Question

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

Answer 1

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]}