One can use (-b/2a, f(-b/2a)), completing the square, graphing, derivative, and other methods for finding the vertex of a parabola. Here's one of many methods.
rewrite: ax²+bx+c as a(x+b/(2a))²+c-b²/(4a) The minimum(when a > 0... if a < 0 it is the maximum... and if a=0 things break since it is not a parabola) value is when x+b/(2a)=0
@@Jack-hd3ov how would there be a pain to differentiate any quadratic? By definition quadratics are in the form ax^2 + bx + c and the derivative is simply 2ax + b? Much easier than finding roots in general
I mean, this is easier. But then you would have to know calculus. Not that it is some arcane knowledge. But, depending on what stage you are, you weren't taught calculus yet(... or ever). In Brazil, they teach you calculus at university. But this problem(finding the vertex of parabola) is given to people at high-school.
@@MithilaMaati that might be in the high school here... or the last year before high school. I also might have overestimated: finding a vertex of a parabola is slightly earlier.
Thank you Sir 🙏 I’ve had to teach myself math this entire college semester. My professor spend 2hrs out of a 3hr45min lecture period teaching us this concept. Was certain I wouldn’t understand it for the final, then watched your video and it clicked instantly with everything I already knew. Much appreciated.
This makes perfect sense since you’re trying to find the mid point of the two points. I’ve always used the partial QF of -(b)/2(a) to solve this, but now I know another way.
Now that's how you make maths so easy. Thanks mister H for your service. Can we expect a video on differentiation, it would be very helpful. Thanks again. :)
When you learn calculus, this becomes much easier, but knowing the basics is essential for learning quadratic equations. Nice video, im a brazilian guy.
Since you already have f(x) in factored form (x-5)(x-1), the easier way to get f(3) is to plug 3 into the factored form (3-5)(3-1) = (-2)(2) = -4. Why easier? The numbers are smaller, and you only have to keep track of two things at a time. 😊
You do know it’s easier to find the y- value by substituting the x- value into the Intercept or factored form, right?!? (3-5)(3-1)= -2(2)=-4. So much easier to do mentally! 🎉❤
u did it so much better and a much faster technique😭 im having my exams tomorrow in math about quadratic equations and functions and deriving qf's! i converted the sf to vertex form then I made table of values and graphed😆
I just took the first derivative with respect to x, set that to 0 and it gives me 3. Then just substitute that into the original equation which gives a value of -4 for y. Of course if you just take the general case for a parabola of y = ax^2 + bx + c then the first derivative is 2ax + b. Set that to 0 and you get 2ax=-b, so x=-b/2a. Substitute a=1, b=-6 and d=5 to get x and then that value into the original equation. Thus x=6/2=3 and y=3^2-6*3+5=-4. Also, that method works with a parabola that doesn't cut the x axis, such as y=x^2-6x+15 (x=3, y=6)...
2 best ways in my opinion: 1st way: Just use the formula that was made by the same way he did it. -b/2a for the x value and -b^2/4a + c for the y value where a, b, and c are 1, -6, and 5. -(-6)/2(1) = 3 (-(-6)^2)/4(1) + 5 = -4 2nd way: Take the derivative and set equal to 0. f(x) = x^2 - 6x + 5 f'(x) = 2x - 6 = 0 2x = 6 x = 3 3^2 - 6(3) + 5 = -4
There are many other ways to find the vertex of this parabola, like: a. Remembering that the coordinates of the vertex are (-b/2a, f(-b/2a)) b. Converting the function into vertex form by completing the square c. Finding the derivative of the function, then setting it equal to zero to find the vertex
Completing the square is more deeply explanatory for understanding the actual shape of the curve, and it is actually quicker. Of course setting the first derivative = 0 is the quickest way to just finding the x coordinate.
Ok, j'ai compris. Je suis heureuse de voir que mes études ressortent petit à petit de l'oubli...et même j'apprends des choses que je n'avais jamais vu dans mes vieux cours.
it is much easier and faster by just simply taking the derivative and make it equal to 0 so 3x-6 = 0 => x = 3 and substitute in the function to get y, which is 3^2-6*3+5 which equals -4. So (3, -4). I just did that in my head even before he started writing on the board
Or you can try rearrange the equation to mimic the form (x-alpha)^2 + beta where alpha and beta are the coordinates of the vertex meaning x^2-6x+5= x^2-6x+9-4= (x-3)^2-4 Hence the point S(3;-4) is the vertex of this parabola
This is why it works. In general, a function shifted horizontally h units to the right, and k units vertically, becomes f(x - h) + k. Write it in vertex form: y = a*(x - h)^2 + k Expand: y = a*(x^2 - 2*x*h + h^2) + k y = a*x^2 - 2*a*h*x + a*h^2 + k Map to the standard form coefficients: y = a*x^2 + b*x + c a = a b = -2*a*h c = a*h^2 + k And from your b-term, you can solve for h, the horizontal shift: h = -b/(2*a).
It's easier imo to just use the derivative to get x f(x) = x^2 -6x +5 f'(x) = 2x -6 = 0 2x = 6 x = 3 y = f(3) = 3^2 -6*3 +5 y = 9 -18 +5 y = -4 Vertex = (3,-4)
My problem with integral calculations, i have no picture, no sense of what im calculate. Then i got the understanding of how many calculate, the understanding of why many make strange mistankes.
Integrals wouldn't help you with this one. You'd need to use the opposite operation of derivatives. Looking for places where the derivative equals zero, is how you can solve for the turning points of functions in general. For a parabola, it's very simple to do. Drop each exponent by 1, and have each original exponent join its coefficient. Simple use of the power rule, d/dx k*x^n = k*n*x^(n-1). d/dx (x^2 - 6*x + 5) = 2*x - 6 Set it equal to zero, and solve for x: 2*x - 6 = 0 x = 3
I use parabolic curves in my artwork. But i never knew it was a math equation to show my work. I honestly don't have a clue as to what im seeing. I would like to understand this in case someone ask.
![ความรักไม่ได้น่ากลัวขนาดนั้น (TRUST ME) - LYKN [ OFFICIAL MV ]](http://i.ytimg.com/vi/9np-lAtCRpQ/mqdefault.jpg)
One can use (-b/2a, f(-b/2a)), completing the square, graphing, derivative, and other methods for finding the vertex of a parabola.
Here's one of many methods.
Bro i was about to say that 😢 but anyways
(-b/2a, -D/4a) where D is discriminant D=b²-4ac
I remember studying parabola vertex
@@RishavPlayz2007 -D/4a
rewrite:
ax²+bx+c
as
a(x+b/(2a))²+c-b²/(4a)
The minimum(when a > 0... if a < 0 it is the maximum... and if a=0 things break since it is not a parabola) value is when x+b/(2a)=0
This is my favorite math short channel of them all... this guy must be a great tutor. Keep teaching the world good sir
Will do! Thank you.
@@mrhtutoring I don't speak English well but your shorts are very simple! As-Salyam from Caucasus
Good luck!
On some real shit, I wholeheartedly love this fucking guy, he’s helped me out a lot during this Engineering process
I like to set the derivative equal to zero to find x and then plug x back into the original to find the y coordinate
@@Jack-hd3ov how would there be a pain to differentiate any quadratic? By definition quadratics are in the form ax^2 + bx + c and the derivative is simply 2ax + b? Much easier than finding roots in general
Same dy/dx = 2x-6
2x-6=0
X = 3
I mean, this is easier. But then you would have to know calculus.
Not that it is some arcane knowledge. But, depending on what stage you are, you weren't taught calculus yet(... or ever).
In Brazil, they teach you calculus at university.
But this problem(finding the vertex of parabola) is given to people at high-school.
@@matheusjahnke8643 find the vertex is taught at the age of 15, or in 10th class in india
@@MithilaMaati that might be in the high school here... or the last year before high school.
I also might have overestimated: finding a vertex of a parabola is slightly earlier.
If this good sir was my maths teacher I would've became Einstein by now
😅😅😅😅😊😊😂
I told my grandson too study the rules and he will understand. He does great now.
You know you're dealing with a teaching pro when comic sans gets used in the title of each problem.
OH MY GOD THIS HELPED SO MUCH THANK YOU!!!
Just put (-b/2a) for the x coordinate and {(4ac-b²)/4a} for the y coordinate when finding the vertex of a parabola.
You can also just use (-b/2a) to find x coordinate and plug it into the function to find y
Exactly what I thought when I heard this - just find the axis of symmetry. Simple.
Nice and clean! Thank you
I love this. One thing I would add to the graph is the y intercept.
Your videos are great! 👍 i would watch your maths lessons!
You could also complete the square, so x²-6x+5=x²-6x+9-4=(x-3)²-4 and its obvius from here on
i was thinking the same!
Obvious.
Fantastic! So clear and straight to the point.
I can't stop watching this guy.
Thank you Sir 🙏 I’ve had to teach myself math this entire college semester. My professor spend 2hrs out of a 3hr45min lecture period teaching us this concept. Was certain I wouldn’t understand it for the final, then watched your video and it clicked instantly with everything I already knew. Much appreciated.
This makes perfect sense since you’re trying to find the mid point of the two points. I’ve always used the partial QF of -(b)/2(a) to solve this, but now I know another way.
Excellent inspirational teaching that broadens a learner's mind through human intuition. Easily Unlock a learner's dormant potential. Thanks
super
Now that's how you make maths so easy. Thanks mister H for your service. Can we expect a video on differentiation, it would be very helpful. Thanks again. :)
Now you can mess with your students by asking them the vertex of a parabola that doesn't cut the X axis in the test 😂. That would be definitely funny.
When you learn calculus, this becomes much easier, but knowing the basics is essential for learning quadratic equations. Nice video, im a brazilian guy.
Surgically perfect👏👏👏
I love how you look and the way you teach😊😊😊
Thank you! 😃
Your videos are great!! Love from India ❤
Glad you like them!
That was some on point explaination sir!
Thanks a lot ❤
Thank you
Just find the axis of symmetry using -b/2a, and plug it in for x
I like to put it into standard form but I’ve never thought to do it like this, nice
This guy is a genius !!!
Math comes so easy for him !!!
It takes someone who is really smart to solve these math problems !!!
Thank you so much
You are a Maths genius!!!
Since you already have f(x) in factored form (x-5)(x-1), the easier way to get f(3) is to plug 3 into the factored form (3-5)(3-1) = (-2)(2) = -4. Why easier? The numbers are smaller, and you only have to keep track of two things at a time. 😊
I have a test tomorrow thanks man this helped a lot
You are amazing teacher my math teacher doesn’t actually teach us
Best of luck!
Egad! This brings back nightmares from HS math 60-odd years ago. I can only wish I had 5% of this man's mathematical savvy. I really envy him.
I didn't think of this solution, so I wound up using calculus. Thanks!
Nice ive literally only ever completed the square but this is cool too
You do know it’s easier to find the y- value by substituting the x- value into the Intercept or factored form, right?!? (3-5)(3-1)= -2(2)=-4. So much easier to do mentally! 🎉❤
We haven't learned this yet in school, do u mind explaining why and how he transformed the equation to 0=(x-5)(x-1) ?
u did it so much better and a much faster technique😭 im having my exams tomorrow in math about quadratic equations and functions and deriving qf's!
i converted the sf to vertex form then I made table of values and graphed😆
Thank you sir
Thank u so much made it so much easier
شكرا لك أستاذي الفاضل 😊❤
Good job as always
Four coordinates needed to draw a graph:
1. x = 0
2. y = 0 -> two coordinates
3. y min/max
I just took the first derivative with respect to x, set that to 0 and it gives me 3. Then just substitute that into the original equation which gives a value of -4 for y.
Of course if you just take the general case for a parabola of y = ax^2 + bx + c then the first derivative is 2ax + b. Set that to 0 and you get 2ax=-b, so x=-b/2a.
Substitute a=1, b=-6 and d=5 to get x and then that value into the original equation. Thus
x=6/2=3 and y=3^2-6*3+5=-4.
Also, that method works with a parabola that doesn't cut the x axis, such as y=x^2-6x+15 (x=3, y=6)...
Im 65 and enjoyed doing that un school. Now kids cant even do basic math forget algebra, calculus, trigonometry!!!
Excellent
2 best ways in my opinion:
1st way: Just use the formula that was made by the same way he did it. -b/2a for the x value and -b^2/4a + c for the y value where a, b, and c are 1, -6, and 5.
-(-6)/2(1) = 3
(-(-6)^2)/4(1) + 5 = -4
2nd way: Take the derivative and set equal to 0.
f(x) = x^2 - 6x + 5
f'(x) = 2x - 6 = 0
2x = 6
x = 3
3^2 - 6(3) + 5 = -4
That's so simple and yet so smart😭
You make concepts that seemed unintelligible to me in Secondary School seem as obvious as breathing. An exceptional talent for teach maths.
Thank you for the very comment. I appreciate it!
Superb
There are many other ways to find the vertex of this parabola, like:
a. Remembering that the coordinates of the vertex are (-b/2a, f(-b/2a))
b. Converting the function into vertex form by completing the square
c. Finding the derivative of the function, then setting it equal to zero to find the vertex
Excellent video
Beautiful
its already in standard form just do (-b/2a, f(-b/2a))
Did you read the comment pinned at the top?
Brilliant 😮
Completing the square is more deeply explanatory for understanding the actual shape of the curve, and it is actually quicker.
Of course setting the first derivative = 0 is the quickest way to just finding the x coordinate.
I can't argue with your comment.
However, the video was just to show another method.
Ok, j'ai compris.
Je suis heureuse de voir que mes études ressortent petit à petit de l'oubli...et même j'apprends des choses que je n'avais jamais vu dans mes vieux cours.
Just differentiate and set that equal to 0.
it is much easier and faster by just simply taking the derivative and make it equal to 0
so 3x-6 = 0 => x = 3 and substitute in the function to get y, which is 3^2-6*3+5 which equals -4. So (3, -4). I just did that in my head even before he started writing on the board
Yeah i learned this is good for using factored form
Or you can try rearrange the equation to mimic the form (x-alpha)^2 + beta where alpha and beta are the coordinates of the vertex meaning
x^2-6x+5= x^2-6x+9-4= (x-3)^2-4
Hence the point S(3;-4) is the vertex of this parabola
I love you sir, wish you could be my tutor
👍👍👍
Sir do you know about JEE Advance Exam conducted in India Please take questions from that exam also
Thanks
This is the smartest Algebra teacher in the world
Good to see a blackboard. The world went to hell with greenboards and white grease pen boards!
You can also do -b/2a. Idk why it works but it does
It works because the first term of the quadratic equation is -b/2a
This is why it works.
In general, a function shifted horizontally h units to the right, and k units vertically, becomes f(x - h) + k.
Write it in vertex form:
y = a*(x - h)^2 + k
Expand:
y = a*(x^2 - 2*x*h + h^2) + k
y = a*x^2 - 2*a*h*x + a*h^2 + k
Map to the standard form coefficients:
y = a*x^2 + b*x + c
a = a
b = -2*a*h
c = a*h^2 + k
And from your b-term, you can solve for h, the horizontal shift:
h = -b/(2*a).
Y- intercept is (0, 5) directly from given quadratic expression.
Nice one. I would like to add that the scale for x and y axes are different. 😅
Y+4=(x-3)^2 then compare with x2 =4ay
Makes so much more sense 10+ years after calculus 😂
What if x^2 has a coefficient of more than one???
It's easier imo to just use the derivative to get x
f(x) = x^2 -6x +5
f'(x) = 2x -6 = 0
2x = 6
x = 3
y = f(3) = 3^2 -6*3 +5
y = 9 -18 +5
y = -4
Vertex = (3,-4)
Te amo señor
What the fk i was never taught this method this is way faster than the vertex formula damn
Great
The best
Thanks ❤
❤
where can i meet u, I WANT TO SAY THANK YOU!!!!!!!!!!!
Needs to be define vertex. Most people don’t know what the meaning of vertex.
Genius
for x equal to zero f(0) equal to 5, so it cut the ordinate axe
Or differentiate and set equal to 0
I'd have gone straight to calculus
Should cross the y axes at (0,5)
Formula is (-b/2a,-D/4a).
İngilizce bilmesem de matematiğin dilinden anladığım için mutluyum sjsjjs
My problem with integral calculations, i have no picture, no sense of what im calculate. Then i got the understanding of how many calculate, the understanding of why many make strange mistankes.
Integrals wouldn't help you with this one. You'd need to use the opposite operation of derivatives.
Looking for places where the derivative equals zero, is how you can solve for the turning points of functions in general.
For a parabola, it's very simple to do. Drop each exponent by 1, and have each original exponent join its coefficient. Simple use of the power rule, d/dx k*x^n = k*n*x^(n-1).
d/dx (x^2 - 6*x + 5) =
2*x - 6
Set it equal to zero, and solve for x:
2*x - 6 = 0
x = 3
🙏🙏
Thank you soo much sir
Most welcome
orr find the derivative
2x - 6
and it equals 0 when x = 3
then do f(3)
thanks
Best analytic geometry tutorial that i have ever seen
Or you can use:
vertex = (-b/2a ; -Δ/4a)
Thxx
You can also use calculus...2x-6=0
What kind of real life situation would this, finding the vertex, come into play? Engineering? Just wondering.
Maximizing profits,minimizing costs, dimensions of a fenced area...
5 at y-axis interception
Yup! 예~
Genuine
I use parabolic curves in my artwork. But i never knew it was a math equation to show my work. I honestly don't have a clue as to what im seeing. I would like to understand this in case someone ask.