How do you find the asymptotes for # f(x)= (x^2 + 1) / (x - 2x^2)#?

2 Answers
Jun 18, 2016

Answer:

You look for extreme values taken by f

Explanation:

For large #x#, only the largest powers remain, and f approaches #x^2/-2x^2 = -1/2# There is a pole at #x - 2x^2 = 0# or #x=1/2#

Jun 18, 2016

Answer:

vertical asymptotes x = 0 , x#=1/2#
horizontal asymptote y#=-1/2#

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: #x-2x^2=0rArrx(1-2x)=0rArrx=0,x=1/2#

#rArrx=0,x=1/2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by #x^2#

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

as #xto+-oo,f(x)to(1+0)/(0-2)#

#rArry=-1/2" is the asymptote"#
graph{(x^2+1)/(x-2x^2) [-10, 10, -5, 5]}