Problem 96 - College Algebra with Modeling and Visualization

Question

Value of a Home In 1999 the value of a house was \(\$ 180,000,\) and in 2009 it was \(\$ 245,000\) (a) Find a linear function \(V\) that approximates the value of the house during year \(x .\) (b) What does the slope of the graph of \(V\) represent? (c) Use \(V\) to estimate the year when the house was worth \(\$ 219,000\)

Step-by-Step Solution

Answer

(a) \(V(x) = 6,500x - 12,284,500\); (b) Slope is \$6,500 per year; (c) Mid-2004.

Step 1: Determining the Linear Function Model

The value of a house increases linearly over time. We have the points (1999, 180,000) and (2009, 245,000). Let \(x\) be the year and \(V(x)\) be the approximate house value. A linear function can be represented as \(V(x) = mx + b\), where "\(m\)" is the slope and "\(b\)" is the y-intercept. We need to first find the slope \(m\) using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\).

Step 2: Calculating the Slope

Using the formula \(m = \frac{245,000 - 180,000}{2009 - 1999}\), the slope \(m\) is \(\frac{65,000}{10} = 6,500\). This means that the value of the house increases by \$6,500 every year.

Step 3: Finding the Y-intercept

To find the y-intercept 'b', use the point (1999, 180,000) and the equation \(V(x) = 6500x + b\). Substitute \(x = 1999\) into the equation: \(180,000 = 6,500 \times 1999 + b\). Solving for \(b\), we find that \(b = -12,284,500\). Thus, the linear function is \(V(x) = 6,500x - 12,284,500\).

Step 4: Interpreting the Slope

The slope of the graph of \(V\), which is \(6,500\), represents the rate of increase in the house value each year. Thus, the house value increases by \$6,500 annually.

Step 5: Estimating the Year for a Given House Value

To estimate when the house was worth \\(219,000, set \(V(x) = 219,000\). That gives the equation: \(219,000 = 6,500x - 12,284,500\). Solving for \(x\), add 12,284,500 to both sides: \(219,000 + 12,284,500 = 6,500x\) \(12,503,500 = 6,500x\) Then, divide by 6,500: \(x = \frac{12,503,500}{6,500} \approx 1924.38\). Since \(x\) represents the year, rounding up, the house was worth \\)219,000 in mid-2004.

Key Concepts

Slope Calculation

Understanding the slope of a line is crucial in analyzing linear functions. In the context of a real-world problem like house valuation, the slope represents the rate of change. To calculate the slope, you use two known points on a line, such as the years and corresponding house values.

The formula to compute the slope (denoted as \(m\)) is:

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

In this function, \((x_1, y_1)\) and \((x_2, y_2)\) are points on the line. Here, these points are \((1999, 180,000)\) and \((2009, 245,000)\). Applying the formula:

• \(m = \frac{245,000 - 180,000}{2009 - 1999} = \frac{65,000}{10} = 6,500\)

The slope of 6,500 means the house's value increases by $6,500 every year. This slope is a way to quantify the change in value over time, highlighting how much more the house is worth annually.

Linear Equation

A linear equation is a powerful tool for modeling relationships where there is a constant rate of change. This type of equation takes the form \( V(x) = mx + b \), where \(m\) is the slope and \(b\) is the y-intercept.

In our exercise, the linear equation representing the house value is:

• \( V(x) = 6,500x - 12,284,500 \)

To derive this equation, we've already calculated the slope, \(m\), as 6,500. The y-intercept, \(b\), is found by substituting one of the given points into the equation and solving for \(b\). For the point \((1999, 180,000)\):

• \(180,000 = 6,500 \times 1999 + b\)
• Solving gives \(b = -12,284,500\)

This equation allows us to predict or estimate the house's value at any given year \(x\). It's a straightforward model of the relationship between time and value.

Y-intercept

The y-intercept is an essential part of a linear equation, representing the point where the line crosses the y-axis. In our equation \( V(x) = 6,500x - 12,284,500 \), the y-intercept is \(-12,284,500\). This value tells us the theoretical value of the house when \(x\), the year, is 0. While this isn't a realistic year for our problem (as it concerns modern home value), it helps form the basis of the equation.

This value is found by solving the equation for \(b\), using one of our known points. Though the interpretation of the y-intercept might not make immediate practical sense in terms of historical analysis, it is crucial for constructing the equation. It essentially shifts the line up or down to match the given data, ensuring the calculated slope fits the actual points provided.