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

1 Answer
Mar 27, 2016

Answer:

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.