pineapplewithmouse
Junior Member
- Joined
- Jun 22, 2021
- Messages
- 52
So because of my curiosity, I wanted to see if I could find by myself with my previous knowledge in math and with trial and error formulas that can calculate the intersect with the x/y axis and the extremum points of polynomial functions.
My findings:
f(x) = ax^0 - a very easy one
f(x) = ax^1+bx^0 - also, a very easy one
f(x) = ax^2+bx^1+cx^0 - the intersect with the y axis was easy of course, and the intersect with the x axis formula I learnt in school, same for the x value of the extremum point: -b/2a. But the y value of the extremum point formula I found myself: (-b^2/4a)+c
And then came the f(x) = ax^3+bx^2+cx^1+dx^0
I really didn't think it will be that hard. After hours of trial and error and trying of thinking of creative ways to find these formulas, this is what I came out with:
The intersect with the y axis - an obvious one
And then came the extremum points... I tried to use the least parameters I could at once so it will be more simple and these are the formulas for a f(x) = ax^3+bx^2+dx^0 function:
X value of the first extremum point: -2b/3a
Y value of the first extremum point: (4b^3/27a^2)+d
X value of the second extremum point: 0
Y value of the second extremum point: 0+d
It alone took a lot of time, and then came the parameter c...
After a really long time of trying I came up just with a Y value of the first extremum point formula:
((4b^3+6√|3*c^3|)/27a^2)+d
And this formula is not perfect either cause it works just when c≠0 and b=0 or when b≠0 and c=0, and I have no idea why.
And the c parameter affects the second extremum point in very weird ways, and all of this is very very difficult.
Can someone help me to understand what am I missing/tell me the formulas? I am really exhausted because of this thing...
And no, the internet didn't help at all (well, just by a coincidence it told me that I should look in the √3 area, which kinda helped), but no formulas like the ones I am trying to find.
My findings:
f(x) = ax^0 - a very easy one
f(x) = ax^1+bx^0 - also, a very easy one
f(x) = ax^2+bx^1+cx^0 - the intersect with the y axis was easy of course, and the intersect with the x axis formula I learnt in school, same for the x value of the extremum point: -b/2a. But the y value of the extremum point formula I found myself: (-b^2/4a)+c
And then came the f(x) = ax^3+bx^2+cx^1+dx^0
I really didn't think it will be that hard. After hours of trial and error and trying of thinking of creative ways to find these formulas, this is what I came out with:
The intersect with the y axis - an obvious one
And then came the extremum points... I tried to use the least parameters I could at once so it will be more simple and these are the formulas for a f(x) = ax^3+bx^2+dx^0 function:
X value of the first extremum point: -2b/3a
Y value of the first extremum point: (4b^3/27a^2)+d
X value of the second extremum point: 0
Y value of the second extremum point: 0+d
It alone took a lot of time, and then came the parameter c...
After a really long time of trying I came up just with a Y value of the first extremum point formula:
((4b^3+6√|3*c^3|)/27a^2)+d
And this formula is not perfect either cause it works just when c≠0 and b=0 or when b≠0 and c=0, and I have no idea why.
And the c parameter affects the second extremum point in very weird ways, and all of this is very very difficult.
Can someone help me to understand what am I missing/tell me the formulas? I am really exhausted because of this thing...
And no, the internet didn't help at all (well, just by a coincidence it told me that I should look in the √3 area, which kinda helped), but no formulas like the ones I am trying to find.