What Are the Asymptotes for This Equation?

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The discussion revolves around finding an equation that has specific asymptotes: x=3 and y=5. One proposed equation was 5x/(3-x), but it was pointed out that this would actually yield asymptotes of x=3 and y=-5. A correct alternative suggested was y = 5x/(x-3), which meets the criteria for the desired asymptotes. The conversation emphasizes the importance of understanding how to manipulate equations to achieve the correct asymptotic behavior. Overall, the focus is on deriving the appropriate functions based on the specified asymptotes.
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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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