How do you identify all asymptotes or holes for #f(x)=(2x-2)/(x^2-2x-3)#?

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Answer 1 · 2016-11-24T06:34:15

The vertical asymptotes are #x=-1# and #x=3#
No slant asymptote
The horizontal asymptote is #y=0#
No holes

Let's factorise the denominator

#x^2-2x-3=(x+1)(x-3)#

The domain of #f(x)# is #D_(f(x))=RR-{-1,3} #

As you cannot divide by #0#, #x!=-1# and #x!=3#

The vertical asymptotes are #x=-1# and #x=3#

The degree of the numerator #<# than the degree of the denominator, thereis no slant asymptote

We calculate the limits of #f(x)#, we take only the terms of highest degree.

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

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

The horizontal asymptote is #y=0#

graph{(2x-2)/(x^2-2x-3) [-8.89, 8.89, -4.444, 4.445]}