# Math Help - graph

1. ## graph

can someone help me find the eqaution of this graph

it has to be in the formatt of

y=e

2. Originally Posted by calem_123
can someone help me find the eqaution of this graph

it has to be in the formatt of

y=e
I don't know about anyone else, but I cant open the spreadsheet.

RonL

3. ## try this

ok well try this one out, its in word

4. ## Don't make multiple posts of the same question

Don't make multiple posts of the same question.

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

RonL

5. ok sorry man

just need it soon

thanks

6. Originally Posted by calem_123
can someone help me find the eqaution of this graph

it has to be in the formatt of

y=e
I have uploaded a corrected plot.

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

7. Originally Posted by calem_123
can someone help me find the eqaution of this graph

it has to be in the formatt of

y=e
Nobody will be able to fine 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:

$
y=e^{f(x)}
$

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

RonL

8. yeah i need an an equation which approximates the graph

$
y=e^{f(x)}
$

yes it has to be in that form

any help wil be greatly appreciated

9. Using the excell solver to fit an equation of the form:

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

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

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

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

RonL

10. Originally Posted by CaptainBlack
Using the excell solver to fit an equation of the form:

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

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

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

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

RonL
Oops, the $a$ term is redundant, this is equivalent to:

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

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

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

(The rational function approximation is better though)

RonL

11. 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

12. Originally Posted by calem_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
Choose three points on the graph $(x_1,y_1),\ (x_2,y_2),\ (x_3,y_3)$, then you have a set of equations:

$y_1=e^{a+bx_1+cx_1^2}$
$y_2=e^{a+bx_2+cx_2^2}$
$y_3=e^{a+bx_3+cx_3^2}$,

now take logs of these equations:

$\log(y_1)=a+bx_1+cx_1^2$
$\log(y_2)=a+bx_2+cx_2^2$
$\log(y_3)=a+bx_3+cx_3^2$.

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

RonL

13. ok thanks a lot

ur a life saver

14. crap, using that way results in the equation being way off

thanks though

15. Originally Posted by calem_123
crap, using that way results in the equation being way off

thanks though

If you have done it right the plot of the equation should pass through
the points you selected though.

RonL