When Tyler was born, his father created a college fund by opening an investment account with $3,500 and an annual interest rate of 2.85%. The current amount of the investment account can be modeled by an exponential function in the form f(x)=a⋅bx

, where x is the number of years after the investment started. Find the function f(x)
.(1 point)
Responses

f(x)=3,500⋅2.85x
f left parenthesis x right parenthesis equals 3,500 times 2.85 superscript x baseline

f(x)=3,500⋅0.715x
f left parenthesis x right parenthesis equals 3,500 times 0.715 superscript x baseline

f(x)=3,500⋅1.285x
f left parenthesis x right parenthesis equals 3,500 times 1.285 superscript x baseline

f(x)=3,500⋅1.0285x

1 answer

To find the function \( f(x) \) that models the amount in the investment account over time, we can use the formula for exponential growth, which is typically given as:

\[ f(x) = a \cdot b^x \]

Where:

  • \( a \) is the initial amount (the principal),
  • \( b \) is the growth factor (1 plus the interest rate),
  • \( x \) is the number of years.

In this case:

  • The initial amount \( a \) is \( 3,500 \).
  • The annual interest rate is \( 2.85% \), which in decimal form is \( 0.0285 \). Therefore, the growth factor \( b \) is \( 1 + 0.0285 = 1.0285 \).

Now we can plug these values into the exponential function:

\[ f(x) = 3,500 \cdot (1.0285)^x \]

Thus, the correct function is:

\[ f(x) = 3,500 \cdot 1.0285^x \]

The correct option from your responses is:

f(x) = 3,500⋅1.0285^x.