Assigining an equation to this graph

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In summary, the conversation is about finding the equation of a graph with a vertical asymptote at x=-1 and a horizontal asymptote at y=1. The equation for this graph is y = \frac{-3}{x+1}+1. The equation was derived using 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. The variable p is a natural number that matches the indexed constants A_{1},...,A_{p}.
  • #1
danago
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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.
 
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  • #2
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
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?
 
  • #4
Okay, so on your graph there are asymptotes at x=-1 and y = 1. This means that as [itex]x\to -1[/itex] then [itex]y\to\pm\infty[/itex]. Now, the horizontal asymptote means that as [itex]x\to + \infty[/itex] then [itex]y\to 1[/itex]. Therefore, the equation of the curve must be something of the form;

[tex]y = \frac{A}{x+1} + 1[/tex]

Can you see why?
 
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  • #5
Hootenanny said:
Therefore, the equation of the curve must be something of the form;

[tex]y = \frac{A}{x+1} + 1[/tex]

Can you see why?

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

[tex] y(x)=\frac{A_{1}x^{p-1}+...+A_{p-1}x-3}{(x+1)^{p}}+1 [/tex]

?

You need to add the conditions

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

Daniel.
 
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  • #6
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 [tex]\frac{A}{x+1}[/tex] approaches zero.

Am i thinking along the right lines?
 
  • #7
Im not quite getting what you said dex :(
 
  • #8
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
So...would i be right in saying that the equation for this graph is:
[tex]y=\frac{-3}{x+1}+1[/tex]
?
 
  • #10
danago said:
So...would i be right in saying that the equation for this graph is:
[tex]y=\frac{-3}{x+1}+1[/tex]
?
That is indeed correct.
 
  • #11
ok thanks for the help :)
 
  • #12
dextercioby said:
No, i don't. But can you see why the equation must have this form

[tex] y(x)=\frac{A_{1}x^{p-1}+...+A_{p-1}x-3}{(x+1)^{p}}+1 [/tex]

?

You need to add the conditions

[itex] p\geq 1 [/itex], 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?

[tex] y(x)=\frac{A_{1}x^{p-1}+...+A_{p-1}x-3}{(x+1)^{p}}+1 [/tex]

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

What does the variable 'p' represent?
 
  • #13
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.
 

Related to Assigining an equation to this graph

1. How do you assign an equation to a graph?

To assign an equation to a graph, you need to first identify the key points on the graph, such as the intercepts, turning points, and slope. Then, you can use these points to create an equation that accurately represents the relationship between the variables in the graph.

2. What is the process for assigning an equation to a graph?

The process for assigning an equation to a graph involves analyzing the key points on the graph, determining the relationship between the variables, and using this information to create an equation that fits the graph.

3. Can you assign an equation to any type of graph?

Yes, you can assign an equation to any type of graph as long as there is a clear relationship between the variables. However, some graphs may be more difficult to assign equations to, such as curved or nonlinear graphs.

4. Are there different methods for assigning equations to graphs?

Yes, there are different methods for assigning equations to graphs, such as using the slope-intercept form or the point-slope form. The method used will depend on the type of graph and the information provided.

5. How do you check if the equation assigned to a graph is accurate?

To check if the equation assigned to a graph is accurate, you can substitute different values into the equation and compare the resulting points to the ones on the graph. If the points match, then the equation is likely accurate.

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