A simplified rational exp, -n - 2, is not undefined at n=-2?

Question

A problem on Khan Academy asks you to simplify the equation

                     (1)   ((−3n^2+ 6n))/3n

(1) in its simplified form is

                      (2)   -n + 2.

Obviously at n = 0 in (1) the equation is undefined.

All of this makes sense to me so far, what is confusing me, however, is that the answer provided said:

The only value of n for which (1) and (2) are undefined is 0.

What is tripping me up is why (2) is not undefined at both n=0 and n=-2.

since,

for n = -2

-n - 2

=

-(-2) - 2

= 0

In the other example problems if a simplified rational expression had the form

            (3)    x - 2

We would have considered the rational equation to have been undefined at

x = 2

since for,

x - 2 = 0

x = 2

Can someone explain why we consider (3) undefined at x = 2,

but not

(2) at n = -2?

Answer 1

The reason a rational function may be undefined at a point is if that value makes the denominator zero. There is no problem, however, with having the numerator be zero (and the denominator be nonzero). Thus $-n+2$ is perfectly well-defined when $n=-2$, and it has a value of $0$ as you showed. Similarly, $x-2$ is defined when $x=2$ and has a value of $0$. Something like $\frac{1}{x-2}$ would not be defined at $x=2$ because you cannot divide by $0$.

Answer 2

I think there is some confusion here, as neither $(3)$ nor $(2)$ are undefined. That is, the rational function $$ x - 2$$ is well-defined at $x = 2$. Indeed, as you have noticed, sticking in $2$ produces the number $0$. This is a perfectly reasonable number, and there is nothing odd about it at all.

Similarly, $-n + 2$ is perfectly well-behaved at $n = -2$. The number $0$ is just another number!

However, in the initial rational function $$\frac{-3n^2 + 6n}{3n},$$ if you tried to plug in $0$ then you would get $$ \frac{0}{0},$$ and what is that supposed to mean? That's why the initial rational function is undefined at $n = 0$.