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Context:
HP Prime hardware: Version C
Software: 2018 02 12 (13441)
CAS version: 1.4.9
OS: V0.050.639

Equation: (X^2 - 25) / (X^2 - 5*X)
Apps: 1) Function and 2) Advanced Graphing

Observation:
1) In Function graph, shows the expected root at X=0.
2) In Advanced Graphing, shows the same plus the discontinuity vertical line at X=-5.
So, shows 2 solutions at x=-5.

Problem?
(1) That's not an equation.

(2) Your comment #2 indicates that you meant the denominator to be X^2+5X (which is the way it was typed into HP's calculator forum and into your HP Prime), not X^2-5X.

(3) Both apps' graphs show the root to be at X=5, not at X=0.

(4) The vertical line at X=-5 is the Advanced Graphing app's way of indicating a removable singularity. It does not indicate another root. Try Y=(X^3-X)/(X^2-1) to see another example of this in action.

(5) Yes, the Function app differs from the Advanced Graphing app. This is intentional. If they were identical, there would be no reason to have both.
Joe, don't be so rude!
Hi dmusic, welcome back! Joe is true, you've been a bit confusing.

(05-01-2018 01:45 AM)Joe Horn Wrote: [ -> ](1) That's not an equation.
True, it's an expression. But like any calculator, you could have infer the missing '=0' Smile

(05-01-2018 01:45 AM)Joe Horn Wrote: [ -> ](2) Your comment #2 indicates that you meant the denominator to be X^2+5X (which is the way it was typed into HP's calculator forum and into your HP Prime), not X^2-5X.
Yes dmusic we think you made a mistake somewhere, but we still understand the question.

For the expression f(x)=(X^2 - 25) / (X^2 - 5*X), there is only 1 root (f(x)=0) at x=-5, not at x=5.
There is a discontinuity at x=5, the calculator does not have any other mean to show it than with a vertical bar, but it is not a root, the limit is 2 (try with x=4.999 or x=5.001).

For the expression f(x)=(X^2 - 25) / (X^2 + 5*X), there is only 1 root (f(x)=0) at x=5, not at x=-5. There is also a discontinuity at x=-5, the limit is also 2 (try x=-4.999 or x=-5.001)



(05-01-2018 01:45 AM)Joe Horn Wrote: [ -> ](3) Both apps' graphs show the root to be at X=5, not at X=0.
Yes dmusic, finding the root of a function means finding f(x)=0, so (X^2 - 25) / (X^2 - 5*X)=0, which is true for (and only for) x^2-25=0 => x^2=25 => x=-5 but not x=5 because of the denominator.
For (X^2 - 25) / (X^2 + 5*X)=0 it is the opposite : x=5 is a root, x=-5 has no value.

(05-01-2018 01:45 AM)Joe Horn Wrote: [ -> ](4) The vertical line at X=-5 is the Advanced Graphing app's way of indicating a removable singularity. It does not indicate another root. Try Y=(X^3-X)/(X^2-1) to see another example of this in action.
There is a discontinuity at x=5 or x=-5 (depending on the expression you used), and the calculator does not have any other mean to show it than with a vertical bar, but it is not a root, the limit is 2 in both cases.

(05-01-2018 01:45 AM)Joe Horn Wrote: [ -> ](5) Yes, the Function app differs from the Advanced Graphing app. This is intentional. If they were identical, there would be no reason to have both.
To be more precise, the advanced app refines the graph around specific values like the discontinuity in your case. It also has a more precise way of plotting functions, find values with cursor, and you can enter any kind of expressions, not only functions, eg. y^2 = 1 - x^2 (see attached pictures)

[attachment=5878] [attachment=5879]
(05-01-2018 08:14 PM)pinkman Wrote: [ -> ]Joe, don't be so rude!

Oh dear... I'm very sorry for sounding rude. Please believe me that my intentions were only to help and to be clear and concise. Therefore you (and/or any other forum members) could help me very much by sending me a PM which points out what I said that was rude, so I can avoid doing that in the future. Thanks in advance.
You always are rude to me Joe. It hurts my feelings and makes me cry... :-(

Why just the other day I cried myself to sleep, tears dripping onto my lumpy, worn out pillow (because you never sent me a new one) and wondered where in my life I went wrong such that Joe feels a need to pick on me so.
Excellent. My work is done here. Big Grin
(05-01-2018 08:14 PM)pinkman Wrote: [ -> ]Joe, don't be so rude!

Rude? I would call his post very informative. But, if Joe is being rude, what label would get OP who asked the question and after that he disappeared?
Sorry for my bad English.

For me, Joe has not been rude, he is responding in a non-didactic way. Horn must be respond in more detail, trying to teach how Pinkman does it.
OMG :-)
(05-01-2018 11:25 PM)Joe Horn Wrote: [ -> ]
(05-01-2018 08:14 PM)pinkman Wrote: [ -> ]Joe, don't be so rude!

Oh dear... I'm very sorry for sounding rude. Please believe me that my intentions were only to help and to be clear and concise. Therefore you (and/or any other forum members) could help me very much by sending me a PM which points out what I said that was rude, so I can avoid doing that in the future. Thanks in advance.

PM sent just now, I copy here my last words: "Be sure I believe you when you say you were not rude."
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