Problem 56 - A Graphical Approach to College Algebra

Question

(Modeling) In Exercises \(53-56\), assume that a linear relationship exists between the two quantities. Depreciation of a Photocopier \(\quad\) A photocopier sold for \(\$ 3000\) in \(2006 .\) Its value in 2014 had depreciated to \(\$ 600 .\) (a) If \(x=0\) represents 2006 and \(x=8\) represents 2014 express the value of the machine, \(y,\) as a linear function of the number of years, \(x,\) after 2006 (b) Graph the function from part (a) in a window \([0,10]\) by \([0,4000] .\) How would you interpret the \(y\) -intercept in terms of this particular situation? (c) Use your calculator to determine the value of the machine in 2010 , and verify your result analytically.

Step-by-Step Solution

Answer

(a) \(y = -300x + 3000\); (b) y-intercept represents \$3000 in 2006; (c) \$1800 in 2010.

Step 1: Determine the slope of the linear function

First, identify the two points representing the years 2006 and 2014, expressed as coordinates: \((0, 3000)\) and \((8, 600)\). The slope of the line, \(m\), is calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Substitute the corresponding values: \(m = \frac{600 - 3000}{8 - 0} = \frac{-2400}{8} = -300\).

Step 2: Write the linear equation

Using the slope-intercept form of a linear equation, \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. We already know the slope \(m = -300\). Since the value of the machine in 2006 (when \(x = 0\)) is \(3000\), this is also the y-intercept. Hence, the equation becomes \(y = -300x + 3000\).

Step 3: Graph the linear function

The function \(y = -300x + 3000\) is graphed in a window with \(x\) spanning from \(0\) to \(10\) and \(y\) from \(0\) to \(4000\). The y-intercept at \((0, 3000)\) symbolizes the initial value of the photocopier in 2006, or when \(x = 0\).

Step 4: Compute the value of the machine in 2010 analytically

Substitute \(x = 4\) (2010) into the linear equation to find \(y\): \(y = -300(4) + 3000 = -1200 + 3000 = 1800\). Therefore, the value of the machine in 2010 is \(\$1800\).

Step 5: Verify the result with a calculator

Use a calculator to substitute \(x = 4\) into the equation \(y = -300x + 3000\). This calculation should also show \(y = 1800\), confirming that the analytical solution is correct.

Key Concepts

Depreciation

Depreciation is the process through which an asset loses its value over time. For example, the photocopier originally purchased for $3000 in 2006 diminished in value to just $600 by 2014. This loss is primarily due to wear and tear, technological advancements, and obsolescence.
Depreciation is often used in accounting to allocate the cost of a tangible asset over its useful life. In a mathematical context, it can be represented as a straight line on a graph, assuming the decrease in value occurs at a constant rate over time.
In our exercise, the story of the photocopier illustrates how depreciation can be captured using linear functions. The decrease from $3000 to $600 over 8 years is uniform, characterized by the straight-line equation derived in the problem.

Slope of a Line

The slope of a line in mathematics tells us how steep the line is and the direction it is heading. Represented by the letter 'm' in the equation of a line, it can be calculated as the change in the y-value divided by the change in the x-value between two points on the line.

• Formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

For our photocopier problem, the slope represents how much value the photocopier loses each year. We find it to be -300, indicating a yearly depreciation of $300.
This negative slope signifies that the photocopier's value is decreasing annually, which is a common scenario when dealing with depreciating assets.

Y-Intercept

The y-intercept of a linear function is the point where the graph crosses the y-axis. It represents the value of 'y' when 'x' is zero.
In the context of our exercise, the y-intercept of $3000 reveals the initial value of the photocopier when it was brand new in 2006.

• The formula used is: \( y = mx + b \), where 'b' is the y-intercept.

Understanding the y-intercept is crucial because it offers insights into the starting point of the circumstance we are modeling. In real-world applications, it often signifies the original condition or initial value of the subject being studied.

Graphing Linear Equations

Graphing linear equations is a powerful visual method to understand relationships between variables. In this instance, by graphing the linear equation \( y = -300x + 3000 \), we can visualize how the value of the photocopier decreases over time.
This specific graph shows a straight diagonal line descending from the y-axis to the right, beginning at (0, 3000) and sloping downwards.
To accurately graph a linear equation, you need:

• The y-intercept, which provides a starting point.
• The slope, which tells us the direction and steepness of the line.

Using these, you can plot additional points on the graph by moving accordingly from the intercept via the slope. This helps to further solidify understanding of how linear functions represent real-life situations like depreciation.