How do you identify all asymptotes or holes for #f(x)=(x^2+11x+18)/(2x+1)#?

1 Answer
Dec 16, 2016

Answer:

There is no hole. We have vertical asymptote at #x=-1/2# and oblique asymptote at #2x-4y+21=0#

Explanation:

Let us examine #f(x)=(x^2+11x+18)/(2x+1)#.

As #x^2+11x+18=x^2+9x+2x+18#

= #x(x+9)+2(x+9)=(x+2)(x+9)#

As a common factor cannot be crossed out from both the numerator and the denominators, we do not have a hole in #f(x)#.

It is also apparent that as #x->-1/2#, #(2x+1)->0#

and as we have #f(x)=(x^2+11x+18)/(2x+1)#, #f(x)->oo#

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

Further, #f(x)=(x^2+11x+18)/(2x+1)#

= #(x/2(2x+1)+21/4(2x+1)+51/4)/(2x+1)#

= #x/2+21/4+51/(4(2x+1)#

Hence, as #x->oo#. #f(x)->x/2+21/4#

and we have an oblique asymptote at #y=x/2+21/4# or #2x-4y+21=0#
graph{(y-(x^2+11x+18)/(2x+1))(x+1/2)(2x-4y+21)=0 [-21.08, 18.92, -5.32, 14.68]}