How do you find the vertical, horizontal or slant asymptotes for #g(x)=(x+7)/(x^2-4)#?

1 Answer
Dec 31, 2016

Answer:

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

Explanation:

We need

#a^2-b^2=(a+b)(a-b)#

Let's factorise the denominator

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

Then,

#g(x)=(x+7)/(x^2-4)=(x+7)/((x+2)(x-2))#

The domain of #g(x)# is #D_g(x)=RR-{-2,2}#

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

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

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

To find the horizontal asymptotes, we calculate #lim g(x)# as #x->+-oo#

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

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

The horizontal asymptote is #y=0#

graph{(y-(x+7)/(x^2-4))(y)=0 [-11.25, 11.26, -5.62, 5.62]}