# Assigining an equation to this graph

1. Nov 22, 2006

### danago

Hi. I was doing a practice exam, and came across this question:

"What is the equation of this graph:
"

If i am given a graph, and i dont know what type of graph it is, is there any way i can find an equation? The question also shows a vertical asymptote at at x=-1, and a horizontal asymptote at y=1.

All help greatly appreciated.

Dan.

2. Nov 22, 2006

### robphy

In general, it's not so easy.... if at all possible.

However, in the special cases, you can do this.. and are probably expected to.

In your example, if it were asymptotic to x=0 and y=0 would it be easier?

3. Nov 24, 2006

### danago

Hmmm not really :( Im not even sure where to start, and how to use the asymptotes.

Should i know the general form of the equation for this type of graph?

4. Nov 24, 2006

### Hootenanny

Staff Emeritus
Okay, so on your graph there are asymptotes at x=-1 and y = 1. This means that as $x\to -1$ then $y\to\pm\infty$. Now, the horizontal asymptote means that as $x\to + \infty$ then $y\to 1$. Therefore, the equation of the curve must be something of the form;

$$y = \frac{A}{x+1} + 1$$

Can you see why?

Last edited: Nov 24, 2006
5. Nov 24, 2006

### dextercioby

No, i don't. But can you see why the equation must have this form

$$y(x)=\frac{A_{1}x^{p-1}+...+A_{p-1}x-3}{(x+1)^{p}}+1$$

?

You need to add the conditions

$p\geq 1$, y(2)=0, the table of sign for y and the requirement that on its domain of definition the first derivative be positive.

Daniel.

Last edited: Nov 24, 2006
6. Nov 24, 2006

### danago

hmmm i think so. The (x + 1) denominator means that when x=-1, the denominator is 0, therefore making y=infinity; and that "1" constant, that comes from the horizontal asymptote, because as x approaches +/- infinity, the $$\frac{A}{x+1}$$ approaches zero.

Am i thinking along the right lines?

7. Nov 24, 2006

### danago

Im not quite getting what you said dex :(

8. Nov 24, 2006

### dextercioby

A minor typo in my first post, p can assume the value 1, which leads to the simple form gave by Hootenay in post #4. The conditions i stated in my previous post have to be checked for the proposed solution by Hootenay.

Daniel.

9. Nov 24, 2006

### danago

So....would i be right in saying that the equation for this graph is:
$$y=\frac{-3}{x+1}+1$$
?

10. Nov 24, 2006

### Hootenanny

Staff Emeritus
That is indeed correct.

11. Nov 24, 2006

### danago

ok thanks for the help :)

12. Nov 26, 2006

### chaoseverlasting

This may sound stupid, but why did you use this equation?

$$y(x)=\frac{A_{1}x^{p-1}+...+A_{p-1}x-3}{(x+1)^{p}}+1$$

I understand the specific version of the equation used but what does this equation mean?

What does the variable 'p' represent?

13. Nov 27, 2006

### dextercioby

p is not a variable, it's just a natural number. It should match the indexed constants A_{1},...,A_{p}.

It's the most general possible solution to the problem.

Daniel.