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

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Answer 1 · 2018-07-12T12:36:20

Vertical asymptote is at $x= +-sqrt5$
Horizontal asymptote: $y=0.5$ , slant asymptote:absent

$f(x)=sqrt(x^4+6x^2+9)/(2x^2-10)$ 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 $x^2-5=0 or x^2=5 or x=+-sqrt5$

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) $:. y= 1/2=0.5$

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

have no slant asymptote.

Vertical asymptote is at $x= +-sqrt5$

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

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

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