can someone help me find the eqaution of this graph

it has to be in the formatt of

y=e

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- Apr 20th 2006, 03:42 AMcalem_123graph
can someone help me find the eqaution of this graph

it has to be in the formatt of

y=e - Apr 20th 2006, 03:47 AMCaptainBlackQuote:

Originally Posted by**calem_123**

RonL - Apr 20th 2006, 03:59 AMcalem_123try this
ok well try this one out, its in word

- Apr 20th 2006, 04:43 AMCaptainBlackDon't make multiple posts of the same question
Don't make multiple posts of the same question.

This is the sub-forum that you will get the best response from. If

a question is urgent post it here, and don't post duplicates in the

other fora.

RonL - Apr 20th 2006, 04:47 AMcalem_123
ok sorry man

just need it soon

thanks - Apr 20th 2006, 04:56 AMCaptainBlackQuote:

Originally Posted by**calem_123**

Try to avoid using proprietary file formats like Excel and Word, Not everyone

has these on their machines. Try to stick to something like .JPG, .PNG

graphics files.

RonL - Apr 20th 2006, 05:07 AMCaptainBlackQuote:

Originally Posted by**calem_123**

**the**equation of the graph.

They will be able to find an equation which approximates the graph.

Now your statement about the required format of the equation is not

clear, I take it to mean that you want an equation in the form:

$\displaystyle

y=e^{f(x)}

$

for some function $\displaystyle f$ of $\displaystyle x$. Is that right?

RonL - Apr 20th 2006, 05:13 AMcalem_123
yeah i need an an equation which approximates the graph

$\displaystyle

y=e^{f(x)}

$

yes it has to be in that form

any help wil be greatly appreciated - Apr 20th 2006, 05:44 AMCaptainBlack
Using the excell solver to fit an equation of the form:

$\displaystyle

y=K e^{a+bx+cx^2}}

$

to a selection of data points from this curve I get:

$\displaystyle

y=25e^{1.098-0.02429x+0.000256x^2}

$

but this is not as good a fit as I would like to see.

RonL - Apr 20th 2006, 07:48 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

$\displaystyle

y=74.95e^{-0.02429x+0.000256x^2}

$

or absorbing the $\displaystyle K$ instead of the $\displaystyle a$ term:

$\displaystyle

y=e^{3.5343-0.02429x+0.000256x^2}

$

(The rational function approximation is better though)

RonL - Apr 20th 2006, 09:55 PMcalem_123
ok thanks very very much

i need to know how to find the equation using algabra

could u explain how to figure it out please, using algabraic methods - Apr 21st 2006, 12:33 AMCaptainBlackQuote:

Originally Posted by**calem_123**

$\displaystyle y_1=e^{a+bx_1+cx_1^2}$

$\displaystyle y_2=e^{a+bx_2+cx_2^2}$

$\displaystyle y_3=e^{a+bx_3+cx_3^2}$,

now take logs of these equations:

$\displaystyle \log(y_1)=a+bx_1+cx_1^2$

$\displaystyle \log(y_2)=a+bx_2+cx_2^2$

$\displaystyle \log(y_3)=a+bx_3+cx_3^2$.

This is a set of three linear simultaneous equations in variables $\displaystyle a,\ b,$ and $\displaystyle c$, which can be solved by the usual methods.

This method is not advised.

RonL - Apr 21st 2006, 02:01 AMcalem_123
ok thanks a lot

ur a life saver - Apr 21st 2006, 07:09 PMcalem_123
crap, using that way results in the equation being way off

thanks though - Apr 21st 2006, 10:10 PMCaptainBlackQuote:

Originally Posted by**calem_123**

If you have done it right the plot of the equation should pass through

the points you selected though.

RonL