NationStates Jolt Archive

http://img100.imageshack.us/img100/3636/math1tb.png

What would the equation:


y = 1 / [g(x)]

do to that graph (represented by g(x))?

Plot the inverses of the points you know, then connect them. Should be accurate enough.

You're gonna have a few asymptotes.

Helped me heaps in my highschool days.

It doesn't calulate anything, but it draws beautiful graphs from functions like yours.
Graphmatica (http://www8.pair.com/ksoft/)

Plot the inverses of the points you know, then connect them. Should be accurate enough.

Ah ha! So you do inverse the points!

You're gonna have a few asymptotes.

Or... you don't invert points?

Or... you don't invert points?No. Any y=1 stays y=1 (1/1=1). Any y=0 becomes y=infinity (1/0=infinity) [and vice versa]. The lowest point on your graph will be (-3,0.5) (1/2=0.5).

You inverse the result of g(x), not mirror their position about y=0

http://skimble.blogspot.com/uploaded_images/un_find_x_lol-744885.jpg

(1/0=infinity)
I haven't done maths for ages, but I am pretty certain that infinity is not so much the answer as there will simply be a break in the graph at a point where something is divided by zero.

In other words, an asymptote (http://en.wikipedia.org/wiki/Asymptote), in a way.

Super Power, I just about died laughing there.

Good show.

I haven't done maths for ages, but I am pretty certain that infinity is not so much the answer as there will simply be a break in the graph at a point where something is divided by zero.

In other words, an asymptote (http://en.wikipedia.org/wiki/Asymptote), in a way.I know what an asymptote is. It's just saying that it equals infinity describes much more eloquently in my opinion what happens at that point and around that point. ie g(x) -> infinity then comes the asymptote which can't be plotted because technically 1/0 is undefined not infinity (but it makes logical (graphical) sense that it would be). And then g(x) comes back from infinity and back to low and more easily plotable numbers.

True, true.
This would be a lot easier if we knew what g(x) is. The points on their own don't seem to make all that much sense to me, and graphmatica just creates a fancy approximation that doesn't actually touch any of those points.
It doesn't have the right feature here, I guess.

No. Any y=1 stays y=1 (1/1=1). Any y=0 becomes y=infinity (1/0=infinity) [and vice versa]. The lowest point on your graph will be (-3,0.5) (1/2=0.5).

You inverse the result of g(x), not mirror their position about y=0

Could you do an example of how I would go about finding the point of the new function by using point (-3, 2)?

True, true.
This would be a lot easier if we knew what g(x) is. The points on their own don't seem to make all that much sense to me, and graphmatica just creates a fancy approximation that doesn't actually touch any of those points.
It doesn't have the right feature here, I guess.

g(x) is that 'tarded function that I graphed.

g(x) is that 'tarded function that I graphed.
The point is that you would need some sort of g = ax + b or something like that, so we can invert that, and then put in values into the inverted formula.

Simply inverting the points won't do the trick I'm afraid, you first have to do the algebra.

The point is that you would need some sort of g = ax + b or something like that, so we can invert that, and then put in values into the inverted formula.

Simply inverting the points won't do the trick I'm afraid, you first have to do the algebra.

so, to get the new point from (-3, 2) I would do this:

2 = 1/ (g(-3)?

etc, etc.?

Not really because you seem to have a uniform function for all x. It does lots of random shit.

Wait, do you just mean find the new individual co-ordinate or the new function of the graph using the co-ordindate (-3,2)?

so, to get the new point from (-3, 2) I would do this:

2 = 1/ (g(-3)?

etc, etc.?Oops. g(-3) = 2
Therefore 1/g(-3) = 1/2

Not really because you seem to have a uniform function for all x. It does lots of random shit.

Wait, do you just mean find the new individual co-ordinate or the new function of the graph using the co-ordindate (-3,2)?

New individual coordinate.

Basically, if you were to apply the equation y = g(x-2) the entire graph would shift 2 units to the right. If you were to apply the equation y = g(2x) to my graph it would be half as long, because (4,0) would become (2,0), etc. (I'm not entirely sure why it shortens instead of lengthens, but that's how we were taught today).

Now here, I'm trying to see what would happen to my graph if I applied the equation y = 1 / [g(x)] to it.

Sorry if you all already knew that, just want to clarify.

Since when does anything/0 = infinity? it's UNDEFINED

New individual coordinate.

Basically, if you were to apply the equation y = g(x-2) the entire graph would shift 2 units to the right. If you were to apply the equation y = g(2x) to my graph it would be half as long, because (4,0) would become (2,0), etc. (I'm not entirely sure why it shortens instead of lengthens, but that's how we were taught today).

Now here, I'm trying to see what would happen to my graph if I applied the equation y = 1 / [g(x)] to it.

Sorry if you all already knew that, just want to clarify.(It shortens, like you said, because you double x. It's like if I count to 100 but only count the even numbers then it takes half the time to count to 100 because I only have to count 50 numbers, not 100. Does that make sense? :confused:

Anyway, 1/g(x) does weird shit to a graph. It makes straight lines curvy, some curvy lines straight, and vaguely mirrors points around y=|x|

And then y=g(1/x) is different again.

Since when does anything/0 = infinity? it's UNDEFINED
But when you look at a graph when the y value aproaches 1/0, the it appears to approach infinity.

Since when does anything/0 = infinity? it's UNDEFINEDStop being so tight arsed. I was trying my best to give a good visual description so that CJLP would know what I was talking about. Undefined is pretty undefining word. It only means something if you know precisely what context it's being used in. And seeing as this is thread to try and explain what is happening to a graph rather than have high level dicussion where using jargon makes the dicussion quicker.

Do you know that a large proportion of people use jargon words to hide their lack of knowledge on a subject, or use a word without being fully able to explain the concept of the word?

Also you just jump in with a single line post, that's going off-topic, and in the process showing that you haven't bothered to read the entirety of less than two page thread, where I've actually already talked about my use of 'infinity', but instead show yourself to be insecure in your own mental faculties that you have to jump in and attempt to demonstrate superior knowledge whenever you can.

Grow up.

:mad:

Stop being so tight arsed. I was trying my best to give a good visual description so that CJLP would know what I was talking about. Undefined is pretty undefining word. It only means something if you know precisely what context it's being used in. And seeing as this is thread to try and explain what is happening to a graph rather than have high level dicussion where using jargon makes the dicussion quicker.

Do you know that a large proportion of people use jargon words to hide their lack of knowledge on a subject, or use a word without being fully able to explain the concept of the word?

Also you just jump in with a single line post, that's going off-topic, and in the process showing that you haven't bothered to read the entirety of less than two page thread, where I've actually already talked about my use of 'infinity', but instead show yourself to be insecure in your own mental faculties that you have to jump in and attempt to demonstrate superior knowledge whenever you can.

Grow up.

:mad:

don't let the trolls bother you. anyway, i knew you meant 1/0 meant an asymptote, so it's all good.

here's my new graph, where arrows = asymptotes.

http://img87.imageshack.us/img87/7992/math7hs.png