To analyze the function \( f(t) = 10,000(1.009)^{12t} \), we can interpret its components:
- The initial amount invested (the principal) is represented by the coefficient outside the parentheses, which is \( 10,000 \).
- The term \( (1.009)^{12t} \) indicates that the money is being compounded monthly, with \( 1.009 \) representing the monthly growth factor.
To find the annual growth rate, we need to convert the monthly growth rate into an annual rate:
- The monthly growth rate is \( 0.009 \) (which corresponds to \( 1.009 - 1 \)).
- To find the effective annual rate (EAR), we use the formula for compounding:
\[ \text{Annual Rate} = (1 + r)^n - 1 \]
where \( r \) is the monthly rate (0.009) and \( n \) is the number of months in a year (12):
\[ \text{Annual Rate} = (1 + 0.009)^{12} - 1 \]
Calculating this:
\[ (1.009)^{12} \approx 1.1136 \quad \text{(approximately)} \] \[ \text{Annual Rate} \approx 1.1136 - 1 \approx 0.1136 \text{ or } 11.36% \]
Thus, the initial amount invested is \( 10,000 \), which grows at an annual rate of approximately \( 11.36% \).
Given the available statements, the most accurate choice is:
The initial amount of money invested is $10,000, which grows at an annual rate of about 11%.