Please, show how to use derivatives to find the vertex of parabola y=4x^2-16x+23?
Well, it is simple. The vertex of parabola, in thi case, represents the minimum point of the function.
We know that we can find out a local extreme point of a function for each critical value of the function. The critical value of the function is the root of the first derivative of that function.
We'll differentiate the given function with respect to x:
f'(x) = 8x - 16
We'll cancel it out:
f'(x) = 0
8x - 16 = 0
8x = 16
x = 16/8
x = 2
The critical value of the function is x = 2, that means that the vertex of the function has the x coordinate, x = 2, and y coordinate will be found replacing x by 2, in the expression of parabola.
f(2) = 4*2^2 - 16*2 + 23
f(2) = 16 - 32 + 23
f(2) = -16 + 23
f(2) = 7
The coordinates of the vertex of the parabola 4x^2-16x+23 are: V(2 ; 7).