Why is x=-2 not a vertical asymptote of f(x)=(x^2-4)/(x+2)

The vertical asymptote of this rational function is none. Why is that? How come it can't be -2?

cuz when you factor using the difference of perfect squares, the bottom and top terms cancel. here ill show you
(x^2-4)/(x+2) = (x+2)(x-2)/(x+2)
then those x+2's cancel and youre left with
f(x)= (x-2) and theres no where wheres it undefined

Because, at x = -2, you have a "hole".

For x > -2, you have a continuous function (equivalent to f(x) =x-2).

For x < -2, you have a continuous function (equivalent to f(x) = x - 2).

But at x = -2, you have effectively 0/0, which is not defined.

x^2 - 4 can be factored into (x-2)(x+2)

Therefore, x + 2 on the top and bottom cancel out, leaving nothing on the bottom. Since they cancel out, -2 cannot be an asymptote. -2 is actually a hole. There are not asymptotes since nothing is left on the bottom.

you can factor the numerator into (x-2)(x+2). notice how the x+2 cancel. therefere, you get the new f(x) to be f(x)= x-2. when drawing the graph, there will be a hole (discontinuity) at x=-2.

Because the numerator factors as (x + 2)(x – 2). The x + 2 cancels leaving f(x) = x – 2 when x ≠ –2 and a hole rather than a vertical asymptote when x = –2.