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

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Answer 1 · 2017-10-25T08:55:33

See below.

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]}

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