Maximum and minimum of a quadratic function

The maximum or minimum of a quadratic function is found at the vertex of its graph¹. Let us find the maximum of $y = -x^2 + 4x - 1$ .

Opening downward

Since the coefficient of $x^2$ is negative, this graph is a downward-opening parabola. The vertex is then the highest point, where the function attains its maximum. There is no lower bound, so no minimum.

Completing the square

\begin{align*} -x^2 + 4x - 1 &= -(x^2 - 4x) - 1 \\ &= -\left((x - 2)^2 - 4\right) - 1 \\ &= -(x - 2)^2 + 3 \end{align*}

So the vertex is $(2, 3)$ . Since $-(x - 2)^2$ is $-1$ times a square, it is always at most $0$ , and it is largest when $x = 2$ . The maximum value $3$ is taken there.

Two routes to the same point

Method	Computation	Result
Completing the square	$-(x-2)^2 + 3$	vertex $(2, 3)$
Vertex formula	$x = -\dfrac{b}{2a} = -\dfrac{4}{-2}$	$x = 2$
Calculus	$y' = -2x + 4 = 0$	$x = 2$ , and $y'' < 0$

Completing the square finds the vertex from the shape of the expression; differentiating finds it from where the slope is $0$ .

Maximum or minimum

Sign of $a$ in $y = a(x - p)^2 + q$	At $x = p$
$a < 0$	maximum $q$
$a > 0$	minimum $q$

Whether it is a maximum or a minimum depends only on the sign of the coefficient of $x^2$ .

A restricted interval

One case deserves care: when $x$ is restricted to an interval. If the vertex lies outside it, the largest and smallest values occur at the endpoints instead. Take this same function on $0 \leq x \leq 1$ .

$x$	$y$	Role
$0$	$-1$	minimum
$1$	$2$	maximum
$2$	$3$	vertex, outside the interval

With a restricted domain, always check whether the vertex is inside it.

An application

To maximize the area of a rectangle of perimeter $20$ : with height $x$ the width is $10 - x$ , so the area is $S = x(10 - x) = -x^2 + 10x$ , whose vertex is at $x = 5$ . The area is greatest, at $25$ , when the rectangle is the $5$ by $5$ square.

The large dot on the graph is the vertex, where the maximum is attained.

Quadratic function, Wikipedia