To write an exponential function [tex]\( y = a \cdot b^x \)[/tex] that represents the given data in the table, let's follow a clear, step-by-step approach:
1. Identify Constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
- First, observe the values of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 0:
[tex]\[
\begin{align*}
x & = 0 \\
y & = 1 \\
\end{align*}
\][/tex]
Substituting [tex]\( x = 0 \)[/tex] into the exponential function [tex]\( y = a \cdot b^0 \)[/tex], we get:
[tex]\[
y = a \cdot 1 = a
\][/tex]
Therefore, [tex]\( a = 1 \)[/tex].
2. Determine the Base [tex]\( b \)[/tex]:
- Use another pair of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values to find [tex]\( b \)[/tex]. Consider [tex]\( x = 1 \)[/tex]:
[tex]\[
\begin{align*}
x & = 1 \\
y & = 3 \\
\end{align*}
\][/tex]
Substitute these values into the function:
[tex]\[
3 = 1 \cdot b^1 \implies b = 3
\][/tex]
Therefore, [tex]\( b = 3 \)[/tex].
3. Formulate the Exponential Function:
Using the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] found:
[tex]\[
y = 1 \cdot 3^x \quad \text{or simply} \quad y = 3^x
\][/tex]
4. Verification:
To ensure our function [tex]\( y = 3^x \)[/tex] fits all given points in the table, let’s check:
[tex]\[
\begin{array}{cc}
x & y = 3^x \\ \hline
-2 & 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \\
-1 & 3^{-1} = \frac{1}{3} \\
0 & 3^0 = 1 \\
1 & 3^1 = 3 \\
2 & 3^2 = 9 \\
\end{array}
\][/tex]
The function [tex]\( y = 3^x \)[/tex] consistently matches all given [tex]\( y \)[/tex] values from the table.
Therefore, the exponential function that represents the data in the table is:
[tex]\[
y = 3^x
\][/tex]
