I'm trying to find removable discontinuities. I'm studying Algebra II for dummies. The problem is: x^3-3x^2-16x-12/x^3-4x^2-11x-6. This factors to (x+2)(x-6)(x+1)/(x+1)(x+1)(x-6). Reducing to x+2/X+1. I thought the only removable discontinuity would be 6. The text lists both 6 and -1. But, since -1is a vertical asymptote even though it was factored out...it hasn't been removed. Am I thinking wrong or is the text wrong?

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I'm trying to find removable discontinuities. I'm studying Algebra II for dummies. The problem is: x^3-3x^2-16x-12/x^3-4x^2-11x-6. This factors to (x+2)(x-6)(x+1)/(x+1)(x+1)(x-6). Reducing to x+2/X+1. I thought the only removable discontinuity would be 6. The text lists both 6 and -1. But, since -1is a vertical asymptote even though it was factored out...it hasn't been removed. Am I thinking wrong or is the text wrong?

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the tect is right. a graph is discontinuous when you have to lift your pencil to continue graphing it, and at a verticle asymptote you have to lift your pencil, therefore it is discontinuous
Okay, but the key word is removable discontinuity. By factoring out 6 it becomes a removable discontinuity. -1 has been factored out, but since it is still in the denominator(it was squared) the discontinuity has not been removed. That is what threw me. I can see the asymptote would be there and the graph would be discontinuous.
well removable discontinuity is when a point or section of a graph is removed, aka a hole, that is when 2 factors are crossed out during simplification

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Right. To simplify my confusion, if the equation were (x+1)/(x+1)^2 the function would reduce to 1/x+1. There would be an asymptote at -1. But, following the apparent logic of the problem in the text...-1 would be considered a removable asymptote because I factored it out in part. I don't see how something can be both a removable discontinuity and an asymptote at the same time.
it is just saying that the point lies on the vertical asymptote. the vertical asymptote means the function wont cross that value, yet a point may lie on it.
Okay, I really appreciate your time. It clears things up a great deal for me.
oo your welcome. i didnt actually learn that till i was in calculus, and thats where i learned that i really like math
Now your just making me feel a bit guilty. I had calc ten years ago and haven't really used math very often. I like it a great deal but I'm out of practice and am working to push my knowledge further. Again, I greatly appreciate your help.
oo your welcome.
I'm out for the night. I'm sure I'll be back in the future.
Is this the expression? \[\frac{x{}^{\wedge}3-3x{}^{\wedge}2-16x-12}{x{}^{\wedge}3-4x{}^{\wedge}2-11x-6}\]
Okay, I'm back. Yes, that is the expression.
The expression simplifies to: \[\frac{2+x}{1+x}\] When x=-1 in the denominator, you have division by zero which is a no no. There seems to be a root at x=-2 Refer to the plot attachment.
1 Attachment
Right, the imaginary line is an asymptote. An asymptote creates a discontinuity. A removable discontinuity is one that can be removed by factoring. My solution agrees with the texts solution until the final answer which lists both 6 and -1 as being removable. My point is that -1 is not removable. I think that is what you are also indicating. According to nctarheelsbball, it is an asymptote but listed as removable because that is where the point lies.
Sorry. I guess I'll have to google "removable discontinuity math"
From: http://www.math-magic.com/algebra/asymptote.htm E. Removable Discontinuities "A removable discontinuity is when you can factor out a term in the numerator and factor out the same term in the denominator, thus canceling each other out." \[ \frac{x{}^{\wedge}3-3x{}^{\wedge}2-16x-12}{x{}^{\wedge}3-4x{}^{\wedge}2-11x-6}==\frac{(x+1) (x+2)(x-6)}{(x+1)(x+1)(x-6)}\] The factors (x+1) and (x-6) cancel out. Perhaps the confusion is that (x+1) is the simplified version of ( x - (-1) ) which equals ( x + 1 )
Well, I don't know if anyone will read this now, and I don't mean to be pedantic. I am very well aware of what a removable discontinuity is, but because of this problem I also googled the definition for greater understanding. I also understand what an asymptote is...the problem is the conflict between names, definitions and this particular rational function: Mainly, that I'm very bloody well aware that (x-6) cancels out: thus fulfilling the very definition of a removable discontinuity. (x+1) cancels out in the numerator and the denominator and thus is removed, however: because it is squared in the denominator (x+1) remains a part of the function and because it remains, it creates an asymptote particularly a verticle asymptote which is the very definition of a non-removable discontinuity. The best answer I've seen so far is nctarheelsball who seems to suggest, in a condensed form: yes, those terms conflict, but both are true.

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