<s>[INST] 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. 
 [/INST] 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.
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