How do you find the vertical, horizontal and slant asymptotes of: #f(x) =(x^2 + 1) / (x - 2x^2)#?

1 Answer
Dec 20, 2016

Answer:

The vertical asymptotes are #x=0# and #x=1/2#
No slant asymptote
The horizontal asymptote is #y=-1/2#

Explanation:

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

The domain of #f(x)# is #D_f(x)=RR-{0,1/2}#

As we cannot divide by #0#, #x!=0# and #x!=1/2#

The vertical asymptotes are #x=0# and #x=1/2#

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

We must calculate the limits as #x->+-oo#, we take the terms of highest degree in the numerator and the denominator

#lim_(x->+-oo)f(x)=lim_(x->+-oo)x^2/(-2x^2)=lim_(x->+-oo)-1/2=-1/2#

The horizontal asymptote is #y=-1/2#

graph{(y-(x^2+1)/(x(1-2x)))(y+1/2)=0 [-28.86, 28.85, -14.43, 14.46]}