Rewrite \(ax^{2} + bx + c\) as \(a(x + p)^{2} + q\) and the maximum or minimum value of the quadratic, and the value of \(x\) where it happens, can be read straight off. This works for any \(a\), including negative \(a\), and it needs no calculator.
Write \(ax^{2} + bx + c = a(x + p)^{2} + q\). A square is never negative, so if \(a > 0\) the quadratic has a MINIMUM value \(q\), and if \(a < 0\) it has a MAXIMUM value \(q\). Either way it occurs when the bracket is zero, at \(x = -p\). The turning point of the graph is \((-p,\ q)\).
Fig. 1: in completed square form the turning point can be read off. y = (x - 2)² - 3 opens upwards, so -3 is its least value, at x = 2; y = 7 - 2(x - 1)² opens downwards, so 7 is its greatest value, at x = 1
Section 01
When a = 1: halve, square, adjust.
In the lab, set \(b\) to 10 and press Make the square. The \(10x\) strip splits into two halves of \(5x\) that wrap around the \(x^{2}\) square, leaving a 5 by 5 corner empty, so \(x^{2} + 10x = (x + 5)^{2} - 25\). With \(-10x\) the same working gives \((x - 5)^{2} - 25\). In symbols: \(x^{2} + bx = \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2}\). Halve \(b\), square the bracket, then take away the square you added.
Once you have \((x + p)^{2} + q\), the bracket is a square, so it is never negative and is zero only when \(x = -p\). The smallest the whole expression can be is therefore \(q\).
Worked example Even coefficient of x
Express \(x^{2} - 10x + 18\) in the form \((x + p)^{2} + q\). Hence state the minimum value of \(x^{2} - 10x + 18\) and the value of \(x\) at which it occurs.
Half of \(-10\) is \(-5\), so \(x^{2} - 10x = (x - 5)^{2} - 25\).
Answer: Minimum value \(-\dfrac{13}{4}\) when \(x = -\dfrac{3}{2}\)
Examiner trap
The sign in the bracket is the sign of \(b\), but the \(x\)-value of the turning point has the OPPOSITE sign: \((x - 5)^{2} - 7\) has its minimum at \(x = 5\), not \(x = -5\). On a non-calculator paper keep fractions such as \(-\frac{13}{4}\) exact; do not round to \(-3.3\).
Section 02
When a is not 1, or is negative.
Take \(a\) out of the \(x^{2}\) and \(x\) terms only, complete the square inside the bracket, then multiply back out the number you subtracted. The constant term stays outside the whole time.
Sign of \(a\) in \(a(x + p)^{2} + q\)
Shape
Value of \(q\)
Where
\(a > 0\)
opens upwards
minimum (least) value
\(x = -p\)
\(a < 0\)
opens downwards
maximum (greatest) value
\(x = -p\)
Worked example Positive a
Express \(3x^{2} + 12x + 5\) in the form \(a(x + b)^{2} + c\), and state its minimum value.
Take out 3 from the first two terms: \(3(x^{2} + 4x) + 5\).
Multiply out the \(-4\): \(3(x + 2)^{2} - 12 + 5 = 3(x + 2)^{2} - 7\).
\(3(x + 2)^{2} \ge 0\) and is zero at \(x = -2\).
Answer: \(3(x + 2)^{2} - 7\); minimum value \(-7\) when \(x = -2\)
Worked example Negative a: a maximum
Express \(5 + 4x - 2x^{2}\) in the form \(p - q(x - r)^{2}\). Hence find the maximum value of \(5 + 4x - 2x^{2}\) and the value of \(x\) at which it occurs.
Take out \(-2\) from the \(x\) terms: \(-2x^{2} + 4x = -2(x^{2} - 2x)\).
When you take \(a\) out, the number you subtract inside the bracket is multiplied by \(a\) on the way out. In \(3\left[(x + 2)^{2} - 4\right] + 5\) the \(-4\) becomes \(-12\), not \(-4\). With a negative \(a\), take out \(-2\) and change BOTH signs inside: \(-2x^{2} + 4x = -2(x^{2} - 2x)\).
Stage 1 · Learn
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Completed square form also works backwards: if you know the maximum or minimum value and where it occurs, you can rebuild the quadratic. It also finds the greatest value of a fraction whose denominator is a quadratic, because the fraction is greatest where the denominator is least.
A check by differentiation. At a turning point the gradient is zero. For \(y = 2x^{2} - 8x + 11\), \(\dfrac{dy}{dx} = 4x - 8 = 0\) gives \(x = 2\), and \(y = 8 - 16 + 11 = 3\), matching \(2(x - 2)^{2} + 3\). You will meet this method properly in the calculus topic; for now completing the square is the main method.
Worked example Rebuild from the minimum
The expression \(x^{2} + px + q\) has a minimum value of 4, which occurs when \(x = -3\). Find \(p\) and \(q\).
The coefficient of \(x^{2}\) is 1 and the minimum is 4 at \(x = -3\), so \(x^{2} + px + q = (x + 3)^{2} + 4\).
Find the greatest value of \(\dfrac{12}{x^{2} - 4x + 7}\) and the value of \(x\) at which it occurs.
Complete the square in the denominator: \(x^{2} - 4x + 7 = (x - 2)^{2} + 3\).
The denominator is always positive and its least value is 3, at \(x = 2\).
A fixed positive numerator divided by the SMALLEST denominator gives the LARGEST fraction: \(\dfrac{12}{3} = 4\).
Answer: Greatest value 4 when \(x = 2\)
Examiner trap
For the fraction, the least value of the denominator gives the GREATEST value of the fraction. Do not report the denominator's least value (3) as the answer, and do not just put \(x = 0\) in to get \(\frac{12}{7}\).
Stage 2 · Exam
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Express \(2x^{2} - 10x + 3\) in the form \(a(x + b)^{2} + c\).
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Find the greatest value of \(1 + 6x - 3x^{2}\) and the value of \(x\) at which it occurs.
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The minimum value of \(x^{2} - 6x + k\) is 4. Find the value of \(k\).
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Find the greatest value of \(\dfrac{20}{x^{2} + 2x + 6}\).
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Turning points read straight from the square.
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