How do you find vertical, horizontal and oblique asymptotes for #1/(x^2-2x-8)#?

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Answer 1 · 2017-02-26T14:43:51

The vertical asymptotes are #x=4# and #x=-2#
The horizontal asymptote is #y=0#
No oblique asymptote

Let's factorise the denominator

#x^2-2x-8=(x-4)(x+2)#

As you canot divide by #0#, #=>#, #x!=4# and #x!=-2#

The vertical asymptotes are #x=4# and #x=-2#

As the degree of the numerator is #<# than the degree of the denominator, there is no oblique asymptote.

#lim_(x->+-oo)1/(x^2-2x-8)=lim_(x->+-oo)1/x^2=0^+#

The horizontal asymptote is #y=0#

graph{(y-1/(x^2-2x-8))(y-10000(x-4))(y-1000(x+2))(y)=0 [-3.85, 7.25, -2.893, 2.66]}