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

Jun 18, 2016

You look for extreme values taken by f

#### Explanation:

For large $x$, only the largest powers remain, and f approaches ${x}^{2} / - 2 {x}^{2} = - \frac{1}{2}$ There is a pole at $x - 2 {x}^{2} = 0$ or $x = \frac{1}{2}$

Jun 18, 2016

vertical asymptotes x = 0 , x$= \frac{1}{2}$
horizontal asymptote y$= - \frac{1}{2}$

#### 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: $x - 2 {x}^{2} = 0 \Rightarrow x \left(1 - 2 x\right) = 0 \Rightarrow x = 0 , x = \frac{1}{2}$

$\Rightarrow x = 0 , x = \frac{1}{2} \text{ are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by ${x}^{2}$

$\frac{{x}^{2} / {x}^{2} + \frac{1}{x} ^ 2}{\frac{x}{x} ^ 2 - \frac{2 {x}^{2}}{x} ^ 2} = \frac{1 + \frac{1}{x} ^ 2}{\frac{1}{x} - 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 + 0}{0 - 2}$

$\Rightarrow y = - \frac{1}{2} \text{ is the asymptote}$
graph{(x^2+1)/(x-2x^2) [-10, 10, -5, 5]}