To determine which exponential equation could represent the given table of values, we must analyze the behavior of each candidate equation and compare its results to the values provided in the table.
The given table of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values is:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
x & 3 & 2 & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline
y & 2.0014 & 2.0041 & 2.0123 & 2.037 & 2.111 & 2.333 & 3 & 5 \\
\hline
\end{array}
\][/tex]
We need to verify each of the four given equations against this data:
1. [tex]\( y = \left(\frac{1}{3}\right)^{x-3} - 3 \)[/tex]
2. [tex]\( y = 3^{-2} - 3 \)[/tex]
3. [tex]\( y = 3^{x-3} + 2 \)[/tex]
4. [tex]\( y = \left(\frac{1}{3}\right)^{x-3} - 2 \)[/tex]
Firstly, let's check the third option:
### Option 3: [tex]\( y = 3^{x-3} + 2 \)[/tex]
We calculate [tex]\( y \)[/tex] for each [tex]\( x \)[/tex] value:
[tex]\[
\begin{aligned}
& x = 3: & y = 3^{3-3} + 2 = 3^0 + 2 = 1 + 2 = 3 \\
& x = 2: & y = 3^{2-3} + 2 = 3^{-1} + 2 = \frac{1}{3} + 2 \approx 2.333 \\
& x = -1: & y = 3^{-1-3} + 2 = 3^{-4} + 2 = \frac{1}{81} + 2 \approx 2.0123 \\
& x = 0: & y = 3^{0-3} + 2 = 3^{-3} + 2 = \frac{1}{27} + 2 \approx 2.037 \\
& x = 1: & y = 3^{1-3} + 2 = 3^{-2} + 2 = \frac{1}{9} + 2 \approx 2.111 \\
& x = 4: & y = 3^{4-3} + 2 = 3^{1} + 2 = 3 + 2 = 5 \\
\end{aligned}
\][/tex]
Comparing these calculated values with the provided table values, we observe that they match closely with a small tolerance (usual in numerical calculations). Therefore, the equation that best fits the data in the table is:
[tex]\[ \boxed{y = 3^{x-3} + 2} \][/tex]
