Question

How do you write an exponential function whose graph passes through (0,0.2) and (4, 51.2)?

Answer 1

`f(x) = 0.2*4^x`

Explanation:

Notice that:

`51.2/0.2 = 256 = 4^4`

So we can write:

`f(x) = 0.2*4^x`

Then:

`f(0) = 0.2*4^0 = 0.2`

`f(4) = 0.2*4^4 = 51.2`

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Note

This is not the only solution.

Instead of `4` we could use any of the other three `4`th roots of `256`, namely:

`-4`, `4i` or `-4i`

So other solutions are:

`f(x) = 0.2*(-4)^x`

`f(x) = 0.2*(4i)^x`

`f(x) = 0.2*(-4i)^x`

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General case

Suppose we want an exponential function that passes through points `(x_1, y_1)` and `(x_2, y_2)`

A general form can be written:

`f(x) = a*b^x`

Then:

`f(x_1) = a*b^(x_1) = y_1`

`f(x_2) = a*b^(x_2) = y_2`

So:

`b^(x_2-x_1) = (a*b^(x_2))/(a*b^(x_1)) = y_2/y_1`

One solution is:

`b = (y_2/y_1)^(1/(x_2-x_1))`

There are other solutions formed by multiplying this by some `(x_2-x_1)`th root of `1` - that is some number of the form:

`cos((2kpi)/(x_2-x_1)) + i sin((2kpi)/(x_2-x_1))`

where `k` is any integer.

If `x_2-x_1` is rational then this results in a finite number of other solutions.

If `x_2-x_1` is irrational then this results in an infinite number of solutions.

Once we have a value for `b` then there is a corresponding value for `a` given by:

`a = y_1/b^(x_1)`