# How do you find the asymptotes for f(x)=(3x^5 + 1) / (2x^6 + 3x -1)?

Jan 12, 2016

$f \left(x\right)$ has horizontal asymptote $y = 0$ and vertical asymptotes
$x = {x}_{1}$ and $x = {x}_{2}$, where:

${x}_{1} \approx 0.3324335502431692846$

${x}_{2} \approx - 1.1414892917449508403$

#### Explanation:

Let $g \left(x\right) = 3 {x}^{5} + 1$ and $h \left(x\right) = 2 {x}^{6} + 3 x - 1$

$f \left(x\right) = \frac{3 {x}^{5} + 1}{2 {x}^{6} + 3 x - 1} = g \frac{x}{h \left(x\right)}$

$g \left(x\right)$ is of degree $5$ and $h \left(x\right)$ is of degree $6$, so:

$f \left(x\right) = g \frac{x}{h \left(x\right)} \to 0$ as $x \to \pm \infty$

So $f \left(x\right)$ has horizontal asymptote $y = 0$

$f \left(x\right)$ will have vertical asymptotes wherever $h \left(x\right) = 0$ and $g \left(x\right) \ne 0$

$h \left(x\right) = 2 {x}^{6} + 3 x - 1$ has $2$ changes of sign, so $1$ or $2$ positive zeros.

$h \left(- x\right) = 2 {x}^{6} - 3 x - 1$ has $1$ change of sign, so $h \left(x\right)$ has one negative zero.

Since the coefficients of $h \left(x\right)$ are Real, any non-Real zeros will occur in Complex conjugate pairs, so there are an even number of non-Real zeros.

So we can deduce that $h \left(x\right)$ has exactly one positive, one negative and four non-Real zeros.

$h \left(x\right) = 0$ has no solution expressible in terms of $n$th roots, but we can find good approximations using Newton's method.

$h ' \left(x\right) = 12 {x}^{5} + 3$

Starting with an initial approximation ${a}_{0}$, iterate using the formula:

${a}_{i + 1} = {a}_{i} - \frac{h \left({a}_{i}\right)}{h ' \left({a}_{i}\right)} = {a}_{i} - \frac{2 {a}_{i}^{6} + 3 {a}_{i} - 1}{12 {a}_{i}^{5} + 3}$

Putting this into a spreadsheet and using initial approximations ${a}_{0} = 1$ and ${a}_{0} = - 1$, I got the following approximations after a few iterations:

${x}_{1} \approx 0.3324335502431692846$

${x}_{2} \approx - 1.1414892917449508403$

Note also that the only Real zero of $g \left(x\right)$ is when:

$x = \sqrt[5]{- \frac{1}{3}} \approx - 0.80274$

So neither of the zeros of $h \left(x\right)$ are zeros of $g \left(x\right)$ and both of the zeros of $h \left(x\right)$ are vertical asymptotes of $f \left(x\right)$.

graph{(3x^5+1)/(2x^6+3x-1) [-10, 10, -5, 5]}