Why does this happen with my graph?


by Nikitin
Tags: graph, happen
Nikitin
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#1
Feb20-12, 01:55 PM
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I was doing some homework, trying to solve an equation by using algebra instead of reading from a graph (like I was supposed to), and I stumbled upon something:

Check out the attachment.

Why does the graph of f(x)=x/x have a T shape, with the vertical line starting at x=0 ? I know the graph isn't kind-of supposed to be defined for x=0, but I can't explain this anyway.
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tiny-tim
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#2
Feb20-12, 02:15 PM
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Hi Nikitin!
Quote Quote by Nikitin View Post
I know the graph isn't kind-of supposed to be defined for x=0, but I can't explain this anyway.
It should just have a tiny hole at x = 0,

since, as you say, it isn't defined there.
I expect that particular graphing site (which one is it?) has a special way of dealing with 0/0

it looks like it's putting it equal to 0
so it joins that point, (0,0), to the rest of the graph by a line.
Nikitin
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#3
Feb20-12, 04:33 PM
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I'm using mathcad. great program.

anyway, I think it's supposed to be like this.. I used this function A=[(x-1)(x-2)^2 +4]/x which had a similar line going vertically from y=0 to y=8.
The line actually represented that for x=0 A could be whatever from 0 to 8, and that actually made sense and was correct. I know my situation is terribly explained, but point is I think the line is supposed to be there.

AlephZero
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Feb20-12, 05:58 PM
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Why does this happen with my graph?


Quote Quote by Nikitin View Post
I know my situation is terribly explained, but point is I think the line is supposed to be there.
What Tiny Tim said.

You shouldn't "believe" anything that is generated by a computer program, unless you understand why it's right or wrong.

Whatever MathCad is doing here, even if it's reinforcing your personal opinion about how math is "supposed to be", that isn't helping you learn what's right.
Nikitin
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#5
Feb21-12, 05:54 AM
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well it's not my personal opinion, if the function in post 3# wouldn't have gone with the vertical line then my answer would be incorrect.
Mark44
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Feb21-12, 06:00 PM
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Quote Quote by Nikitin View Post
anyway, I think it's supposed to be like this.. I used this function A=[(x-1)(x-2)^2 +4]/x which had a similar line going vertically from y=0 to y=8.
The line actually represented that for x=0 A could be whatever from 0 to 8, and that actually made sense and was correct. I know my situation is terribly explained, but point is I think the line is supposed to be there.
A is undefined when x = 0. There should NOT be a vertical line.

Your function A is identical to y = x2 - 5x + 8, except that there is a "hole" in the graph of A at the point (0, 8). The graph of y = x2 - 5x + 8 is a parabola that opens up.

Quote Quote by Nikitin View Post
well it's not my personal opinion, if the function in post 3# wouldn't have gone with the vertical line then my answer would be incorrect.
NemoReally
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May28-12, 11:21 AM
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Quote Quote by tiny-tim View Post
Hi Nikitin!

It should just have a tiny hole at x = 0,

since, as you say, it isn't defined there.
I expect that particular graphing site (which one is it?) has a special way of dealing with 0/0

it looks like it's putting it equal to 0
so it joins that point, (0,0), to the rest of the graph by a line.
By default, Mathcad defines 0/0 as 0. There is an option to turn it off, thus allowing the user to catch the error and deal with it according to the nature of the problem.

NR

from the main menu:
Tools \ Worksheet Options ...
This brings a dialog box with several tabs, choose the 'Calculation' tab and clear the '0/0=0' checkbox.
tiny-tim
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May28-12, 11:43 AM
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Quote Quote by NemoReally View Post
By default, Mathcad defines 0/0 as 0. There is an option to turn it off
ahh!
thanks, NemoReally
HallsofIvy
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May28-12, 12:30 PM
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However, it is very likely that "round off error" for numbers very close to 0 will cause difficutly with things like f(x)/g(x) where f(x) and g(x) are very close to 0 themselves.


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