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?

$(1)$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.
