Vertical asymptotes occur at the values of #x# for which the function is undefined. In this case if #x=2#
#sqrt(x^2+2)/(3x-6)=sqrt(x^2+2)/(3(2)-6)=sqrt(x^2+2)/0# ( undefined division by 0).
So the line #x=2# is a vertical asymptote:
We now need to see what happens as #x# increases without bound in the positive direction and the negative direction.
As #x-> oo# both numerator and denominator are positive and approach infinity, so the limit is 0.
#lim_(x->oo)sqrt(x^2+2)/(3x-6)=0#
As #x-> -oo# the numerator is positive and the denominator is negative and both approach infinity, so the limit is 0.
#lim_(x->-oo)sqrt(x^2+2)/(3x-6)=0#
This shows that the x axis is a horizontal asymptote. So the line:
#y=0# is an asymptote.
Graph:
graph{(sqrt(x^2+2))/(3x-6) [-14.23, 14.25, -7.12, 7.11]}
