Question

How do you write an equation in the form f(x) = kx^n for the direct variation functions given f(0.3) = 45 and n = 2?

Answer 1

Function is `f(x)=500x^2`

Explanation:

As `f(0.3)=45` and function is of the form `f(x)=kx^n` and `n=2`

Hence `f(0.3)=k(0.3)^2=45`

or `0.09k=45`

and `k=45/0.09=45xx100/9=500`

Hence function is `f(x)=500x^2`

Answer 2

`color(blue)(f(x)=500x^2)`

Explanation:

Direct variation is given as:

`y prop kx^n`

Where `bbk` is the constant of variation.

To find `bbk`:

From `f(0.3)=45`, `n=2`

`:.`

`45=k(0.3)^2`

`k=45/(0.3)^2=500`

So we have:

`f(x)=500x^2`