Question about vertical asymptotes

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The rational function f(x) = [(x+2)(x-5)]/[(x-3)(x+1)] has vertical asymptotes at x=3 and x=-1, determined by setting the denominator to zero. The confusion arose from the misunderstanding that one could simply switch signs in the denominator to find these values. The horizontal asymptote at y=1 is derived from the leading coefficients of the numerator and denominator as x approaches infinity. Clarification was provided that the "+" and "-" signs in the asymptote equations represent positive and negative values, not operations. Overall, the discussion emphasizes the correct method for identifying vertical and horizontal asymptotes in rational functions.
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This is a example in my book

a rational function like f(x) = [(x+2)(x-5)]/[(x-3)(x+1)] has vertical asymptotes at x=3 and x=(-1)

Is all you do is switch the addition and subtraction signs in [(x-3)(x+1)]?

The same equation has a horizontal asymptotes at y=1, which i can't seem to understand where it came from

The is the best i could explain this, thank you for reading this
 
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Johnnycab said:
The same equation has a vertical asymptotes at y=1, which i can't seem to understand where it came from

Do you mean, a horizontal asymptote? Here, this rough sketch might help: http://archives.math.utk.edu/visual.calculus/1/horizontal.5/index.html" .
 
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-sorry iam unable to view the flash pictures, because i can't install macromedia plugin (wont verify)
-could someone give me a written explanation please, iam still lost
 
Johnnycab said:
This is a example in my book

a rational function like f(x) = [(x+2)(x-5)]/[(x-3)(x+1)] has vertical asymptotes at x=3 and x=(-1)

Is all you do is switch the addition and subtraction signs in [(x-3)(x+1)]?

The same equation has a horizontal asymptotes at y=1, which i can't seem to understand where it came from

The is the best i could explain this, thank you for reading this

No, you don't ever just arbitrarily "switch signs"! A rational function has vertical asymptotes where the denominator is 0 (and the numerator isn't). The denominator of f(x) is (x-3)(x+1) which equals 0 when x- 3= 0 or x+ 1= 0. To solve x- 3= 0, add 3 to both sides: x- 3+ 3= x= 0+ 3= 3. To solve x+ 1= 0, subtract 1 from both sides: x+ 1- 1= x= 0- 1= -1. See? I didn't "switch signs" anywhere!

I feel I should also point out that the "+" and "-" signs in
x= +3 and x= -1 are not "addition" and "subtraction" signs- they are "positive" and "negative" signs. The difference is subtle but important. (TI calculators have separate "subtraction" and "negative" keys for that reason.)
 
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thanks hallsofivy it makes sense now i see what your talking about, thanks for your help
 

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