Modeling Exponential Growth and Decay with A=P(1+r)^1

Question

A) It's an initial Deposit of $200 is made into an account that pays 4.2% annual interests. Write a model to represent the balance in the account after t years.

B) You buy a used car for $7000 And it depreciates in value at a rate of 6% per year. Write a model to represent the value of the car after t years.

Answer 1

An exponential function is defined as a function of the form f(x) =ab^x, where a and b are constants. Eg f(x)=2x3^x is an exponential function.

b > 1, it is a growth function as with x incresing f(x) increases.

If b<1, f(x) decreases with x increasing, it is a decay function.

Let's denote amount at the end of t years as A(t)

Here initial principal in dollars, p =200, interest compounded annually = 4.2% = 0.42

A(0) = p =200

At the end of 1 year, amount, A(1)=p(1+r)

At the end of 2 years amount A(2) =(principal at the beginning of the year)x(1+r)= p(1+r)(1+r)=p(1+r)²

Similarly Amount at the end of t years,A(t)=p(1+r)^t=200(1+0.042)^t

i.e. A(t)=200x1.042^t which is an exponential growth function,.

B)

For depreciation, let v(t) be the value at the end of t year,

v(1)=700(1-6%)=700x.94

At the end of 2nd year, 6% depreciation happens over the value at the beginning of the year,

i.e. v(2)= (700x0.94)x0.94 = 700x0.94²

Going on v(t)=700x0.94^t, which is an exponential decay function.

Answer 2

a=p(1+r) p-(p*.06)=new price

200(1+.042)= in 1 yr; $208.40

7000-(7000*.06)=7000-420=6580 1yr. repeat

6580-(6580*.06)=6580-394.80=6185.20 2nd yr. repeat

Answer 3

A or otherwise expressed as f(t) = P(1+.042)^t where P is your initial value of 200

A= 200(1.042)^t

A= 7000(1-.06)^t

A= 7000(.94)tt