Why does completing the square and solving for $y$ for $y = ax^2 + bx +c$ give the coordinates of the vertex?

Question

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?

Answer 1

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.

Answer 2

The vertex question is the same as you are finding a point that is at the maximum infimum or the minimum supermum of a function. In the case of a parabola, you can use $a^2 \geqslant 0 (a \in \mathbb R)$ So, using the perfect square, you will get the alternative form of $ax^2+bx+c$, $a(x-h)^2+k$，which indicates one inequity:for$a>0,a(x-h)^2+k \geqslant k$ and for $a<0, a(x-h)^2+k \leqslant k$. Hence, you find the vertex of the parabola, $(h,k)$.