How do you find the asymptotes for #y=(2x^2+3)/(x^2-6)#?

1 Answer
Aug 7, 2016

Answer:

vertical asymptotes at #x=±sqrt6#
horizontal asymptote at y = 2

Explanation:

The denominator of y cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #x^2-6=0rArrx^2=6rArrx=±sqrt6#

#rArrx=-sqrt6" and " x=sqrt6" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" (a constant)"#

divide the terms on the numerator/denominator by the highest power of x, that is #x^2#

#((2x^2)/x^2+3/x^2)/(x^2/x^2-6/x^2)=(2+3/x^2)/(1-6/x^2)#

as #xto+-oo,yto(2+0)/(1-0)#

#rArry=2" is the asymptote"#
graph{(2x^2+3)/(x^2-6) [-10, 10, -5, 5]}