Question

How do you find the asymptotes for `f(x)=(2x)/(sqrt (9x^2-4))`?

Answer 1

Asymptotes: f(x) = y = `+-2/3` and `x=+-2/3`..

Explanation:

Let y = f(x). It is defined for `9x^2-4>0`. So, `|x|>2/3`.

`yto+-oo` as `xto+-2/3`. So, `x=+-2/3` are asymptotes..

Inversely, solve for x.
`x=(2y)/sqrt(9y^2-4)`.
Interestingly the form is the same. So, `|y|>2/3`.

`xto+-oo` as `yto+-2/3`. So, `y=+-2/3` are asymptotes..

The graph comprises four branches symmetrical about the origin in four regions:

Both x and `y > 2/3` in the first quadrant, `x<-2/3` and `y > 2/3` in the second quadrant, both x and `y < -2/3` in the third quadrant and `x > 2/3` and `y < -2/3` in the fourth quadrant.

As a whole, the graph is outside of `|x| <=2/3 and |y|<=2/3`.

In this formula, `f^(-1)-=f`, like f(x) = 1/x. Indeed, interesting.