Another approach is to factor and rewrite it as y = 3(x+2)^2 - 7 I like how the answer can be immediately seen in this form. The vertex occurs at x+2=0 and y=-7
Excellent, so easy to learn , thank you! Wish I had math presented this way 45 years ago when in school! Learning again and enjoying this logic now is such a refreshing exercise for depression
or take the first derivative of the given parabola and set it equal to zero and evaluate (this works because of the minima of the parabola that is the vertex)
Since this is a parabola that opens upwards or downwards, you can also take the first derivative and the slope at the point where the vertex is will be zero. y= 3x²+12+5 Differentiate w.r.t x dy/dx = 6x+12 = 0 (the derivative is the slope of a function) 6x + 12 = 0 x= -2 Put the value of x back in the original equation y= 3(-2)² + 12(-2) + 5 y= -7 Hence the vertex is (-2,-7)
Professor, I figured it out differently, x = -b / 2ac , sos, x= -12/ (2*3*5) , x = -12/ 30, x incinclitude -2/ 5. incentitude, uclide, -.4. y=-b(a) + c where c is the inetitude of y sound value. y= -12 (3) + 5; y = -36 + 7, y= -37 + 7 + 1, -30 + 1, -29. x= -.4, y = -29, the x and y value for the vertex of eq.
This is rote memorization. Something we were not supposed to do, years ago. This only works because a) calculus says dy/dx = 0 gives a minimum or maximum value for a parabola WHEN and ONLY WHEN the parabola is not oriented on a skewed axis and there is an axis of symmetry parallel to the y axis. How to obtain the vertex of a SKEWED axis of symmetry parabola?
It also works to find the inflection point of a cubic, y=a*x^3 + b*x^2 + c*x + d. -b/(3*a) gives the x-position of the inflection point. In general, -b/(n*a) gives the x-position of the central point of a polynomial of degree n.
@@mrhtutoring both methods are nice. To be honest, I've been doing so much with measure theory and functional analysis that seeing things like this is nice again. I kinda miss this type of math honestly
@@mrhtutoring actually, doesn't this only work with parabolas? A dervative is more generalized method I would assert. Still I'll teach your method to my kids when they get to algebra:)
First derivative: Hold my beer.
you are right. But it is limit. not vertex. It is a parabola.
@@mishelreyes-z5o If you get first derivative and solve for y'=0, you get x = -b/2a
What if it is a rightward opening parabola(y² = +- 4ax)
@@ntard LOL, but it isn’t.
@@ntard we take dx/dy instead of dy/dx
Another approach is to factor and rewrite it as y = 3(x+2)^2 - 7
I like how the answer can be immediately seen in this form. The vertex occurs at x+2=0 and y=-7
Fantastic! Thank you, Professor!
Learned more from you than my college professor❤ thanks for making it easy to understand I really appreciate this🌸
Excellent, so easy to learn , thank you! Wish I had math presented this way 45 years ago when in school! Learning again and enjoying this logic now is such a refreshing exercise for depression
or take the first derivative of the given parabola and set it equal to zero and evaluate (this works because of the minima of the parabola that is the vertex)
works everytime??? would be so much eazier
wait, derivative is 6x+12 we equal to 0 means x=2. not -2 as in the video
@@seekingCK 6*x + 12 = 0, solves for x as x = -2. The 12 becomes negative, when you shuffle it to the opposite side.
@@carultch damn. Im sorry i was way to tired i think. Shame on me sry. Thx for the tipp tho, way faster that way
@@seekingCK I made the same mistake. LOL
Instead of plugging in you can use synthetic division, whatever the remainder is that’s ur min/max.
yeah, you can get the tangent of the equation, and you get also, another form of vertex of a parabola.
Thank you so much worlds best teacher
Since this is a parabola that opens upwards or downwards, you can also take the first derivative and the slope at the point where the vertex is will be zero.
y= 3x²+12+5
Differentiate w.r.t x
dy/dx = 6x+12 = 0 (the derivative is the slope of a function)
6x + 12 = 0
x= -2
Put the value of x back in the original equation
y= 3(-2)² + 12(-2) + 5
y= -7
Hence the vertex is (-2,-7)
Derivation Still seems to be much easier to Remember
What? This has to be satire lol, derivation is years more advanced. Although I'd use derivation for anything of larger degree then that of a quadratic
@@captainkibbles3502 Yes. Thanks derivation I do not have to Remember any formulas about polynoms
you are on track. but it is acceleration within the parabolic distance vs. change in motion. Physics! and or Calculus.
In the next video, show a figure that represents the result, this helps a lot with understanding.
Thank you
That surely is the easiest way to get the solution. Great work !
I think it’s better to either differentiate or factorise. These methods are much easier to remember than a formula and apply to other situations.
Another way to find the y of a vertex is using the equation c - (b^2 /4a).
-Δ/4a
miles better than my actual algebra teacher this saved me
Wow! Than you!
good way to find the vertex,untill you see the question required to use completing the square to find the vertex
Great work
y=ax^2+bx+c Xs=-b/2a
y=3x^2+12x+5 Xs= -12/(2•3)
Xs=-2 (put in the first equation)
y=3•(-2)^2+12•(-2)+5y=-7
The vertex of the parabola is
v=(-2/-7)
Professor, I figured it out differently,
x = -b / 2ac , sos, x= -12/ (2*3*5) , x = -12/ 30, x incinclitude -2/ 5. incentitude, uclide, -.4.
y=-b(a) + c where c is the inetitude of y sound value.
y= -12 (3) + 5; y = -36 + 7, y= -37 + 7 + 1, -30 + 1, -29.
x= -.4, y = -29, the x and y value for the vertex of eq.
Someone please find this man.
I want to thank him
You switched from two different markers, a white board, and being silent, to using a chalkboard and talking!
Hahaaha!
Thank you so much
Thank you sir!
Ti 84 with a min or max function is even easier.
This is rote memorization. Something we were not supposed to do, years ago. This only works because a) calculus says dy/dx = 0 gives a minimum or maximum value for a parabola WHEN and ONLY WHEN the parabola is not oriented on a skewed axis and there is an axis of symmetry parallel to the y axis. How to obtain the vertex of a SKEWED axis of symmetry parabola?
You used a lot of words to say "when y is a function of x".
Nice trick 😊😊
It also works to find the inflection point of a cubic, y=a*x^3 + b*x^2 + c*x + d.
-b/(3*a) gives the x-position of the inflection point.
In general, -b/(n*a) gives the x-position of the central point of a polynomial of degree n.
Those who are watching mrhtutoring shorts from India 👇
I think he mean inflection ( not vertex)
How did he get y = -7?
Thanku sir
Happy April Fools Day. 😅😅
(-b/2a,4ac-b^2/4a)
-2,-7
why do you have no comments
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dy/dx = 0
6x + 12 =0
x = -2
y = -7
I set the 1st order derivative to 0
😁🤪👍👋
First dervative?????
Yes, however it may require more steps
@@mrhtutoring both methods are nice. To be honest, I've been doing so much with measure theory and functional analysis that seeing things like this is nice again.
I kinda miss this type of math honestly
@@mrhtutoring actually, doesn't this only work with parabolas? A dervative is more generalized method I would assert.
Still I'll teach your method to my kids when they get to algebra:)