Question

How do you find the asymptotes for `y=(2x^2 + 3)/(x^2 - 6)`?

Answer 1

Three asymptotes are `x=sqrt6`, `x=-sqrt6` and `y=2`.

Explanation:

Vertical asymptotes can be found by factorizing function in the denominator i.e. here `x^2-6`. As such here `x+sqrt6=0` and `x-sqrt6=0` or `x=-sqrt6=` and `x=sqrt6` are two such asymptotes.

Location of the horizontal asymptote is determined by looking at the degrees of the numerator and denominator.

If degree of numerator is less (than denominator), `y=0` is the asymptote. If degree of numerator is greater (than denominator), there is no horizontal asymptote (but if degree of numerator is just one degree higher there is a slanting asymptote).

In case the degrees are same, as in the instant case, let `a` be the ratios of the coefficient of highest degrees in numerator and denominator, then asymptote is at `y=a`.

Here as the ratio is `(2x^2)/x^2` or `2`, asymptote is at `y=2`.