How to calculate the vertex of the curve y=2x^2-4x+4 using derivatives?
The vertex of the curve is the point where the curve has a point of minima or maxima.
Here y = 2x^2 - 4x + 4
To find the x-coordinate of the extreme point we solve for x in y'=0.
y' = 4x - 4
4x - 4 = 0
=> x = 1
At x = 1, y = 2 - 4 + 4 = 2
The vertex is (1, 2)
The vertex of the given parabola represents the extreme point of the function.
To determine the extreme point, we'll have to calculate the critical value of the function.
The critical value of the function is the zero of the 1st derivative.
y' = 4x-4
We'll cancel the 1st derivative:
4x-4 = 0
4x = 4 => x = 1
f(1) = 2 - 4 + 4
f(1) = 2
The vertex of the parabola is the minimum point of the function and it's coordinates are (1,2).