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
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. :)
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. 😊
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
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 want you to be my schools math teacher my math teacher in my school is very bs and no one understands her english it will really help me and my class if you become our math teacher she keeps targeting random people for no reason
@@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.
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! 🎉❤
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
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)
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.
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)...
You'd be dealing with a parabola with imaginary roots and hence does not cut the x-axis. Best way to deal with this is probably differential calculus. There are other ways. For instance, x^2+1=0. If we let f(x)=x^2+1 => f'(x)=2x, which is the derivative of f(x); we locate the stationary points. Thus, when f'(x)=0 => 2x=0 => x=0. Substituting x=0 in f(x) => f(0)=1. Hence vertex is (0,1).
@@gaopinghu7332 No problem at all. Just remember that the vertex is a stationary point of the given parabola and at stationary points, dy/dx=0 or f'(x)=0, as there's no change in the gradient at this point. So, this knowledge prompts us to think of the derivative and lets us apply it the way you've seen in the example I gave.
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.
