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

Jul 12, 2018

Vertical asymptote is at $x = \pm \sqrt{5}$
Horizontal asymptote: $y = 0.5$, slant asymptote:absent

#### Explanation:

$f \left(x\right) = \frac{\sqrt{{x}^{4} + 6 {x}^{2} + 9}}{2 {x}^{2} - 10}$ or

f(x)=sqrt((x^2+3)^2)/(2( x^2-5) or

f(x)=((x^2+3))/(2( x^2-5)

Vertical asymptote is at ${x}^{2} - 5 = 0 \mathmr{and} {x}^{2} = 5 \mathmr{and} x = \pm \sqrt{5}$

Horizontal asymptote: The degree of numerator and denominator

is same i.e $2$ , so we have a horizontal asymptote at

$y =$ (numerator's leading coefficient) / (denominator's leading

coefficient) $\therefore y = \frac{1}{2} = 0.5$

Slant asymptote: Since numerator's degree is not greater , we

have no slant asymptote.

Vertical asymptote is at $x = \pm \sqrt{5}$

Horizontal asymptote: $y = 0.5$, slant asymptote:absent

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