asymptote

UrbanXrisis

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

arildno

Horizontal (right-hand side) asymptote: [tex]\lim_{x\to\infty}f(x)[/tex]
Horizontal (left-hand side) asymptote: [tex]\lim_{x\to-\infty}f(x)[/tex]

What does the derivative of a function tell you about the function's behaviour?

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?

Pyrrhus

are you familiar with limits?

arildno

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)

UrbanXrisis

So how would I find the equation for each horizontal asymptote of: f(x)=(9x^2-36)/(x^2-9)

arildno

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:
[tex]9x^{2}-36\approx9x^{2}[/tex]
(the relative error is tiny)
Similarly:
[tex]x^{2}-9\approx{x^{2}}[/tex]
Hence:
[tex]f(x)=\frac{9x^{2}-36}{x^{2}-9}\approx\frac{9x^{2}}{x^{2}}=9[/tex]
(for huge positive x's)
Hence, the horizontal right-hand asymptote is L(x)=9

UrbanXrisis

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?

arildno

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: [tex]3x^{2}-7x+14\approx3x^{2}[/tex] when x is huge?
Why?
[tex]\frac{3x^{2}}{7x}=\frac{3}{7}x[/tex] 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.

UrbanXrisis

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?

arildno

I quote myself:
In particular, how is information of whether a function is increasing or decreasing given by the values of the derivative?

Tide

Analyze the behavior of the derivative of the function. If it's positive then the function is increasing.

UrbanXrisis

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?

Tide

Yes, the function is increasing when f '(x) > 0.

arildno

NO!!
You find the intervals where f'(x) is greater than zero, not the intervals where f'(x) is increasing..

UrbanXrisis

f`(x)=(90x)/(x^2-9)^2

this is positive from [0,3) and (3,infinity)

is this correct?

arildno

Yes; assuming your expression for f'(x) is correct.