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

1 Answer
Feb 19, 2016

Answer:

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#.