Look at this table:

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
1 & 8 \\
\hline
2 & 16 \\
\hline
3 & 32 \\
\hline
4 & 64 \\
\hline
5 & 128 \\
\hline
\end{tabular}

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

[tex]\[ y = \ \square \][/tex]



Answer :

Looking at the given table:

[tex]\[ \begin{tabular}{|c|c|} \hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
1 & 8 \\
\hline
2 & 16 \\
\hline
3 & 32 \\
\hline
4 & 64 \\
\hline
5 & 128 \\
\hline
\end{tabular}
\][/tex]

We can see a clear pattern: each value of \( y \) is double the previous value of \( y \). This indicates the relationship between \( x \) and \( y \) is exponential in nature, rather than linear.

An exponential function is typically of the form:

[tex]\[ y = a \cdot b^x \][/tex]

To determine the values of the coefficients \( a \) and \( b \), we use the given data points.

From the table, we can see:
[tex]\[ y(1) = 8 \\ y(2) = 16 \\ y(3) = 32 \\ y(4) = 64 \\ y(5) = 128 \][/tex]

To find \( b \):
Taking the ratio of consecutive \( y \)-values to determine the base \( b \):

[tex]\[ \frac{y(2)}{y(1)} = \frac{16}{8} = 2 \][/tex]

Thus, \( b = 2 \).

To find \( a \):
Using the first data point \( (x = 1, y = 8) \):

[tex]\[ 8 = a \cdot 2^1 \Rightarrow a = \frac{8}{2} = 4 \][/tex]

So, the exponential function that models the data is:

[tex]\[ \boxed{y = 4 \cdot 2^x} \][/tex]