How do you identify all asymptotes or holes and intercepts for #f(x)=(x^3-1)/((2(x^2-1))#?

1 Answer
Nov 25, 2016

Answer:

There is a hole at #x=1#
A vertical asymptote is #x=-1#
The slant asymptote is #y=x/2#
There is no horizontal asymptote.

Explanation:

Let's factorise the denominator

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

Let's factorise the numerator
#x^3-1=(x-1)(x^2+x+1)#

Therefore,

#f(x)=(x^3-1)/(2(x+1)(x-1))=(cancel(x-1)(x^2+x+1))/(2(x+1)cancel(x-1))#

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

There is a hole at #x=1#

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

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

A vertical asymptote is #x=-1#

As the degree of the numerator is #># than the degree of the denominator, we expect a slant asymptote.

Let's do a long division

#color(white)(aaaa)##x^2+x+1##color(white)(aaaa)##∣##2x+2#

#color(white)(aaaa)##x^2+x##color(white)(aaaaaaa)##∣##x/2#

#color(white)(aaaaa)##0+0+1#

So, #(x^2+x+1)/(2(x+1))=x/2+1/(2x+2)#

The slant asymptote is #y=x/2#

To calculate the limit, we take the terms of highest degree.

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

There is no horizontal asymptote.

graph{(y-(x^2+x+1)/(2x+2))(y-x/2)=0 [-5.542, 5.553, -2.783, 2.766]}