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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 :)

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  • $\begingroup$ Can you use calculus? $\endgroup$ – SalmonKiller May 19 '15 at 19:46
  • $\begingroup$ @SalmonKiller What do you mean by that? $\endgroup$ – Ash May 19 '15 at 19:49
  • $\begingroup$ Do you have to use limit ideas to show asymptotes? And btw, is it logarithm base 10 or the natural log? $\endgroup$ – SalmonKiller May 19 '15 at 19:50
  • $\begingroup$ @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. $\endgroup$ – Ash May 19 '15 at 19:53
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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 $0 \le x \le 1$

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  • $\begingroup$ What about $h$? $\endgroup$ – SalmonKiller May 19 '15 at 19:57
  • $\begingroup$ $h(x) = f(g(x))$ $\endgroup$ – WW1 May 19 '15 at 19:59
  • $\begingroup$ @WW1 does f(x) have a vertical asymptote at x=-3? I'm going by a graph of the function. $\endgroup$ – Ash May 19 '15 at 20:02
  • $\begingroup$ @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. $\endgroup$ – SalmonKiller May 19 '15 at 20:03
  • $\begingroup$ yep, $x=-3$ is the only solution to $x^2+6x+9=0$ $\endgroup$ – WW1 May 19 '15 at 20:03
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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.

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