Question

Writing Linear and Exponential EquationsAn investment is initially worth $9,900. Write an equation representing the value of this investment V after t years ineach of the following situations.V=a) The value decreases by 11% per yearV=b) The value decreases by $896 per yearV=c) The value increases by 6% per yearV=d) The value increases by $743 per year

Answer 1

We have an initial value of $9,900.

We have to model the following situations.

a) The value decreases by 11% per year.

We can start expressing this as:

`V(t)=V(t-1)-0.11\cdot V(t-1)=0.89\cdot V(t-1)`

This is a recursive function, as V(t) is function of V(t-1). We can find the explicit function as:

`\begin{gathered} V(t)=0.89\cdot V(t-1) \\ V(t)=0.89(0.89\cdot V(t-1))=0.89^2\cdot V(t-2) \\ \Rightarrow V(t)=0.89^tV(t-t)=0.89^t\cdot V(0) \\ V(t)=9900\cdot0.89^t \end{gathered}`

b) The value decreases by $896 per year

This can be modeled as:

`\begin{gathered} V(t)=V(0)-896\cdot t \\ V(t)=9900-896t \end{gathered}`

c) The value increases by 6% per year

This will have similarities with the function in the point a).

We can model this as:

`\begin{gathered} V(t)=1.06\cdot V(t-1) \\ V(t)=1.06^t\cdot V(0) \\ V(t)=9900\cdot1.06^t \end{gathered}`

d) The value increases by $743 per year

This will be a linear model, as it is the model in point b).

`V(t)=9900+743t`

Answer:

a) V = 9900*0.89^t

b) V = 9900 - 896t

c) V = 9900*1.06^t

d) V = 9900 + 743t