How do you find the vertical, horizontal or slant asymptotes for #2 - 3/x^2#?

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Answer 1 · 2016-11-28T08:12:57

The vertical asymptote is #x=0#
No slant asymptote
The horizontal asymptote is #y=2#

Let #f(x)=2-3/x^2#

The domain of #f(x)# is #RR-{0}#

As you cannot divide by #0#, #x!=0#

So, the vertical asymptote is #x=0#

As the degree of the numerator is #=# th the degree of the denominator, there is no slant asymptote.

#lim_(x->+-oo)f(x)=im_(x->+-oo)(2-3/x^2)=2#

The horizontal asymptote is #y=2#

graph{(y-(2-3/x^2))(y-2)=0 [-7.9, 7.9, -3.95, 3.95]}