Use derivatives to find the vertex of parabola y=x^2-4x+7?
Take the derivative:
y' = 2x - 4
Given that the original equation is concave up, we are looking for an absolute/local minimum. At a minimum, the slope is 0. So lets set the derivative equal to 0 and solve for x
0 = 2x - 4
x = 4/2 = 2.
The vertex is at point x = 2. You can plug this back into the original to get the y value.
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).