How to find the asymptotes for #(3x^2) / (x^2-4)#?

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Answer 1 · 2018-08-11T22:13:18

#(3x^2)/(x^2-4)# has horizontal asymptote #y=3# and vertical asymptotes #x=-2# and #x=2#

Note that:

#lim_(x->+-oo) (3x^2)/(x^2-4) = lim_(x->+-oo) 3/(1-4/x^2) = 3#

So #y=3# is a horizontal asymptote.

Also:

#x^2-4 = (x-2)(x+2)#

So the denominator is zero when #x=+-2# and the numerator is non-zero at those values.

So #(3x^2)/(x^2-4)# has vertical asymptotes at #x=-2# and #x=2#

graph{(y-(3x^2)/(x^2-4)) = 0 [-10.09, 9.91, -3.6, 6.4]}