Question

An exponential relationship y = f(x) has a l-unit percent change of 12% with f(0) = 28.i. Define a formula for function f modeling this relationship. Be sure to use function notation, and be sure to use the correct function name.

Answer 1

`f(x)=28*(1.12)^x`

The percent change is 12%, meaning after each interval x the new value is 112% times the previous value. This can be expressed as (for x = 1)

`\text{new value=28}*(112)/(100)`

when x = 2 the above value becomes the "previous value" and therefore, the new value will be

`\text{new value =(28}*(112)/(100))*(112)/(100)`

when the above becomes the previous value the new value then will be

`\text{new value =((28}*(112)/(100))*(112)/(100))*(112)/(100)`

and so on until we have x number of 122/100 =1.12 terms. Then the new value for any x can be written as

`\text{new value for any x=28(}(112)/(100))^x`

or

`f(x)=28(1.12)^x`