<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] 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 $y = x + 3$.
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