# How to get the horizontal asymptote.

1. Sep 18, 2004

### 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

2. Sep 18, 2004

### arildno

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?

3. Sep 18, 2004

### 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?

4. Sep 18, 2004

### Pyrrhus

are you familiar with limits?

5. Sep 18, 2004

### 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)

6. Sep 18, 2004

### UrbanXrisis

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

7. Sep 18, 2004

### 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:
$$9x^{2}-36\approx9x^{2}$$
(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

8. Sep 18, 2004

### 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?

9. Sep 18, 2004

### 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: $$3x^{2}-7x+14\approx3x^{2}$$ 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.

10. Sep 18, 2004

### 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?

11. Sep 18, 2004

### arildno

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

12. Sep 18, 2004

### Tide

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

13. Sep 18, 2004

### 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?

14. Sep 18, 2004

### Tide

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

15. Sep 18, 2004

### arildno

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

16. Sep 18, 2004

### UrbanXrisis

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

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

is this correct?

17. Sep 18, 2004

### arildno

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

18. Sep 18, 2004

### UrbanXrisis

I actually cheated and used a graphing calculator to solve for when (90x)/(x^2-9)^2 is positive. How would I solve it without a graphing calculator?

19. Sep 18, 2004

### arildno

Well, the denominator is always non-negative (why?)
Hence, only the sign of the numerator is of importance (why?)

20. Sep 18, 2004

### UrbanXrisis

okay, I got a good understanding of it all. Thanks for the help & advice.