# How do you find vertical, horizontal and oblique asymptotes for (x^4 -81) /(x^3+3x^2 - x - 3)?

Feb 20, 2017

Invisible hole at (-3, -27/2). Slant asymptote : $x - y + 3 = 0$. Vertical asymptotes : $\uparrow x = \pm 1 \downarrow$. See graphs, The third is not to scale but gets graph for great $| y |$ near.

#### Explanation:

$y = \frac{{x}^{4} - {3}^{4}}{\left(x + 3\right) \left({x}^{2} - 1\right)}$

$= \left(\frac{x + 3}{x + 3}\right) \frac{\left(x - 3\right) \left({x}^{2} + 9\right)}{\left(x - 1\right) \left(x + 1\right)}$.

Sans the hole at $H \left(- 3 , - \frac{27}{2}\right)$,

$y = \frac{\left(x - 3\right) \left({x}^{2} + 9\right)}{\left(x - 1\right) \left(x + 1\right)}$
graph{(x-3)(1+10/(x^2-1)) [-164, 164.1, -82.1, 82]}

graph{((x-3)(1+10/(x^2-1))-y)(y-x-3)(x-1+.01y)(x+1-.01y)=0 [-164, 164.1, -82.1, 82]}

graph{(x-3)(1+10/(x^2-1)) [-16, 16, -160, 160]}

Feb 21, 2017

We have vertical asymptote at $x = 1$ and $x = - 1$ and oblique asymptote at $y = x - 3$

#### Explanation:

First factorize numerator and denominators

${x}^{4} - 81 = {\left({x}^{2}\right)}^{2} - {9}^{2} = \left({x}^{2} - 9\right) \left({x}^{2} + 9\right) = \left(x + 3\right) \left(x - 3\right) \left({x}^{2} + 9\right)$

and ${x}^{3} + 3 {x}^{2} - x - 3 = {x}^{2} \left(x + 3\right) - 1 \left(x + 3\right) = \left({x}^{2} - 1\right) \left(x + 3\right) = \left(x + 1\right) \left(x - 1\right) \left(x + 3\right)$

Hence $f \left(x\right) = \frac{{x}^{4} - 81}{{x}^{3} + 3 {x}^{2} - x - 3} = \frac{\left(x + 3\right) \left(x - 3\right) \left({x}^{2} + 9\right)}{\left(x + 1\right) \left(x - 1\right) \left(x + 3\right)}$

= $\frac{\left(x - 3\right) \left({x}^{2} + 9\right)}{\left(x + 1\right) \left(x - 1\right)} = \frac{{x}^{3} - 3 {x}^{2} + 9 x - 27}{{x}^{2} - 1}$

Observe that although we have cancelled out $x + 3$, we have an undefined function $\frac{{x}^{4} - 81}{{x}^{3} + 3 {x}^{2} - x - 3}$ at $x = - 3$, though close to $x = - 3$, on either side, $f \left(x\right)$ is defined. As such we have a hole at $x = - 3$

Further at $x \to 1$ as also for ${x}_{\succ} 1$, $f \left(x\right) \to \pm \infty$ depending on whether we approach limits from left or right and hence

we have vertical asymptotes at $x = 1$ and $x = - 1$

Here as degree of polynomial in numerator is just one more than that of denominator, we do not have a horizontal asymptote, but we do have a oblique or slant asymptote.

As $f \left(x\right) = \frac{{x}^{4} - 81}{{x}^{3} + 3 {x}^{2} - x - 3} = x - 3 + \frac{10 {x}^{2} - 90}{{x}^{3} + 3 {x}^{2} - x - 3}$

= $x - 3 + \frac{\frac{10}{x} - \frac{90}{x} ^ 3}{1 + \frac{3}{x} - \frac{1}{x} ^ 2 - \frac{3}{x} ^ 3}$

and as $x \to \pm \infty$, $y \to x - 3$

we have an oblique or slant asymptote $y = x - 3$
graph{(x^4-81)/(x^3+3x^2-x-3) [-20, 20, -40, 40]}