Problem 62 - Introductory Algebra for College Students

Question

A car was purchased for \(\$ 22,500\). The value of the car decreases by \(\$ 3200\) per year for the first seven years. Write a function \(V\) that describes the value of the car after \(x\) years, where \(0 \leq x \leq 7 .\) Then find and interpret \(V(3)\).

Step-by-Step Solution

Answer

The function describing the value of the car after \(x\) years is given by \(V = 22500 - 3200x\), and the worth of the car after 3 years is $15,100.

Step 1: Write the Depreciation Function

Firstly, we need to understand that the value of the car decreases by $3200 each year. Therefore, the function representing the value of the car \(V\) after \(x\) years from the time of purchase will be \(V = 22500 - 3200x\). This equation simply shows the initial value of the car ($22500) and how it decreases each year by $3200 times the number of years (\(x\)).

Step 2: Calculate and Interpret V(3)

Then, we need to find the value of the car after 3 years. We do this by substituting \(x = 3\) into our equation. Therefore, \(V(3) = 22500 - 3200 * 3 = 15100 .\) This means that after 3 years, the value of the car is $15100. It represents the value of the car in dollar amount after it has depreciated for 3 years.

Key Concepts

Depreciation Function

When we talk about a depreciation function, we're discussing a mathematical way to represent how an item's value decreases over time. In the context of a car, various factors such as wear and tear, age, and new models entering the market can reduce the value of a vehicle. The depreciation function is a straightforward tool that helps us predict the future value of the car given its current worth and the annual rate of depreciation.

Consider the example of a car that was bought for \(22,500. Its value decreases linearly by \)3,200 every year. To encapsulate this in a function, we can establish a linear equation: V(x) = 22500 - 3200x, where V represents the car's value after x years. This function reflects a linear depreciation because the amount of value the car loses each year is constant, which is one of the simplest forms of a depreciation function.

Linear Equations

In algebra, a linear equation is a statement of equality between two expressions that forms a straight line when graphed on a coordinate plane. These types of equations often come in the form y = mx + b, where m represents the slope of the line and b is the y-intercept or the point where the line crosses the y-axis.

In the case of the car depreciation problem, we can relate this to our function V(x) = 22500 - 3200x. Here, -3200 is the slope, which tells us that the car's value is dropping by \(3200 each year. The y-intercept, in this instance, is the initial value of the car, which is \)22,500. This equation is an example of how linear equations can be used in practical situations to model relationships between two variables. In this problem, the variables are the time in years and the value of the car.

Function Interpretation

The process of function interpretation involves making sense of the numerical results we obtain from a function and explaining them in the context of the problem at hand. To interpret the function V(x) = 22500 - 3200x, we need to plug in different values of x and analyze the results.

For instance, when we calculate V(3), this provides us with the car's value after 3 years. By substituting x with 3, we get V(3) = 22500 - 3200*3 = 15100. Interpreting this result, we understand that after 3 years, the car would be worth $15,100. This interpretation gives us insight into the car's depreciation and can be used by owners to make informed decisions about selling or trading the vehicle. Interpreting functions is an essential skill in algebra, allowing us to apply mathematical models to real-life scenarios and understand their implications.