4 Methods to Find Minimum Value From Quadratic Equation
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- Опубліковано 10 вер 2024
- For Guidance Contact : anil.anilkhandelwal@gmail.com
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You could also use differentiation to do this as well, no? There's a 5th method.
if f(x) = x^2 - 8x + 12
then f'(x) = 2x-8
turning points are when rate of change is 0 so set f'(x) to 0:
2x-8 = 0
2x = 8
x = 4
plug back into original equation
f(4) = 4^2 - 8(4) +12
= 16 - 32 + 12
= -16 + 12
= -4
Thus, minimum value is -4.
Thanks. That is correct. The students who are doing Functions now will take calculus in the next Semester.
Therefore, I am not using derivatives.
I agree totally that the differentiation method could be used, but at ordinary level math students have not done differentiation.
After using- b/2a to find x, one can use 4ac-b^2/4a to find y. Note it is 4ac-b^2/all over 4a.
You sir are a tru hero! From everyone i watched talking about this topic.. no one mentioned this method.. thank you!
can you please tell the trick to find maxima ? In the same equation..?
and 4 is the maximum value? please answer
I love it when he says, "Do you understand?" LMFAOO
Is that a JoJo reference, bro
Timestamp please
@@justyourtypicalhamster5046 stfu
@@justyourtypicalhamster5046 idgaf
Yeah right, I do as well
Thank you sir, please help me find the minimum value of y=2x^2-8x+10
2 steps for vertex of a quadratic equation ax²+bx+c
1. Use -b/2a you will get the x of the vertex.
2. Then plug that x into the original equation to get y. Easy.
or just use c-(b^2/4a)
Sir i want to know how the coordinates of vertex came (4, -4)
We can find with method
Let f(x)=x^2-8x+12
By using formula for x
x=-b/2a
Here a=1 b=-8 and c = 12
So, x = -(-8)/2(1)
x=4
Now finding y
Put x=4 in equation above
Y=(4)^2-8(4)+12
Y=16-32+12
Y=28-32
Y=-4
Therefore (x,y)=(4,-4)
Thanks A lot Sir I've exams after 1 months this help me a lot. Love you sir.
This awesome others have been confusing me all over youtube but Here you gave me what I'm looking for straight to the point ,thanks
Sir There is 5'th method also .Put y = ax^2+bx +c then find dy/dx = 2ax +b and equate it to zero. This gives x =--b/2a .Then substitute this value of x in y . DrRahul Rohtak Haryana India
Sir Anil Kumar, sorry to correct you but the partial factor is wrong because x=0 & x=8 are not the x-intercepts of the parabola function f(x) = x^2 - 8x + 12, the correct x-intercepts are x=2 & x=6 from intercept form of parabola function f(x) = (x - 2)(x - 6).
x = (2+6)/2 = 4, x=4 is the axis of symmetry. Evaluate f(x): f(4) = (4)^2 - 8(4) + 12 = -4 or f(4) = (4 - 2)(4 - 6) = -4. So the minimum value of the parabola is -4. It just coincidentally happened that when you solved axis of symmetry x = (0+8)/2 = 4 for partial factor and therefore conclude minimum value is -4 which you based on your first solved vertex (4, -4). :)
I have used different method, called Partial Factoring. Many trinomials cannot be factored. If you notice, I have factored only the first two terms and therefore it is called partial factor. Your solution is correct, and is generally used. Partial Factoring is an efficient method, good to know. Please go through the video again and see if that provides value.
Thanks.
Yes, I believe your video provided value on your audience and I learn something new from you. Thank you very much.
It just a different method of getting the same result as you got. So, the method is correct.
Any way, follow the method you like.
Thanks
Thanks you very much Sir
You know nothing
I have a question is: If y=∣x−3∣+(3−x)^2 − 5, what is the value of x - y when y is the minimum?
At 6:48, the graph intercepts the X axis at (2,0) and (6,0) and the axis of symmetry is the line x=4.
How do you solve the same question with the Eqn X2 + 4x + 2.....it's coming -2+√2
Thank you sir for your help
Thanks for appreciation and your valuable time. All the Best!
2nd method is not used correctly. this method could only be used when the graph passes through the origin. in this equations case it doesnot.
Watch some videos to understand the second method. Thanks: ua-cam.com/play/PLJ-ma5dJyAqoylQuxhRGpFRGYQAFqMvcK.html
@@MathematicsTutor i disagree with the graph that you have made. the y intercept of this graph should be at y=12 . and its image should be at x=8 making the line of symmetry at x=4.
@@sadeedwaseem8124 Sir, I have drawn exactly what you are saying. However, your interpretation is different. The horizontal line is y = 12 and not the x-axis. So, please look at it again. Thanks
@@sadeedwaseem8124 anil kumar sir is saying right just check first .
It helped me, but this is not the same method we learned.
Sorry but why do we get different answers with the different methods, one is postive and the other negative which one is the actual answer
Thank you!
Thank you
Sir maximum and minimum value of a polynomial Kahi value nikal neka formula hai na ye?.......agar ye Max n min value of a quadratic polynomial ka nikal neke formula hai to.........Max n min value of a cubic Polynomial n higher order Polynomial ka value kese nikalte hai......sir plzz batana😧😭
Have you done calculus. Then the first derivative is zero at maximum/minimum.
Here is a Playlist with some examples: ua-cam.com/video/oLWWh31SpNI/v-deo.html
Here is a video with few examples with Boundary Conditions: ua-cam.com/video/bLESE2kkwMc/v-deo.html
Let me know if that helps
@@MathematicsTutor sir mujhe calculus ka Max n min nikalna atahai......paar mere syllabus me-General properties of polynomial,graphical representation of a polynomial,maximum and minimum value of a polynomial ye sab hai........ab sir mujhe ye janna tha ki cubic polynomial ka Max n min kese nikalta hai....
@@beinghuman5345 Post a picture of your questions on my email: anil.anilkhandelwal@gmail.com
I will make a video for you on this topic in Hindi
Thanks
SOmething else useful: the y-coordinate of the vertex will be c-(b^2/4a)
Good
Thank you sir,.Sir, would you please give a lecture on maximum and minimum value pertaining to probability...
i had no lit bit of knowledge about maxima and minima....so please suggest.....basic video to learn maxima and minima
Thanks, here are few videos to start with: Vertex represents maximum or a minimum for the given quadratic equation. If the parabola opens up we have minimum and if the parabola opens down we will have a maximum. Parabola opens up id the leading coefficient is positive (3 x^2). It opens down if the leading coefficient is negative (-3x^2). Here is how you can find vertex from different forms of quadratic equations. ua-cam.com/video/zQWBo6ThAO4/v-deo.html
Here is video to state max/min values. ua-cam.com/video/OlX0OJc0c8k/v-deo.html
Hope that helps
thanks tmrw i will watch your videos afrer wards. ....i tell my level ....thanks anil kumar
Good teaching sir.
Nice sir
Thank you sir!
edit: you are the savior, and you remind me of my former math teacher
Partial factor, second step x(x - 8) + 12 not able to understand how x in two points is equals 12 and value of x 0.
How?
Thanks and I hope that will help many others to understand.
y = x(x - 8) + 12. For x = 0 and and for x = 8 y = 12. Substitute the value and check.
Axis of symmetry is midway between these two x values
Thesefor y = (0 + 8)/2 is the equation of axis of symmetry on which the turning point will lie.
Hope that is better.
Thanks
yes. thanks
i wanna cryy
Why?
What is the maximum and minimum value of x square - x + 7
F(x)=(X+2)^2 how can i solve this problem
Thank u sir for this
Thanks for appreciation and your time.
Mad helpful
Sir pz explain in differetation method.
How did u write vertex.
amazing, thank you.
It helped
-4 squard is +16 sir you have put it -16
Thanks. Please note the -4^2 = -16 and (-4)^2 = 16
Minimum value of x2 - 2x + 5x
Min at x = 1, y = 4
@@MathematicsTutor Which way?
@@rantimadhu1048 Brah, do your homework, lol
16x2 - 8x + 1 minimum value
y = (4x - 1)^2
Minimum at x = 1/4, y = 0
Thanks
@@MathematicsTutor thanks 🙏🏻🙏🏻
But why is the minimum to the right solution -4 ? Is it wrong ?
f(x) = (x - 4)^2 - 4. Parabola opens upwards, so we have minimum defined by the vertex.
Minimum of -4 is at x = 4. or f(4) = -4.
Hope that is clear.
First Method enough.remaining methods confusion to students.
My advice Anil sir...u tell one method and direct and concept..dont tell methods ..that negative effect on ur carrer..be silent and energy stored inside U...be happy sir
Thank you sir
Thanks for appreciation
Thanks m8!
how you find a vertex
Here is my best video to find the vertex from different forms of Quadratic Equations.
ua-cam.com/video/cSZKrGdOlbQ/v-deo.html
Thanks for the request
Minimum value of polynomial 4x2-6x+1
At x = 3/4 find the value of y. That should be the minimum value. Try one of the methods discussed in this video and find the solution.
Let me know if that works. Thanks
4x3+9x-12 slove it please
9x
really helped thx m8
👍
"Do You understand"😂😂
X^2-5x+=0 please anyone solve it? I’m cryinggg I have no time
X2-5X = X(X-5)
Therefore, x=0 or 5.
@@ismartgaming8528 wow I commented that 3 months ago in term 1😳 Thank you anyway😄!!
@@Ahtohallan. it's basic sense that humans should help each other right? :) Anyway, I'm glad that I helped you.
@@ismartgaming8528 appreciated🥺✨
s
Wtf?
Chutiyapa
thank you!