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0 0
1 0
2 0
3 10
4 83
5 170
6 300
7 500
8 750
9 1150
10 1500
11 1750
12 1800
Can anyone give me the formula for this plot? The numbers are not absolute, as they were taken off a small graph, so a formula that would aproximate this curve is ok. THANKS!
asympyotes 0, 1800
This is the curve for which I need an equation:
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Here is a silly answer that uses operators that are probably disallowed in algebra in this fashion for some reason or other.
y=0(x<3) + (3 ≤ x<11)(1800/(11-3))(x-3) + (x ≥ 11)1800
igloo myrtilles fourmis
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The tanh(x) function has two horizontal asymptotes, it could be adapted possibly.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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arctan(x) also has two horizontal asymptotes.
But I think that these functions cannot be adapted for this graph, because it is not symmetric.
IPBLE: Increasing Performance By Lowering Expectations.
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Why not this for solution?:
if:3<x<12 y=((1/logx^3)+1)*x^3
kdria posted that meaning to reply to this topic, but accidentally starting a new one instead.
You can reply using the 'post reply' button. ;)
Why did the vector cross the road?
It wanted to be normal.
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arctan(x) also has two horizontal asymptotes.
But I think that these functions cannot be adapted for this graph, because it is not symmetric.
Yes I quote him! You can use arctan(x) by creating your own function
For example try to plot a function like this:
2*atan(x-3)+3
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krassi_holmz wrote:arctan(x) also has two horizontal asymptotes.
But I think that these functions cannot be adapted for this graph, because it is not symmetric.Yes I quote him! You can use arctan(x) by creating your own function
For example try to plot a function like this:2*atan(x-3)+3
Or wait also tanh(x) is good. I think is perfect for your case
Try to plot it
tanh(x-3/5)+1
Last edited by seerj (2005-12-29 03:21:16)
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Yes, yes.
And seerj's curves are close.
IPBLE: Increasing Performance By Lowering Expectations.
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