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

1 Answer
Dec 17, 2016

Answer:

There is no hole. We have vertical asymptote at #x=-1# and oblique asymptote at #y=2x+1#

Explanation:

Let us examine #g(x)=(2x^2+3x-1)/(x+1)#.

As for the #x=-1#, although denominator #x+1=0# but #2x^2+3x-1!=0#, there is no common factor between numerator and denominator and hence we do not have any hole.

It is also apparent that as #x->-1#, #(2x^2+3x-1)/(x+1)->oo#

and as we have #f(x)=(x^2+11x+18)/(2x+1)#, #f(x)->oo# or #f(x)->-oo#, as we approach #x=-1# from negative or positive side.

Hence, we have a vertical asymptote at #x=-1/2#.

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

As such, #f(x)=(2x^2+3x-1)/(x+1)=2x+1-2/(x+1)#

Hence, as #x->oo#. #f(x)->2x+1#

and we have an oblique asymptote at #y=2x+1#
graph{(2x^2+3x-1)/(x+1) [-10, 10, -5, 5]}