Look at this table:

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-6 & -3 \\
\hline
-5 & -4 \\
\hline
-4 & -5 \\
\hline
-3 & -6 \\
\hline
-2 & -7 \\
\hline
\end{tabular}

Write a linear [tex]$(y=mx+b)$[/tex], quadratic [tex]$\left(y=ax^2\right)$[/tex], or exponential [tex]$\left(y=a(b)^x\right)$[/tex] function that models the data.

[tex]$y=$[/tex]

[tex]$\square$[/tex]



Answer :

Sure! Let's analyze the provided data to determine the best-fitting function.

The data points given are:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -6 & -3 \\ \hline -5 & -4 \\ \hline -4 & -5 \\ \hline -3 & -6 \\ \hline -2 & -7 \\ \hline \end{array} \][/tex]

To model this data, we look for a potential linear relationship [tex]\( y = mx + b \)[/tex].

The linear function is determined by the coefficients [tex]\( m \)[/tex] (slope) and [tex]\( b \)[/tex] (y-intercept). From the analysis:

The slope [tex]\( m = -1.0000000000000004 \approx -1 \)[/tex]

The intercept [tex]\( b = -8.999999999999998 \approx -9 \)[/tex]

Thus, the linear function that best models the data is:

[tex]\[ y = -1x - 9 \][/tex]

So, the function that models the data is:

[tex]\[ y = -x - 9 \][/tex]