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

Suppose that $88,000 is invested at 6% 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)^{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 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 $106,780 (Round to the nearest dollar as needed.) The amount of money in the account at t = 5 years is $122,226 (Round to the nearest dollar as needed.)

Solution

Suppose that $88,000 is invested at \( 6\frac{1}{2} \% \) interest, compounded quarterly. a) The function for the amount to which the investment grows after \( t \) years is \[ A(t) = 88000(1.01625)^{4t} \] b) To graph the function, we would visually choose the graph that represents this exponential function with a starting point at \( t = 0 \). The correct graph would show an upward curve starting from $88,000. c) Find the amount of money in the account at \( t = 0, 3, \) and \( 5 \) years. For \( t = 0 \), \[ A(0) = 88000(1.01625)^{0} = 88000 \] So, the amount of money at \( t = 0 \) years is $88,000. For \( t = 3 \), \[ A(3) = 88000(1.01625)^{12} \approx 106,780 \] So, the amount of money at \( t = 3 \) years is $106,780. For \( t = 5 \), \[ A(5) = 88000(1.01625)^{20} \approx 122,226 \] So, the amount of money at \( t = 5 \) years is $122,226. d) To find when the amount of money in the account will reach $300,000, we solve: \[ 300,000 = 88000(1.01625)^{4t} \] Divide both sides by 88,000: \[ \frac{300,000}{88,000} = (1.01625)^{4t} \] Simplify: \[ 3.4091 \approx (1.01625)^{4t} \] Take the natural logarithm of both sides: \[ \ln(3.4091) \approx 4t \ln(1.01625) \] Solve for \( t \): \[ t \approx \frac{\ln(3.4091)}{4 \ln(1.01625)} \approx 17.67 \] So, the amount in the account will reach $300,000 after approximately 17.67 years.