Problem 29 - College Algebra and Calculus: An Applied Approach

Question

A business purchases a piece of equipment for $$\$ 875$$. After 5 years the equipment will have no value. Write a linear equation giving the value \(V\) of the equipment during the 5 years.

Step-by-Step Solution

Answer

The linear equation representing the value of the equipment over time is \(V(t)=-175t+875\)

Step 1: Determine Initial Value

The initial value of the equipment is given in the problem as $$875. This will be the equipment’s value at year 0, so this is the y-intercept of the linear equation we will create. This makes it \(V(0)=875\).

Step 2: Determine Final Value

The problem states that after 5 years, the equipment will have no value. This means the value of the equipment at year 5 is 0. So, \(V(5)=0\).

Step 3: Formulate Linear Equation

Now, we have two points (0,875) and (5,0) on the line which represents the value of the equipment over time. To determine the equation, we will use the slope-intercept form of a line, which is \(y=mx+b\), where \(m\) is the slope and \(b\) is the y-intercept. We already know that \(b=875\). The slope \(m\) can be calculated as \(m = (y_2-y_1) / (x_2-x_1)\). Substituting \(x_1=0\), \(y_1=875\), \(x_2=5\) and \(y_2=0\), we get \(m = (0-875) / (5-0) = -175\). So, the linear equation representing the value of the equipment over time is \(V(t)=-175t+875\), where \(t\) is the time in years.

Key Concepts

Understanding the Slope-Intercept Form

Linear equations are fundamental in algebra and are frequently used to represent real-world situations.
One of the most widely used forms of linear equations is the slope-intercept form.
This can be written as \[y = mx + b\].
Let's break this down:

• \(y\) represents the dependent variable which we are trying to find or predict.
• \(x\) is the independent variable, which usually represents time or another measurable factor.
• \(m\) is the slope of the line, which tells us how steep the line is.
• \(b\) is the y-intercept, which shows where the line crosses the y-axis.

In our exercise, the slope-intercept form is used to model the depreciation of equipment over time.
This form makes it easier to understand the relationship between time and the value of the equipment.

Identifying the Y-intercept

The y-intercept is a crucial component in linear equations as it provides the starting point of the line on the graph.
In simple terms, it is the value of \(y\) when \(x = 0\).
In the exercise, the initial value of the equipment, which is \(875, acts as the y-intercept.
This is because it represents the equipment's value right at the start.

• This means when no time has passed (\(t = 0\)), the equipment is valued at \)875.

Understanding the y-intercept helps in grasping where the equation starts on the graph.
This is essential for drawing the line that corresponds to the equation and figuring out the line's behavior over other values of \(x\).

Performing Slope Calculation

The slope in a linear equation indicates how the dependent variable changes as the independent variable changes.
It is calculated by measuring the change in \(y\) over the change in \(x\).
For the exercise given, we calculate the slope \(m\) using the two points provided: (0, 875) and (5, 0).
These points show the equipment's value at the beginning and at the end of its useful life.
The formula for calculating the slope \(m\) is:
\[m = \frac{y_2 - y_1}{x_2 - x_1}\]
Substitute the values:

• \(y_1 = 875\), \(x_1 = 0\)
• \(y_2 = 0\), \(x_2 = 5\)
• Slope \(m = \frac{0 - 875}{5 - 0} = -175\)

The negative slope implies that the equipment's value is decreasing over time, which matches the real-world scenario of depreciation.
The slope gives us a rate of depreciation, indicating how much value the equipment loses per year.