How do you find the asymptotes for $h(x)=(x^2-4)/ x$ ?

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Answer 1 · 2016-11-15T14:52:22

The vertical asymptote is $x=0$
The slant asymptote is $y=x$
No horizontal asymptote

As you cannot divide by $0$ ,

$x!=0$

So $x=0$ is a vertical asymptote.

The degree of the numerator is $>$ the degree of the denominator, so we expect a slant asymptote.

Let's simplify the $h(x)$

$h(x)=(x^2-4)/x=x-4/x$

Therefore, $y=x$ is a slant asymptote.

$lim_(x->+-oo)h(x)=lim_(x->+-oo)x=+-oo$

graph{(y-(x^2-4)/x)(y-x)=0 [-11.25, 11.25, -5.63, 5.62]}

How do you find the asymptotes for h(x)=(x^2-4)/ xh(x)=(x^2-4)/ x?