Horizontal and Vertical Asymptotes of functions

Question

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".

If someone could please help me with this, that would be appreciated :)

Answer 1

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 $0 \le x \le 1$

Answer 2

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.