The vertex of a parabola and completing the square

The graph of a quadratic function (a parabola) turns around at its tip, the vertex. Let us find the vertex of $y = x^2 - 2x + 3$.

To do this we complete the square. Half of the coefficient $-2$ of the linear term is $-1$, and we use that $(x - 1)^2 = x^2 - 2x + 1$ contains $x^2 - 2x$.

Since $(x - 1)^2$ is a square, it is always at least $0$, and it is smallest when $x = 1$, where $y = 2$. So the vertex is $(1, 2)$, and because $a = 1 > 0$ the parabola opens upward, taking its minimum value $2$ at $x = 1$.

The vertical line $x = 1$ through the vertex is the axis of the parabola. For example $x = 0$ and $x = 2$ both give $y = 3$: points equally far to the left and right of the axis have the same height.

The form $(x - 1)^2 + 2$ shows that this graph is $y = x^2$ shifted $1$ to the right and $2$ up; indeed the vertex has moved from the origin $(0, 0)$ to $(1, 2)$. In general $y = a(x - p)^2 + q$ is $y = ax^2$ shifted $p$ right and $q$ up, with vertex $(p, q)$ and axis $x = p$. The two large dots on the graph are the vertex $(0, 0)$ of $y = x^2$ and the vertex $(1, 2)$ of $y = x^2 - 2x + 3$.