# How do you find the asymptotes for g(x) = (x^4 -81) /(x^3+3x^2 - x - 3)?

Feb 22, 2018

Vertical asymptotes at $x = - 1$and $x = 1$ and oblique or slanting asymptote at $y = x$. Wealso have a hole at $x = - 3$.

#### Explanation:

Let us first factorize the numerator ${x}^{4} - 81$ and denominator ${x}^{3} + 3 {x}^{2} - x - 3$

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

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

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

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

Observe that when we have $x + 3 = 0$ or $x - 1 = 0$ or $x + 1 = 0$, $g \left(x\right)$ is not defined. However, as $x + 3$ appears both in numerator and denominator, this cancels out and we can arrive at limiting value of $f \left(x\right)$ when $x + 3 = 0$ or $x = - 3$.

Hence, we have a hole at $x = - 3$.

Further, when $x \to 1$ or $x \to - 1$ $g \left(x\right) \to \pm \infty$ depending on whether we approach these values from left or right.

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

Further as degree of numerator is just one more that of denominator, we should have an oblique or slanting asymptote.

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

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

Hence, we have an oblique or slanting asymptote at $y = x$

graph{(x^4-81)/(x^3+3x^2-x-3) [-84, 76, -33, 47]}