Question

How do you determine the formula for the graph of an exponential function given (0,2) and (3,1)?

Answer 1

I found: `y(x)=2e^(-0.231x)`

Explanation:

Let us start with the general form of an exponential function:
`y(x)=Ae^(kx)`
where `A` and `k` are two constants we need to evaluate.
We use the coordinates of the points into the general expression to find the two constants:

First : `x=0 and y=2` getting:
`y(0)=Ae^(0k)` that must be equal to `2`
or:
`Ae^(0k)=2`
`Ae^(0)=2`
`A*1=2`
so that `A=2`

Second : `x=3 and y=1` getting:
`y(3)=Ae^(3k)` that must be equal to `1`;
we know from the previous step that `A=2` so:
`y(3)=2e^(3k)=1`
and so:
`2e^(3k)=1`
`e^(3k)=1/2`
take the natural log of both sides:
`ln(e^(3k))=ln(1/2)`
rearranging:
`3k=ln(1/2)`
and:
`k=-0.231`

Finally our function will be:

`y(x)=2e^(-0.231x)`

Graphically:
graph{2e^(-0.231x) [-10, 10, -5, 5]}