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

Question

2068 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-26T13:45:08

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

Notice that:

$51.2/0.2 = 256 = 4^4$

So we can write:

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

Then:

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

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

$color(white)()$
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:

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

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

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

$color(white)()$
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:

$a = y_1/b^(x_1)$