Question

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

Answer 1

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`

Answer 2

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]}