To find the exponential regression equation that fits the given data, we'll use the general form for an exponential function:
[tex]\[ y = a \cdot b^x \][/tex]
Given the steps required to derive this form:
1. Fit the Data to Exponential Model: We are using the data points [tex]\((x, y)\)[/tex] provided in the table. These points are:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & 4 \\
\hline
2 & 9 \\
\hline
3 & 27 \\
\hline
4 & 90 \\
\hline
5 & 250 \\
\hline
6 & 580 \\
\hline
\end{array}
\][/tex]
2. Transform the Exponential Model: To use linear regression, we can take the natural logarithm of both sides:
[tex]\[ \ln(y) = \ln(a) + x \cdot \ln(b) \][/tex]
This transformation linearizes the relationship between [tex]\(x\)[/tex] and [tex]\(\ln(y)\)[/tex].
3. Perform Linear Regression: We perform linear regression on the data points [tex]\((x, \ln(y))\)[/tex] to find the coefficients [tex]\(\ln(a)\)[/tex] and [tex]\(\ln(b)\)[/tex].
4. Exponentiate the Coefficients: Once we have [tex]\(\ln(a)\)[/tex] and [tex]\(\ln(b)\)[/tex], we exponentiate these to return to the original exponential form:
[tex]\[ a = e^{\ln(a)} \][/tex]
[tex]\[ b = e^{\ln(b)} \][/tex]
Following these steps, we find the coefficients to be approximately:
[tex]\[ a \approx 1.31 \][/tex]
[tex]\[ b \approx 2.80 \][/tex]
Therefore, the exponential regression equation that fits the given data is:
[tex]\[ y = 1.31 \cdot (2.80^x) \][/tex]
Looking at the provided answer choices:
A. [tex]\( y = 2.80(1.31^x) \)[/tex]
B. [tex]\( y = 39.16x^2 - 169.38x + 158.9 \)[/tex]
C. [tex]\( y = 104.74x - 206.6 \)[/tex]
D. [tex]\( y = 1.31 \cdot (2.80^x) \)[/tex]
The correct answer is:
[tex]\[ \boxed{y = 1.31 \cdot (2.80^x)} \][/tex]
So, the correct answer is option D.
