Assigining an equation to this graph

[danago]

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

"What is the equation of this graph:
http://img96.imageshack.us/img96/8089/untitled1copyhe1.jpg "[/URL]

If i am given a graph, and i don't 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.

Thanks in advance,
Dan.

 

[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?

 

[danago]

Hmmm not really :( I am 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?

 

[Hootenanny]

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?

 

[dextercioby]

Therefore, the equation of the curve must be something of the form;

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

Can you see why?

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.

 

[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?

 

[danago]

Im not quite getting what you said dex :(

 

[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.

 

[danago]

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

 

[Hootenanny]

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

That is indeed correct.

 

[danago]

ok thanks for the help :)

 

[chaoseverlasting]

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.

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?

 

[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.