I found another way to justify the third fact: a parabola of the form ax^2+bx+c is just a translation of the parabola ax^2 (completing the square makes this explicit) whose turning point is at x=0 and whose value at x=1 is a.

@zeyads.el-gendy4227

2 жыл бұрын

Nice

@Mateusz-Maciejewski

2 жыл бұрын

Yes, I just wanted to write that in order to prove the last part it's better to use the vertex form y=a(x-p)^2+q and this is what you've written about translating the parabola y=ax^2. Btw, when I read "fact you probably didn't know" I thought "are you kidding", but I really didn't know the second fact about b.

@bitti1975

2 жыл бұрын

Yes, when so many things cancel out, it's a good hint that there is an easier way. Still I wonder how to make this rigorous while still keeping it simple?

That mic(pen) drop at the end though! Really interesting facts, thank you, Dr. Peyam

@alwysrite

2 жыл бұрын

Dr Peyam, some advice - once you drop the "mic" you do not speak any more !

@AlphaNumeric123

2 жыл бұрын

@@alwysrite Well I suppose you could say this is a pen drop, distinct from a mic drop, because he was finished writing on the board and so drop the pen. What would be real classy is to drop the pen after you're done writing, and then drop the mic after you're done talking. In fact, I suggest we make this a cultural norm with everything we do.

@alwysrite

2 жыл бұрын

@@AlphaNumeric123 ; )

Another cool fact: a affects concavity, c affects the y-intercept but what does b affect. It turns out that changing b translates the parabola along the path y=-ax^2+c.

this is wonderful! When I'm helping people with quadratic formula, I sometimes like to show them where -b/(2a) and ±sqrt(Δ)/(2a) lie in the graph. The former is the x-coordinate of the vertex and axis of symmetry, while the latter is the distance from the axis of symmetry to each x-intercept.

@antoniopedrofalcaolopesmor6095

2 жыл бұрын

Wow, that's awesome, I never thought of the quadratic formula this way rather than just memorizing and using it mechanically, but you are entirely and absolutely right, thanks for sharing it.

I'm watching it again and now I see ... Sending a kiss for such a straight line 😅 I love you dear Peyam

@ClearerThanMud

2 жыл бұрын

That was a very impressive tangent line; I marveled at it too!

@user-wu8yq1rb9t

2 жыл бұрын

@@ClearerThanMud me too 😀

@antoniopedrofalcaolopesmor6095

2 жыл бұрын

@@ClearerThanMud Indeed, same for me

Simple yet interesting. Thanks sir for making my day. I would have never thought of parabolas in this manner. Looking forward to more such cool videos from you.

Hi Dr. Peyam, Thnx! Reading off parameter 'b' directly from the graph is new to me. Reading off parameter 'a' can be explained in an another way: Any graph can be obtained by transforming the standard graph (with relation R(x,y) or function y=f(x)). Transformations are: scaling, translating and/ or rotating. e.g. any (non-rotated) ellipse is obtained by scaling and translating the standard circle x^2+y^2=1. If you translate the function y=f(x) (or relation R(x,y)) along (a,b) you get y-b = f(x-a) (or R(x-a,y-b)). If you scale y=f(x) (or R(x,y)) with (alpha,beta) you get y/beta = f(x/alpha) (or R(x/alpha,y/beta)). The parabola y = ax^2 + bx + c is obtained by scaling the standard parabola y=x^2 with (1,a) and then translating it. So reading off 'a' has become obvious now. TJ

Great, as usual 😊😊

Excellent explanation Dr Peyam . Thanks .

That is really cool! Awesome job!

Love the video! And also the pen drop in the end🔥

Very interesting! Thank you for the facts :)

That is great! Thank you for this video

Excellent - thank you!

Very nice. A practical problem is that constructing the tangent line on a drawing for determining b could be a bit imprecise. For greater precision you could also use the fact that b is also equal to (f(1)-f(-1))/2 or more general b=(f(x_1)-f(-x_1))/(2 x_1). On a given graph it is also not easy to determine where exactly the maximum is situated, which makes determining the coefficient a imprecise. The following fact gives you a more precise result: a=(f(1)+f(-1))/2-c or more general: a=((f(x_1)+f(-x_1))-2c)/(2 x_1^2)

@bendriver3242

2 жыл бұрын

Nice. Your result for b follows because the tangent to a parabola at any point P is always parallel to any chord joining points on the parabola which are equally spaced horizontally on either side of P.

Dude dropped the chalk like a boss.

It feel enlighten. Many "simple" facts connected after watching the video.

Actually I feel like buying a big chunky notebook and write down such amazing stuff lol

bloody brilliant

The behaviour of the first coefficient is very obvious in my opinion. The slope of the tangent is new to me. Thanks for that.

@chessandmathguy

2 жыл бұрын

The behavior might be obvious... The clear cut proof, I'm not sure if it's as obvious.

@easymathematik

2 жыл бұрын

@@chessandmathguy The proof is very trivial.

Nice video. Excellent tangent.

A much more elegant method is to think of the parabola as the path of a projectile in frictionless air. In that context the ax^2= (1/2) * (2a) * x^2 term represents the vertical distance the projectile falls below the tangent line. At the maximum, i e. the topmost point of the trajectory the tangent line is horizontal so that in unit time the distance fallen equals (1/2)*(2*a) *(1^2) = a. QED.

@MichaelRothwell1

2 жыл бұрын

Yes indeed, where the 1st derivative is zero, the changes are due to the 2nd derivative (2a).

Lovely. Thanks for sharing 👍

Beautiful. Just beautiful.

Yhay was a quite nice video about parabolas. It is interesting that you think you know all about them, somebody genius like Prof. Peyam can open your eyes a little bit more! Thank you!

There’s always something new to learn on a simple parabola, thanks👍😘

Thank you, this keeps my brain going!

Yes, I did not know ! Thankyou.

Interesting fact about the linear coefficient b that I didnt know. Please, keep posting these facts. They are useful to many of us.

@MrBeen992

2 жыл бұрын

@래오산 that would give c, not b. I think you meant f ' (0). No, it never crossed my mind to calculate the slope at the y intercept. Have you ?

wow this was fascinating.

I did like this! And I subscribed! This may be useful in position vs time with uniform acceleration graphing in physics class

Wow thats pretty cool!!

Thank you.

Fabulous......fact... Thanks❤❤❤

Damn the quadratic coefficient thing is neat. Dr. Peyam if you don't mind who is the original Twitter source you found this from?

Ok, that pen drop was really Badass

When you want to say anything could happen but differently : 0:09

I wonder what other things some of us still don't know about parabolas. Great vid.

This is so cool

Shoulda been some gangster music at the end.

Interesting!

I first read the title as “a parabola fact you *parabobly* didn’t know”

@drpeyam

2 жыл бұрын

Haha

much content such quality, wow

Great stuff! And a very interesting view of that equation and its coefficients! (Is there something equivalent for the cubic form, ax^3+bx^2...?) I'am a Maths teacher and even didn't know this. I will teach it to my students.

@easymathematik

2 жыл бұрын

Some properties are easily to generalize. Consider a n degree polynomial a_n * x^n + ... + a_1 * x + a_0 The slope of the tangent in P ( 0 / a_0) is a_1 For quadratic polynomials we get P (0/c) and slope b For linear polynomials y=mx+ b we get of coursre P(0/b) and slope m. Polynomials of higher degree are very oscillating and it is very hard to classify them. Coefficients of order higher > 1 affects extremal points and turning points. If you calculate d^k /x^k p( x_critical) you get some wild terms involving polynomials in the coefficients. For example the slope of the tangent in the turning point of a cubic ax^3 + bx^2 + cx + d is (3ac - b^2 )/ (3a) Where you can see in this case the similarity to the quadratic. Here it is easy because a cubic has exactly one turning point. For cubic you have for example the following nice property. Let x_1, x_2, x_3 be three real roots of f(x) = ax^3 + bx^2 + cx + d Consider following tangent line: x_0 = (x_i + x_j)/2, i,j = 1,2,3 and i and j distinct. y = f ' ( x_0 ) (x - x_0) + f(x_0) What is the zero of this tangent? It is the remaining zero. :)

Nice to learn something new about parabolas. But I think it’s even nicer: y=bx+c is the tangent line at x=0, a is the difference between the parabola and the tangent line at unit distance from the tangent point

@drpeyam

2 жыл бұрын

That’s nice too!

upvoted for the kiss !

I did not only know this, I've actually taught this already several times to my students. :) And even more: For a line, the equation for the slope is (Delta y) / (Delta x). For a parabola, the equation is quite similar: a = (Delta y) / (Delta x)². You only have to pay attention that you start at the vertex of the parabola.

These results are very intuitive if we consider the Taylor expansion of the quadratic f(x)=ax²+bx+c. Since f is a quadratic, its second order Taylor polynomial (about x = d) will be exact: f(x) = f(d) + f'(d)(x-d) + f''(d)/2×(x-d)² Note that f'(x) = 2ax+b, f''(x) = 2a, so, substituting just for f''(d), we get: f(x) = f(d) + f'(d)(x-d) + a(x-d)² For c and b, we just consider d = 0, i.e. the second order Maclaurin polynomial, which is f(x) rewritten slightly as f(x) = c+bx+ax², to get c = f(0) (the y-intercept) and b = f'(0) (the gradient at the y-intercept). Now let's think about the value of a. Notice that the first two terms of the Taylor expansion, f(d) + f'(d)(x-d), give the best linear approximation to f(x) at x=d, i.e. the tangent at that point. So quadratic term in the Taylor expansion, a(x-d)², gives the amount by which the curve lies above the tangent. So if x=d±1, this amount is a(±1)²=a. In other words, a is the amount by which the quadratic lies above its tangent at a point at a value of x that is 1 to either side of that point. This applies, for example, at the y-intercept, but becomes simpler where the first derivative vanishes, i.e. at the vertex (where x=d0=-b/(2a), though we don't need the value for this argument). So about the vertex x=d0, we get simply f(x) = f(d0) + a(x-d0)², in particular f(d0±1) = f(d0) + a(±1)² = f(d0) + a, and so a is the amount by which the quadratic lies above its vertex at a value of x that is 1 to either side of the vertex. Although we certainly don't need Taylor series to prove these results, I think they give a nice insight into what's going on.

"a parabola might or might not change your life" Climb up to the roof, run, jump, the trajectory is a parabola, and it will change your life, in ___ seconds

@drpeyam

2 жыл бұрын

Hahaha

@fabiopilnik827

2 жыл бұрын

So far I´ve noticed the universe begins with two quantities, call them the Yin and the Yang, seperated by a distance of a constant c. Velocity and not necessarily geometry is endogenous to gravity: stationary masses cannot afford gravity. I´m holding my breath with great pause before Dr. Peyam gives me what the "a" stands for.

can u show an example?

@drpeyam

2 жыл бұрын

x^2 + 5x -7

Hello Dr. Peyam. *Marker Drop* ....Nice

I did discovered this in junior high And i have been using this for all math and physics problems for a while now

You want the min or max of the parabola y = ax^2 + bx + c. The parabola's extrema is located at x = -b/(2a). To get the the value of the extrema, you can compute the y value which results from plugging -b/(2a) in for x. But, there is a short cut. Instead of plugging in x = -b/(2a) into the full equation y = ax^2 + bx + c, you can instead plug x = -b/(2a) into y = (1/2)bx + c. Why it works: y = ax^2 + bx + c at ax = -b/(2a) -> y = a(-b/(2a))^2 + b(-b/(2a)) + c -> y = a(b^2/(4a^2)) - b^2/(2a) + c -> y = b^2/(4a) - b^2/(2a) + c -> y = b^2/(4a) - 2b^2/(4a) + c -> y = -b^2/(4a) + c -> y = (1/2)b(-b/(2a)) + c which is y = (1/2)bx + c at x = -b/(2a). ■

Such a great pen drop there.

Wow, that was neat

Something I find amusing is that the third fact - about a - can also be translated into the language of discrete derivatives: that is, the discrete derivative at the vertex is a.

Clever!

Yay! A new parabola fact!

Good to know

4:10 keeping it real, Dr. Peyam!

What a tangent man.... People may solve Riemann hypothesis but such a tangent is happens once in century....

Not gonna lie, I thought "yeah right, something I've never heard about parabolas". Liked and subscribed sir.

The way I would describe that third fact is to say that A is the slope of the secant line through the vertex that has a width of one

@sharpnova2

2 жыл бұрын

you could think of it as kind of like a unit secant line

@sharpnova2

2 жыл бұрын

but actually maybe it would be more appropriate to reserve that term for a secant line that had a length of one

I read the title as, "A probability fact you probably didn't know" and thought that it was a nice wordplay...

You can speak faster than I can think!!! Brilliant!!!!!!!!!!!

Thanks for the video. While watching the video I had that idea, "a" is half the second derivative of the function.

now do the Carlyle Circle!

#3 is quite a nice property, never thought of the first coefficïent like that.

I like the mic drop

nice

Thanks

@drpeyam

2 жыл бұрын

Thanks so much!!!! I really appreciate the super thanks, it means a lot to me

@nellwackwitz

2 жыл бұрын

@@drpeyam I adore your videos Dr. Peyam! When I feel overtired from teaching, I know I can watch any video that you have made, and then I can grade all night and face my wonderful students with a smile! It’s all because of you! ❤️😄

@drpeyam

2 жыл бұрын

Awwwwww thank you!!! This warms my heart 😌

I also saw the post a few days ago as you retweeted it. @chessematics

First time I have seen a pen drop!

Can you explain in more detail, maybe in another video, why the b value is the slope of the tangent line evaluated at 0? It doesn't make intuitive sense to me.

@drpeyam

2 жыл бұрын

The definition of derivative *is* slope of tangent line

@BobbyMack

2 жыл бұрын

@@drpeyam Why is the b-value the tangent evaluated at 0? Why not evaluated at some other point?

@drpeyam

2 жыл бұрын

You could do it at another point, but the answer would be more complicated, not just “b”

Wow! I used this way of explaining all the time in straight lines, but I hadn't realized it for the 'b'. I think I'm gonna use this! The a could be a bit hard for 2nd graders, (with b/2a for the already).

I have literally been looking for this information forever. Now a b and c mean something (mostly b)

Life changed ... hehehehe ... nice ... cheers.

Good job, may I suggest that when you do a proof you have them already generated on white board and reveal them one line at a time. And drive home the point when setting x= zero. Dr. P great job, i'd like to see something like it for cubics or quartics. Have a wonderful day sir.

@drpeyam

2 жыл бұрын

I’ve tried that, people didn’t like it! But thank you :)

Share the twitter post next time as well, please =-)

Sweet

Thats fucking nuts

Another Interesting Fact: It's just a jump to the left, And then a step to the riiiiiiiggght Put your hands on your hips, And then bring your knees in tiiiiiiiiggght...

Thank u Always some new views with u! Very good but note that it can be explain by only translation of: y=ax² graph to considered y=ax²+bx+c..Graph . To say translation of sommit vertex S=(0;0) to S'=(-b/a; ∆/4a) by vector translaton=[-b/a; ∆/4a] All proprieties of sommit vertex S are herited by S' after translation M (1;a) is translated to M' [-b/a+1;(∆/4a)+a)] We don't need the values calculated neither in your post nor i calculated... We only observe: as the sommit vertex S goes vertically = a unit ordinate when it abscissa goes horizontally =1 unit so S' as well as do

the fist fact is well known. the second: the slope was nice. and three is trivial (take the normal parabola and stretch it by a.) so the point 1/1 gets stretched to 1/a. End of proof. This fact actually only helps in the vertex form. In the form ax^2 +bx +c its useless, unless you first calculate the vertex using Mitternacht (Midnight) formulas. And last but not least: I would cut my students points for drawing a parabola which is not even a function ;-) Hey: This is an instructional video. At least set some standards for the quality of visualization.

Great video! Though I knew all of this already, it's nothing new, but still helpful to know

why hadn't I encountered teacher like you

@drpeyam

2 жыл бұрын

Awwwww thanks!!!

hi dr. aren't you originally from Tabriz - Iran?

@drpeyam

2 жыл бұрын

I’m from Iran but not from Tabriz

@ensiehbehzadian5061

2 жыл бұрын

@@drpeyam glad to have you as my compatriot Dr.

I subed!!!I really don't have to finish d lesson....Just the way math shld b taught

Thank you! Great presentation. A lapel microphone would be better.

@drpeyam

2 жыл бұрын

I forgot mine that day

Why is the abscissa of the vertex minus b/2a when that point is on the positive x side of the graph? (2:06)

@drpeyam

2 жыл бұрын

In that picture a is negative

@coolbombe

2 жыл бұрын

I don't understand why that point on the x axis is minus b/2a.

@coolbombe

2 жыл бұрын

Never mind. The maximum value of ax2+bx+c occurs at y'=0. So 2ax+b=0, x=minus b/2a. That was not explained.

@drpeyam

2 жыл бұрын

It’s taken as a given

Don’t tell what I probably don’t know about parabolas… I was married to one once. Things went south between us though when she became overly hyperbolic.

This ought to be known as the abc's of a parabola.

HOw exciting!

THankyou!! this was really interesting.... i wonder why this is not in school books??

@drpeyam

2 жыл бұрын

Thank you! And not sure either!!

Next on Peyam's list: "a 2*2 fact you probably didn't know". And after that: "2+2 fact you probably didn't know". But what is really surprising is that he managed not to copy from or pick into his cheat sheet for over 4 minutes. Is this the same Peyam who cannot solve a simple integral without copying from his notes?

@drpeyam

2 жыл бұрын

I had my notes in front of me 😝

One think I like to argue, (yes, I have no life), is if you throw a ball up "Does the ball ever stop at the top of the throw?" I argue that at no point does the ball 'stop' it is either going up or coming down (more probably going up AND down at the same time). But you cannot plot ANY two points where there is zero difference between the points on any scale, throw the ball up and it goes up and down but never STOPS, it is ether going up or down. I argue that even though it changes direction, the ball never 'stops' it never has zero velocity. Do you think I am right? or should I just try to get out more :)

@TheDRAGONFLITE

2 жыл бұрын

you should consider that fact that the velocity is the derivative of position with respect to time and that it does in fact equal zero at the top of a parabola. The acceleration is always a constant but the velocity does in fact stop for a instant.

@Darryl_Frost

2 жыл бұрын

@@TheDRAGONFLITE That is true, but you can say the same for any position on the parabola, if you take one data point, its velocity is zero. But for any delta t there will be no two instants of the same value. Also, every ball no matter how 'stiff' even a steel ball has some elasticity, So I expect the top edge of the ball and the bottom edge of the ball out not be going up and down together, you could have the top going down and the bottom still going up. It would be hard to see with a steel ball, but I bet you would see that effect with a big blow up beach ball (like a coke ball). But of course if it changes direction it must go through zero, but it cannot be zero for any period of time. We probably both need to get out more LOL.. And thanks for your response.

@colleenforrest7936

2 жыл бұрын

If you are measuring "between two points" the ball does not stop. You ate measuring the "change" in velocity between those two points, and since only one point, the point at the top of the arc will, will have velocity 0 and the other cannot, the change in velocity will be a non zero value But if you look at just that one point at the top of the apex, that's the only point in the ball's journey where v=0. One picosecond more or one picosecond less and the velocity is no longer 0. Your distance could be 1/infinity picoseconds away from the apex and still have a non zero velocity. It is only at the instant of the apex that v=0 and only at that one point.

@Darryl_Frost

2 жыл бұрын

@@colleenforrest7936 you make an excellent point.. thanks.

You are right! I don't know a.

Insta like after the pen drop

Mwaa!

👍👏👏👏