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]}