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Look at this table:

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-6 & 324 \\
\hline
-5 & 225 \\
\hline
-4 & 144 \\
\hline
-3 & 81 \\
\hline
-2 & 36 \\
\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]\[ y = \ \square \][/tex]


Sagot :

To determine which type of function best models the data, let's analyze the table provided:

[tex]\[ \begin{tabular}{|c|c|} \hline x & y \\ \hline -6 & 324 \\ \hline -5 & 225 \\ \hline -4 & 144 \\ \hline -3 & 81 \\ \hline -2 & 36 \\ \hline \end{tabular} \][/tex]

We need to determine whether the data fits a linear, quadratic, or exponential model. Let's start testing the quadratic function [tex]\( y = ax^2 \)[/tex].

Consider the first data point [tex]\( (-6, 324) \)[/tex]:

[tex]\[ y = ax^2 \implies 324 = a(-6)^2 \implies 324 = 36a \implies a = 9 \][/tex]

Next, let's verify this [tex]\( a \)[/tex] value with other points:

For [tex]\( x = -5 \)[/tex]:

[tex]\[ y = 9(-5)^2 = 9 \times 25 = 225 \][/tex]

For [tex]\( x = -4 \)[/tex]:

[tex]\[ y = 9(-4)^2 = 9 \times 16 = 144 \][/tex]

For [tex]\( x = -3 \)[/tex]:

[tex]\[ y = 9(-3)^2 = 9 \times 9 = 81 \][/tex]

For [tex]\( x = -2 \)[/tex]:

[tex]\[ y = 9(-2)^2 = 9 \times 4 = 36 \][/tex]

All data points perfectly fit the quadratic equation [tex]\( y = 9x^2 \)[/tex]. Thus, the function that models the data is:

[tex]\[ y = 9x^2 \][/tex]
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