A house's value, originally bought for$311000 in 2017, has been growing at a rate of \displaystyle{h(t) = 6(1.18)^t} thousand dollars per years, where t is the number of years since 2017 calculator checkpoint: h(2)=8.3544. A. how much has the value of the house grown from 2017 to 2023? round to the three decimal place

B. how much is the value of the house projected to grow from 2023 to 2033? round to the three decimal place

C. find the specific antiderivative that models the value of the house in thousand dollars t years since 2017





### Part A: How much has the value of the house grown from 2017 to 2023?

The house's growth rate is given by $ h(t) = 6(1.18)^t $, where $ t $ is the number of years since 2017. We need to find the total growth from 2017 to 2023.

To do this, we'll calculate the value of the house in 2023 (i.e., $ t = 6 $) and compare it with the value in 2017 (i.e., $ t = 0 $).

\[
h(6) = 6(1.18)^6
\]
\[
h(0) = 6(1.18)^0 = 6
\]

The growth from 2017 to 2023 is given by:
\[
\text{Growth from 2017 to 2023} = h(6) - h(0)
\]

Let me calculate this value.The value of the house has grown by approximately $ 10.197 $ thousand dollars from 2017 to 2023.

### Part B: How much is the value of the house projected to grow from 2023 to 2033?

To calculate the projected growth from 2023 to 2033, we need to calculate the values at $ t = 16 $ (which represents 2033) and $ t = 6 $ (which represents 2023), then find the difference.

\[
h(16) = 6(1.18)^{16}
\]
\[
h(6) = 6(1.18)^6
\]

The projected growth from 2023 to 2033 is given by:
\[
\text{Growth from 2023 to 2033} = h(16) - h(6)
\]

Let's calculate this.It seems like I can’t do more advanced data analysis right now. Please try again later.

You can calculate the projected growth from 2023 to 2033 using the same method as before by plugging the values into the formula for $ h(t) $.

### Part C: Find the antiderivative of the growth function $ h(t) = 6(1.18)^t $

The antiderivative (or indefinite integral) of the function $ h(t) = 6(1.18)^t $ will give the total value of the house in thousand dollars over time.

To find the antiderivative:

\[
\int 6(1.18)^t \, dt
\]

We can apply the rule for integrating exponential functions:

\[
\int a^t \, dt = \frac{a^t}{\ln(a)}
\]

So,

\[
\int 6(1.18)^t \, dt = \frac{6(1.18)^t}{\ln(1.18)} + C
\]

where $ C $ is the constant of integration. This gives the specific antiderivative that models the value of the house.

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Would you like me to explain any part in more detail or try another calculation? Here are some follow-up questions to expand on this topic:

1. How can you calculate future values of a home based on its rate of appreciation?
2. How does the integration of the growth function reflect the total value over time?
3. What other financial applications use similar exponential growth models?
4. How would the model change if the growth rate varied over time?
5. How does compounding growth influence long-term projections in real estate?

**Tip:** When working with exponential growth, small changes in the rate can lead to large differences in long-term projections!

A house's value, originally bought for $311,000 in 2017, has been growing at a rate of h(t) = 6(1.18)^t thousand dollars per year, where t is the number of years since 2017. A. How much has the value of the house grown from 2017 to 2023? Round to three decimal places. B. How much is the value of the house projected to grow from 2023 to 2033? Round to three decimal places. C. Find the specific antiderivative that models the value of the house in thousand dollars t years since 2017.

Analyze how a house's value grows using exponential functions. Solve for the increase from 2017 to 2023, project growth to 2033, and find the antiderivative to model the house's value over time. Suitable for high school students studying calculus and exponential growth models.

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