The vertex of a parabola and completing the square

The graph of a quadratic function turns around at its tip, the vertex. Let us find the vertex of $y = x^2 - 2x + 3$ by completing the square¹.

Completing 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 $x^2 - 2x$ .

\begin{align*} x^2 - 2x + 3 &= (x^2 - 2x + 1) - 1 + 3 \\ &= (x - 1)^2 + 2 \end{align*}

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 there.

The axis

The vertical line $x = 1$ through the vertex is the axis of the parabola. Points equally far to the left and right of the axis have the same height.

$x$	$y$
$0$	$3$
$1$	$2$
$2$	$3$

Vertex form

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 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 general formula

Completing the square on $y = ax^2 + bx + c$ gives the following.

y = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}

So the $x$ -coordinate of the vertex is always $-\dfrac{b}{2a}$ . Here that is $1$ , matching what we found. The $y$ -coordinate follows by substituting back.

What the sign of $a$ decides

Sign of $a$	Opening	The vertex is
$a > 0$	upward	a minimum
$a < 0$	downward	a maximum

Either way, the vertex is the point that answers questions about largest and smallest values. The sign of its $y$ -coordinate also decides whether the parabola meets the $x$ -axis: here the vertex sits above the axis and the parabola opens upward, so the graph never crosses the $x$ -axis at all.

The two large dots on the graph are the vertex of $y = x^2$ and the vertex of $y = x^2 - 2x + 3$ .

Completing the square, Wikipedia