Completing the square: the vertex, the sketch and the range

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Quadratics open Paper 1, Pure Mathematics 1, which every AS and A Level candidate sits.

You met quadratics at IGCSE; here the aim is speed and exactness, because almost every later Paper 1 topic (functions, coordinate geometry, calculus) leans on this algebra.

Completing the square rewrites a quadratic so that $x$ appears only once, inside a squared bracket.

That one change tells you the vertex of the graph, its line of symmetry and the least or greatest value of the expression, with no calculus at all.

The case $a = 1$

Halve the coefficient of $x$ to get the number in the bracket, then subtract its square so that nothing has changed:

x^2 + bx + c = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2 + c.

For example, $x^2 - 6x + 11$ : half of $-6$ is $-3$ , so $(x - 3)^2$ gives $x^2 - 6x + 9$ , and we need $2$ more.

So $x^2 - 6x + 11 = (x - 3)^2 + 2$ .

Expand your answer in your head to check it: this catches most slips.

When $a$ is not 1

Take $a$ out of the $x^2$ and $x$ terms only, complete the square inside, then multiply back out:

ax^2 + bx + c = a\left[x^2 + \tfrac{b}{a}x\right] + c = a\left[\left(x + \tfrac{b}{2a}\right)^2 - \tfrac{b^2}{4a^2}\right] + c.

The constant inside the square bracket gets multiplied by $a$ when you remove the bracket.

Forgetting that factor is the classic slip.

When $a$ is negative

Papers often ask for a form such as $q - a(x + p)^2$ with $a > 0$ .

Take out $-a$ , and watch two signs: the $x$ coefficient inside the bracket changes sign, and the constant you subtract inside becomes an addition outside.

Write the square-bracket line out in full; it is where the marks are lost.

Reading the graph from the form

Suppose $y = a(x + p)^2 + q$ .

The squared bracket is never negative, and it is zero only when $x = -p$ .

If $a > 0$ , the graph is U-shaped. Its least value is $q$ , at $x = -p$ , so the vertex (a minimum point) is $(-p, q)$ .

If $a < 0$ , the graph is n-shaped. Its greatest value is $q$ , at $x = -p$ , and the vertex is a maximum point.

The line of symmetry is $x = -p$ .

The range of the function follows at once: $y \geqslant q$ when $a > 0$ , $y \leqslant q$ when $a < 0$ .

Notice the sign: $(x - 3)^2 + 2$ has its vertex at $x = +3$ , because the bracket is zero when $x = 3$ .

Sketching a quadratic

A sketch on these papers shows its key features as values, not a scale drawing.

For $y = ax^2 + bx + c$ , mark:

1. the shape (U for $a > 0$ , n for $a < 0$ );

2. the vertex, from the completed square;

3. the $y$ -intercept, $(0, c)$ ;

4. the $x$ -intercepts, if any, found by solving $y = 0$ . The completed square gives them straight away: set the bracket form to zero and square-root.

The diagram below shows all four features for $y = 2x^2 - 8x + 3 = 2(x - 2)^2 - 5$ (Worked example 2).

The U-shaped graph of y = 2x² − 8x + 3 with its minimum point (2, −5), a dashed guide from the vertex to the axes showing the line of symmetry x = 2, the y-intercept (0, 3) and the x-intercepts at about (0.419, 0) and (3.58, 0), which are exactly 2 − ½√10 and 2 + ½√10.

How the marks work

B1: a correct fact or value, earned whatever method you used. M1: a correct method applied to this question, with your numbers in; a slip later does not lose it. A1: a correct answer or step, which needs the M1 before it.

M1* then DM1: the DM1 is only available if the starred M1 was earned. B2 for one item means both marks with no error, one mark with one error.

AG: answer given; every step to the printed result must be on the page. FT: credit for correctly using your own earlier wrong answer. OE: any equivalent form is fine. AWRT: any answer which rounds to the stated value.

A correct answer with no working earns at most a special-case mark (SC), often nothing.

Marks once earned are kept, so a harmless extra line after a correct answer costs nothing, except where a final answer must stand alone, such as a final inequality.

Key terms

Completed-square form: a quadratic written as $a(x + p)^2 + q$ , in which $x$ appears once, inside the squared bracket.

Vertex: the turning point of the graph of a quadratic, at $(-p, q)$ for $y = a(x + p)^2 + q$ ; a minimum point when $a > 0$ and a maximum point when $a < 0$ .

Line of symmetry: the vertical line $x = -p$ through the vertex.

Least (or greatest) value: the value $q$ of the expression at the vertex.

Worked example 1: the case $a = 1$

Express $x^2 + 10x + 18$ in the form $(x + a)^2 + b$ , where $a$ and $b$ are constants, and state the coordinates of the vertex of $y = x^2 + 10x + 18$ . [3]

1. Half of 10 is 5: $(x + 5)^2 = x^2 + 10x + 25$ , and $18 - 25 = -7$ .

2. $x^2 + 10x + 18 = (x + 5)^2 - 7$ , so $a = 5$ B1, $b = -7$ B1

3. The bracket is zero when $x = -5$ , so the vertex is $(-5, -7)$ B1 FT (from your $a$ and $b$ )

Answer: $(x + 5)^2 - 7$ ; vertex $(-5, -7)$ .

Check: $(x + 5)^2 - 7 = x^2 + 10x + 25 - 7$ , which is the original.

Worked example 2: a leading coefficient, then the sketch

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

(b) Hence find the exact $x$ -coordinates of the points where $y = 2x^2 - 8x + 3$ crosses the $x$ -axis, and sketch the graph. [4]

1. (a) $2x^2 - 8x + 3 = 2\left[x^2 - 4x\right] + 3 = 2\left[(x - 2)^2 - 4\right] + 3$

2. $= 2(x - 2)^2 - 8 + 3 = 2(x - 2)^2 - 5$ B2 (all three of $a = 2$ , $b = -2$ , $c = -5$ ; one error earns B1)

3. (b) $2(x - 2)^2 - 5 = 0 \Rightarrow (x - 2)^2 = \tfrac{5}{2}$ M1 (the completed square from (a) used)

4. $x - 2 = \pm\sqrt{\tfrac{5}{2}} = \pm\tfrac{\sqrt{10}}{2}$ , so $x = 2 \pm \tfrac{1}{2}\sqrt{10}$ A1

5. Sketch: U-shape, vertex $(2, -5)$ B1; $y$ -intercept $(0, 3)$ and both $x$ -intercepts marked B1

Answer: $x = 2 - \tfrac{1}{2}\sqrt{10}$ and $x = 2 + \tfrac{1}{2}\sqrt{10}$ (about 0.419 and 3.58); the sketch is the diagram above.

Both roots sit symmetrically about $x = 2$ , which is a quick check.

The second sketch shows an n-shaped graph, for Worked example 3.

The n-shaped graph of y = 5 + 8x − 2x² with its maximum point (2, 13), a dashed guide from the vertex to the axes, the y-intercept (0, 5) and the x-intercepts at about (−0.550, 0) and (4.55, 0).

Worked example 3: a negative $x^2$ coefficient

Express $5 + 8x - 2x^2$ in the form $a - b(x - c)^2$ , where $a$ , $b$ and $c$ are positive constants.

Hence state the greatest value of $5 + 8x - 2x^2$ and the range of $f(x) = 5 + 8x - 2x^2$ for $x \in \mathbb{R}$ . [4]

1. $5 + 8x - 2x^2 = -2\left[x^2 - 4x\right] + 5$ (the sign of the $x$ term changes inside)

2. $= -2\left[(x - 2)^2 - 4\right] + 5 = -2(x - 2)^2 + 8 + 5$ (here $-2 \times -4 = +8$ )

3. $= 13 - 2(x - 2)^2$ B2 ( $a = 13$ , $b = 2$ , $c = 2$ ; one error earns B1)

4. The squared term is never negative and is subtracted, so the greatest value is 13, when $x = 2$ B1 FT

5. Range: $f(x) \leqslant 13$ B1 FT

Answer: $13 - 2(x - 2)^2$ ; greatest value 13; range $f(x) \leqslant 13$ .

The diagram above confirms it: the maximum point is $(2, 13)$ .

Worked example 4: an exam-length chain using the form

(a) Express $4x^2 - 12x + 13$ in the form $(2x - a)^2 + b$ . [2]

(b) Hence show that $4x^2 - 12x + 13 > 0$ for all real values of $x$ . [2]

1. (a) $(2x - a)^2 = 4x^2 - 4ax + a^2$ , so $4a = 12$ and $a = 3$ B1

2. $(2x - 3)^2 = 4x^2 - 12x + 9$ , so $b = 13 - 9 = 4$ , giving $(2x - 3)^2 + 4$ B1

3. (b) $(2x - 3)^2 \geqslant 0$ for every real $x$ M1

4. so $(2x - 3)^2 + 4 \geqslant 4 > 0$ , and therefore $4x^2 - 12x + 13 > 0$ for all real $x$ A1 AG

5. (c) The fraction is greatest when its positive denominator is least, which is 4 at $2x - 3 = 0$ M1

6. Greatest value $\tfrac{8}{4} = 2$ , when $x = \tfrac{3}{2}$ A1

Answer: (a) $(2x - 3)^2 + 4$ (b) shown (c) 2, at $x = \tfrac{3}{2}$ .

Part (b) needs the reason in words; a few trial values of $x$ prove nothing.

Part (c) works only because the denominator is always positive, which (b) has just shown.

Exam tip

The papers ask for a stated form such as $a(x + b)^2 + c$ or $a - b(x + c)^2$ ; give the constants in exactly that form and check by expanding.

Later parts often use the form silently, for a vertex, a range, an inverse or an equation, so look back at it before you start a later part from scratch.

Common mistakes

Sign slips with a negative $x^2$ coefficient. Taking out $-2$ from $-2x^2 + 8x$ gives $-2(x^2 - 4x)$ , and the $-4$ inside becomes $+8$ outside. Write the square-bracket line in full and expand your answer back to check.

The factor $a$ not applied to the subtracted square. $2\left[(x - 2)^2 - 4\right] + 3$ is $2(x - 2)^2 - 8 + 3$ , not $2(x - 2)^2 - 4 + 3$ .

Vertex sign reversed. $(x - 3)^2 + 2$ has its vertex at $(3, 2)$ , not $(-3, 2)$ .

Range not read from the form. Once you have $13 - 2(x - 2)^2$ , the range $f(x) \leqslant 13$ is immediate; examiners reported candidates missing it. Write a range with $f(x)$ or $y$ , never with $x$ .

Answering a different question. If the vertex is asked for, give both coordinates; if a sketch is asked for, label the vertex and both intercepts as values.

Check your understanding

1. Express $x^2 - 14x + 40$ in the form $(x - a)^2 - b$ , where $a$ and $b$ are constants. [2]

2. (a) Express $3x^2 + 18x + 20$ in the form $a(x + b)^2 + c$ . [2]
(b) Hence state the coordinates of the minimum point of $y = 3x^2 + 18x + 20$ and the equation of its line of symmetry. [2]

3. Show that $2x^2 - 10x + 17 > 0$ for all real values of $x$ . [3]

4. It is given that $f(x) = 7 - 6x - x^2$ for $x \in \mathbb{R}$ .
(a) Express $f(x)$ in the form $a - (x + b)^2$ . [2]
(b) State the greatest value of $f(x)$ and the range of $f$ . [2]
(c) Hence solve the equation $f(x) = 0$ . [2]

5. The curve $y = 2x^2 + px + q$ has its minimum point at $(3, -5)$ . Find the values of the constants $p$ and $q$ . [4]

6. Express $9x^2 + 12x + 7$ in the form $(3x + a)^2 + b$ . Hence find the least value of $9x^2 + 12x + 7$ and the value of $x$ at which it occurs. [4]

7. It is given that $y = x^2 - 2kx + k^2 + 2k - 3$ , where $k$ is a constant.
(a) Express $y$ in the form $(x - a)^2 + b$ , giving $a$ and $b$ in terms of $k$ . [2]
(b) Hence show that, whatever the value of $k$ , the vertex of the curve lies on the line $y = 2x - 3$ . [2]
(c) Find the set of values of $k$ for which $x^2 - 2kx + k^2 + 2k - 3 > 0$ for all real $x$ . [2]

Answers

1. $(x - 7)^2 - 49 + 40$ M1; $= (x - 7)^2 - 9$ , so $a = 7$ , $b = 9$ A1

2. (a) $3\left[(x + 3)^2 - 9\right] + 20 = 3(x + 3)^2 - 7$ B2 ( $a = 3$ , $b = 3$ , $c = -7$ ; one error B1)
(b) Minimum point $(-3, -7)$ B1 FT; line of symmetry $x = -3$ B1 FT

3. $2x^2 - 10x + 17 = 2\left[(x - \tfrac{5}{2})^2 - \tfrac{25}{4}\right] + 17$ M1; $= 2(x - \tfrac{5}{2})^2 + \tfrac{9}{2}$ A1; $(x - \tfrac{5}{2})^2 \geqslant 0$ , so the expression is at least $\tfrac{9}{2}$ , which is positive, for all real $x$ A1 AG (the reason must be stated)

4. (a) $7 - (x^2 + 6x) = 7 - \left[(x + 3)^2 - 9\right] = 16 - (x + 3)^2$ B1 ( $a = 16$ ) B1 ( $b = 3$ )
(b) Greatest value 16 B1 FT; range $f(x) \leqslant 16$ B1 FT
(c) $(x + 3)^2 = 16$ , $x + 3 = \pm 4$ M1 (the form from (a) used); $x = 1$ or $x = -7$ A1

5. Vertex form $y = 2(x - 3)^2 - 5$ M1 (vertex $(3, -5)$ put into $a(x + b)^2 + c$ with $a = 2$ ); $= 2(x^2 - 6x + 9) - 5$ M1 (expanded); $= 2x^2 - 12x + 13$ A1; $p = -12$ , $q = 13$ A1. Or: the line of symmetry $x = -\tfrac{p}{4} = 3$ gives $p = -12$ (M1 A1), then $18 - 36 + q = -5$ gives $q = 13$ (M1 A1).

6. $(3x + a)^2 = 9x^2 + 6ax + a^2$ , so $6a = 12$ , $a = 2$ B1; $b = 7 - 4 = 3$ , giving $(3x + 2)^2 + 3$ B1; least value $3$ B1 FT; when $3x + 2 = 0$ , $x = -\tfrac{2}{3}$ B1 FT

7. (a) $(x - k)^2 - k^2 + k^2 + 2k - 3$ M1; $= (x - k)^2 + 2k - 3$ , so $a = k$ , $b = 2k - 3$ A1
(b) The vertex is $(k, 2k - 3)$ B1 FT; at $x = k$ the line gives $y = 2k - 3$ , the same as the vertex's $y$ -coordinate, so the vertex lies on $y = 2x - 3$ for every $k$ A1 AG (the comparison must be stated)
(c) The least value $2k - 3$ must be positive M1; $k > \tfrac{3}{2}$ A1

Completing the square: the vertex, the sketch and the range

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