Vtyjkyuk

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
$$
fracx^2+x-6x-2 = frac(x+3)(x-2)x-2 = x+3.
$$
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.

You're right that
$$
y = fracx^2+x-6x-2
$$
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
$$
y = fracx^2+x-6x-2
$$
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
$$
y = begincases
fracx^2+x-6x-2, & x neq 2,\
5, & x = 2.\
endcases
$$
Appending this value makes it continuous everywhere, and makes it identically equal to $y = x + 3$.

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.

Function $xmapstofracx^2+x-6x-2=frac(x+3)(x-2)x-2$ is defined on $mathbb R-2$.

Function $xmapsto x+3$ 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.

An answer for you after learning complex analysis in the future

Let $f(z)=fracz^2-z+6z-2$ and $g(z)=z-3$.

$f(z)$ has a domain of $mathbb Csetminus2$.

$g(z)$ has a domain of $mathbb C$.

Since $f(z)-g(z)=0$ for every $zinmathbb Csetminus2$, by the principle of analytic continuation, $g$ is the unique analytic continuation of $f$. This explains why the graphs look the same.

To elaborate on rbird's answer, if you have a $0/0$ singularity, you can use L'Hopital's rule to find the solution:

$$
lim_xrightarrow c fracf(x)g(x)= lim_xrightarrow c fracf'(x)g'(x)
$$

and so

$$
lim_xrightarrow2 fracx^2 +x-6x-2 = lim_xrightarrow2frac2x+11
$$

which equals 5 QED