Question: Suppose that $83,000 is invested at \(4\frac{1}{2}\%\) interest, compounded quarterly. a) Find…

Suppose that $83,000 is invested at \(4\frac{1}{2}\%\) 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 \(t = 0\), 4, 7, and 10 years.

d) When will the amount of money in the account reach $100,000?

a) The function for the amount to which the investment grows after \(t\) years is \(A(t) = 83,000(1.01125)^{4t}\).

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

b) Graph the function. Choose the correct answer below. Use the viewing window [0, 50, 0, 500,000] with Xscl = 5 and Yscl = 50,000.

[Graph options: A, B, C]

c) Find the amount of money in the account at \(t = 0\), 4, 7, and 10 years.

The amount of money in the account at \(t = 0\) years is \$.

(Round to the nearest dollar as needed.)

Solution

To solve this problem, let’s address each part: a) The function for the amount to which the investment grows after \( t \) years is given by: \[ A(t) = 83,000(1.01125)^{4t} \] b) To graph the function, it is necessary to choose the correct plot. From the options A, B, and C, the correct graph is one that shows an exponential growth curve starting at $83,000. c) Find the amount of money in the account at \( t = 0, 4, 7, \) and \( 10 \) years. For \( t = 0 \): \[ A(0) = 83,000(1.01125)^{4 \times 0} = 83,000 \] For \( t = 4 \): \[ A(4) = 83,000(1.01125)^{16} \] Calculating this, \[ A(4) \approx 83,000 \times 1.1911 \approx 98,856 \] For \( t = 7 \): \[ A(7) = 83,000(1.01125)^{28} \] This evaluates to approximately, \[ A(7) \approx 83,000 \times 1.3553 \approx 112,493 \] For \( t = 10 \): \[ A(10) = 83,000(1.01125)^{40} \] This evaluates to approximately, \[ A(10) \approx 83,000 \times 1.5416 \approx 127,953 \] d) When will the amount of money in the account reach $100,000? We need to solve for \( t \) in the equation: \[ 83,000(1.01125)^{4t} = 100,000 \] First, divide both sides by 83,000, \[ (1.01125)^{4t} = \frac{100,000}{83,000} \approx 1.2048 \] Take the natural logarithm of both sides, \[ 4t \cdot \ln(1.01125) = \ln(1.2048) \] Solving for \( t \), \[ t = \frac{\ln(1.2048)}{4 \cdot \ln(1.01125)} \] Calculating this gives approximately \( t \approx 4.7 \) years.