How do you find the horizontal asymptote for #(2x^2)/(x^2-4)#?
1 Answer
Jun 5, 2016
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]}