If anyone is having a hard way doing this to get the local minimum or maximum (coordinates of the vertex) go check prof Dave’s videos on differentiation. If you take the first derivative of the function and set the answer = 0 and solve for X you get the x value of the local minimum or maximum you’re looking for. When you plug that into the original function for x you get the y value of the coordinate of the vertex. Thank you prof Dave for learning me all of this.
I felt slightly overwhelmed during this video for the first time in the series but actually got all the comprehension questions right on the first try. That boosted my confidence for sure.
I haven't finished the video yet, but I see a lot of the comments are saying that this video is confusing and not as helpful as others have been. I'd just like to share what I hope will be a helpful method I've been using. Professor Dave Is always my first choice. I go through his courses in order, and don't move on to the next video till I feel I've mastered that videos content. In doing this, I've found it helpful to find the corresponding video on the subject from The Organic Chemistry Tutor TH-cam channel. If I'm still lost, I turn to Khan academy's channel, and lastly, a specific Google search for the "Lumen Learning" site along with the topic. For example: google "Lumen Learning graphing parabola" I can't find a way to search on the website, but Lumen Learning has a text book layout of a lot of concepts as well as interactive graphs and input sections for working out examples. These other resources have been incredible secondary sources when I'm just not getting it, so I hope it can be for someone else as well.
when “a” multiplies with “b²/4a²”, the resultant will be “b²/4a”. To compensate for this new term we will add a (-b²/4a) term to make it 0. BEFORE *multiplying* with *a* : a( x² + b/a x + b²/4a² ) + c -b²/4a AFTER *multiplying* with *a* : ax² + bx + *b²/4a* + c *-b²/4a* AFTER *subtracting* *(-b²/4a)* from *b²/4a* : ax² + bx + c (standard form)
@Teiwaz111 1 год назад (изменено): Yeah, that one is a bit tricky and I had to think about it too. You add b²/4a² to the term in parenthesis, but you haven't yet added anything to the rest of the equation (the "+ c" part), you still gotta do that. And to figure out what you need to add to the "+ c" part you solve "a (b²/4a²)" part --> ab²/4a², or b²/4a. As it is a FUNCTION you have to SUBTRACT that term from everything else to not change the value of the expression itself. Hope that clarified it. [edit] For further clarification: If you have f(x) = 7 You could also say f(x) = a + b Which would still be true if f.e. a = 3 and b = 4 If you add something to a though f(x) = (a + 2) + b Your f(x) suddenly would become f(x) = (3 + 2) + 4 f(x) = 9 To still comply to f(x) = 7 you gotta do f(x) = (3 + 2) + (4 - 2) f(x) = 5 + 2 f(x) = 7
Hi Professor Dave, thanks for the lesson, I have a question (sorry for bad English): In order to go from vertex form back to standard form, do I use the (x +- h)^2 = (x^2 +- 2xh + h^2) formula, and then multiply every term by a? Also, is it correct to say (based on the formula) that to go to standard form from vertex, a = a, b = -2ah, c = ah^2+k ? Thanks in advanced, I missed this lesson because of the virus.
Professor to find vector form from parabolic form in strategy 1 you have added b^2/4a into the parentheses that has been multiplied by a ,But whereas outside u have subtracted b^2/4a instead of ab^2/4a
@Proffesor Dave Explains ihave been seeing your many videos on many topics and saw many comments like Jesus and all I wanna say you obviously are not Jesus but for the students that are struggling with a topic or subject you are a type of for them really 😂 Thank you 👍
It's a perfect square so it can be expressed as the square of the binomial, note the exponent outside the parentheses. Check my earlier tutorial on completing the square if it seems confusing.
Yeah, that one is a bit tricky and I had to think about it too. You add b²/4a² to the term in parenthesis, but you haven't yet added anything to the rest of the equation (the "+ c" part), you still gotta do that. And to figure out what you need to add to the "+ c" part you solve "a (b²/4a²)" part --> ab²/4a², or b²/4a. As it is a FUNCTION you have to SUBTRACT that term from everything else to not change the value of the expression itself. Hope that clarified it. [edit] For further clarification: If you have f(x) = 7 You could also say f(x) = a + b Which would still be true if f.e. a = 3 and b = 4 If you add something to a though f(x) = (a + 2) + b Your f(x) suddenly would become f(x) = (3 + 2) + 4 f(x) = 9 To still comply to f(x) = 7 you gotta do f(x) = (3 + 2) + (4 - 2) f(x) = 5 + 2 f(x) = 7 [/edit]
Use the given quadratic equation in explaining substitute it like in "a use the -x because its the given" don't just fucking use a because not all understand how everything happens like me
If you are confused about the quadratic formula, visit my tutorial where I explain that. Each tutorial is about a very specific topic and assumes knowledge of the previous tutorials.
@@miguelantonlobaton5667 if you expand the line with the first "=", there will be a +2 extra compared to the function. So the -2 is there to balance it.
If anyone is having a hard way doing this to get the local minimum or maximum (coordinates of the vertex) go check prof Dave’s videos on differentiation. If you take the first derivative of the function and set the answer = 0 and solve for X you get the x value of the local minimum or maximum you’re looking for. When you plug that into the original function for x you get the y value of the coordinate of the vertex. Thank you prof Dave for learning me all of this.
I think they should understand this way first before learning calculus
I felt slightly overwhelmed during this video for the first time in the series but actually got all the comprehension questions right on the first try. That boosted my confidence for sure.
can you help me with the second one (╥﹏╥).
Thank you for your Lesson
as always, thank you professor dave for the math knowledge
I finally got this fast-pace topic after watching hours of lectures on TH-cam
That odd satisfaction from the intro song doesn`t get old.
If you're stuck on this one try starting with problems 2 and 3 before 1. 1 is trickier because of the 2x^2 and - 4x
Great stuff!
Thx again! 😀
6:07 , how did you plot the x intercept values? Prof. Could you xplain it further?
I have solved using quadratic formula but the values are in root, prof.
Thanks for the vid pls upload moree you also got my subscription
If only we were thought math this way in schools... Thank u sooo much sir you are a star
I haven't finished the video yet, but I see a lot of the comments are saying that this video is confusing and not as helpful as others have been. I'd just like to share what I hope will be a helpful method I've been using. Professor Dave Is always my first choice. I go through his courses in order, and don't move on to the next video till I feel I've mastered that videos content. In doing this, I've found it helpful to find the corresponding video on the subject from The Organic Chemistry Tutor TH-cam channel. If I'm still lost, I turn to Khan academy's channel, and lastly, a specific Google search for the "Lumen Learning" site along with the topic. For example: google "Lumen Learning graphing parabola"
I can't find a way to search on the website, but Lumen Learning has a text book layout of a lot of concepts as well as interactive graphs and input sections for working out examples. These other resources have been incredible secondary sources when I'm just not getting it, so I hope it can be for someone else as well.
3:24 why don't we subtract a times -b^2/4a since its being multiplied by a
when “a” multiplies with “b²/4a²”, the resultant will be “b²/4a”.
To compensate for this new term we will add a (-b²/4a) term to make it 0.
BEFORE *multiplying* with *a* :
a( x² + b/a x + b²/4a² ) + c -b²/4a
AFTER *multiplying* with *a* :
ax² + bx + *b²/4a* + c *-b²/4a*
AFTER *subtracting* *(-b²/4a)* from *b²/4a* :
ax² + bx + c (standard form)
What is length of the parabola curve between any two points
What is line equations for that length
How they are correlated
Why the a in c-b²/4a isn't squared?
@Teiwaz111 1 год назад (изменено):
Yeah, that one is a bit tricky and I had to think about it too.
You add b²/4a² to the term in parenthesis, but you haven't yet added anything to the rest of the equation (the "+ c" part), you still gotta do that.
And to figure out what you need to add to the "+ c" part you solve "a (b²/4a²)" part --> ab²/4a², or b²/4a.
As it is a FUNCTION you have to SUBTRACT that term from everything else to not change the value of the expression itself.
Hope that clarified it.
[edit]
For further clarification:
If you have
f(x) = 7
You could also say
f(x) = a + b
Which would still be true if f.e. a = 3 and b = 4
If you add something to a though
f(x) = (a + 2) + b
Your f(x) suddenly would become
f(x) = (3 + 2) + 4
f(x) = 9
To still comply to f(x) = 7 you gotta do
f(x) = (3 + 2) + (4 - 2)
f(x) = 5 + 2
f(x) = 7
hello bit confused there the standard forms of parabola like y2=4ax how can relate both of them ? sorry it may seems a dum question !
Hi Professor Dave, thanks for the lesson, I have a question (sorry for bad English):
In order to go from vertex form back to standard form, do I use the (x +- h)^2 = (x^2 +- 2xh + h^2) formula, and then multiply every term by a? Also, is it correct to say (based on the formula) that to go to standard form from vertex, a = a, b = -2ah, c = ah^2+k ?
Thanks in advanced, I missed this lesson because of the virus.
when i saw x^2 and derivatives in an algbera II course, i was like, what, i thought this was algebra II?
superbb!!
How does 1 in c became 2 in completing the square manually
Thank you so much, you need more views, and subscribers
i agree, tell all your friends!
Will do!
Want to recommend you to every friend I know. though I don't thin they're gonna watch
3.34 Why did we express x + b/2a as x - (-b/2a)? What is the reason?
Because it's the same expression, but he did to make sure that you will know where does the positive sighn came from.
Done this lesson.
1:07 I didn't understand what the word "derivatives" was doing there on the bottom right.
oh it just means we know how to graph simple transformations of basic graphs
I see, thanks for the clarification.
Professor to find vector form from parabolic form in strategy 1 you have added b^2/4a into the parentheses that has been multiplied by a ,But whereas outside u have subtracted b^2/4a instead of ab^2/4a
@Proffesor Dave Explains ihave been seeing your many videos on many topics and saw many comments like Jesus and all I wanna say you obviously are not Jesus but for the students that are struggling with a topic or subject you are a type of for them really 😂
Thank you 👍
Great feeling when I understand something 3 years after they tried to teach it to me in school
how do come up with h ( -2)\2 (-1)=-1
-2\-2=-1.
correct me if i'm wrong
maby its +1.
sorry im still trying to figure it out
3:25 what happened there? how does he get from (x^2 + bx/a + b^2/4a^2) to (x + b/2a)?
It's a perfect square so it can be expressed as the square of the binomial, note the exponent outside the parentheses. Check my earlier tutorial on completing the square if it seems confusing.
Yeah, that one is a bit tricky and I had to think about it too.
You add b²/4a² to the term in parenthesis, but you haven't yet added anything to the rest of the equation (the "+ c" part), you still gotta do that.
And to figure out what you need to add to the "+ c" part you solve "a (b²/4a²)" part --> ab²/4a², or b²/4a.
As it is a FUNCTION you have to SUBTRACT that term from everything else to not change the value of the expression itself.
Hope that clarified it.
[edit]
For further clarification:
If you have
f(x) = 7
You could also say
f(x) = a + b
Which would still be true if f.e. a = 3 and b = 4
If you add something to a though
f(x) = (a + 2) + b
Your f(x) suddenly would become
f(x) = (3 + 2) + 4
f(x) = 9
To still comply to f(x) = 7 you gotta do
f(x) = (3 + 2) + (4 - 2)
f(x) = 5 + 2
f(x) = 7
[/edit]
❤❤❤
what is vertex form????
Great...😎
7:45 in problem 2,isn't it ((x-6)^2)-4
nah then you would have -12x
Use the given quadratic equation in explaining substitute it like in "a use the -x because its the given" don't just fucking use a because not all understand how everything happens like me
If you are confused about the quadratic formula, visit my tutorial where I explain that. Each tutorial is about a very specific topic and assumes knowledge of the previous tutorials.
At -3:53 why add 2 instead of 1
Isn't the first vertex (1,4)? How did you get the value "3"?
the k term is the y coordinate, that's 3
@@ProfessorDaveExplains Ooooh. I see. Thank you so much!!
this one was confusing as hell
The answer for the first comprehension it's incorrect. The correct answer is vertex form 2(x-1)^2+4 and vertex are 1 and 4
If you expand that you'll find you're incorrect. It's correct as shown.
@@ProfessorDaveExplains can u please give me the steps as I'm not getting an correct answer only for the first comprehension?
@@girishsingh2535 Here are the steps;
f(x)=2x^2 - 4x + 5
= 2(x^2 - 2x + 1) + 5 - 2
= 2(x - 1)^2 + 3
so, vertex (h,k) = (1,3)
@@Mike_Toreno hi, why did u minus 2 on the side of 5? While 1 on the left side? I thought it should be the same? Pls enlighten mee
@@miguelantonlobaton5667 if you expand the line with the first "=", there will be a +2 extra compared to the function. So the -2 is there to balance it.
a1
This video is not helpful like circles and eclipses
How?
@any.user.allowed. because he explains everything incompletely 😮😮😮
the expression is slightly incorrect
❤❤❤