Answer :
To determine which exponential function matches the values given in the table, we analyze the behavior of each function and compare it with the table.
Given Values:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & 16 \\ \hline -1 & 8 \\ \hline 0 & 4 \\ \hline 1 & 2 \\ \hline 2 & 1 \\ \hline \end{array} \][/tex]
Possible Functions:
1. [tex]\( f(x) = \frac{1}{2} (4)^x \)[/tex]
2. [tex]\( f(x) = 4 (4)^x \)[/tex]
3. [tex]\( f(x) = 4 \left(\frac{1}{2}\right)^x \)[/tex]
4. [tex]\( f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^x \)[/tex]
### Checking Each Function:
1. [tex]\( f(x) = \frac{1}{2} (4)^x \)[/tex]:
[tex]\[ \begin{align*} x = -2, & \quad f(x) = \frac{1}{2} (4)^{-2} = \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{32} \\ x = -1, & \quad f(x) = \frac{1}{2} (4)^{-1} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \\ x = 0, & \quad f(x) = \frac{1}{2} (4)^{0} = \frac{1}{2} \cdot 1 = \frac{1}{2} \\ x = 1, & \quad f(x) = \frac{1}{2} (4)^{1} = \frac{1}{2} \cdot 4 = 2 \\ x = 2, & \quad f(x) = \frac{1}{2} (4)^{2} = \frac{1}{2} \cdot 16 = 8 \\ \end{align*} These values do not match the table. 2. \( f(x) = 4 (4)^x \): \[ \begin{align*} x = -2, & \quad f(x) = 4 (4)^{-2} = 4 \cdot \frac{1}{16} = \frac{1}{4} \\ x = -1, & \quad f(x) = 4 (4)^{-1} = 4 \cdot \frac{1}{4} = 1 \\ x = 0, & \quad f(x) = 4 (4)^{0} = 4 \cdot 1 = 4 \\ x = 1, & \quad f(x) = 4 (4)^{1} = 4 \cdot 4 = 16 \\ x = 2, & \quad f(x) = 4 (4)^{2} = 4 \cdot 16 = 64 \\ \end{align*} These values do not match the table. 3. \( f(x) = 4 \left(\frac{1}{2}\right)^x \): \[ \begin{align*} x = -2, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{-2} = 4 \cdot 4 = 16 \\ x = -1, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{-1} = 4 \cdot 2 = 8 \\ x = 0, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{0} = 4 \cdot 1 = 4 \\ x = 1, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{1} = 4 \cdot \frac{1}{2} = 2 \\ x = 2, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{2} = 4 \cdot \frac{1}{4} = 1 \\ \end{align*} These values match the table. 4. \( f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^x \): \[ \begin{align*} x = -2, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{-2} = \frac{1}{2} \cdot 4 = 2 \\ x = -1, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{-1} = \frac{1}{2} \cdot 2 = 1 \\ x = 0, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{0} = \frac{1}{2} \cdot 1 = \frac{1}{2} \\ x = 1, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{1} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \\ x = 2, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{2} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \\ \end{align*} These values do not match the table. After checking these functions, we see that the values in the table match exactly with the function: \[ f(x) = 4 \left(\frac{1}{2}\right)^x \][/tex]
Thus, the exponential function represented by the values in the table is:
\[
f(x) = 4 \left(\frac{1}{2}\right)^x
\
Given Values:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & 16 \\ \hline -1 & 8 \\ \hline 0 & 4 \\ \hline 1 & 2 \\ \hline 2 & 1 \\ \hline \end{array} \][/tex]
Possible Functions:
1. [tex]\( f(x) = \frac{1}{2} (4)^x \)[/tex]
2. [tex]\( f(x) = 4 (4)^x \)[/tex]
3. [tex]\( f(x) = 4 \left(\frac{1}{2}\right)^x \)[/tex]
4. [tex]\( f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^x \)[/tex]
### Checking Each Function:
1. [tex]\( f(x) = \frac{1}{2} (4)^x \)[/tex]:
[tex]\[ \begin{align*} x = -2, & \quad f(x) = \frac{1}{2} (4)^{-2} = \frac{1}{2} \cdot \frac{1}{16} = \frac{1}{32} \\ x = -1, & \quad f(x) = \frac{1}{2} (4)^{-1} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \\ x = 0, & \quad f(x) = \frac{1}{2} (4)^{0} = \frac{1}{2} \cdot 1 = \frac{1}{2} \\ x = 1, & \quad f(x) = \frac{1}{2} (4)^{1} = \frac{1}{2} \cdot 4 = 2 \\ x = 2, & \quad f(x) = \frac{1}{2} (4)^{2} = \frac{1}{2} \cdot 16 = 8 \\ \end{align*} These values do not match the table. 2. \( f(x) = 4 (4)^x \): \[ \begin{align*} x = -2, & \quad f(x) = 4 (4)^{-2} = 4 \cdot \frac{1}{16} = \frac{1}{4} \\ x = -1, & \quad f(x) = 4 (4)^{-1} = 4 \cdot \frac{1}{4} = 1 \\ x = 0, & \quad f(x) = 4 (4)^{0} = 4 \cdot 1 = 4 \\ x = 1, & \quad f(x) = 4 (4)^{1} = 4 \cdot 4 = 16 \\ x = 2, & \quad f(x) = 4 (4)^{2} = 4 \cdot 16 = 64 \\ \end{align*} These values do not match the table. 3. \( f(x) = 4 \left(\frac{1}{2}\right)^x \): \[ \begin{align*} x = -2, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{-2} = 4 \cdot 4 = 16 \\ x = -1, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{-1} = 4 \cdot 2 = 8 \\ x = 0, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{0} = 4 \cdot 1 = 4 \\ x = 1, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{1} = 4 \cdot \frac{1}{2} = 2 \\ x = 2, & \quad f(x) = 4 \left(\frac{1}{2}\right)^{2} = 4 \cdot \frac{1}{4} = 1 \\ \end{align*} These values match the table. 4. \( f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^x \): \[ \begin{align*} x = -2, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{-2} = \frac{1}{2} \cdot 4 = 2 \\ x = -1, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{-1} = \frac{1}{2} \cdot 2 = 1 \\ x = 0, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{0} = \frac{1}{2} \cdot 1 = \frac{1}{2} \\ x = 1, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{1} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \\ x = 2, & \quad f(x) = \frac{1}{2} \left(\frac{1}{2}\right)^{2} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8} \\ \end{align*} These values do not match the table. After checking these functions, we see that the values in the table match exactly with the function: \[ f(x) = 4 \left(\frac{1}{2}\right)^x \][/tex]
Thus, the exponential function represented by the values in the table is:
\[
f(x) = 4 \left(\frac{1}{2}\right)^x
\
