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

Dec 20, 2016

The vertical asymptotes are $x = 0$ and $x = \frac{1}{2}$
No slant asymptote
The horizontal asymptote is $y = - \frac{1}{2}$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + 1}{x - 2 {x}^{2}} = \frac{{x}^{2} + 1}{x \left(1 - 2 x\right)}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{0 , \frac{1}{2}\right\}$

As we cannot divide by $0$, $x \ne 0$ and $x \ne \frac{1}{2}$

The vertical asymptotes are $x = 0$ and $x = \frac{1}{2}$

As the degree of the numerator $=$ to the degree of the denominator, there is no slant asymptote.

We must calculate the limits as $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} {x}^{2} / \left(- 2 {x}^{2}\right) = {\lim}_{x \to \pm \infty} - \frac{1}{2} = - \frac{1}{2}$

The horizontal asymptote is $y = - \frac{1}{2}$

graph{(y-(x^2+1)/(x(1-2x)))(y+1/2)=0 [-28.86, 28.85, -14.43, 14.46]}