To determine the exponential function that best fits the given data points, we will find an exponential function of the form:
[tex]\[ y = a \cdot e^{b \cdot x} \][/tex]
Given the data points:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 1 & 2 & 3 & 4 & 5 \\
\hline
y & 3 & 7 & 15 & 33 & 85 \\
\hline
\end{array}
\][/tex]
Here's a step-by-step approach to find the coefficients [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
1. Transforming the data: To use linear regression to determine the best-fit exponential function, we take the natural logarithm of the [tex]\(y\)[/tex]-values. This transforms the problem to a linear form, as taking the log of both sides gives:
[tex]\[ \ln(y) = \ln(a) + bx \][/tex]
2. Finding the linear relationship: Perform linear regression on the transformed data [tex]\((x, \ln(y))\)[/tex] to find the best-fit line. The slope of this line will be [tex]\(b\)[/tex], and the intercept will be [tex]\(\ln(a)\)[/tex].
3. Solving for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
- The intercept of the linear model in the logarithmic space is [tex]\(\ln(a)\)[/tex], so [tex]\( a = e^{\text{intercept}} \)[/tex].
- The slope of the line in the logarithmic space is directly [tex]\(b\)[/tex].
After performing these steps with the given data, we find:
- [tex]\( a \approx 1.306 \)[/tex]
- [tex]\( b \approx 0.824 \)[/tex]
4. Constructing the exponential function: With the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] determined, we can now write the exponential function as:
[tex]\[ y = 1.306 \cdot e^{0.824x} \][/tex]
This function represents the best fit for the given data points.
Thus, the final answer for the exponential function is:
[tex]\[ y = 1.306 \cdot e^{0.824x} \][/tex]
