What Are the Asymptotes for This Equation?

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Homework Help Overview

The discussion revolves around identifying possible equations that exhibit specific asymptotes, namely x=3 and y=5, as well as x=-2 and y=3. Participants are exploring the characteristics of these asymptotes in the context of rational functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants suggest manipulating known functions, such as y = 1/x, to derive new equations that meet the asymptotic criteria. There is also a discussion about the validity of proposed equations and their corresponding asymptotes.

Discussion Status

Some participants have offered potential equations and are questioning their correctness in relation to the specified asymptotes. There is an ongoing evaluation of different approaches, with some suggestions being acknowledged as incorrect while others are considered viable.

Contextual Notes

Participants are navigating through the constraints of the problem, including the need to ensure that the proposed equations accurately reflect the desired asymptotic behavior. There is a mention of previous errors in function signs that affect the outcome.

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Homework Statement


Make an example of a possible equation whose graph got the asymptotes:
a)
x=3 and y=5
b)
x=-2 and y=3


Homework Equations





The Attempt at a Solution


I have no idea but I'm ready to hear those kind of solutions you might suggest me
 
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you know the asymptote of y = 1/x ? Try moving that graph around
 
willem2 said:
you know the asymptote of y = 1/x ? Try moving that graph around

I've found out a possible solution if we state that x=3 and y=5, then we can say that
x-a=0
x-3=0
x=3
then we can put our value in and put it under our fraction so we now got:

( )/3-x

Nothing in the parantese:
Now we need the y asymptote:

y=5
f(0)=5 to make this statement true:
We need to say that:
(5x/x)/(3/x - x/x)
x-> eternity
so 3/x will be close to zero and in math = 0
so we got 5/1 = 5 this equals true. So our possible equation for our asymptotes:

5x/(3-x).

Is there some errors in my solving this equation or is this a possible answer?
 
Edit: Had a wrong function with the wrong sign; mrkuo is correct.

Another one which came to mind first was 1/(x-3) + 5. This is just a shifting of the graph of 1/x.
 
Last edited:
Unfortunately,

y=5x/(3-x)

will not produce the asymptotes x=3 and y=5, rather, x=3 and y= negative 5.

Use y = 5x/(x-3) instead. Previous poster's answer works just as well.
 

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