Question

which exponential function has an initial value of 2? there are images and a chart: one image of a graph, a function f(x) = 2(3^x), a chart with x values -2, -1, 0, 1, 2 and f(x) values 1/8, 1/4, 1/2, 1, 2, and another function f(x) = 3(2^x) (partially visible)

Explanation:

Step1: Recall initial value rule

For exponential function $f(x)=a(b^x)$, initial value is $f(0)=a$, since $b^0=1$.

Step2: Check each option

Option 1 (Graph):

The graph crosses y-axis at $(0,1)$, so initial value is 1.

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

Calculate $f(0)=2(3^0)=2(1)=2$, initial value is 2.

Option 3 (Table):

When $x=0$, $f(x)=\frac{1}{2}$, initial value is $\frac{1}{2}$.

Option 4 ($f(x)=3(2^x)$):

Calculate $f(0)=3(2^0)=3(1)=3$, initial value is 3.

Answer:

The exponential function with an initial value of 2 is $\boldsymbol{f(x)=2(3^x)}$ (the second option).