Question
In year 1, a house is worth 500000. Due to a recession, the house’s value drops to 410000 in year 6.
Write the rule to model the value of the house.
Write the rule to model the value of the house.
Answers
GPT-4o mini
To model the value of the house over time, we need to establish a function that shows how the house’s value decreases from year 1 to year 6. We have the following points:
- At Year 1: \( V(1) = 500000 \)
- At Year 6: \( V(6) = 410000 \)
We can use a linear function to model the value of the house since we have two points. A linear function can be expressed in the form:
\[
V(t) = mt + b
\]
where \( V(t) \) is the value of the house at year \( t \), \( m \) is the slope (the rate of change), and \( b \) is the y-intercept (initial value at \( t = 1 \)).
1. **Find the slope (m):**
The slope \( m \) can be calculated as follows:
\[
m = \frac{V(6) - V(1)}{6 - 1} = \frac{410000 - 500000}{6 - 1} = \frac{-90000}{5} = -18000
\]
2. **Find the initial value (b):**
Since the initial value \( V(1) = 500000 \) (at year 1), we can directly use this as the y-intercept.
Now we can plug \( m \) and \( b \) into the linear function:
\[
V(t) = -18000t + b
\]
To determine \( b \), we first rewrite it in terms of \( t \) at \( t = 1 \) to calculate:
\[
V(1) = -18000(1) + b = 500000 \\
-18000 + b = 500000 \\
b = 500000 + 18000 \\
b = 518000
\]
3. **Final model:**
Now we plug this value back in to get the function:
The rule modeling the value of the house is:
\[
V(t) = -18000t + 518000
\]
Keep in mind that this model is linear and valid from Year 1 onwards, but the specific behavior of house values may vary according to other factors not incorporated into this simple model.
- At Year 1: \( V(1) = 500000 \)
- At Year 6: \( V(6) = 410000 \)
We can use a linear function to model the value of the house since we have two points. A linear function can be expressed in the form:
\[
V(t) = mt + b
\]
where \( V(t) \) is the value of the house at year \( t \), \( m \) is the slope (the rate of change), and \( b \) is the y-intercept (initial value at \( t = 1 \)).
1. **Find the slope (m):**
The slope \( m \) can be calculated as follows:
\[
m = \frac{V(6) - V(1)}{6 - 1} = \frac{410000 - 500000}{6 - 1} = \frac{-90000}{5} = -18000
\]
2. **Find the initial value (b):**
Since the initial value \( V(1) = 500000 \) (at year 1), we can directly use this as the y-intercept.
Now we can plug \( m \) and \( b \) into the linear function:
\[
V(t) = -18000t + b
\]
To determine \( b \), we first rewrite it in terms of \( t \) at \( t = 1 \) to calculate:
\[
V(1) = -18000(1) + b = 500000 \\
-18000 + b = 500000 \\
b = 500000 + 18000 \\
b = 518000
\]
3. **Final model:**
Now we plug this value back in to get the function:
The rule modeling the value of the house is:
\[
V(t) = -18000t + 518000
\]
Keep in mind that this model is linear and valid from Year 1 onwards, but the specific behavior of house values may vary according to other factors not incorporated into this simple model.