Could that be done via exporting/importing the coordinates/area to/from an external script?]]>

I am able to shade a polar-function circle centred at the origin (eg r(t)=1). I am also able to construct a relation circle (eg (x-2)^2+(y-2)^2=1) away from the origin, but then Insert Shading is greyed out. How do I shade a circle not centred at (0,0)?

Thanks for any guidance!

]]>Thanks.

Me, not being able to use your programs settings!

Actually there are 2 Real roots (2 & 3) and 2 Irrational roots (1.8189364... & 3.61800635...)

If there's any way I can help please let me know (but remember I'm 83 and slowing down!)

Regards

Pete]]>

Hi. I am using the Beta version 4.5, build 584. When I input x(t)=re(3-2i) and y(t)=im(3-2i) as a parametric function nothing appears on the graph. I would have expected to see some sort of marker at x = 3 and y = -2. Is this a problem at your end or am I doing something wrong?

chris

]]>I actually need someone to find bugs in Graph. Otherwise they don't get fixed. And you do provide some interesting test cases.

]]>If you've time & patience would you check this equation out for me please.

It's a X^4... equation which can only give 4, 2 or 0 roots.

On the Graph it gives 3 !!

The equation is : x^4 - 9.9999x^3 + 35.9992x^2 - 54.9979x + 29.9982

I'm NOT deliberately looking for problems in your program - honest!!!

Regards

Pete

I'd like to be able to have x and y axis numbering as "1/2", or "2/3", etc.. Is it possible to do this? If not, is it possible to set a number of decimal places to x and y numbering such as having "1/3" be shown as "0.33". Either way could you consider these for the next version of Graph?

Thanks

chris

]]>I recently ran my program (Find factors) with

x^4 - 10x^3 + 36x^2 - 55x + 31

and it came up with Q"No solutions" (Real or irrational)

so naturally I went to Graph.

Initially it seemed to concur but (very briefly) I saw a very narrow "spike"

but I was not able to capture it again.

It was asymptotic, and almost vertical.

Would you be so kind and see if you can find it ?

It appeared to be around x = 1.01652...

A close equation with the constant as 30 gives :

(x - 2)(x - 3)

2 approximate (irrational?) roots

x1 = +1.381966011250 f(x1) = +0.000000000000239808173319034

x2 = +1.381966011251 f(x2) = -0.00000000000200017780116468

x1 = +3.618033988749 f(x1) = -0.00000000000200017780116468

x2 = +3.618033988750 f(x2) = +0.00000000000019984014443252

Regards

Pete

https://www.padowan.dk/doc/english/EvaluateDialog.html]]>