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krassi_holmz
2005-12-30 08:25:16

Yes, yes.
And seerj's curves are close.

seerj
2005-12-30 02:16:45

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

seerj
2005-12-30 02:12:30

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

mathsyperson
2005-12-29 23:43:13

kdria041 wrote:

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.

krassi_holmz
2005-12-28 23:46:03

arctan(x) also has two horizontal asymptotes.
But I think that these functions cannot be adapted for this graph, because it is not symmetric.

MathsIsFun
2005-12-28 17:00:31

The tanh(x) function has two horizontal asymptotes, it could be adapted possibly.

John E. Franklin
2005-12-28 11:01:48

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

kdria041
2005-12-28 08:16:37

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: