How do you find the vertical, horizontal or slant asymptotes for #(2x+4)/(x^2-3x-4) #?

1 Answer
Dec 9, 2016

Answer:

The vertical asymptotes are #x=-1# and #x=4#
No slant asymptote
The horizontal asymptote is #y=0#

Explanation:

Let's factorise the denominator

#x^2-3x-4=(x+1)(x-4)#

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

The domain of #f(x)# is #D_f(x)=RR-{-1,4} #

As we cannot divide by #0#, #x!=-1# and #x!=4#

The vertical asymptotes are #x=-1# and #x=4#

The degree of the numerator is #<# than the degree of the denominator, so there is no slant asymptote

To calculate the limits as #x->+-oo#, we take the terms of highest degree in the numerator and the denominator.

#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+4)/(x^2-3x-4) [-11.25, 11.25, -5.625, 5.625]}