Question: Suppose that $88,000 is invested at 6 1/2% interest, compounded quarterly. a) Find the function…

Suppose that $88,000 is invested at 6 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) = 88000(1.01625)^t. (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 XscI = 5 and YscI = 50,000.

c) Find the amount of money in the account at t = 0, 3, 5, and 10 years. The amount of money in the account at t = 0 years is $88000. (Round to the nearest dollar as needed.) The amount of money in the account at t = 3 years is $. (Round to the nearest dollar as needed.)

Solution

The problem involves compounding interest. Let’s go through each part step by step. a) Finding the function for the amount after \( t \) years: The function given is: \[ A(t) = 88000(1.01625)^{4t} \] This represents the compounding formula where the interest is compounded quarterly. b) Graph the function: The correct choice (from the options visible) is graph B. It shows an exponential growth which is characteristic of compound interest. c) Find the amount of money in the account at \( t = 0, 3, 5, \) and \( 10 \) years: 1. At \( t = 0 \): \[ A(0) = 88000(1.01625)^{4 \times 0} = 88000 \] 2. At \( t = 3 \): \[ A(3) = 88000(1.01625)^{4 \times 3} \] Calculate \( (1.01625)^{12} \) and then multiply by 88000. 3. At \( t = 5 \): \[ A(5) = 88000(1.01625)^{4 \times 5} \] Calculate \( (1.01625)^{20} \) and then multiply by 88000. 4. At \( t = 10 \): \[ A(10) = 88000(1.01625)^{4 \times 10} \] Calculate \( (1.01625)^{40} \) and then multiply by 88000. d) When will the amount of money reach $300,000? Set up the equation: \[ 88000(1.01625)^{4t} = 300000 \] Divide both sides by 88000: \[ (1.01625)^{4t} = \frac{300000}{88000} \] Calculate \( \frac{300000}{88000} \), then use logarithms to solve for \( t \): \[ 4t = \frac{\log\left(\frac{300000}{88000}\right)}{\log(1.01625)} \] Solve for \( t \). This will give you when the amount reaches $300,000.