Question

which exponential function has an initial value of 2?
○ $f(x) = 2(3^x)$
graph of an exponential curve on a coordinate plane with grid
○
○ $f(x) = 3(2^x)$
table with columns x and f(x), first row under x is empty, under f(x) is 1

Explanation:

Step1: Recall the form of exponential function

The general form of an exponential function is \( f(x) = a(b^x) \), where \( a \) is the initial value (the value of the function when \( x = 0 \)) and \( b \) is the base (\( b>0, b
eq1 \)).

Step2: Find the initial value of each function

• For \( f(x)=2(3^x) \): Substitute \( x = 0 \) into the function. We know that \( 3^0=1 \), so \( f(0)=2\times(3^0)=2\times1 = 2 \).
• For \( f(x)=3(2^x) \): Substitute \( x = 0 \) into the function. We know that \( 2^0 = 1 \), so \( f(0)=3\times(2^0)=3\times1=3 \).

Answer:

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