Why $(h,k)$ in equation $y= a(x-h)^2 +k$ is the vertex of a parabola?

Question

As in the title , I know how to convert normal explicit equation to a vertex form equation by completing the square . But what is the reasoning behind why $(h,k)$ must be the vertex , but not other parts of the graphs ?

Answer 1

A straight parabola (i.e., one with its directrix parallel to the $\;y$- axis) has its maximal or minimal point at its vertex (depending on its leading coefficient's sign), and

$$f(x)=y=(x-h)^2+k\implies f'(x)=2(x-h)=0\iff \color{red}{x=h}$$

and then

$$f(h)=k$$

so the vertex is indeed at $\;(h,k)\;$ , and it is a minimum point iff

$$f''(h)=2>0\;\;\color{green}\checkmark$$

Answer 2

Because the vertex of a parabola is the point $(\bar{x},\bar{y})$ such that $\bar{y}$ is maximum (if $a < 0$) or minimum (if $a > 0$), and these situations both happen for $\bar{x} = h$, which ultimately leads to $\bar{y} = k$. You can also think of the symmetry of the parabola in the $y$-axis. Meaning, find the roots of $a(x-h)^2+k$ and take their arithmetic mean. You'll get $h$.