How do you find the vertical, horizontal or slant asymptotes for #f(x)=sqrt(x^2+2)/(3x-6)#?

1 Answer
Oct 25, 2017

Answer:

See below.

Explanation:

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]}