Answered

Elias purchases a home for [tex]$\$[/tex]38,900[tex]$. The value of the home, in thousands of dollars, since his purchase is shown in the table.

\begin{tabular}{|c|c|}
\hline
\begin{tabular}{c}
Years since \\
Purchase
\end{tabular} & \begin{tabular}{c}
Value \\
(thousands of $[/tex]\[tex]$)
\end{tabular} \\
\hline
0 & 38.9 \\
\hline
5 & 62.4 \\
\hline
10 & 89.3 \\
\hline
15 & 145.2 \\
\hline
20 & 210.8 \\
\hline
25 & 326.5 \\
\hline
\end{tabular}

Find an exponential function that models the data. Round numerical values to the nearest hundredth. Let $x$ be the number of years since the purchase.

The function $f(x) = \square \cdot \square^x$ models the data. Use the model to predict the home's value.

After 12 years, the home's value will be about $\$[/tex] [tex]$\square$[/tex] thousand.

After 35 years, the home's value will be about [tex]$\$[/tex][tex]$ $[/tex]\square$ thousand.



Answer :

To determine an exponential function that models the given data of the home's value over time, we follow these detailed steps:

1. Understanding the data: The table gives us the value of the home (in thousands of dollars) at different time points (years since purchase).

2. Exponential Model: We assess that the home value can be modeled by an exponential function of the form:
[tex]\[ f(x) = a \cdot e^{bx} \][/tex]
where [tex]\( x \)[/tex] is the number of years since the purchase, [tex]\( a \)[/tex] is the initial value of the home in thousands of dollars, and [tex]\( b \)[/tex] is the exponential growth rate.

3. Converting the data for linear regression:
To find [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we take the natural logarithm of the value data to linearize it:
[tex]\[ \ln(y) = \ln(a \cdot e^{bx}) = \ln(a) + bx \][/tex]
This transforms our exponential relationship into a linear form:
[tex]\[ \ln(y) = bx + \ln(a) \][/tex]

4. Performing linear regression:
Using the given data:
[tex]\[ x = [0, 5, 10, 15, 20, 25] \][/tex]
[tex]\[ y = [38.9, 62.4, 89.3, 145.2, 210.8, 326.5] \][/tex]
We perform linear regression on the transformed data [tex]\( (\ln(y)) \)[/tex] to find the slope [tex]\( b \)[/tex] and the intercept [tex]\( \ln(a) \)[/tex].

5. Extracting [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
By solving the linear regression:
- [tex]\( a \)[/tex] is found by exponentiating the intercept.
- [tex]\( b \)[/tex] is the slope.

6. Fitting the Model:
After performing the calculation, we get:
[tex]\[ a \approx 39.59 \][/tex]
[tex]\[ b \approx 0.08 \][/tex]
Thus, the exponential function that models the data is:
[tex]\[ f(x) \approx 39.59 \cdot e^{0.08x} \][/tex]

7. Predicting the home value after 12 and 35 years:
Using the model [tex]\( f(x) = 39.59 \cdot e^{0.08x} \)[/tex]:
- After 12 years:
[tex]\[ f(12) \approx 39.59 \cdot e^{0.08 \cdot 12} \approx 109.05 \text{ thousand dollars} \][/tex]
- After 35 years:
[tex]\[ f(35) \approx 39.59 \cdot e^{0.08 \cdot 35} \approx 760.28 \text{ thousand dollars} \][/tex]

Summarizing our findings for the question:

The function [tex]\( f(x) \approx 39.59 \cdot e^{0.08x} \)[/tex] models the data.

- After 12 years, the home's value will be about [tex]\( \$109.05 \)[/tex] thousand.
- After 35 years, the home's value will be about [tex]\( \$760.28 \)[/tex] thousand.