I have a math question

[marciosmg]

zexen_lowe said:

ctk495 said:
y=mx.c

Or atlest that's what I think.

That's for lineal equations

marciosmg said:
y = ax2 + bx +c

xv = -b/2a
yv = -(b2 - 4ac)/4a

v = vertix (in your pics case xv = yv = 0)

NOw, for every (x,y) you have, you throw it in the y = ... equation to find a, b and c

Then you will get the whole y = ... eqaution which is the formula of that parabole (I don´t the name in english - that shape)

That's right, but here's the easier form

a= (y-yv)/(x-xv)^2

b= -2a*xv

c= yv + a*xv^2

Maybe you are right, but I have been doing t my way for so long that your way seems weird to me. So, I will stick to mine, thank you very mcuh.