Problem 64 - A Graphical Approach to Precalculus with Limits

Question

Average Wages The average hourly wage (adjusted to 1982 dollars) was \(\$ 7.66\) in 1990 and \(\$ 9.16\) in 2017. (a) Find a point-slope form of a line that passes through the points \((1990,7.66)\) and \((2017,9.16)\) (b) Interpret the slope. (c) Use the equation from part (a) to approximate the hourly wage in \(2009 .\) Compare it with the actual value of \(\$ 8.27\)

Step-by-Step Solution

Answer

(a) \(y - 7.66 = 0.0556(x - 1990)\); (b) Wage increases \$0.0556/year; (c) Predicted 2009 wage is \$8.72, actual \$8.27.

Step 1: Identify the Points

We are given two points that describe average hourly wages: \((1990, 7.66)\) and \((2017, 9.16)\). These represent the years and corresponding average wages.

Step 2: Calculate the Slope

The slope \(m\) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For our points, this is \(m = \frac{9.16 - 7.66}{2017 - 1990} = \frac{1.5}{27} \approx 0.0556\).

Step 3: Write the Point-Slope Form Equation

The point-slope form of a line is \(y - y_1 = m(x - x_1)\). Using point \((1990, 7.66)\) and slope \(0.0556\), the equation is \(y - 7.66 = 0.0556(x - 1990)\).

Step 4: Interpret the Slope

The slope \(0.0556\) indicates that the average hourly wage increased by approximately \$0.0556 per year from 1990 to 2017, adjusted to 1982 dollars.

Step 5: Use the Equation to Find Wage in 2009

Substitute \(x = 2009\) into the equation from Step 3: \(y - 7.66 = 0.0556(2009 - 1990)\). This simplifies to \(y - 7.66 = 0.0556 \times 19\), so \(y = 7.66 + 1.0564 = 8.7164\).

Step 6: Compare with Actual Wage

The predicted wage in 2009 is \(\\(8.72\) while the actual wage was \(\\)8.27\). The prediction is higher than the actual value by about \$0.45.

Key Concepts

Point-Slope Form

The point-slope form is a method used to define a straight line when we know a point on the line and the slope of the line. It's a simple yet powerful tool that can help easily write the equation of a line. The general formula is:

• \( y - y_1 = m(x - x_1) \)

In this formula:

• \( m \) stands for the slope of the line.
• \((x_1, y_1)\) is a specific point on the line.

In our exercise, the specific point is \((1990, 7.66)\), representing the year 1990 and an average wage of \(\$7.66\). The slope \( m \) we calculated is approximately \(0.0556\). Using these, we form the point-slope equation as:

• \( y - 7.66 = 0.0556(x - 1990) \)

This means for any year \( x \), you can plug it into the equation to find the expected average wage \( y \) for that year.

Slope Interpretation

Understanding the slope is crucial as it tells us about the change that occurs in one variable compared to the other. The slope is essentially the rate of change:

• It is calculated using: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
• The slope from our data is \( 0.0556 \).

What does this mean in our context?

• From 1990 to 2017, the average hourly wage, adjusted to 1982 dollars, increased by approximately \(\$0.0556\) each year.

This information provides insights into wage trends over time, allowing us to predict or analyze wage patterns.

Equation of a Line

Creating the equation of a line allows us to describe a linear relationship between two variables. For our exercise on average wages, the equation gives a mathematical representation of how the wage has evolved over the years. Using point-slope form, the equation derived is:

• \( y - 7.66 = 0.0556(x - 1990) \)

Once we have this equation, we can rearrange it into slope-intercept form for easier interpretation or comparison:

• \( y = 0.0556x - 110.8844 + 7.66 \)
• Simplified, this becomes \( y = 0.0556x - 103.2244 \)

This equation can now be used to predict average wages at any given year by substituting the year for \( x \). For example, substituting 2009 for \( x \) gives an approximate wage prediction of \( \\(8.72 \), slightly higher than the actual recorded wage of \( \\)8.27 \). This discrepancy might highlight external factors not captured by this simple linear model.