vertex formula of a cubic curve

I will show you how to derive the vertex formula of a cubic curve. This is an introduction to calculus derivatives. Notice how this vertex formula looks quite similar to the quadratic formula. enjoy!
You can also check out part1 here: • how to find the vertex...
For more calculus tutorials, check out my new channel ‪@bprpcalculusbasics‬
👉 / justcalculus
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[@Koisheep]

I was like "he's not gonna set C=c right?" He did.

[@agfd5659]

madlad

[@PontusLarsson1]

It does not matter in this case, since we are strictly talking about cubic polynomials, but it should be noted that f''(x) = 0 does not in general mean the function has a saddle point. Consider for example f(x) = x^4.

[@blackpenredpen]

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

[@Ocklepod]

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

[@PontusLarsson1]

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.

[@hybmnzz2658]

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

[@MarcoLiedekerken]

Something extra: The left 3 graphs have an "a" which is positive. The right 3 graphs have an "a" which is negative.

[@DylanCJ9]

That's common sense.

[@matthewlee5677]

Wow man, you really are dedicated. You remade the whole video because of your somewhat failing microphone. Props to you, you're great!

[@blackpenredpen]

Thank you!!!!!!!!

[@Abdega]

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

[@HrushikeshNaik1650763n73]

You explain graphs very well. Can you do a video seperately just on graphs for various functions.

[@Cloud88Skywalker]

The local min is the point where the positive square root is used. The local max is the point for the negative square root. Because, for x = (-b ± √(b^2-3ac))/3a; y" = 6a ((-b ± √(b^2-3ac))/3a) + 2b = ± 2√(b^2-3ac)

[@hassanakhtar5112]

Excellent way ....Liked And Subscribeddddd...Best Of Luck

[@goodnessokoronkwo6773]

I didn't know math. I always want to skip math but now math is now important to me, I want to get to understand this formula but you're fast in teaching.

[@trueriver1950]

Another way to find which vertex is min and which is max is to consider the first derivative (gradient) where the curve changes from concave up to concave down. At this point the second derivative is zero, so using your expression for y" we get x=-b/3a. It's easy to remember as it is the equation we get if we ignore the square root in the vertex eqn. So set x=-b/3a in y' we get a number that is + or 0 or - If it is "+" the local max must be the right hand one If it is zero it's a saddle If it is "-" then the vertex on the left is the max

[@user-mx8sj1nc6v]

It is also nice to look at the 6 possibilities of a parabolla to the x axis .... and conclude the 3rd degree polinomial shape "above" them.

[@5767chinmay]

You should make more videos on Algebra topics 👍

[@blackpenredpen]

I plan to make the cubic formula soon

[@ssdd9911]

when?

[@fauxpassant]

@@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?

[@alexvarjabedian9207]

Wow I derived this!!!

[@diaries8568]

Love you sir🔥❤️

[@SebastienPatriote]

But if a is positive, the smallest x will be a local maximum, and vice-versa, why the need for the sexond derivative at all?

[@deepakjindal9874]

9:26 it not only depends upon the discriminant but also the value of b and c individually........ For First two graphs of having local minima and local maxima....... We must have also this conditions that:- 1) b should not be zero while 2) c could be zero And for last two graphs of strictly increasing or strictly decreasing We may can also say that :- 1)b should be zero while 2) c should be lesser than zero

[@davidseed2939]

Did you notice, that the equation for x can be rewritten to a form that shows that y”/2 is equal to the discriminant

[@mike4ty4]

The interesting question: 1. Are there "natural" equations in which a relevant parameter is x = (-b +/- sqrt(b^2 - 2ac))/(2a) ? 2. Are there, more generally, "natural" settings in which for some fixed N, or perhaps even N and M, both positive integers, a relevant parameter is x = (-b +/- sqrt(b^2 - Nac))/(Ma) i.e. either perhaps N = M or not? Let's explore "generalized quadratic formulae". :g:

[@trollme.trollmehard.9524]

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

[@tarat.techhh]

vintage sweater nice one mate

[@ghiabaghiaba4195]

What is the defrences betwine y- and dy/dx ? It s look like the same am I coorect

[@carultch]

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").

[@torresfan1143]

Was that the opening song of doraemon at the start of the video ?

[@willie333b]

Somehow I figured out a similar thing myself in high school

[@OonHan]

Firstly, great video. Secondly, the vertex formula is *very* similar to the quadratic formula

[@OonHan]

in fact, the derivative of the cubic is quadratic

[@OonHan]

SADDLE POINT = INFLECTION

[@muhammadwibowo1661]

Oon Han ikr

[@OonHan]

lol

[@muhammadwibowo1661]

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

[@ieatgarbage8771]

3ax^2+2bx+c=0 wow, that already looks like the quadratic formula (-2b±sqrt[4b^2-12ac])/6a (-b±sqrt[b^2-3ac])/3a And now I can’t even type this shid xd

[@mihaiciorobitca3343]

can be the x of vertex negative ?

[@reetasingh1679]

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

[@mihaiciorobitca3343]

Reeta Singh thanks

[@justinlewtp]

Never thought "algebra-student friendly" goes well with Calculus :)

[@lightofamphy6764]

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.

[@megacarlosloki]

At the first derivative, how a slope can be a second power function? A slope isnt a line? Sorry any english error, I am brazilian and dont speek very well. Also I love the channel.

[@megacarlosloki]

Entendi

[@waishingtseung6930]

3rd and 4th comp…Ans:bb=3ac

[@ameraibrahim1138]

Damn I randomly made that equation too today 😭😭😭

[@RajAdityaNag]

The world: Hmmm, since this guy is an asian he must be better than the rest of us in math Me(also am asian): I couldn't agree more with the world.

[@mikyjet065]

Why do you need the second derivative, can't you just figure it out based on if the A coefficient is negative or positive? Now correct me if I am wrong but, if the A coefficient is positive than the first vertex (from left) is going to be a local maximum. And when it's negative the first vertex (from right) is going to be a local minimum.

[@carultch]

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.

[@dewman7477]

Looks very similar to quadratic formula

[@OonHan]

if you dare, do the vertex formula of a quartic!!! ;)

[@OonHan]

Zacharie Etienne lol makes sense

[@pacolibre5411]

Will the answer ever not be -b/3a?

[@carultch]

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.

[@particleonazock2246]

9:01 Classic old timer. ISN'T IT???

[@noobmaster-dm7tu]

What is with the Doraemon music

[@pleindespoir]

The lowest pair of graphs cannot be a function of a 3rd grade polynomial function - ist HAS TO HAVE at least a saddlepoint or inflection point. The 2nd derivative must become zero at any value on the x-axis. If not, there is no cubic function. Am I right ?

[@thecubeur33]

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

[@thecubeur33]

Consider x^3+x !

[@pleindespoir]

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 :(

[@nestoreleuteriopaivabendo5415]

@@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.

[@hakimchulan]

Doraemon!!

[@OonHan]

oreo

[@dr.rahulgupta7573]

Plz put all zebras in one cage .DrRahul Rohtak Haryana India

[@muhammadwibowo1661]

Its too family friendly hahahaha

[@neuralwarp]

Oh gosh, the content was great, but I'd love to work with you on your English pronunciation skills.