Problem 92 - College Algebra with Modeling and Visualization

Question

VISA Cards Annual transactions on VISA cards increased from \(\$ 400\) billion in 2002 to \(\$ 635\) billion in \(2007 . \quad\) (Source: CardWeb.) (a) Find a linear function \(V\) that models the data. (b) Estimate when this number was between \(\$ 450\) billion and \(\$ 540\) billion.

Step-by-Step Solution

Answer

(a) \( V(t) = 47t + 400 \) (b) The transactions were between \( \$450 \) and \( \$540 \) billion from early 2003 to late 2004.

Step 1: Define the Linear Function

We start by defining the years as variables. Let 2002 be the year zero, so 2007 will be year 5. We want to find a linear function \( V(t) = mt + b \) that fits the data points (0, 400) and (5, 635). The formula for a linear function is \( V(t) = mt + b \).

Step 2: Calculate the Slope

The slope \( m \) is calculated as \[ m = \frac{V(5) - V(0)}{5 - 0} = \frac{635 - 400}{5} = \frac{235}{5} = 47. \]

Step 3: Find the y-intercept

To find \( b \), use one of the points. Using (0, 400):\[ V(0) = 47(0) + b = 400, \]\[ b = 400. \]

Step 4: Write the Equation of the Linear Model

Substituting \( m \) and \( b \) into the linear function gives:\[ V(t) = 47t + 400. \]

Step 5: Solve for Time Interval When Transactions Are Between 450 and 540 Billion

Use the linear equation. For \( V(t) = 450 \):\[ 47t + 400 = 450, \]\[ 47t = 50, \]\[ t = \frac{50}{47} \approx 1.06. \]For \( V(t) = 540 \):\[ 47t + 400 = 540, \]\[ 47t = 140, \]\[ t = \frac{140}{47} \approx 2.98. \]

Step 6: Determine the Years Corresponding to the Solutions

Since each year corresponds to a value of \( t \) (with year zero as 2002), when \( t = 1.06 \) it corresponds to early 2003 and when \( t = 2.98 \) it is late 2004.

Key Concepts

Linear Modeling

Linear modeling is a powerful tool that aids in understanding and predicting trends between two variables. In the context of this exercise, we are modeling the annual transactions on VISA cards over the years. Linear models, like the one used here, are represented by functions that illustrate a relationship in the form: \( V(t) = mt + b \), where \( t \) is the time variable (years since 2002). The first step in linear modeling is identifying your data points. For our example, these are \((0, 400)\) and \((5, 635)\), signifying \(2002\) and \(2007\) respectively.
These models allow us to estimate and make predictions about data trends easily by providing a straight-line graph representing the relationship. They are particularly handy when we suspect a linear association, as they simplify complex data into a clearer form.

Slope Calculation

The slope of a line in linear models is crucial as it indicates the rate of change between the variables. In this exercise, to find the slope \( m \), we apply the formula \( m = \frac{V(5) - V(0)}{5 - 0} \). This means we calculate the difference in the transaction values and divide it by the difference in years. The result here is 47. This means that annually, the transactions increase by \( \$47 \) billion.
Slope is fundamentally important because it tells us how steep the line of the graph is and, in practical terms, it shows how quickly or slowly the transaction amount is changing each year. Understanding slopes can be quite powerful, allowing one to predict future trends or even find potential anomalies in data by looking at historical changes.

Y-intercept

The y-intercept is the value of the linear function when \( t = 0 \). It represents the initial value or starting point before any changes occur in a linear relationship. In this problem, we find the y-intercept \( b \) by substituting \( t = 0 \) into our equation \( V(t) = 47t + b \), which provides \( V(0) = 400 \), suggesting \( b = 400 \).
Thus, the y-intercept shows the starting transactions in \( 2002 \), which was \( \$400 \) billion. Understanding the y-intercept provides context on where the data started in a given set, and by extension, helps foresee how it might change if those conditions remained constant over time.

Estimating Values

Estimating values using a linear model is beneficial for interpreting data outside the specific points initially provided. This model helps us determine when the transaction amounts on VISA cards were between \( \\(450 \) billion and \( \\)540 \) billion. We set \( V(t) = 450 \) and \( V(t) = 540 \) in our linear equation \( V(t) = 47t + 400 \).
Solving these gives \( t = 1.06 \) and \( t = 2.98 \). Since we started counting years from \( 2002 \), \( t = 1.06 \) represents early \( 2003 \) and \( t = 2.98 \) indicates the end of \( 2004 \). This means between these times, the transactions transformed critically, reaching those amounts. Estimating values in this way provides a tangible approach to forecasting and data analysis, supporting strategic and informed decision-making.