Question about vertical asymptotes

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Discussion Overview

The discussion revolves around understanding vertical and horizontal asymptotes in the context of a specific rational function, f(x) = [(x+2)(x-5)]/[(x-3)(x+1)]. Participants explore how to identify vertical asymptotes and clarify misconceptions regarding the signs in the function's denominator.

Discussion Character

  • Conceptual clarification, Technical explanation, Homework-related

Main Points Raised

  • One participant states that the rational function has vertical asymptotes at x=3 and x=-1, questioning if the signs in the denominator are simply switched to find these values.
  • Another participant corrects the misunderstanding, explaining that vertical asymptotes occur where the denominator equals zero, not by switching signs.
  • A participant expresses confusion about horizontal asymptotes, mistakenly referring to them as vertical asymptotes and seeking clarification.
  • A later reply emphasizes the distinction between positive/negative signs and addition/subtraction signs in the context of solving for asymptotes.
  • One participant acknowledges the clarification provided by another, indicating improved understanding.

Areas of Agreement / Disagreement

There is no consensus on the initial misunderstanding regarding the identification of vertical asymptotes, but participants engage in clarifying the concepts involved. The discussion remains unresolved regarding the participant's confusion about horizontal asymptotes.

Contextual Notes

Participants express limitations in understanding the graphical representation of asymptotes due to technical issues with viewing external resources. The discussion includes varying levels of familiarity with mathematical terminology and concepts.

Johnnycab
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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
 
Last edited:
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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" .
 
Last edited by a moderator:
-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.)
 
Last edited by a moderator:
thanks hallsofivy it makes sense now i see what your talking about, thanks for your help
 

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