How do you find the vertical, horizontal or slant asymptotes for #f(x)=(6x) / sqrt( x^2 - 3)#?

1 Answer
A. S. Adikesavan
Mar 27, 2016

Vertical asymptotes#x=+-sqrt3#.
Horizontal asymptotes: #y=+-6#.

Explanation:

#y=(6x)/sqrt(x^2-3)#.
For real y, #|x|>sqrt3#.
As #xto+-sqrt3, yto+-oo#.
So, #x==-sqrt3# are asymptotes.

Inversely,
#x=(sqrt3 y)/sqrt(y^2-36)#.
For real x, #|y|>6#.
As #yto+-6, xto+-oo#.
So, #y=+-6# are asymptotes.

The graph splits into four branches, in the four quadrants.
The graph is exterior to the region about the origin, bounded by the vertical and horizontal asymptotes, #x=+-sqrt3# and #y=+-6#.

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