Why isn't there a vertical asymptote at $x = 2$ for $y = \frac{x - 2}{x^2 - 3x +2}$? 
Why isn't there a vertical asymptote at $x = 2$ for $y = \frac{x -
 2}{x^2 - 3x +2}$?

If you factorize the denominator, you get: $(x-1)(x-2)$. So, when $x = 1$ or $2$, the denominator will be $0$. But I noticed that when I put this function into a graphing program, there is only a vertical asymptote at $x = 1$ and the function seems to be continuous at $x = 2$.
I can also see that the $(x - 2)$ will cancel with the numerator, but I was just wondering on a deeper level why this means that only $x = 1$ is undefined.
 A: We have
$$y = \frac{x-2}{x^2-3x+2}$$
You can check that at $x=1$, we face the following problem
$$y(1) = -\frac{1}{0}$$
So the function is not defined at $x=1$. Similarily
$$y(2) = \frac{0}{0}$$
So the function is not defined at $x=2$ either.
How do we differentiate an asymptote from a removable discontinuity$^{1}$? We need to find the limits. In this case, we have
\begin{align*}
\lim_{x\to 2} y &=\lim_{x\to 2}\frac{x-2}{x^2-3x+2}\\
 &=\lim_{x\to 2}\frac{x-2}{(x-2)(x-1)}\\
&=\lim_{x\to 2}\frac{1}{x-1}\\
&=1
\end{align*}
so there is no asymptote at $x=2$, but rather a "hole". We can fill it in, and in fact make $y$ continuous there by simply defining $y(2)=1$.
However, at $x=1$, as you note, there is an asymptote.
$1:$ [the function] is discontinuous there, but the function can be redefined so that it can be continuous at that point. J.M.
A: The function is not defined at $x = 2$ because of the 0/0 that appears there, but the limit there still exists and so the function does not have an infinity-singularity that would lead to an asymptote.
A: The function is:
$$f(x)=\frac{x-2}{x^2-3x+2}$$
Observe that:
$$\lim_{x\rightarrow 1^-}{f(x)}=-\frac{1}{0}=-\infty $$
and,
$$\lim_{x\rightarrow 1^+}{f(x)}=\frac{1}{0}=\infty $$
And that
$$\lim_{x=2}{f(x)}=\frac{0}{0}=(undefined!) $$
We know through L'Hôspital's Rule that:
$$ If \lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}=\frac{0}{0}$$
$$then$$
$$\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}=\lim_{x\rightarrow a}{\frac{f'(x)}{g'(x)}}$$
$$\lim_{x\rightarrow 2}{\frac{x-2}{x^2-3x+2}}=\lim_{x\rightarrow 2}{\frac{(x-2)'}{(x^2-3x+2)'}}=\lim_{x\rightarrow 2}{\frac{1}{2x-3}}=\frac{1}{2(2)-3}=\frac{1}{4-3}=\frac{1}{1}=1$$
This shows that though $f(2)$ is undefined, the limit is defined. However, for $f(1)$, the limit cannot be rationally defined because $\Big(\big[x\rightarrow 1^{\pm}\big],\big[f(x)\rightarrow \pm\infty\big]\Big)$
Usually, for graphing softwares, $x$-values which are not defined, but have a limit, are graphed as "holes" perhaps missing a pixel or so. These $x$ values are usually either:


*

*Values which are not defined on $f(x)$, but have a existing limit.

*Values which can be computed after "cancelling out" something.


In our case, $f(2)$ is both!
Explaining that there is no asymptote (on graphing software) at $x=2$; if you zoom in enough, the pixels will become messy/non-existant/forming an odd shape (such as a 2-sided-square or circle...), depending on which graphing software you use and its programming.
