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

Question

Caleb Alexander · Accepted Answer

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

Christian Antunez · Answer

You look for extreme values taken by f

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#

Christian Antunez · Answer

undefined To find the asymptotes of $ f(x) = \frac{x^2 + 1}{x - 2x^2} $, first identify any vertical asymptotes by setting the denominator equal to zero and solving for $ x $. Then, identify any horizontal asymptotes by examining the behavior of the function as $ x $ approaches positive or negative infinity. Finally, determine any oblique asymptotes by performing polynomial division if the degree of the numerator is one more than the degree of the denominator.