Writing a quadratic expression of the form x squared plus bx plus c in the completed-square form (x plus p) squared plus q, and using it to identify the turning point of the parabola.

What this dot point is asking

The SQA wants you to rewrite a quadratic expression $x^2 + bx + c$ in the completed-square form $(x + p)^2 + q$, and to use that form to state the turning point and minimum (or maximum) value of the parabola. At National 5 the coefficient of $x^2$ is $1$, which keeps the method clean.

What completing the square does

A quadratic such as $x^2 + bx + c$ can always be rewritten as a single squared bracket plus a constant: $(x + p)^2 + q$. This completed-square form is useful because the squared bracket is never negative, so the smallest the whole expression can be is $q$, reached when the bracket is zero. That tells you the turning point of the graph without any calculus.

The method

The key idea is that $(x + p)^2 = x^2 + 2px + p^2$. Comparing this with $x^2 + bx$, the coefficient of $x$ tells you $2p = b$, so $p = \dfrac{b}{2}$. The square then carries an unwanted $+p^2$, which you subtract back off.

Reading the turning point

In completed-square form $y = (x + p)^2 + q$, the bracket is zero when $x = -p$, and at that point $y = q$. Because a squared term is never negative, this is the lowest point of an upward parabola.

The line of symmetry of the parabola passes vertically through the turning point, so for $y = (x - 3)^2 - 4$ it is the line $x = 3$. This is useful for sketching: once you have the turning point and you know whether the curve opens up or down, you can plot the y-intercept (the value of $y$ when $x = 0$) and reflect it in the line of symmetry to get a second point quickly.

Why the method works

The whole trick rests on the identity $(x + p)^2 = x^2 + 2px + p^2$. The middle term of the expanded bracket is $2px$, so to match the $bx$ in $x^2 + bx$ you need $2p = b$, that is $p = \tfrac{b}{2}$. Squaring that half then produces a constant $p^2$ that was never in the original expression, so you subtract it straight back off. Seeing why you halve and subtract makes the steps memorable rather than mechanical, and it stops the most common error of forgetting the correction.

Examples in context

Completing the square answers "what is the best possible value" questions. If the cost of producing $x$ thousand items, in thousands of pounds, is modelled by $C = x^2 - 8x + 25$, completing the square gives $C = (x - 4)^2 + 9$. The cost is least when the bracket is zero, at $x = 4$ thousand items, where the minimum cost is $\pounds 9$ thousand. The completed-square form reads the optimum straight off without trial and error.

Try this

Q1. Express $x^2 + 4x + 1$ in the form $(x + p)^2 + q$. [2 marks]

• Cue. $(x + 2)^2 - 3$.

Q2. Express $x^2 - 2x - 6$ in completed-square form. [2 marks]

• Cue. $(x - 1)^2 - 7$.

Q3. State the turning point of $y = x^2 + 8x + 10$. [2 marks]

• Cue. $(x + 4)^2 - 6$, so the turning point is $(-4, -6)$.