Express a quadratic in completed-square form and use it to find the vertex, the maximum or minimum value, and the line of symmetry

What this dot point is asking

SEAB wants you to rewrite a quadratic $ax^2 + bx + c$ in completed-square form $a(x - h)^2 + k$, because that form hands you the vertex $(h, k)$, the maximum or minimum value, and the line of symmetry $x = h$ at a glance. It is the single most useful manipulation for understanding a parabola.

The answer

Why completed-square form is useful

A squared term is never negative. So in $a(x - h)^2 + k$, the bracket contributes its least value $0$ when $x = h$, and the whole expression equals $k$ there. If $a > 0$ the parabola opens upward and $k$ is the minimum; if $a < 0$ it opens downward and $k$ is the maximum.

Completing the square when a equals 1

For $x^2 + bx + c$, halve the coefficient of $x$, square it, add and subtract it:

The vertex is at $x = -\dfrac{b}{2}$.

Completing the square when a is not 1

Factor $a$ out of the $x$ terms first, complete the square inside the bracket, then expand the constant back out:

Reading off the key features

From $a(x - h)^2 + k$: the vertex (turning point) is $(h, k)$; the line of symmetry is the vertical line $x = h$; the maximum or minimum value is $k$, and which it is depends on the sign of $a$.

Sketching from the completed-square form

The completed-square form gives everything needed for a quick sketch: plot the vertex $(h, k)$, draw the axis of symmetry through it, note whether the curve opens up or down from the sign of $a$, and mark the $y$-intercept by setting $x = 0$ in the original. The shape follows without a table of values.

Solving equations by completing the square

Completing the square also solves a quadratic equation: rearranging $a(x - h)^2 + k = 0$ to $(x - h)^2 = -\dfrac{k}{a}$ and taking the square root gives the roots directly. This is the manipulation that derives the quadratic formula itself.

Examples in context

Example 1. Maximum height of a projectile. A height model $h = -5t^2 + 20t + 2$ completes to $h = -5(t - 2)^2 + 22$, so the greatest height is $22$ metres at $t = 2$ seconds, found instantly without calculus.

Example 2. Minimising cost. A cost function $C = 3x^2 - 12x + 50$ completes to $3(x - 2)^2 + 38$, showing the least cost is $38$ at $x = 2$ units, the kind of optimisation a quadratic model captures directly.

Try this

Q1. Express $x^2 + 8x + 3$ in the form $(x + a)^2 + b$. [2 marks]

• Cue. Half of $8$ is $4$: $(x + 4)^2 - 16 + 3 = (x + 4)^2 - 13$.

Q2. State the minimum value of $(x - 5)^2 + 7$ and where it occurs. [1 mark]

• Cue. Minimum value $7$ at $x = 5$.

Q3. Express $2x^2 - 12x + 1$ in completed-square form. [3 marks]

• Cue. $2(x^2 - 6x) + 1 = 2[(x - 3)^2 - 9] + 1 = 2(x - 3)^2 - 17$.