Use derivatives to find the vertex of parabola y=x^2-4x+7?

The vertex of parabola represents an extreme point. In
this case the parabola is convex and the vertex is it's minimum
point.

Any extreme of a function can be determined using
derivatives.

We'll determine f'(x) =
y':

f'(x) = 2x - 4

We'll
cancel f'(x):

2x - 4 = 0 => x - 2 = 0 => x =
2

The critical value of the function is x = 2. The extreme
of the function will be calculated replacing x, into the expression of f(x), by the
critical value found earlier.

f(2) = 2^2 - 4*2 +
7

f(2) = 4 - 8 + 7

f(2) =
3

The vertex of parabola is represented by
the pair of coordinates: (2 , 3).