Vertex of Parabola Using X-Intercepts

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.

Superb

thanks

Makes so much more sense 10+ years after calculus 😂

Genuine

That's so simple and yet so smart😭

I’m in sixth grade…

Y+4=(x-3)^2 then compare with x2 =4ay

You are a Maths genius!!!

Excellent inspirational teaching that broadens a learner's mind through human intuition. Easily Unlock a learner's dormant potential. Thanks

I feel like I'm going to cry rn! finally I can replay this over and over until it's stuck with me

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😆

Wow im flabbergasted

Te amo señor

Thanks ❤

Thank you sir

Grade 1 problems 😂😂😂😂😂

One thing prof,how do you you it open upward, you plot upward first. I mean with finding vertex and x intercept we can. But how you figure out first that it is pointing upward

Thank you

How about general form to standard form sir

Im 65 and enjoyed doing that un school. Now kids cant even do basic math forget algebra, calculus, trigonometry!!!

Yeah i learned this is good for using factored form

Me encantó, lo resolví sin saber parábolas 😂😅😔🥺😭

5 at y-axis interception

Fantastic! So clear and straight to the point.

Excellent

Needs to be define vertex. Most people don’t know what the meaning of vertex.

This is the smartest Algebra teacher in the world

F(x)=ax^2+bx+c
When a>0
There is one minimum
point of F(x).
When a<0
There is one maximum
Point of F(x).
Minimum/maximum
point(x,y):
x=-b/(2a)
y=F[-b/(2a)]#😅

The best

I still never seem to understand why the equation of a parabola is equated to zero

Easy way derivative of f( x)=2x-6=0 then x=3 f(3)= -4

Good to see a blackboard. The world went to hell with greenboards and white grease pen boards!

Thank you soo much sir

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

❤

What kind of real life situation would this, finding the vertex, come into play? Engineering? Just wondering.

Can use derivative, slope as 0

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! 🎉❤

Formula is (-b/2a,-D/4a).

When you learn calculus, this becomes much easier, but knowing the basics is essential for learning quadratic equations. Nice video, im a brazilian guy.

orr find the derivative
2x - 6
and it equals 0 when x = 3
then do f(3)

I can't stop watching this guy.

😂😂😂👏🏿👏🏿👏🏿👍🏿

Wont work when positive/negative definite

Where can I use this in my daily life

Should cross the y axes at (0,5)

Why not just find where f'(x)=0 then plug that into f(x) or is that too advanced for the level you are teaching?

(X^2-6x)+5
(X^2-6+9)+5-9
(X-3)^2 -4
x=3,-4