Question

which exponential function has an initial value of 2? (there is a table, a function expression $f(x)=3(2^x)$, and a graph in the problem, but the ocr text here only retains the main question text)

Explanation:

Step1: Define initial value rule

For exponential function $f(x)=a(b^x)$, initial value is $f(0)=a$. We need $a=2$.

Step2: Check table function

Find $f(0)$ from table: $f(0)=\frac{1}{2}
eq 2$.

Step3: Check given formula

Calculate $f(0)$ for $f(x)=3(2^x)$:
$f(0)=3(2^0)=3(1)=3
eq 2$.

Step4: Check graph function

Find $y$-intercept (initial value) of graph: when $x=0$, $y=2$, which matches the required initial value.

Answer:

The exponential function represented by the bottom graph (with y-intercept at 2) has an initial value of 2.