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]}

How do you find the vertical, horizontal or slant asymptotes for 2 - 3/x^22 - 3/x^2?