# Why is there a vertical asymptote at x = 3 but g(x) does not have a vertical asymptote at x = 3?

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 $x = 3$

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

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

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.
• What did you get when you factored everything in $g(x)$? Did any $(x - 3)$ factors cancel? Commented Oct 16, 2014 at 19:02
• You should have gotten: $$g(x) = \frac{-2(x^2 - 6x + 9)}{4(x^2 - 2x - 3)} = \frac{-(x - 3)^2}{2(x + 1)(x - 3)}$$ Commented Oct 16, 2014 at 19:26
we have $f(x)=\frac{x^2-7x+4}{4(x-3)(x+2)}$ and $g(x)=-\frac{1}{2}\frac{(x-3)^2}{(x+3)(x+1)}$