Question

which exponential function has an initial value of 2?

x	f(x)
-1	1/4
0	1/2
1	1
2	2

graph of an exponential function

$f(x) = 3(2^x)$

$f(x) = 2(3^x)$

Explanation:

Step1: Recall initial value definition

The initial value of an exponential function \( f(x) = a(b^x) \) is the value when \( x = 0 \), since \( b^0 = 1 \), so \( f(0)=a \).

Step2: Analyze the table

For the table, when \( x = 0 \), \( f(0)=\frac{1}{2} \), so initial value is \( \frac{1}{2} \), not 2.

Step3: Analyze the graph

Visually, the graph's y - intercept (when \( x = 0 \)) appears to be 1 (estimating from the grid), not 2.

Step4: Analyze \( f(x)=3(2^x) \)

For \( f(x)=3(2^x) \), when \( x = 0 \), \( f(0)=3(2^0)=3(1) = 3 \), initial value is 3, not 2.

Step5: Analyze \( f(x)=2(3^x) \)

For \( f(x)=2(3^x) \), when \( x = 0 \), \( f(0)=2(3^0)=2(1)=2 \), so initial value is 2.

Answer:

\( f(x) = 2(3^x) \)