Problem 26 - College Algebra Essentials

Question

A new car worth \(\$ 45,000\) is depreciating in value by \(\$ 5000\) per year. a. Write a formula that models the car's value, \(y,\) in dollars, after \(x\) years. b. Use the formula from part (a) to determine after how many years the car's value will be \(\$ 10,000\). c. Graph the formula from part (a) in the first quadrant of a rectangular coordinate system. Then show your solution to part (b) on the graph.

Step-by-Step Solution

Answer

a. The depreciation is represented by the model \(y = 45000 - 5000x\). b. The value of the car would depreciate to \(\$10000\) after 7 years. c. The graph of the model, with a marked point at the seventh year, can help to visualize the depreciation process and particular years' values.

Step 1: Establish the Depreciation Model

The car's value decreases by \(\$5000\) annually. Therefore, the formula to calculate the remaining value (\(y\)) of the car after \(x\) years can be stated as \(y = 45000 - 5000x\).

Step 2: Find the Year When Car Value Decreases to \$10000

To determine the year number when the car's value will descend to \(\$10000\), we need to solve the equation \(45000 - 5000x = 10000\). From this, we obtain \(x = 7\), which means that the car would be worth \(\$10000\) after 7 years.

Step 3: Graph the Depreciation Model and Mark the Specific Year

Plotting the model expression \(y = 45000 - 5000x\) on the coordinate plane depicts the linear depreciation rate. The x-axis stands for the years while the y-axis replicates the car's price. To illustrate the solution from step 2, we should mark the 7th year on the x-axis and draw a straight line from this point to intersect the graph. The intersecting point on the y-axis reflects the \($10000\) value.

Key Concepts

Depreciation Model

When we talk about linear depreciation, we are referring to a consistent loss of value over time. In the context of our example, we have a car bought at \( \\(45,000 \) and it loses \( \\)5,000 \) in value every year. This loss can be modeled with a linear equation. The general formula for linear depreciation is expressed as:

• \( y = C - Dx \)

where \( C \) is the initial cost (\( \\(45,000 \) in our case) and \( D \) is the depreciation per time period (\( \\)5,000 \) per year). \( x \) is the time in years. For our car, the formula becomes \( y = 45000 - 5000x \).
Linear depreciation models are vital in predicting how long an asset remains useful or attractive economically. They help both sellers and buyers understand value loss over specified time frames.

Graphing Linear Equations

Graphing a linear equation such as \( y = 45000 - 5000x \) helps visualize the relationship between time and the car's value. Here’s a simplified method to plot this on a graph:

• Start by labeling your axes: the horizontal (x-axis) for time in years, and the vertical (y-axis) for the car's value in dollars.
• Identify key points using the formula. For instance, when \( x = 0 \), \( y = 45000 \), and when \( x = 7 \) (from our solution step), \( y = 10000 \).

Connecting these points will form a straight line, illustrating the consistent value loss each passing year due to depreciation.
By graphing linear equations, it becomes easier to:

• Spot overall trends, and
• Forecast future values or decisions at a glance.

Using a graph enables a visual understanding that complements numerical solutions.

Solving Linear Equations

Solving a linear equation involves finding the value of the variable that makes the equation true. In our car depreciation model \( 45000 - 5000x = 10000 \), we need to solve for \( x \), which represents the time in years. Here's how to solve it:

• First, subtract \( 10000 \) from \( 45000 \.\) This simplifies to \( 5000x = 35000 \).
• Then, divide both sides by \( 5000 \) to isolate \( x \), resulting in \( x = 7 \).

This tells us that after 7 years, the car's value will depreciate to \( \$10,000 \).
Solving linear equations like these is crucial in scenarios where predictions or precise time calculations are necessary. It involves simple algebraic manipulations, aiming to balance equations to uncover the unknown element, which in practical terms can greatly aid in personal or commercial decision-making.