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

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SUMMARY

The vertical asymptote of the function f(x) = (5 - x^2)/(x - 3) is definitively located at x = 3. This is due to the denominator becoming zero at this point, rendering the function undefined. The discussion also clarifies that the function g(x) = (9 - x^2)/(x - 3) does not have a vertical asymptote at x = 3, as it is not equivalent to f(x). Additionally, the relationship between vertical asymptotes and discontinuities is explored, confirming that a vertical asymptote indicates a point of discontinuity.

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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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