Problem 2 - Algebra Form and Function

Question

A car's value \(t\) years after it is purchased is given by \(V(t)=18,000-1700 t\). How long does it take for the car's value to drop to \(\$ 2000 ?\)

Step-by-Step Solution

Answer

Answer: It takes approximately 9.41 years for the car's value to drop to $2000.

Step 1: Write down the given information and target

We are given the value function of the car as \(V(t) = 18000 - 1700t\). The target value is when \(V(t) = 2000\). Now, we will set \(V(t)\) equal to \(2000\) and solve for \(t\).

Step 2: Set up the equation

First, set the equation \(2000 = 18000 - 1700t\). Our goal is to isolate \(t\) and solve for its value.

Step 3: Solve the equation for \(t\)

To solve for \(t\), follow these steps: 1. Subtract 18000 from both sides: \(-16000 = -1700t\). 2. Divide both sides by -1700: \(t = \frac{-16000}{-1700}\).

Step 4: Calculate the value of \(t\)

Now, we can calculate the value of \(t\) by dividing the numbers: \(t = \frac{-16000}{-1700} = 9.41\).

Step 5: Interpret the result

It takes approximately 9.41 years for the car's value to drop to $2000.

Key Concepts

Value Function

The value function is a mathematical expression used to represent the value of an asset over time. In this exercise, the value function for a car is given by \(V(t) = 18,000 - 1700t\). Here, \(V(t)\) represents the car's value \(t\) years after purchase.

• The term \(18,000\) is the initial value of the car at \(t = 0\) (i.e., when it is brand new).
• The coefficient \(-1700\) represents the annual depreciation, or the amount the car's value decreases each year.

Understanding the components of the value function is crucial for predicting how the car's worth will change over time. It allows us to calculate the future value of the car for any given \(t\). In this scenario, the value function is linear, meaning that the car’s value decreases at a constant rate annually.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. In this problem, we need to determine how many years \(t\) it takes for the car's value to drop to \(2000.To do this, we solve the equation \(2000 = 18000 - 1700t\) by isolating \(t\):

• First, subtract \(18,000\) from both sides to get \(-16,000 = -1700t\).
• Then, divide both sides by \(-1700\) to calculate \(t\): \[t = \frac{-16000}{-1700} = 9.41\]

By performing these steps, we've solved the equation, finding that it takes approximately 9.41 years for the car's value to decrease to \)2000. This procedure demonstrates the power of algebraic manipulation, helping to figure out specific time frames or conditions related to financial investments and depreciation.

Depreciation

Depreciation refers to the decrease in value of an asset over time. For items like cars, this is an important concept because it affects the resale value and investment decisions. In this exercise, the car depreciates at a rate of $1700 per year, as indicated by the equation's slope in the value function.

• Types of Depreciation: The problem focuses on straight-line depreciation, which assumes a constant rate of decrease in value over time.
• Importance: Understanding depreciation helps individuals plan for future expenses, decide when to sell or replace assets, and manage budgets effectively.

Depreciation also plays a key role in accounting and tax calculations, as it affects the financial statements and determines the book value of assets. In practical terms, being aware of how quickly a car depreciates can influence buyers' decisions, such as when purchasing a vehicle or calculating its long-term ownership costs.