Let's look at the given data points:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2 & -8 \\
\hline
3 & -18 \\
\hline
4 & -32 \\
\hline
5 & -50 \\
\hline
6 & -72 \\
\hline
\end{array}
\][/tex]
We need to determine which type of function, linear [tex]\((y = mx + b)\)[/tex], quadratic [tex]\((y = ax^2)\)[/tex], or exponential [tex]\((y = a(b)^x)\)[/tex], best fits this data.
After analyzing the patterns within the table:
1. Linear Fit:
- Linear equations follow the form [tex]\(y = mx + b\)[/tex]. Checking the first differences of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] can give us insight. If these first differences are not constant, a linear model is not suitable.
2. Quadratic Fit:
- Quadratic equations follow the form [tex]\(y = ax^2 + bx + c\)[/tex]. In this problem, we assume a simplified form without the linear term, i.e., [tex]\(y = ax^2\)[/tex]. We have checked the fit of a quadratic function.
3. Exponential Fit:
- Exponential equations follow the form [tex]\(y = a(b)^x\)[/tex]. Checking the ratios of consecutive [tex]\(y\)[/tex]-values can give us insight. If these ratios are not constant, an exponential model is not suitable.
Given that we are provided the coefficients for a quadratic function [tex]\(y = ax^2\)[/tex]:
After analyzing the data, it was found that the quadratic function fits well. The equation for the quadratic fit is:
[tex]\[
y = -2x^2
\][/tex]
This conclusion comes from determining the quadratic coefficients. The coefficient [tex]\(a\)[/tex] is approximately [tex]\(-2\)[/tex], and the coefficients [tex]\(b\)[/tex] and [tex]\(c\)[/tex] were not significant given the near-zero values taken from the analysis. Therefore:
[tex]\[
\boxed{-2x^2}
\][/tex]
This quadratic equation [tex]\(y = -2x^2\)[/tex] best models the given data set.
