Where is the Vertical Asymptote of f(x) = (5 - x^2)/(x - 3)?

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Discussion Overview

The discussion revolves around identifying the vertical asymptote of the function f(x) = (5 - x^2)/(x - 3). Participants are seeking clarification on the concept of vertical asymptotes and their relationship to discontinuities, as well as the steps involved in determining them.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Conceptual clarification

Main Points Raised

  • Some participants assert that the vertical asymptote is at x = 3, citing that the denominator becomes zero at this point, leading to an undefined function.
  • Others question the nature of discontinuity at x = 3, asking if it can be classified as such when the denominator is zero.
  • There is a discussion about the function g(x) = (9 - x^2)/(x - 3), with some participants stating it does not have a vertical asymptote, while others reference the asymptote of f(x) in relation to it.
  • One participant expresses confusion regarding the attribution of the vertical asymptote to the correct function, seeking clarification on the statements made by others.
  • Questions are raised about the relationship between discontinuity and vertical asymptotes, specifically whether both terms can be used interchangeably or if they have distinct meanings.

Areas of Agreement / Disagreement

Participants generally agree that x = 3 is a vertical asymptote for f(x), but there is disagreement regarding the classification of g(x) and the definitions of discontinuity and vertical asymptotes. The discussion remains unresolved on these points.

Contextual Notes

There are limitations in the discussion regarding the definitions of discontinuity and vertical asymptotes, as well as the specific conditions under which these concepts apply. Some mathematical steps and reasoning are not fully explored.

mathdad
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Find the vertical asymptote of f(x) = (5 - x^2)/(x - 3). I need the steps not the solution.
 
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Vertical asymptote is x = 3 ... why?
 
RTCNTC said:
Find the vertical asymptote of f(x) = (5 - x^2)/(x - 3). I need the steps not the solution.

$g(x) = \dfrac{9-x^{2}}{x-3}$

NOT a Vertical Asymptote. Why?
 
skeeter said:
Vertical asymptote is x = 3 ... why?

At x = 3, the denominator of the function has a zero and becomes undefined.

Can we also say there is discontinuity at x = 3?
 
tkhunny said:
$g(x) = \dfrac{9-x^{2}}{x-3}$

NOT a Vertical Asymptote. Why?

According to Skeeter, x = 3 is a vertical asymptote.

Let f be a function.

Let g be a function.

Correct me if I am wrong.

Say we have a rational function y = f/g.

If y = 0/g, is this also discontinuity or does discontinuity apply only when the denominator of a function is 0?

If y = f/0, is this a vertical asymptote?

Are the words discontinuity and vertical asymptote related?
 
RTCNTC said:
tkhunny said:
$g(x) = \dfrac{9-x^{2}}{x-3}$
NOT a Vertical Asymptote. Why?

According to Skeeter, x = 3 is a vertical asymptote.

Where did I say that?

I said $f(x) = \dfrac{5 - x^2}{x-3}$ had the vertical asymptote at $x=3$, not the function cited by tkhunny.
 
skeeter said:
Where did I say that?

I said $f(x) = \dfrac{5 - x^2}{x-3}$ had the vertical asymptote at $x=3$, not the function cited by tkhunny.

Sorry about the confusion.
 

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