To determine which type of function best fits the given data, we need to analyze the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. Given the data points:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -6 \\
\hline
2 & -18 \\
\hline
3 & -54 \\
\hline
4 & -162 \\
\hline
5 & -486 \\
\hline
\end{array}
\][/tex]
We will consider the possibility of an exponential function of the form [tex]\( y = a \cdot b^x \)[/tex].
1. Assume an Exponential Model: Since the given data points are negative, we take their absolute values for easier computations and analysis, though the result will incorporate the negativity in the concluding steps.
2. Transform the Exponential Model:
Taking the natural logarithm of both sides of the equation [tex]\( y = a \cdot b^x \)[/tex], we get:
[tex]\[
\log(y) = \log(a) + x \cdot \log(b)
\][/tex]
This converts our exponential function into a linear form where [tex]\(\log(y)\)[/tex] is a linear function of [tex]\( x \)[/tex].
3. Fit a Line:
Using the logarithmic relationships of the data points:
[tex]\[
\log(|y|) = \log(a) + x \cdot \log(b)
\][/tex]
The fitting coefficients [tex]\(\log(a)\)[/tex] and [tex]\(\log(b)\)[/tex] are determined to be such that:
[tex]\[
\log(a) \approx 0.693
\][/tex]
and
[tex]\[
\log(b) \approx 1.099
\][/tex]
4. Exponentiate the Coefficients:
Converting back to the coefficients of our original model:
[tex]\[
a = e^{0.693} \approx 2
\][/tex]
and
[tex]\[
b = e^{1.099} \approx 3
\][/tex]
So, the exponential function that models this data is:
[tex]\[
y = -2 \cdot 3^x
\][/tex]
Thus, the final model representing the given data is:
[tex]\[
y = -2(3)^x
\][/tex]
