The coordinates of the vertex are $(h, k)$ where $h = \frac{-b}{2a}$ and $k=\frac{-b^2+4ac}{4a}$. Since the axis of symmetry passes through the vertex, the x-coordinate of the vertex is the midpoint of the x-intercepts. This fact provides a rather intuitive way of deriving the x-coordinate of the vertex by taking the average of the zeros of a quadratic.
The two zeros (roots) of a quadratic are $\frac{-b\pm\sqrt{b^2 -4ac}}{2a}$ and the resulting sum will be $\frac{-b}{a}$. Then to get the average just divide by 2 and the average is $\frac{-b}{2a}$
Is there some intuitive way, similar to the above, to show why manipulating the form of a quadratic by completing the square gives the coordinates of a vertex?
Best Answer
The vertex of the parabola $a x^2$ is at 0.
The parabola $a (x-h)^2+k$ is the same parabola but translated $h$ units to the right and $k$ units up. So the $x$ coordinate of the vertex is now at $h$.
So whenever you write your parabola in the form $a (x-h)^2+k$, (e.g. by completing the square) you automatically see where the vertex is.