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Sagot :
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.
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.
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