# How do you find the asymptotes for y= (sqrt(x^4 + 6x^2 + 9)) / (2x^2 - 10)?

Aug 29, 2016

Vertical asymptotes: $x = \pm \sqrt{5}$
Horizontal asymptotes: $y = \frac{1}{2}$

#### Explanation:

Start by factoring everything.

$y = \frac{\sqrt{\left({x}^{2} + 3\right) \left({x}^{2} + 3\right)}}{2 \left({x}^{2} - 5\right)}$

We can now cancel the sqrt:

$y = \frac{{x}^{2} + 3}{2 \left({x}^{2} - 5\right)}$

We are now left with a basic rational function.

Since the degree of the numerator and of the denominator are equal, we will concern ourselves with the two types of asymptotes found in this kind of rational function.

Vertical Asymptotes:

These occur when the denominator equals $0$ (which renders the function undefined).

Their placement can be determined by setting the denominator to $0$ and then solving for $x$.

$2 \left({x}^{2} - 5\right) = 0$

${x}^{2} = 5$

$x = \pm \sqrt{5}$

$\therefore$There will be vertical asymptotes at $x = \sqrt{5}$ and $x = - \sqrt{5}$.

Horizontal asymptotes:

Horizontal asymptotes, as their name suggests, are horizontal, or parallel to the x axis.

Their placement will depend on the highest degree of the numerator and the denominator. If the degree of the numerator is less than that of the denominator there will be a horizontal asymptote that lies on the line $y = 0$, or in other words, directly on the $x$-axis.

If the degree of the numerator is equal to that of the denominator, the asymptote will occur at the ratio between the coefficients of the term with the highest degree.

If the degree of the numerator is larger than that of the denominator, there will be no horizontal asymptote, but there will be an oblique asymptote. However, for reasons of simplicity, I will not talk about those in this answer.

What is our case? Well, the highest degree in the numerator is $2$ and the highest degree in the denominator is $2$. We will have a horizontal asymptote at the ratio between the coefficients of the numerator and the denominator.

The coefficient of the numerator is $1$. The coefficient of the denominator is $2$.

Hence, the equation of the horizontal asymptote is $y = \frac{1}{2}$.

Hopefully this helps!