How to know if this irrational function has no asymptotes?

In summary, the function F(x) = x+1-3sqrt((x-1)/(ax+1)) has two asymptotes when a>0 and one vertical asymptote when a<0. However, there is no value of a that will result in no asymptotes. The only possible value for a is 0, which would result in an oblique asymptote. However, this function does not have an oblique asymptote. Instead, it has an asymptotic direction in the form of a parabolic branch. The domain of this function is dependent on the value of a.
  • #1
Jeanclaud
16
0
1. The problem statement, all variables and given/known dat
F(x)=x+1-3sqrt((x-1)/(ax+1))
For which value of a ,(c) has no asymptote?

Homework Equations

The Attempt at a Solution


I know if a>0 then (c) will have 2 asymptote
And if a<o then (c) will have 1 vertical asymptote.
But I can't find any value of a so that (c) has no asymptote. Please give me some hints thank you.

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  • #2
What is leftover after you have eliminated a>0 and a<0? There is only one value possible.
 
  • #3
RUber said:
What is leftover after you have eliminated a>0 and a<0? There is only one value possible.
a=0? But still it has an O.A.
 
  • #4
I don't see what you are saying? Your denominator would be (0x+1) = 1.
 
  • #5
I see now...you meant Oblique Asymptote.
I don't think this would have one.
By scale, yes -- the trend is in the direction of y = x as x gets large, but you can't argue that the function value gets closer to the line in the limit.
 
  • #6
Jeanclaud said:
a=0? But still it has an O.A.
If a = 1, the part inside the radical approaches 1 for large values of |x|. That will give you an oblique asymptote.
RUber said:
I don't see what you are saying? Your denominator would be (0x+1) = 1.
The OP hasn't been very clear, but I think the asymptotes he's referring to are one that's oblique and one that's vertical.
 
  • #7
RUber said:
I see now...you meant Oblique Asymptote.
I don't think this would have one.
By scale, yes -- the trend is in the direction of y = x as x gets large, but you can't argue that the function value gets closer to the line in the limit.
Then it will have only an asymptotic direction (parabolic branch). Thank you.
 
  • #8
Jeanclaud said:
Then it will have only an asymptotic direction (parabolic branch). Thank you.
Determine the domain of this function. That, of course, depends upon the value of a .
 

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