# Finding the domain and asymototes of a function

• Sep 12th 2012, 07:01 PM
Kathrynm77
Finding the domain and asymototes of a function
Hi..I'm not completely clueless on this problem I would just like to see if my answers are correct. The problem is: let f be the function given by f(x)=x/sqrt((x^2)-4)
A) find the domain of f
B) find the vertical asymototes of f
C) find the horizontal asymototes of f
When I attempted this problem algebraic ally (which is what the directions said) I got the answer of (-infinity,-2),(2,infinity) for the domain, vertical asymototes at x=2 and x=-2 and the horizontal asymptote to be y=o because the degree of the denominator is bigger than the numerator...however thinking twice about those rules I now think that only applies to rationalized functions..which this is not...annd when I graphed it on my calculator I got a corner hyperbola with what appeared to be a vertical asymptote at x=I and horizontal asymototes at y=1 and y= -1....please help!! Thanks!-Kathryn
• Sep 12th 2012, 07:10 PM
SworD
Re: Finding the domain and asymototes of a function
Your thinking is on the right track but there is just one small mistake:

Quote:

because the degree of the denominator is bigger than the numerator
The degrees are in fact equal, because of the square root. In the denominator, as x approaches infinity or negative infinity, the $\displaystyle -4$ term matters less and less and eventually not at all, as it gets swamped by the increasingly large $\displaystyle x^2$ term, which is square rooted. So essentially, as x approaches positive infinity, the function behaves like $\displaystyle \frac{x}{\sqrt{x^2}} = \frac{x}{x} = 1$. At negative infinity, this is similar, except the numerator will be negative while the denominator will always be positive, so it approaches -1. This accounts for the horizontal asymptotes.

And there are actually vertical asymptotes at $\displaystyle x = \pm2$, you were correct there.
• Sep 12th 2012, 07:12 PM
MarkFL
Re: Finding the domain and asymototes of a function
You did fine for the domain and the vertical asymptotes, for the horizontal asymptotes, consider:

$\displaystyle \lim_{x\to-\infty}f(x)$

$\displaystyle \lim_{x\to\infty}f(x)$

What do you find when you evaluate these limits?
• Sep 12th 2012, 07:28 PM
MaxJasper
Re: Finding the domain and asymototes of a function
$\displaystyle \lim_{x\to \infty } \, f(x) \to +1$

$\displaystyle \lim_{x\to -\infty } \, f(x) \to -1$
• Sep 13th 2012, 05:21 PM
Kathrynm77
Re: Finding the domain and asymototes of a function
Oh ok..it makes sense now to use limits..thanks so much!!