How do you find the vertical, horizontal and slant asymptotes of: #f(x)=1/(x^2-2x-8 )#?

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Answer 1 · 2017-01-21T12:44:39

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

Let's factorise the denominator

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

So,

#f(x)=1/(x^2-2x-8)=1/((x+2)(x-4))#

As you cannot divide by #0#, #x!=-2# and #x!=4#

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

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

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

The horizontal asymptote is #y=0#

graph{(y-1/(x^2-2x-8))(y)(y-50x-100)(y-50x+200)=0 [-9.84, 10.18, -5.175, 4.825]}