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).