Question

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

Answer 1

Vertical asymptotes are `x=sqrt5` and `x=-sqrt5` and horizontal asymptote is `y=1/2`

Explanation:

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

or `x^2-5=0` i.e. `x=sqrt5` and `x=-sqrt5`

this is so as when `x->+-sqrt5`, `sqrt(x^4+6x^2+9)/(2x^2-10)->+-oo`

For horizontal asymptote, observe that degree of numerator and denominator both is `2` - we take `sqrtx^4` as `x^2` - hence, we have a horizontal asymptote,

as `lim_(x->oo)sqrt(x^4+6x^2+9)/(2x^2-10)`

= `lim_(x->oo)sqrt(1+6/x^2+9/x^4)/(2-10/x^2)`

= `1/2`

Hence, horizontal asymptote is `y=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]}