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

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Answer 1 · 2016-12-20T06:44:25

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

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

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]}

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