# How do you find the asymptotes for (2x^6 + 6x^3 )/( 4x^6 + 3x^3)?

Jun 30, 2018

HA at $y = \frac{1}{2}$

VA at $x = 0$ and $x = \sqrt[3]{- \frac{3}{4}}$

#### Explanation:

When we want to find the horizontal asymptotes, we want to analyze the highest degrees on the top and bottom.

$\frac{2 {x}^{6}}{4 {x}^{6}}$

Since we have the same degree on the top and bottom, the horizontal asymptote just simplifies to the coefficients. We're left with

$\frac{2}{4}$, or $y = \frac{1}{2}$

For our vertical asymptote, we want to think about what value(s) make our function defined.

We can set the denominator equal to zero. We get

$4 {x}^{6} + 3 {x}^{3} = 0$

We can factor our an ${x}^{3}$ to get

${x}^{3} \left(4 {x}^{3} + 3\right) = 0$

Setting both factors equal to zero, we get

$\textcolor{s t e e l b l u e}{x = 0}$ and

$4 {x}^{3} = - 3$

${x}^{3} = - \frac{3}{4}$

$\textcolor{s t e e l b l u e}{x = \sqrt[3]{- \frac{3}{4}}}$

These are our vertical asymptotes.

Hope this helps!