Is the hole and horizontal asymptote the same?
y=(x+2)(x-1)
---------
(x^2-1)

Question

Is the hole and horizontal asymptote the same?
y=(x+2)(x-1)
    ---------
    (x^2-1)

amistre64 · Answer

no, they are 2 different things

amistre64 · Answer

they are caused by similar problems tho

sh3lsh · Answer

:/ What?

sh3lsh · Answer

I thought if you crossed out the same factor on top, the (x-1), the hole is 1. Then, x/x would equal 1.

sh3lsh · Answer

Therefore, the horizontal asymptote and hole would be the same.. What am I doing wrong?

amistre64 · Answer

the hole is a point that looks like its been erased; this happens in the middle of the graph

a horisontal asymp is what happens to the graph at the ends of infinity

having troubles with system freezing ....

sh3lsh · Answer

could you tell me what I am doing incorrectly then? 
Yeah, me too.

amistre64 · Answer

(x+2)(x-1)     (x+2)(x-1)
--------- =  ---------- = (x+2)\(x+1)
  (x^2-1)        (x-1)(x+1)

the simplified version is NOT the original, it is an equivalent with different properties

sh3lsh · Answer

That's what I had. Doesn't that mean that it's all 1?

amistre64 · Answer

when x=1 in the equivalent, we determnine the "hole" that the original has at x=1;  the point (1,3/2)

amistre64 · Answer

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