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\begin{tabular}{|c|c|}
\hline [tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline -1 & 18 \\
\hline 0 & 6 \\
\hline 1 & 2 \\
\hline 2 & [tex]$\frac{2}{3}$[/tex] \\
\hline
\end{tabular}

What is the decay factor of the exponential function represented by the table?

A. [tex]$\frac{1}{3}$[/tex]
B. [tex]$\frac{2}{3}$[/tex]
C. 2
D. 6


Sagot :

To find the decay factor of the exponential function represented by the table, we need to calculate the ratios of consecutive function values ([tex]\(f(x)\)[/tex]) as [tex]\(x\)[/tex] increases by 1 unit. This is because, in an exponential function of the form [tex]\(f(x) = a \cdot b^x\)[/tex], where [tex]\(b\)[/tex] is the decay factor, the ratio of successive function values is constant.

Given the table:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -1 & 18 \\ \hline 0 & 6 \\ \hline 1 & 2 \\ \hline 2 & \frac{2}{3} \\ \hline \end{tabular} \][/tex]

1. Calculate the ratio of [tex]\(f(0)\)[/tex] to [tex]\(f(-1)\)[/tex]:
[tex]\[ \frac{f(0)}{f(-1)} = \frac{6}{18} = \frac{1}{3} \][/tex]

2. Calculate the ratio of [tex]\(f(1)\)[/tex] to [tex]\(f(0)\)[/tex]:
[tex]\[ \frac{f(1)}{f(0)} = \frac{2}{6} = \frac{1}{3} \][/tex]

3. Calculate the ratio of [tex]\(f(2)\)[/tex] to [tex]\(f(1)\)[/tex]:
[tex]\[ \frac{f(2)}{f(1)} = \frac{2/3}{2} = \frac{2/3}{2} = \frac{1}{3} \][/tex]

As you can see, the ratio of consecutive [tex]\(f(x)\)[/tex] values is consistently [tex]\(\frac{1}{3}\)[/tex].

Therefore, the decay factor of the exponential function represented by the table is [tex]\(\frac{1}{3}\)[/tex]. Among the answer choices, the correct one is:
[tex]\[ \boxed{\frac{1}{3}} \][/tex]
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