How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=(6x^2+2x-1 )/ (x^2-1)#?

Question

924 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-16T16:32:32

The vertical asymptotes are #x=1# and #x=-1#
The horizontal asymptote is #y=6#
No oblique asymptote

The denominator is #(x^2-1)=(x-1)(x+1)#

As we cannot divde by #0#, #x!=+-1#

Therefore the vertical asymptotes are #x=1# and #x=-1#

As the degree of the numerator #=# degree of the denominator, we don't expect a slant asymptote.

For the limit, we take the term with the highest coefficient

#lim_(x->+-oo)f(x)=lim_(x->+-oo)(6x^2)/x^2=6#

The horizontal asymptote is #y=6#

graph{(y-(6x^2+2x-1)/(x^2-1))(y-6)=0 [-12.83, 12.49, -1.84, 10.83]}