Question

Determine what type of model best fits the given situation:

The value of a vehicle decreased by 20% each year.

Answer 1

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P =
`(20)/(100) * P = 0.2P`

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P =
`(20)/(100) * 0.8P = (0.2)(0.8P)`

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

`=P(0.8)^(2)`-------------------------(2)

Similarly after 3 years, this value
`P(0.8)^(2)` will again be decreased by 20 %

so 20% of
`P(0.8)^(2) (20)/(100) * P(0.8)^(2) = (0.2)P(0.8)^(2)`

So after 3 years value is decreased by
`(0.2)P(0.8)^(2)`

so value after 3 years will be =
`P(0.8)^(2) - (0.2)P(0.8)^(2)`

taking
`P(0.8)^(2)` common out we get
`P(0.8)^(2)(1-0.2)`

`P(0.8)^(2)(0.8)`

`P(0.8)^(3)`-----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or
`P(0.8)^(1)`

value after 2 years is
`P(0.8)^(2)`

value after 3 years is
`P(0.8)^(3)`

so value after x years will be
`P(0.8)^(x)` ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

`y = P(0.8)^(x)` where y is the value after x years

so thats the final answer