#1
Anyone know how to linearize a graph? I tried squaring the X data, but it didn't work right?
#2
Eh? More in depth please: What's the dependent variable, what's the independent variable and what function relates the two?
#3
Quote by LordBishek
Eh? More in depth please: What's the dependent variable, what's the independent variable and what function relates the two?

The X variables are on the left
0 - 60
1 - 65
2 - 70
3 - 79
4 - 90
5 - 100
6 - 125
6.2 - 135

EDIT: IV is X, DV is Y.
Last edited by Harmonicer at Dec 6, 2009,
#4
Quote by Harmonicer
The X variables are on the left
0 - 60
1 - 65
2 - 70
3 - 79
4 - 90
5 - 100
6 - 125
6.2 - 135

EDIT: IV is X, DV is Y.


Oh, so you're plotting the graph empirically? There's no equation that you know of to describe the line?
#5
Quote by LordBishek
Oh, so you're plotting the graph empirically? There's no equation that you know of to describe the line?

Yeah, its empirical...I don't know the equation.
#6
Quote by Harmonicer
Yeah, its empirical...I don't know the equation.


Hmm, I can't help you then man, I'm sorry. It's been ages since I looked at this stuff. If this is from statistics, I presume "linearisation" refers to finding a line of best fit or something? I'm pretty sure you can find the equation for that relatively easily.

Tell you what, bang the values in Excel, create a scatter graph (WITHOUT trendline) and then add a trendline to the data. If you right click the trend line and select the properties, that can give you the equation of the trend line. But that's the maximum I'm able to do for you at this point dude, my bad.
#7
Quote by LordBishek
Hmm, I can't help you then man, I'm sorry. It's been ages since I looked at this stuff. If this is from statistics, I presume "linearisation" refers to finding a line of best fit or something? I'm pretty sure you can find the equation for that relatively easily.

Tell you what, bang the values in Excel, create a scatter graph (WITHOUT trendline) and then add a trendline to the data. If you right click the trend line and select the properties, that can give you the equation of the trend line. But that's the maximum I'm able to do for you at this point dude, my bad.

Thats ok...and I need to actually modify the data so that the line becomes straight, rather than find a best fit line.
#8
Quote by Harmonicer
Thats ok...and I need to actually modify the data so that the line becomes straight, rather than find a best fit line.


What, in ANY way?

Man, there are dozens of solutions - an infinite number, in fact. Do you have any points you KNOW you have to keep constant?
#9
Quote by LordBishek
What, in ANY way?

Man, there are dozens of solutions - an infinite number, in fact. Do you have any points you KNOW you have to keep constant?

I'm not totally sure exactly...like I said, my teacher said the squaring the X variables would result in it being linear, but that didn't work for some reason.
#10
What I'd do in your situation is plug it into your TI 89 calc (or similiar) or excel and get a best fit line equation, then plug in the x variable in order to modify the data to the straight line.

But I don't remember a lot of stats. The only thing that squaring the x variable reminds me of is correlative data.

Sorry bud.
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#11
It gets them a little more linear, but not a great deal.

I'm outta ideas man, I gotta run anyway. Sorry I couldna be more help
#12
looks pretty linear after the modification to me.

it's pretty obvious that you have some form of y = x^2 dataset

what exactly is the data from?
Attachments:
xy.png
xy2.png
#13
You have a pre-calculus final tomorrow don't you? I do.
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#14
Here ya go man. If you're that clueless odds are your instructor is too.
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#15
I, personally, would use the method of Least Squares.
that is, if we assume a line y = ax + b then we must minimize the sum
Sigma || ax +b - y||^2

and if the points x form a matrix A and the y's form a matrix Y then the solution is

x = (tran(A)*A)^{-1} *tran(A) * Y

where tran(A) denotes the transpose of A.