# What are the asymptotes of f(x)=-x/((x^2-8)(5x+2)) ?

May 27, 2017

vertical asymptotes: x = +- 2sqrt(2); x = -2/5
horizontal asymptotes: $y = 0$

#### Explanation:

Given: $- \frac{x}{\left({x}^{2} - 8\right) \left(5 x + 2\right)}$

Factor the difference of squares in the denominator ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right) :$

$\left({x}^{2} - {\left(\sqrt{8}\right)}^{2}\right) = \left(x - \sqrt{8}\right) \left(x + \sqrt{8}\right)$

$\sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$

Therefore:
$- \frac{x}{\left({x}^{2} - 8\right) \left(5 x + 2\right)} = - \frac{x}{\left(x - 2 \sqrt{2}\right) \left(x + 2 \sqrt{2}\right) \left(5 x + 2\right)}$

Find vertical asymptotes $D \left(x\right) = 0$:

vertical asymptotes: x = +- 2sqrt(2); x = -2/5

Find horizontal asymptotes using Precalculus for a rational function: $\frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n} + \ldots}{{b}_{m} {x}^{m} + \ldots}$

When $n < m$ horizontal asymptote is $y = 0$

Using Calculus to find horizontal asymptotes:

$\lim x \to \infty - \frac{x}{\left({x}^{2} - 8\right) \left(5 x + 2\right)} =$

$\lim x \to \infty - \frac{x}{5 {x}^{3} + 2 {x}^{2} - 40 x + 16}$

Divide all by the largest $x$ in the denominator:

$\lim x \to \infty \frac{- \frac{x}{x} ^ 3}{\frac{5 {x}^{3}}{x} ^ 3 + \frac{2 {x}^{2}}{x} ^ 3 - \frac{40 x}{x} ^ 3 + \frac{16}{x} ^ 3}$

$\lim x \to \infty \frac{- \frac{1}{x} ^ 2}{5 + \frac{2}{x} - \frac{40}{x} ^ 2 + \frac{16}{x} ^ 3} = \frac{0}{5} = 0$