How do you find the asymptotes for $f(x) = (3 - x) / x^2$ ?

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Answer 1 · 2016-11-07T08:06:51

The vertical asymptote is $x=0$
The horizontal asymptote is $y=0$

As you cannot divide by 0, the vertical asymptote is $x=0$

There is no slant asymptote as the degree of the numerator is less than the degree of the denominator:

$lim_(n rarr -oo )(3-x)/x^2=lim_(n rarr -oo )-x/x^2=lim_(n rarr -oo )-1/x=0^+$

$lim_(n rarr +oo )(3-x)/x^2=lim_(n rarr +oo )-x/x^2=lim_(n rarr +oo )-1/x=0^-$

The horizontal asymptote is $y=0$
graph{(3-x)/x^2 [-10, 10, -5, 5]}

How do you find the asymptotes for f(x) = (3 - x) / x^2f(x) = (3 - x) / x^2?