# Horizontal and Vertical Asymptotes of functions

So I'm completing a chart analyzing the different properties of three different functions: $f(x)=\log(x^2+6x+9), g(x)=\sqrt{x^2 -1}$ (sorry not sure how to do square roots on here), $h(x)=f(x)(g(x))$

it asks for the horizontal and vertical asymptotes. However, I am unsure of how to tell whether or not these kind of equations would have them. I know that vertical asymptotes you set the denominator equal to zero (but here I do not see any rational functions). Horizontal asymptotes you divide the "leading terms".

• Can you use calculus? May 19, 2015 at 19:46
• @SalmonKiller What do you mean by that?
– Ash
May 19, 2015 at 19:49
• Do you have to use limit ideas to show asymptotes? And btw, is it logarithm base 10 or the natural log? May 19, 2015 at 19:50
• @salmonKiller I don't think that is what it is. It's just taking the equation and the graph and looking for the details through them. I just know that you can use algebra to find the asymptotes, but I am not sure how to tell if whether or not these functions would have asymptotes. I would presume that it would be the natural log.
– Ash
May 19, 2015 at 19:53

no horizontal asymptotes exist.

$log(x)$ has domain $x>0$ and an asymptote at $x=0$

so $f(x)$ will have an asymptote where $x^2+6x+9=0$

$g(x)$ has domain $-1\le x \le +1$ and range $0 \le y \le 1$ but no assymptotes.

$h(x) = f(g(x))$ also has domain $-1\le x \le +1$ it has no asymptotes because $x^2+6x+9>0$ whenever $0 \le x \le 1$

• What about $h$? May 19, 2015 at 19:57
• $h(x) = f(g(x))$
– WW1
May 19, 2015 at 19:59
• @WW1 does f(x) have a vertical asymptote at x=-3? I'm going by a graph of the function.
– Ash
May 19, 2015 at 20:02
• @Ash As I said in my answer, $x=-3$ is a root of the quadratic $x^2+6x+9$, so yeah, it is a vertical asymptote. May 19, 2015 at 20:03
• yep, $x=-3$ is the only solution to $x^2+6x+9=0$
– WW1
May 19, 2015 at 20:03

Well, I would start by looking for the horizontal asymptotes. For $f$, there is no horizontal asymptote, since the quadratic inside grows larger and larger as $x$ grows and as $x$ decreases. For $g$, the same argument as for $f$. For vertical asymptotes of $f$, we need to look at where it is unbounded. The natural is unbounded at 0, so $x^2+6x+9 = 0$. So there is an asymptote at $x=-3$ since that's the only solution. For $g$, it exists everywhere on it's domain and is never unbounded. For $h(x) = f(g(x))$, we can say that $g(x) \neq -3$, so $h$ has no asymptotes.