# How do you find vertical, horizontal and oblique asymptotes for (sqrt(x^4 + 6x^2 + 9)) / (2x^2 - 10)?

Mar 21, 2018

Vertical asymptotes are $x = \sqrt{5}$ and $x = - \sqrt{5}$ and horizontal asymptote is $y = \frac{1}{2}$

#### Explanation:

Vertical asymptotes are given by putting denominator as $0$ i.e. $2 {x}^{2} - 10 = 0$

or ${x}^{2} - 5 = 0$ i.e. $x = \sqrt{5}$ and $x = - \sqrt{5}$

this is so as when $x \to \pm \sqrt{5}$, $\frac{\sqrt{{x}^{4} + 6 {x}^{2} + 9}}{2 {x}^{2} - 10} \to \pm \infty$

For horizontal asymptote, observe that degree of numerator and denominator both is $2$ - we take ${\sqrt{x}}^{4}$ as ${x}^{2}$ - hence, we have a horizontal asymptote,

as ${\lim}_{x \to \infty} \frac{\sqrt{{x}^{4} + 6 {x}^{2} + 9}}{2 {x}^{2} - 10}$

= ${\lim}_{x \to \infty} \frac{\sqrt{1 + \frac{6}{x} ^ 2 + \frac{9}{x} ^ 4}}{2 - \frac{10}{x} ^ 2}$

= $\frac{1}{2}$

Hence, horizontal asymptote is $y = \frac{1}{2}$

There is no oblique or slnting asymptote. For this we should have degree of numerator just one more than that of denominator.

graph{sqrt(x^4+6x^2+9)/(2x^2-10) [-10, 10, -5, 5]}