How do you identify all asymptotes or holes for #f(x)=(-2x^3-12x^2-16x)/(x^3+x^2-2x)#?

Question

1266 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-13T10:49:31

The vertical asymptote is #x=1#
The holes are when #x=0# and #x=-2#

The horizontal asymptote is #y=-2#

Let's factorise the numerator

#-2x^3-12x^2-16x=-2x(x^2+6x+8)#
#=-2x(x+2)(x+4)#

Lets factorise the denominator

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

So,

#f(x)=(-2x^3-12x^2-16x)/(x^3+x^2-2x)#

#=(-2cancelxcancel(x+2)(x+4))/(cancelxcancel(x+2)(x-1))#

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

There are holes when #x=0# and #x=-2#

As you cannot divide by #0#, #x!=1#

So, the vertical asymptote is #x=1#

Now, we calculate the limits as #x->oo#

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

The horizontal asymptote is #y=-2#

graph{(y-(-2(x+4))/(x-1))=0 [-41.1, 41.1, -20.56, 20.56]}