How do you find the Vertical, Horizontal, and Oblique Asymptote given #((x-3)(9x+4))/(x^2-4)#?

1 Answer
Jan 26, 2017

Answer:

The vertical asymptotes are #x=2# and #x=-2#
The horizontal asymptote is #y=9#
No slant asymptote

Explanation:

Let #f(x)=((x-3)(9x+4))/(x^2-4)#

Let's factorise the denominator

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

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

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

As the degree of the numerator is #=# the degree of the denominator, there is no slant asymptote

#lim_(x->+-oo)f(x)=lim_(x->+-oo)(9x^2)/x^2=9#

The horizontal asymptote is #y=9#

graph{(y-((x-3)(9x+4)/(x^2-4)))(y-9)=0 [-41.1, 41.1, -20.55, 20.57]}