Why is the graph of $y=(x^2+x-6)/(x-2)$ the same as the graph of $y=x+3$, and a continuous function? 
Why is the graph of $y=(x^2+x-6)/(x-2)$ the same as the graph of $y=x+3$, and a continuous function?

I understand that this is because
$$
\frac{x^2+x-6}{x-2} = \frac{(x+3)(x-2)}{x-2} = x+3.
$$
But my problem with this is that in $(x-2)$, when $x=2$ the denominator will be $0$. So shouldn't there be a break here, because if $y=(x^2+x-6)/(x-2)$, then $y$ should then be undefined? So why can $y=(x^2+x-6)/(x-2)$ still be considered continuous and equal to $y=x+3$?
Please try to explain this simply. I'm still just beginning Khan Academy calculus. Thank you. 
 A: Their graphs are not the same, nor is $f(x) = (x^2+x-6)/(x-2)$ continuous at $x=2$.
Indeed, as you say $f(x) = (x^2+x-6)/(x-2)$ is undefined at $x=2$ and therefore has a sort of a "puncture" at $x = 2$.
In particular, $f$ cannot be continuous at $x=2$, since the definition of continuity at a point requires the function to be defined at that point.
A: You're right that
$$
y = \frac{x^2+x-6}{x-2}
$$
has a discontinuity at $x = 2$, so as it is, it isn't truly continuous. However, you've shown that this is equal to $y = x + 3$ for every $x \neq 2$, and by looking at the graph of
$$
y = \frac{x^2+x-6}{x-2}
$$
it makes sense that at $x = 2$, we should have $y = 5$ (when you come to learn about limits, this will make sense and can be proved formally). So, what we should actually do is write
$$
y = \begin{cases} 
       \frac{x^2+x-6}{x-2}, & x \neq 2,\\
       5, & x = 2.\\
    \end{cases}
$$
Appending this value makes it continuous everywhere, and makes it identically equal to $y = x + 3$.
A: Function $x\mapsto\frac{x^2+x-6}{x-2}=\frac{(x+3)(x-2)}{x-2}$ is defined on $\mathbb R-\{2\}$.
Function $x\mapsto x+3$ is defined on $\mathbb R$.
So the functions are not the same, but the second function restricted to set $\mathbb R-\{2\}$ is the same as the first. 
The first function is continuous at any element of $\mathbb R-\{2\}$ hence is a continuous function.
We cannot say that it is continuous at $2$ because it is not defined there, but what we can say is that it can be extended to a continuous function on $\mathbb R$ (which is the second function).
For a related question that I posed myself see here.
A: To elaborate on rbird's answer, if you have a $0/0$ singularity, you can use L'Hopital's rule to find the solution:
$$
\lim_{x\rightarrow c} \frac{f(x)}{g(x)}= \lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}
$$
and so
$$
\lim_{x\rightarrow2} \frac{x^2 +x-6}{x-2} = \lim_{x\rightarrow2}\frac{2x+1}{1}
$$
which equals 5 QED
A: An answer for you after learning complex analysis in the future
Let $f(z)=\frac{z^2-z+6}{z-2}$ and $g(z)=z-3$.
$f(z)$ has a domain of $\mathbb C\setminus\{2\}$.
$g(z)$ has a domain of $\mathbb C$.
Since $f(z)-g(z)=0$ for every $z\in\mathbb C\setminus\{2\}$, by the principle of analytic continuation, $g$ is the unique analytic continuation of $f$. This explains why the graphs look the same.
