Problem 125 - College Algebra with Modeling and Visualization

Question

Super BowlAds The table lists the cost in millions of dollars for a 30 -second Super Bowl commercial for selected years. $$ \begin{array}{lccccc} \text { Year } & 1990 & 1994 & 1998 & 2004 & 2008 \\ \hline \text { Cost } & 0.8 & 1.2 & 1.6 & 2.3 & 2.7 \end{array} $$ Source: MSNBC. (a) Find a linear function \(f\) that models the data. (b) Estimate the cost in 1987 and compare the estimate to the actual value of \(\$ 0.6\) million. Did your estimate involve interpolation or extrapolation? (c) Use \(f\) to predict the year when the cost could reach \(\$ 3.2\) million.

Step-by-Step Solution

Answer

(a) \(f(x) = 0.1056x - 209.248\), (b) Estimate = \$0.4832 million, extrapolation, (c) Year: 2109.

Step 1: Choose Two Points for the Linear Model

To create a linear function, we need to determine the slope using two data points from the table. Let's select (1990, 0.8) and (2008, 2.7) as our points to calculate the slope.

Step 2: Calculate the Slope

The slope \(m\) is calculated using the formula \(m = \frac{y_2-y_1}{x_2-x_1}\). Substitute the chosen points into the equation: \(m = \frac{2.7 - 0.8}{2008 - 1990} = \frac{1.9}{18} \approx 0.1056\).

Step 3: Determine the Linear Function

Using point-slope form \(y - y_1 = m(x - x_1)\), substitute one of the points (1990, 0.8) and the slope \(m = 0.1056\): \(y - 0.8 = 0.1056(x - 1990)\). Solve for \(y\) to get \(f(x) = 0.1056x - 209.248\).

Step 4: Estimate Cost for 1987

Substitute \(x = 1987\) into the linear function: \(f(1987) = 0.1056(1987) - 209.248 \approx 0.4832\). The actual value is 0.6 million. Since we're estimating beyond the available years, this is extrapolation.

Step 5: Predict the Year Cost Reaches $3.2 Million

Set the function equal to 3.2 and solve for \(x\): \(3.2 = 0.1056x - 209.248\). Solving this gives \(x \approx 2108.86\). Therefore, the cost will likely reach \$3.2 million around the year 2109.

Key Concepts

Slope Calculation

Understanding the slope of a line is fundamental in working with linear functions. The slope tells us how much the y-value changes for a unit change in the x-value. It's often represented by the letter "m" in the equation of a line. To calculate the slope using two points on a graph, we use the formula:

• \( m = \frac{y_2-y_1}{x_2-x_1} \)

In our exercise, the points (1990, 0.8) and (2008, 2.7) are chosen to find the slope. Using these points, substitute them into the formula:

• \( m = \frac{2.7 - 0.8}{2008 - 1990} \)
• This simplifies to \( m = \frac{1.9}{18} \approx 0.1056 \)

So, the slope of our linear model is approximately 0.1056, which means that each year the cost increases by about $0.1056 million.

Extrapolation

Extrapolation is the process of using a mathematical model to predict unknown values by extending an existing data trend. In our case, we used extrapolation to estimate the cost of a Super Bowl ad in 1987, which is outside our given data range. By substituting the value for the year 1987 into the linear equation:

• \( f(1987) = 0.1056 \times 1987 - 209.248 \)
• We calculated approximately \( \\(0.4832 \) million

Since 1987 is a year before the provided data points (1990–2008), we call this prediction an extrapolation. Extrapolations can be less accurate because they involve predicting outside the range of known data. The actual cost in 1987 was \\)0.6 million, showing a slight difference from our estimate.

Point-Slope Form

Point-slope form is a way to express the equation of a line. It's particularly useful when we know the slope of the line and one point through which the line passes. The formula for point-slope form is:

• \( y - y_1 = m(x - x_1) \)

Here, \((x_1, y_1)\) is a point on the line, and \(m\) is the slope. In our step-by-step solution, we used the point (1990, 0.8) with the calculated slope \(m = 0.1056\). By substituting these values into the formula, we obtained:

• \( y - 0.8 = 0.1056(x - 1990) \)

Rearranging the equation gives us the linear function:

• \( f(x) = 0.1056x - 209.248 \)

Point-slope form is ideal for quickly forming a linear function when raw data points are available. It simplifies to the slope-intercept form \(y = mx + b\), making interpretations like predicting values easier.