Sure, let's walk through the problem step by step.
### Part (a): Finding the Function
To determine the growth of an investment with compounding interest, we use the compound interest formula:
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Given:
- The principal amount [tex]\( P = 68,000 \)[/tex] dollars.
- The annual interest rate [tex]\( r = 3.5\% = 3.5 / 100 = 0.035 \)[/tex] (in decimal).
- The interest is compounded quarterly, which means [tex]\( n = 4 \)[/tex].
Plugging these values into the formula, we get:
[tex]\[ A(t) = 68000 \left(1 + \frac{0.035}{4}\right)^{4t} \][/tex]
[tex]\[ A(t) = 68000 \left(1 + 0.00875\right)^{4t} \][/tex]
[tex]\[ A(t) = 68000 (1.00875)^{4t} \][/tex]
### Part (b): Calculating the Amount after different years
Using the function [tex]\( A(t) = 68000 (1.00875)^{4t} \)[/tex], we will find the amount of money after [tex]\( t = 0, 4, 5, \)[/tex] and [tex]\( 10 \)[/tex] years.
1. When [tex]\( t = 0 \)[/tex] years:
[tex]\[ A(0) = 68000 (1.00875)^{4 \cdot 0} = 68000 (1.00875)^0 = 68000 \times 1 = 68000.0 \][/tex]
2. When [tex]\( t = 4 \)[/tex] years:
[tex]\[ A(4) = 68000 (1.00875)^{4 \times 4} = 68000 (1.00875)^{16} \approx 78171.00156549881 \][/tex]
3. When [tex]\( t = 5 \)[/tex] years:
[tex]\[ A(5) = 68000 (1.00875)^{4 \times 5} = 68000 (1.00875)^{20} \approx 80943.10635621526 \][/tex]
4. When [tex]\( t = 10 \)[/tex] years:
[tex]\[ A(10) = 68000 (1.00875)^{4 \times 10} = 68000 (1.00875)^{40} \approx 96349.80097931727 \][/tex]
So, the amounts are:
- After [tex]\( 0 \)[/tex] years: \[tex]$68,000.00
- After \( 4 \) years: \$[/tex]78,171.00
- After [tex]\( 5 \)[/tex] years: \[tex]$80,943.11
- After \( 10 \) years: \$[/tex]96,349.80
