# How do you find the vertical, horizontal or slant asymptotes for (4+x^4)/(x^2-x^4)?

Apr 27, 2018

Vertical asymptotes at $x = - 1$, $x = 0$ and $x = 1$ and horizontal asymptote at $y = - 1$.

#### Explanation:

We have vertical asymptotes for $f \left(x\right)$ at $x = a$, when ${\lim}_{x \to a} f \left(x\right) = \pm \infty$. If $f \left(x\right)$ is an algebraic function, which happens for values of $x$, for which denominator is $0$.

Here in denominator, we have ${x}^{2} - {x}^{4} = {x}^{2} \left(1 - {x}^{2}\right)$ and this is equal to zero when $x = 0 , 1$ or $- 1$.

Hence, we have vertical asymptotes at $x = - 1$, $x = 0$ and $x = 1$.

We have horizontal asymptote when ${\lim}_{x \to \pm \infty} f \left(x\right) = k$, a constant. This hapens in algebraic functions, when degree of numerator is equal to that of denominator, which is the case here.

Here ${\lim}_{x \to \pm \infty} \frac{4 + {x}^{4}}{{x}^{2} - {x}^{4}}$

= ${\lim}_{x \to \pm \infty} \frac{\frac{4}{x} ^ 4 + 1}{\frac{1}{x} ^ 2 - 1}$

= $\frac{1}{- 1} = - 1$

Hence we have a horizontal asymptote $y = - 1$

Slant asymptote is there when degree of numerator is just one more than that of denominator. Here we do not have a slant asymptote.

graph{(4+x^4)/(x^2-x^4) [-10, 10, -10, 25]}

As an example of slant asymptote consider $\frac{4 + {x}^{5}}{{x}^{2} - {x}^{4}}$

${\lim}_{x \to \pm \infty} \frac{4 + {x}^{5}}{{x}^{2} - {x}^{4}}$

= ${\lim}_{x \to \pm \infty} \frac{\frac{4}{x} ^ 4 + x}{\frac{1}{x} ^ 2 - 1}$

= $\frac{x}{- 1} = - x$ and slant asymptote is $y = - x$ or $x + y = 0$

graph{(4+x^5)/(x^2-x^4) [-10, 10, -10, 25]}

Graphs not drawn to scale.