How do you identify all asymptotes or holes for #f(x)=(-2x^2-6x-4)/(x^2+x)#?

1 Answer
Feb 25, 2017

Answer:

The vertical asymptote is #x=0#
The horizontal asymptote is #y=-2#
There is a hole at #x=-1#
No slant asymptote

Explanation:

Let's factorise the denominator and the numerator

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

#-2x^2-6x-4=-2(x^2+3x+2)=-2(x+1)(x+2)#

Therefore,

#f(x)=(-2x^2-6x-4)/(x^2+x)=(-2cancel(x+1)(x+2))/(xcancel(x+1))#

There is a hole at #x=-1#

As we cannot divide by #0#, #x!=0#

The vertical asymptote is #x=0#

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

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

The horizontal asymptote is #y=-2#

graph{(y+2(x+2)/(x))(y+2)=0 [-20.28, 20.27, -10.14, 10.14]}