# How do you find the asympototes for y = (-8x^3 - 2x + 12)/( 7x^5 + 9x + 12)?

Jul 28, 2015

The horizontal asymptote is $y = 0$.

There is a vertical asymptote $x = a$, where $a \cong - 0.89261052$ is the only real root of $7 {x}^{5} + 9 x + 12 = 0$.

#### Explanation:

Since the degree of the denominator is greater than the numerator, we find $y \to 0$ as $x \to \pm \infty$. So $y = 0$ is a horizontal asymptote.

$- 8 {x}^{3} - 2 x + 12 = 0$ has a single real root at $x \cong 1.0720202$

$7 {x}^{5} + 9 x + 12 = 0$ has a single real root at $x \cong - 0.89261052$

In general, the roots of quintics cannot be expressed in terms of ordinary radicals. In this particular case, the quintic is almost in Bring-Jerrard normal form, so it's relatively easy (believe me) to express the root in terms of the Bring Radical:

Let ${x}_{1} = {\left(\frac{7}{9}\right)}^{\frac{1}{4}} x$

Then:

$0 = 7 {x}^{5} + 9 x + 12$

$= 7 {\left({\left(\frac{9}{7}\right)}^{\frac{1}{4}} {x}_{1}\right)}^{5} + 9 {\left(\frac{9}{7}\right)}^{\frac{1}{4}} {x}^{1} + 12$

$= 9 {\left(\frac{9}{7}\right)}^{\frac{1}{4}} {x}_{1}^{5} + 9 {\left(\frac{9}{7}\right)}^{\frac{1}{4}} {x}_{1} + 12$

Divide through by $9 {\left(\frac{9}{7}\right)}^{\frac{1}{4}}$ to get:

${x}_{1}^{5} + {x}_{1} + \frac{4}{3} {\left(\frac{7}{9}\right)}^{\frac{1}{4}} = 0$

So ${x}_{1} = B R \left(\frac{4}{3} {\left(\frac{7}{9}\right)}^{\frac{1}{4}}\right)$

and $x = {\left(\frac{9}{7}\right)}^{\frac{1}{4}} \cdot {x}_{1} = {\left(\frac{9}{7}\right)}^{\frac{1}{4}} B R \left(\frac{4}{3} {\left(\frac{7}{9}\right)}^{\frac{1}{4}}\right)$