How do you find the vertical, horizontal or slant asymptotes for #(x + 1)/(x^2 + 5x − 6)#?

1 Answer
Dec 29, 2016

Answer:

The vertical asymptotes are #x=1# and #x=6#
No slant asymptote
The horizontal asymptote is #y=0#

Explanation:

We start by factorising the denominator

#x^2+5x-6=(x-1)(x+6)#

Let #f(x)=(x+1)/((x-1)(x+6))#

The domain of #f(x)# is #D_f(x)=RR-{1,-6} #

As we cannot divide by #0#, #x!=1# and #x!=-6#

So,

The vertical asymptotes are #x=1# and #x=6#

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

For the horizontal asymptote, we calculate the #lim# of #f(x)# as #x->+-oo#

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

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

Therefore,

The horizontal asymptote is #y=0#

graph{(y-(x+1)/(x^2+5x-6))(y)=0 [-10, 10, -5, 5]}