Do quartics have any shadow functions? Examples?

Question

Do quartics have any shadow functions? Examples? <I tried graphing quartics that I guess are shadow functions of each other but they don't look like how I thought they would based on shadow functions of cubics at all>

anonymous · Answer

Is this something to do with finding complex roots of a polynomial?

anonymous · Answer

Yes. =D I'm quite unsure about how the graphs turned out here as they don't seem to reflect each other(if they even have to), and hence I do not know what to make of the shadow generating function. T_T

anonymous · Answer

Hmm....sort of reversed and reflected....

anonymous · Answer

Well, normally when I do this for cubic functions, the shadow generating function would be a linear line draw through the intersections of the pair of shadow functions. But for this one, they won't make up a straight line, seems like the shadow generating function has to be a quadratic for quartics?

anonymous · Answer

Not even sure if they are a pair of shadow functions... = =''

anonymous · Answer

I hadn't really heard about shadow functions before:

If you had a cubic (x+2)(x-(3+2i))(x-(3-2i)  then how would you get the shadow function?

anonymous · Answer

That screenshot looks good, sort of rotated out of the page....

anonymous · Answer

Sort of like multiplying by i.....

anonymous · Answer

What i've found out, for cubic functions, is that the shadow function for (x+2)(x-(3+2i))(x-(3-2i)  is (x+2)(x-(3+2))(x-(3-1)), which is (x+2)(x-5)(x-2)
So, it seems like pulling the "i" off the other function

anonymous · Answer

sorry (x-(3-2))*****

anonymous · Answer

I guess you have to see what happens with quadratic and cubic, then extend the principle to quartic...

anonymous · Answer

I am not familiar with the "rules" of this exercise, I would have to do a study and see.....

anonymous · Answer

It seems like trying to do an Argand diagram but in the same plane as the usual Cartesian.....

anonymous · Answer

So does your screenshot identify the complex roots or not?

anonymous · Answer

Yep, so when I graphed it, I had to expand, say, (x-(3+2i))(x-(3-2i)) to 
x^1-6x+9^2+2^2 first, otherwise the geogebra wouldn't let me graph anything = =''

anonymous · Answer

no they don't

anonymous · Answer

I don't know what to say, really.

Maybe draw some diagrams for quartics with 2 real and 2 complex roots, find those roots in the usual way and mark them on the usual Cartesian diagram as if it was an Argand diagram and see what turns up....

anonymous · Answer

Actually, I am not even entirely sure of the point of the whole exercise at all..:-)

anonymous · Answer

Thank you anyways. =)
I think I've kinda found out what to do now ;D

anonymous · Answer

OK, good luck!