How do you find the horizontal asymptote for #(2x^2)/(x^2-4)#?

1 Answer
Jun 5, 2016

Answer:

vertical asymptotes x = ± 2
horizontal asymptote y = 2

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: #x^2-4=0rArr(x-2)(x+2)=0rArrx=±2#

#rArrx=-2,x=2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by #x^2#

#((2x^2)/x^2)/(x^2/x^2-4/x^2)=2/(1-4/x^2)#

as #xto+-oo,f(x)to2/(1-0)#

#rArry=2" is the asymptote"#
graph{(2x^2)/(x^2-4) [-10, 10, -5, 5]}