vertex formula of a cubic curve 

Ah, that's right!! Thankfully this is true for cubic.

how do we know it's always true for all cubic polynomials there are?

Assuming you have some familiarity with calculus, the simplest ways to do this is either to check whether the second derivative changes sign in a neighborhood of the critical point, or by using the general derivative test. For a proof of the former, let f(x) = ax^3 + bx^2 + cx + d. We know that f(x) has a critical point x* that satisfies f'(x*) = 0 and f''(x*) = 0. Consequently, for ε > 0, we get f''(x* + ε) = 6a(x* + ε) + 2b = 6ax* + 2b + 6aε = f''(x*) + 6aε = 6aε, f''(x* - ε) = 6a(x* - ε) + 2b = 6ax* + 2b - 6aε = f''(x*) - 6aε = -6aε, which means that the second derivative does change sign in the neighborhood of the critical point i.e. x* is an inflection point. Thus, x* is a stationary inflection point, or in other words a saddle point. And we are done.

Nice. It is analagous to how f'(x) = 0 does not always imply a local min/max.

That's common sense.

Thank you!!!!!!!!

He could have just dubbed it over in post but blackpenredpen doesn’t do things halfway!

madlad

I plan to make the cubic formula soon

when?

@@blackpenredpen hmmm... will you be using the method to convert any cubic equation into the form x^3 + ax + b = 0 and solving from there?

Are you still around? I know the answer to this question.

The bigger question is whether or not there *is* a local maximum and local minimum to the cubic. You are correct that *if* there are two turning points of a cubic equation, that the sign on A will tell you whether the local maximum comes before the local minimum for positive, or vice versa for negative. But, in the event that there are no real values of x at which the cubic has a local extreme point, then it's a moot point to find the vertices of the cubic. The sign on A ultimately tells you whether concave-down curvature comes before concave-up curvature for positive, or vice-versa for negative. You get the "roller coaster cubic" with two turning points, when the derivative at the inflection point, is of the opposite sign to the 3rd derivative (as indicated by the sign on the A-term). When the inflection point coincides with a stationary point, or when the 3rd derivative is of the same sign as the derivative at the inflection point, there are no local extreme points.

in fact, the derivative of the cubic is quadratic

SADDLE POINT = INFLECTION

Oon Han ikr

lol

Oon Han i've seen your video. Good job kid.

Two different notations. I assume you mean y', and not y-. y' is Lagrange's notation (pronounced y prime), and dy/dx is Leibnitz's notation (pronounced "dy dx", or "dy on dx").

Entendi

Had no idea what the hell calculus was going into this, still don't know, but I was able to grasp what a derivative is and how it could be useful for these equations with the help of these videos. So for sure, it is algebra-student friendly.

Mihai Ciorobitca The x coordinate of the vertex, you mean? Of course it can...

Reeta Singh thanks

The inflection point will always be at x=-b/(3*a). This is the first step in Cardono's method for simplifying any cubic to become a depressed cubic (one with no x^2 term). You shift the cubic either to the left or right, so that the b-term disappears. This will mean shifting it by a distance of b/(3*a). You then can divide through by the a-term, and reduce it so there are only two coefficients to uniquely determine the solutions. They usually call them p and q in the textbooks. So you reduce the cubic from the form of: 0 = a*x^3 + b*x^2 + c*x + d to: 0 = t^3 + p*t + q You do this by letting t = x - b/(3*a) for your change of variables, to eliminate the x^2 term. Now you only require two inputs to the cubic formula, instead of four.

Zacharie Etienne lol makes sense

No you aren't, not all second degree polynomials have real roots ;) Fullofhope :)

Consider x^3+x !

OK - thank you ! A point of inflection stays, but slope needn't to become zero. That makes the graphs looking like the lowest pair. My hope to be a genious is destroyed again :(

@@pleindespoir never think about yourself like that. You just overcome the doubt by putting it on the table and asking for it. There's nothing wrong with it, in fact it is good for anyone who is trying to be a genius. Genius is 99% hard work, only 1% talent.