Question: Suppose that $88,000 is invested at \(6\frac{1}{2}\)% interest, compounded quarterly. a) Find…

Suppose that $88,000 is invested at \(6\frac{1}{2}\)% interest, compounded quarterly.

a) Find the function for the amount to which the investment grows after \(t\) years.

b) Graph the function.

c) Find the amount of money in the account at \(t = 0\), 3, 5, and 10 years.

d) When will the amount of money in the account reach $300,000?

a) The function for the amount to which the investment grows after \(t\) years is \(A(t) = \boxed{\quad}\).

(Simplify your answer. Type an expression using \(t\) as the variable.)

Solution

Suppose that $88,000 is invested at \(6 \frac{1}{2}\%\) interest, compounded quarterly. a) To find the function for the amount to which the investment grows after \( t \) years, we use the compound interest formula: \[ A(t) = P \left( 1 + \frac{r}{n} \right)^{nt} \] where: - \( P = 88,000 \) is the principal amount, - \( r = 0.065 \) (6.5% as a decimal) is the annual interest rate, - \( n = 4 \) is the number of times interest is compounded per year (quarterly), - \( t \) is the time in years. Substituting the values, we have: \[ A(t) = 88000 \left( 1 + \frac{0.065}{4} \right)^{4t} \] Simplify the expression: \[ A(t) = 88000 \left( 1 + 0.01625 \right)^{4t} \] \[ A(t) = 88000 \left( 1.01625 \right)^{4t} \] This is the function for the amount after \( t \) years.