sh3lsh
  • sh3lsh
Is the hole and horizontal asymptote the same? y=(x+2)(x-1) --------- (x^2-1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
no, they are 2 different things
amistre64
  • amistre64
they are caused by similar problems tho
sh3lsh
  • sh3lsh
:/ What?

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sh3lsh
  • sh3lsh
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
  • sh3lsh
Therefore, the horizontal asymptote and hole would be the same.. What am I doing wrong?
amistre64
  • amistre64
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
  • sh3lsh
could you tell me what I am doing incorrectly then? Yeah, me too.
amistre64
  • amistre64
(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
  • sh3lsh
That's what I had. Doesn't that mean that it's all 1?
amistre64
  • amistre64
when x=1 in the equivalent, we determnine the "hole" that the original has at x=1; the point (1,3/2)
amistre64
  • amistre64
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amistre64
  • amistre64
|dw:1332707328171:dw|
amistre64
  • amistre64
graph might not be according to the function, but it shows the difference between a hole and HA

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