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?

2 Answers
Jun 26, 2018

Function is f(x)=500x^2f(x)=500x2

Explanation:

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

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

or 0.09k=450.09k=45

and k=45/0.09=45xx100/9=500k=450.09=45×1009=500

Hence function is f(x)=500x^2f(x)=500x2

Jun 26, 2018

color(blue)(f(x)=500x^2)f(x)=500x2

Explanation:

Direct variation is given as:

y prop kx^nykxn

Where bbkk is the constant of variation.

To find bbkk:

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

:.

45=k(0.3)^2

k=45/(0.3)^2=500

So we have:

f(x)=500x^2