Problem 83 - College Algebra with Modeling and Visualization

Question

In 2002 sales of premium online music totaled \(\$ 1.6\) billion. In 2005 this revenue reached \(\$ 3.6\) billion. (A) Find a point-slope form of the line passing through \((2002,1.6)\) and \((2005,3.6) .\) Interpret the slope. (B) Use the equation to estimate projected sales in2008. Did you use interpolation or extrapolation? (C) Find the slope-intercept form of this line.

Step-by-Step Solution

Answer

(A) Point-slope: \(y - 1.6 = 0.6667(x - 2002)\). Slope = 0.6667. (B) Projected sales in 2008: $5.6 billion (extrapolation). (C) Slope-intercept: \(y = 0.6667x - 1332.6\).

Step 1: Find the Slope of the Line

To find the slope (m) of the line passing through the points \(2002, 1.6\) and \(2005, 3.6\), we use the formula for slope, which is \((y_2 - y_1) / (x_2 - x_1)\). Here, \((x_1, y_1) = (2002, 1.6)\) and \((x_2, y_2) = (2005, 3.6)\). Therefore, \m = (3.6 - 1.6) / (2005 - 2002) = 2.0 / 3 = 0.6667.\The slope of 0.6667 indicates that for each year, the revenue increases by $0.6667 billion.

Step 2: Find the Point-Slope Form

The point-slope form of a line is given by \(y - y_1 = m(x - x_1)\). Now that we have the slope \(m = 0.6667\) and the point \(x_1, y_1 = (2002, 1.6)\), we can substitute these values into the formula: \y - 1.6 = 0.6667(x - 2002).This is the equation in point-slope form.

Step 3: Estimate Projected Sales in 2008

To find the projected sales for the year 2008 using the point-slope form, substitute \(x = 2008\): \y - 1.6 = 0.6667(2008 - 2002)=> y - 1.6 = 0.6667(6)=> y - 1.6 = 4=> y = 5.6.Thus, the projected sales in 2008 is $5.6 billion. Since 2008 is beyond the given data range, we used extrapolation.

Step 4: Find the Slope-Intercept Form

The slope-intercept form of the equation is \(y = mx + b\). From the point-slope form \(y - 1.6 = 0.6667(x - 2002)\), expand and simplify: \y = 0.6667x - 0.6667 \times 2002 + 1.6=> y = 0.6667x - 1334.2 + 1.6=> y = 0.6667x - 1332.6.Thus, the equation in slope-intercept form is \y = 0.6667x - 1332.6\.

Key Concepts

Point-Slope Form

The point-slope form is very helpful when you know a point on the line and the slope. It is represented by the formula \(y - y_1 = m(x - x_1)\). Here, \((x_1, y_1)\) is a known point on the line, and \(m\) is the slope.

In our exercise, the points provided are \((2002, 1.6)\) and \((2005, 3.6)\), which indicate years and sales in billions. First, we calculated the slope \(m\) to find how much sales increase each year:

• Formula for slope: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
• Substituting the given values: \(m = \frac{3.6 - 1.6}{2005 - 2002} = \frac{2}{3} = 0.6667\)

This shows that revenue increases by \(0.6667\) billion dollars annually. Then, using the point-slope formula, the equation becomes \(y - 1.6 = 0.6667(x - 2002)\), where if you have any future year, you can plug it into this formula to estimate revenues.

Slope-Intercept Form

Slope-intercept form is another way to represent the equation of a line, which is useful for understanding how variables are connected. It is typically shown as \(y = mx + b\). In this formula, \(m\) is the slope, and \(b\) is the y-intercept.

From the point-slope form \(y - 1.6 = 0.6667(x - 2002)\), we transformed to slope-intercept form by expanding the equation:

• \(y = 0.6667x - 0.6667 \times 2002 + 1.6\)
• This simplifies to: \(y = 0.6667x - 1332.6\)

The slope-intercept form \(y = 0.6667x - 1332.6\) makes it easy to identify both the rate at which sales are growing yearly (0.6667 billion dollars) and the y-intercept. This form is particularly useful for quickly estimating sales at different years by replacing \(x\) with the desired year.

Extrapolation

Extrapolation is a method used to estimate values beyond the original range of data, based on an observed linear trend. It allows predictions into future data points.

In our example, we needed to estimate the sales for 2008, though only data from 2002 and 2005 was given. By extending the trend line, we computed the future sales amount:

• Using point-slope: substitute 2008 for \(x\)
• \(y - 1.6 = 0.6667(2008 - 2002)\)
• Solving gives: \(y = 5.6\)

Because 2008 is outside the original data range, this prediction is termed 'extrapolation.' While useful, extrapolated values should be considered with caution, as they rely heavily on the assumption that the observed trend continues unchanged.