## asymptote

f(x)=(9x^2-36)/(x^2-9)

I forgot how to get the hotizontal asymptote. Is it f(x)=0?
When I do that, there is no way to solve it

Also, how do I find the intervals for when f is increasing?

I'm not asking for the answer, but the equation, I totally forgot

thanks
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 Recognitions: Gold Member Homework Help Science Advisor Horizontal (right-hand side) asymptote: $$\lim_{x\to\infty}f(x)$$ Horizontal (left-hand side) asymptote: $$\lim_{x\to-\infty}f(x)$$ What does the derivative of a function tell you about the function's behaviour?
 I'm even more confused. I know that getting the vertical asymptote is setting the denominaor equal to zero and solving for it but what is it for the horizontal?

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## asymptote

are you familiar with limits?

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 Quote by UrbanXrisis I'm even more confused. I know that getting the vertical asymptote is setting the denominaor equal to zero and solving for it but what is it for the horizontal?
What value L do f(x) approach when x goes to infinity?
The line g(x)=L (parallell to the x-axis!) is called the horizontal asymptote to f(x)
 So how would I find the equation for each horizontal asymptote of: f(x)=(9x^2-36)/(x^2-9)
 Recognitions: Gold Member Homework Help Science Advisor Think this way: Let x be a huge positive number. Then, surely, 9x^2 must be a lot larger than 36. Similarly, x^2 must be a lot larger than 9. Hence, you don't make a big mistake by setting: $$9x^{2}-36\approx9x^{2}$$ (the relative error is tiny) Similarly: $$x^{2}-9\approx{x^{2}}$$ Hence: $$f(x)=\frac{9x^{2}-36}{x^{2}-9}\approx\frac{9x^{2}}{x^{2}}=9$$ (for huge positive x's) Hence, the horizontal right-hand asymptote is L(x)=9
 I understand your process of thinking however, is there a specific formula that can calculate that number? I know that to obtain the vertical asymptote, it's by setting the denominator to zero. e.g. f(x)=(9x^2-36)/(x^2-9) x^2-9=0 x=+3,-3 so the equations for the vertical asymptote is x=3 and x=-3 how do I do this with the horizontal asymptote?
 Recognitions: Gold Member Homework Help Science Advisor No, there is in general no foolproof method in determining limit values (in your case, to get the horizontal asymptotes) There exist a rigourous method which in principle tells you whether a chosen number is a limit or not (i.e, if you've made a right (or wrong!) guess) The reasoning I gave you, however, is sufficient to determine the horizontal assymptotes you'll meet. So here's a method, if you like: 1. Think of x as a huge number. Practical meaning: If x appears in a sum (or difference) with a constant, discard that constant. Further, retain only the highest "power" of x in a sum where x appears in multiple terms. For example: $$3x^{2}-7x+14\approx3x^{2}$$ when x is huge? Why? $$\frac{3x^{2}}{7x}=\frac{3}{7}x$$ which is huge since x is huge. That is the magnitude (always positive!) of the term "3x^2" is much bigger than the magnitude of "-7x". 2. These simplifications should be enough to find the asymptote.
 I understand the idea now. My second question addressed finding the intervals for when f is increasing. f(x)=(9x^2-36)/(x^2-9) how do I find the intervals for when f is increasing?

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I quote myself:
 Quote by arildno What does the derivative of a function tell you about the function's behaviour?
In particular, how is information of whether a function is increasing or decreasing given by the values of the derivative?
 Recognitions: Homework Help Science Advisor Analyze the behavior of the derivative of the function. If it's positive then the function is increasing.
 So find the derivative of: f(x)=(9x^2-36)/(x^2-9) then find the intervals of when it is increasing and that will tell me when the function is increasing. Correct?

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