I am trying to find the formula to find the extremum points of a ax^3+bx^2+cx+d type function

[pineapplewithmouse]

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.

 

[Deleted member 4993]

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.

Cardano's equations for roots of cubic equation - Google it.

 

[pineapplewithmouse]

1. I didn't try to find the formulas for the roots
2. I didn't understand so well what the equation means

 

[Dr.Peterson]

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.

To find extrema, we use calculus, and it's not hard that way (just solving a quadratic).

You appear to be trying to do it with algebra only; can you show how you obtained your results?

 

[pineapplewithmouse]

Sure I guess.
I do not remember how I found the x value of the left extremum point, but I do remember how I found the y value:
I took the equation ax^3+bx^2=y, and did a chart in which I wrote every y value of the left extremum point with different values for the parameters a and b each time.

Then I saw that there is a pattern:
The numbers in the second column are 4 times smaller than in the first column, the numbers in the third column are 9 times smaller than in the first column, the numbers in the forth column are 16 times smaller than in the first column and the same for the fifth column. I quickly understood that they are square numbers, and the formula had to do something with it.
Then I saw almost a similar pattern but in rows:
The numbers in the second rows are 8 times larger than in the first row, the numbers in the third rows are 27 times larger than in the first row etc.
And these numbers were cubic numbers, so the formula had to do something with it too.
And after some thinking and trying, I came up with the formula 4b^3/27a^2.

When I tried to add the parameter c into the formula, it caused problems, but after a little googling I found that it has something with √3.
And after a lot of pure guesses and a little bit of thinking, I came up with the formula ((4b^3+6√|3*c^3|)/27a^2)+d.
Which unfortunately, doesn't work perfectly, because for some reason it works only when c≠0 and b=0 or when b≠0 and c=0, and I have no idea why.

 

[Dr.Peterson]

I do not remember how I found the x value of the left extremum point, but I do remember how I found the y value:
I took the equation ax^3+bx^2=y, and did a chart in which I wrote every y value of the left extremum point with different values for the parameters a and b each time.

Then I saw that there is a pattern:

Looking for patterns in numbers can sometimes work, but it is not nearly as effective as doing actual algebra.

I first want to point out that if you have found the x-coordinate of a point on a graph, all you need to do to find the y-coordinate is to "plug in" the x in the equation. It is not necessary to separately look for a pattern in y.

Here is what we would do using calculus:

Given [imath]y = ax^3+bx^2+cx+d[/imath], the derivative, which gives the slope of the curve, is [imath]y' = 3ax^2+2bx+c[/imath]. At any maximum or minimum (and sometimes at other points), this will be zero; so to find the extrema, we just solve [imath]3ax^2+2bx+c = 0[/imath] for x. Using the quadratic formula, [math]x=\frac{-2b\pm\sqrt{4b^2-12ac}}{6a}=\frac{-b\pm\sqrt{b^2-3ac}}{3a}[/math]
I would not bother substituting this into the equation for y to get a formula for the y-coordinate, but just evaluate it and put the resulting number into the equation.

Applying this to the special case you considered, [imath]f(x) = ax^3+bx^2+d[/imath], so that c=0, we get
[math]x=\frac{-b\pm\sqrt{b^2}}{3a}=\frac{-b\pm b}{3a}=0\text{ or }\frac{-2b}{3a}[/math]which is just what you found. And substituting this into the formula, we get
[math]y=a\left(\frac{-2b}{3a}\right)^3+b\left(\frac{-2b}{3a}\right)^2+d=\frac{-8ab^3}{27a^3}+\frac{4b^3}{9a^2}+d=\frac{-8b^3+12b^3}{27a^2}+d=\frac{4b^3}{27a^2}+d[/math]which is what you found by guessing.

 

[JeffM]

As Dr. Peterson explained, you are retracing the history of mathematics in the 16th and 17th century.

The quadratic formula, which was known long before 1500, gives the roots of a quadratic function, and the location of the extremum falls midway between the roots. During the 16th century, the formulas for finding the roots of arbitrary cubic and quartic functions were developed in a master feat of algebra. Not until the 19th century was it shown that no such formula exists for arbitrary polynomials of degree greater than 4.

A method for finding extrema (calculus) was developed in the 17th century. In essence, finding the locations of extrema of a polynomial of degree n involves finding the roots of a related polynomial of degree (n - 1) called the derivative.

To find the locations of the extrema of

[math]p(x) = \sum_{j=0}^n a_jx^{(n-j)}, \text { where } a_0 \ne 0[/math]
you need to find the roots of the derivative, which equals

[math]\sum_{j=0}^{n-1} (n-j)a_jx^{\{(n-1)-j\}}.[/math]
For example, the derivative of

[math]ax^3 + bx^2 + cx + d[/math]
is

[math]3ax^2 + 2bx + c.[/math]
The roots of that derivative are

[math]x = \dfrac{-2b \pm \sqrt{4b^2 - 12ac}}{6a} = \dfrac{- b \pm \sqrt{b^2 - 3ac}}{3a}[/math]
If [imath]b^2 - 3ac \le 0[/imath], the cubic has no finite extrema.

 

[Deleted member 4993]

......calculate the intercept with the x/y axis

I suggested Cardano's formula because you wanted to find x-intercepts of a cubic.

 

[pineapplewithmouse]

I first want to point out that if you have found the x-coordinate of a point on a graph, all you need to do to find the y-coordinate is to "plug in" the x in the equation. It is not necessary to separately look for a pattern in y.

I know that, but I just wondered if there is an equation for that.

And I still didn't learnt calculus in school, but I wanted to try starting learning it myself with the help of the internet, probably it will make my life easier while trying finding answers to questions like that.
Thank you