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^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#

Jun 26, 2018

#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#