Answer :
To determine which table represents an exponential function of the form [tex]\( y = b^x \)[/tex] with [tex]\( 0 < b < 1 \)[/tex], let's analyze each table carefully.
### First Table Analysis:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & \frac{1}{27} \\ \hline -2 & \frac{1}{9} \\ \hline -2 & \frac{1}{3} \\ \hline 0 & 1 \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 27 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]: [tex]\( \frac{1/9}{1/27} = 3 \)[/tex]
- For [tex]\( x = -2 \)[/tex] to [tex]\( x = 0 \)[/tex]: [tex]\( \frac{1}{1/9} = 9 \)[/tex]
- For [tex]\( x = 0 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\( \frac{9}{1} = 9 \)[/tex]
- For [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\( \frac{27}{9} = 3 \)[/tex]
We can see that the values don't maintain a consistent ratio, suggesting it is not an exponential function of the desired form.
### Second Table Analysis:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 27 \\ \hline -2 & 9 \\ \hline -1 & 3 \\ \hline 0 & 1 \\ \hline 1 & \frac{1}{3} \\ \hline 2 & \frac{1}{9} \\ \hline 3 & \frac{1}{27} \\ \hline \end{array} \][/tex]
- For [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]: [tex]\( \frac{9}{27} = \frac{1}{3} \)[/tex]
- For [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]: [tex]\( \frac{3}{9} = \frac{1}{3} \)[/tex]
- For [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]: [tex]\( \frac{1}{3} = \frac{1}{3} \)[/tex]
- For [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]: [tex]\( \frac{1/3}{1} = \frac{1}{3} \)[/tex]
- For [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\( \frac{1/9}{1/3} = \frac{1}{3} \)[/tex]
- For [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\( \frac{1/27}{1/9} = \frac{1}{3} \)[/tex]
Each ratio [tex]\( \frac{y_{n+1}}{y_{n}} \)[/tex] is [tex]\( \frac{1}{3} \)[/tex], which is consistent and within the range [tex]\( 0 < b < 1 \)[/tex].
Therefore, the second table represents an exponential function of the form [tex]\( y = b^x \)[/tex] with [tex]\( 0 < b < 1 \)[/tex].
The table number is [tex]\( \boxed{2} \)[/tex].
### First Table Analysis:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & \frac{1}{27} \\ \hline -2 & \frac{1}{9} \\ \hline -2 & \frac{1}{3} \\ \hline 0 & 1 \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 27 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]: [tex]\( \frac{1/9}{1/27} = 3 \)[/tex]
- For [tex]\( x = -2 \)[/tex] to [tex]\( x = 0 \)[/tex]: [tex]\( \frac{1}{1/9} = 9 \)[/tex]
- For [tex]\( x = 0 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\( \frac{9}{1} = 9 \)[/tex]
- For [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\( \frac{27}{9} = 3 \)[/tex]
We can see that the values don't maintain a consistent ratio, suggesting it is not an exponential function of the desired form.
### Second Table Analysis:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 27 \\ \hline -2 & 9 \\ \hline -1 & 3 \\ \hline 0 & 1 \\ \hline 1 & \frac{1}{3} \\ \hline 2 & \frac{1}{9} \\ \hline 3 & \frac{1}{27} \\ \hline \end{array} \][/tex]
- For [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]: [tex]\( \frac{9}{27} = \frac{1}{3} \)[/tex]
- For [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]: [tex]\( \frac{3}{9} = \frac{1}{3} \)[/tex]
- For [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]: [tex]\( \frac{1}{3} = \frac{1}{3} \)[/tex]
- For [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]: [tex]\( \frac{1/3}{1} = \frac{1}{3} \)[/tex]
- For [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\( \frac{1/9}{1/3} = \frac{1}{3} \)[/tex]
- For [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\( \frac{1/27}{1/9} = \frac{1}{3} \)[/tex]
Each ratio [tex]\( \frac{y_{n+1}}{y_{n}} \)[/tex] is [tex]\( \frac{1}{3} \)[/tex], which is consistent and within the range [tex]\( 0 < b < 1 \)[/tex].
Therefore, the second table represents an exponential function of the form [tex]\( y = b^x \)[/tex] with [tex]\( 0 < b < 1 \)[/tex].
The table number is [tex]\( \boxed{2} \)[/tex].
