts and the resulting interest income over one year. By organizing the data into ordered pairs

• (500, 35)
• (1000, 70)
• (2000, 140)
• (3500, 245)

we turn raw data into a presentable form that makes trends easier to spot. Analyzing these ordered pairs, we notice each increase in investment corresponds to a proportional increase in interest earned. Identifying these patterns is essential in data analysis. It helps in understanding the nature of the data, which in this case shows a clear linear relationship.

Graphing Functions

Graphing is a powerful tool in mathematics to visualize relationships between variables. Here, we're tasked with making a line graph from the dataset showing investment versus interest income. Start by plotting each data point on a coordinate plane. The x-axis represents investments, and the y-axis represents interest. Connect these points with a line.

Why Graphing is Helpful:

• It provides a visual representation of the data, making it easier to identify trends.
• Graphing helps decide if the relation is linear or nonlinear. In this instance, connecting the open dots on our graph forms a straight line.
• It offers a quick way to predict future values, by extending the line.

Here, every sequential point appears to maintain an equal slope, confirming the function's linearity. By graphing, we find that the function describing the interest earned from investments is indeed linear, reflecting a constant rate of change.