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The question was an explanation for why $f(x)$ has a vertical asymptote at $x = 3$, but $g(x)$ does not. Mention limits within the answer.

Perhaps I'm solving them wrong but it looks like there is an asymptote in both. Also I would imagine that limits come into play because the values approach the asymptote at $x = 3$

$$f(x) = \frac{x^2 - 7x + 4}{4x^2 - 4x - 24}$$

$$g(x) = \frac{-2x^2 + 12x - 18}{4x^2 - 8x - 12}$$

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2 Answers 2

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Hint: Factor everything, and look for any $(x - 3)$ factors. If such a factor is found in both the numerator and denominator so that they cancel, then the limit as $x \to 3$ exists so that the discontinuity at $x = 3$ is a hole, not an asymptote. Otherwise, if the $(x - 3)$ factor is in the denominator but not the numerator, then the discontinuity must indeed be a vertical asymptote.

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    $\begingroup$ Thank you very much. This is a very helpful answer! When I solved it, there looked to be a vertical asymptote in both f(x) and g(x) so I'm not sure how g(x) doesn't have the asymptote at x=3. $\endgroup$
    – androidguy
    Commented Oct 16, 2014 at 18:50
  • $\begingroup$ What did you get when you factored everything in $g(x)$? Did any $(x - 3)$ factors cancel? $\endgroup$
    – Adriano
    Commented Oct 16, 2014 at 19:02
  • $\begingroup$ The same as the answer below this. g(x) = -1/2 ((x-3)^2) / (x+3)(x+1) $\endgroup$
    – androidguy
    Commented Oct 16, 2014 at 19:10
  • $\begingroup$ You should have gotten: $$ g(x) = \frac{-2(x^2 - 6x + 9)}{4(x^2 - 2x - 3)} = \frac{-(x - 3)^2}{2(x + 1)(x - 3)} $$ $\endgroup$
    – Adriano
    Commented Oct 16, 2014 at 19:26
  • $\begingroup$ Oh okay, so then the (x-3)'s can cancel so g(x) would not have the vertical asymptote? $\endgroup$
    – androidguy
    Commented Oct 16, 2014 at 19:41
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we have $f(x)=\frac{x^2-7x+4}{4(x-3)(x+2)}$ and $g(x)=-\frac{1}{2}\frac{(x-3)^2}{(x+3)(x+1)}$

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