Are the graphs of these two functions equal to each other? The functions are: $y=\frac{x^2-4}{x+2}$ and $(x+2)y=x^2-4$.
I've seen this problem some time ago, and the official answer was that they are not.
My question is: Is that really true? 
The functions obviously misbehave when $x = -2$, but aren't both of them indeterminate forms at that point? Why are they different?
 A: $(1)$The first function is undefined at $x = -2$, 
$(2)$ the second equation is defined at $x = -2$:
$$(x + 2) y = x^2 - 4 \iff xy + 2y = x^2 - 4\tag{2}$$ It's graph includes the entire line $x = -2$. At $x = -2$, all values of y are defined, so every point lying on the line $x = -2$: each of the form $(-2, y)$ are included in the graph of function (2). Not so with the first equation.
ADDED:
Just to see how well Wolfram Alpha took on the challenge:
Graph of Equation $(1)$: (It fails to show the omission at $x = -2$) But it does add: 


Graph of Equation $(2)$:

Note: The pair of graphs included here do not match in terms of the scaling of the axes, so the line $y = x - 2$ looks sloped differently in one graph than in the other.
A: In the second equation, when $x=-2$, there is no restriction whatsoever on $y$. The graph of the second equation includes the entire vertical line $x=-2$.
A: Forget functions for a moment.  An equation is a sentence, and the graph of an equation (in two variables) is the set of pairs $(x,y)$ which satisfy the sentence.  So $(1, -\frac{1}{3})$ is in the graph of the equation $y = \frac{x^2 -4}{x+2}$ and $(x+2)y = x^2-4$.  You can see this by plugging in $x = 1$ and $y = -1/3$ -- each sentence is true.
The point $(-2, 0)$ lies in the graph of the second equation, but not in the first, because $\frac{0}{0}$ is a nonsense phrase (i.e. "undefined").  It doesn't make sense to call nonsense like $y = \frac{0}{0}$ true or false.  Moreover, $(-2, 3)$ and $(-2, 4)$ also satisfy the second sentence.  In fact, ANY pair $(-2, y)$ satisfies the second equation, so all those points lie on the graph of the second equation.
Now let's return to functions.  The first equation is the graph of a function.  The second equation is not, because it fails the vertical line test.  To be really nit-picky, a function is a rule which assigns to each element of the domain a unique element of the range.  So an equation is not a function.
Finally, "indeterminate forms" concern limits, which don't seem immediately relevant here.
