Completing the square, the vertex and sketching

What you need to be able to do

What you should already know

The graph of a quadratic: the parabola

The graph of every quadratic \(y=ax^2+bx+c\) is a parabola. Four features describe it completely:

• Shape: \(a>0\) gives a ∪ (it opens upwards); \(a<0\) gives a ∩. The bigger \(|a|\), the narrower the curve.
• Vertex (turning point): the lowest point of a ∪, the highest point of a ∩.
• Line of symmetry: the vertical line through the vertex. Points the same distance either side of it have the same \(y\).
• Intercepts: it crosses the \(y\)-axis at \((0,\,c)\), and the \(x\)-axis at the roots (if there are any).

Here is the idea this whole page rests on. \(y=(x-2)^2-3\) is the graph of \(y=x^2\) moved 2 right and 3 down: the bracket shifts it sideways, the number outside shifts it up or down. So its vertex is where the vertex of \(y=x^2\) lands: \((2,\,-3)\). If we can rewrite any quadratic in that shape, the vertex is in front of us. Rewriting it is called completing the square. (Moving graphs in general is on the page Transformations in the Functions chapter.)

Three ways to write the same quadratic

Every quadratic can be written in three forms. They are the same curve; each form makes a different fact easy to read off.

Not every quadratic factorises with whole numbers (\(x^2-4x+1\) doesn't), but every quadratic can be completed. That is why Paper 1 leans on completing the square so heavily: it turns up inside functions (range, inverse), coordinate geometry (circles) and transformations questions as well.

Why completing the square works

Completing the square: the method, case by case

The method changes slightly with the form of the quadratic. Learn the general steps, then the seven forms Paper 1 actually uses.

1. Make the \(x^2\) coefficient 1 inside a bracket: if \(a\ne1\), take \(a\) out of the \(x^2\) and \(x\) terms only (not the constant).
2. Halve the coefficient of \(x\) (inside the bracket). That half goes in the square: \((x+\text{half})^2\).
3. Subtract the half squared, because \((x+\text{half})^2\) has added it.
4. Multiply back out by \(a\) and collect the numbers: you get \(a(x+p)^2+q\).
5. Check by expanding your answer: it must give back the original quadratic. This takes 20 seconds and catches every sign slip.

Two methods on the same question

There are two standard ways to complete the square, and both are correct. The second, expanding and comparing coefficients, makes fewer slips with awkward forms like \((4x+a)^2+b\). Here is one question done both ways: express \(3x^2-12x+7\) in the form \(p(x+q)^2+r\).

Reading the vertex, the least or greatest value and the line of symmetry

Once you have \(y = a(x+p)^2+q\), everything about the turning point is visible. The key fact: a square is never negative, \((x+p)^2 \ge 0\), and it equals 0 only when \(x = -p\).

Sketching a quadratic

Using the completed square to prove something is always positive

Because \((x+p)^2\ge0\) for every \(x\), \(a(x+p)^2+q\) with \(a>0\) is always \(\ge q\). If \(q>0\), the expression is always positive: it never equals zero, so the equation has no real roots.

Key facts at a glance

• Every quadratic graph is a parabola: ∪ if \(a>0\), ∩ if \(a<0\).
• \(x^2+bx=\left(x+\tfrac b2\right)^2-\left(\tfrac b2\right)^2\): halve, square, subtract.
• \(a(x+p)^2+q\): vertex \((-p,\,q)\), line of symmetry \(x=-p\), least (\(a>0\)) or greatest (\(a<0\)) value \(q\).
• A square is \(\ge0\), so \(a(x+p)^2+q\) with \(a>0\) and \(q>0\) is always positive.
• Always expand your answer to check it.

How this is examined

Completing the square turns up in almost every Paper 1, often inside a Functions question (range, inverse) or a circle. The June 2021 report found that candidates who expanded and compared coefficients made fewer slips with forms like \((4x+a)^2+b\). Proving an expression is always positive by trying values earns nothing (November 2021 report).

Schemes award B2 for all three values (B1 for two). They also say "condone correct values in an expression of the correct form, even if missing the square on the bracket", so if the question lists \(p, q, r\), state them as well as writing the expression.

The November 2022 report on a sketch like this: weaker answers "plotted points, plotted a graph with a maximum point or thought that the coordinates of the stationary point were where the graph crossed the axes". A sketch is a shape plus labelled key points, not a plot.

Mark-scheme focus

Express … in the form … (the form decides the marks)

Credited

• \(2[(x-2)^2+3]\) B1 B1: one mark for \(a=2\), one for \(b=3\)

Not credited

• \(2(x-2)^2+6\): equal to the quadratic, but not the form asked for: scores B1 B0
• \(2[(x+2)^2+3]\): the sign in the bracket

Why

Read the brackets in the question before you start. Here the 2 multiplies everything, so it comes out of all three terms: 2[x² − 4x + 7].

From: 9709/12/F/M/22 Q5(a).

Find the coordinates of the minimum point

Credited

• completing the square as far as \((x-4)^2+\dots\) M1, or \(\dfrac{dy}{dx}=2x-8=0\), or \(x=-\dfrac{b}{2a}\)
• \((4,\ -11)\) A1

Not credited

• using the quadratic formula: \(\dfrac{8\pm\sqrt{64-20}}{2}\) then averaging the roots: M0 A0
• \((-4,\ -11)\): the sign trap

Why

Any of the three methods for the turning point is accepted, but the roots of \(y=0\) are not a method for the vertex. If a question says "by completing the square", only the completed square earns the M1.

From: 9709/12/O/N/23 Q6(a).

Increasing for all x: argue from the completed square

Credited

• (a) \(5(y-3)^2+5\) B1 B1
• (b) \(f'(x)=5x^4-30x^2+50\) B1
• \(=5(x^2-3)^2+5\), using (a) with \(y=x^2\) M1
• \(>0\) for all \(x\), so f is increasing A1

Not credited

• checking \(f'(x)>0\) at a few values of \(x\)
• "increasing" with no reason, or the reason with no conclusion

Why

The November 2021 report: "showing a function is positive for a selection of x values does not prove it is positive for all values of x." The square is never negative, so 5(x² − 3)² + 5 ≥ 5 > 0 covers every x.

Also accepted

\(b^2-4ac<0\) for \(5y^2-30y+50\) together with one value of \(f'(x)\) that is positive.

From: 9709/13/O/N/21 Q3.

Worked examples

Example 1 · Completed square, then no real roots, then exact roots (Paper 1, 5 marks) ★★ Medium

1 Read it with me what is actually happening?

One quadratic, three jobs. First rewrite it in completed-square form. Then use that form twice: to see which values of \(k\) the quadratic can never reach, and to solve an equation exactly.

• Hencepart (b) must use part (a): the completed square shows the least value straight away
• no real rootsthe equation \(9x^2-36x+8=k\) has no solution: \(k\) is below every value the quadratic can take
• exact rootsleave surds in: \(2+\frac{\sqrt{13}}{3}\), not 3.20

2 What is it asking for? the target, part by part

• (a)\(p,\ q,\ r\)2 marks
• (b)set of values of \(k\)1 mark
• (c)exact roots2 marks

3 What do we know? everything the question gives us

• \(9x^2-36x+8\)the quadratic
• \(=k\)(b)
• \(=-15\)(c)

• in completed-square form, the least value of the quadratic is the constant at the end

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a) Find\(p(x+q)^2+r\)
  • Usetake 9 out of the x terms, halve −4, subtract, multiply back
• (b) Find\(k\) with no real roots
  • Use\(9(x-2)^2\ge0\), so the quadratic is never below \(r\)
    • Needsr from (a)
• (c) Findexact roots
  • Useset \(9(x-2)^2-28=-15\), isolate the square, square-root (±)
  • Trapforgetting ± gives only one root

Why is the answer to (b) a range of k, not one value?

Every \(k\) below the least value \(-28\) is unreachable, so there are infinitely many: \(k<-28\).

Should (b) be k < −28 or k ≤ −28?

At \(k=-28\) there IS a root (\(x=2\)), a repeated one. "No real roots" needs strictly below: \(k<-28\).

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \(9(x^2-4x)+8 = 9[(x-2)^2-4]+8 = \mathbf{9(x-2)^2-28}\) B2
2. (b) \(9(x-2)^2\ge0\), so \(9x^2-36x+8\ge-28\). No real roots when \(\mathbf{k<-28}\) B1FT
3. (c) \(9(x-2)^2-28 = -15\Rightarrow (x-2)^2 = \tfrac{13}{9}\) M1
4. \(x-2 = \pm\tfrac{\sqrt{13}}{3}\Rightarrow x = \mathbf{2\pm\tfrac{\sqrt{13}}{3}}\) (or \(\tfrac{6\pm\sqrt{13}}{3}\)) A1

The scheme also accepts the formula in (c): \(x = \tfrac{36\pm\sqrt{36^2-4\times9\times23}}{18}\). It allows "ISW if decimals given after the exact answers", but decimals alone lose the A1.

6 Check it does the answer make sense?

Expand (a): \(9(x^2-4x+4)-28 = 9x^2-36x+8\) ✓. In (c), \(-15 > -28\), so roots must exist ✓; \(2\pm1.20\) gives 3.20 and 0.80, and \(9(3.2)^2-36(3.2)+8 \approx -15\) ✓.

Example 2 · A negative x² coefficient and the vertex (Paper 1, 3 marks) ★ Easy

1 Read it with me what is actually happening?

The curve is an upside-down parabola (the \(x^2\) term is negative). We rewrite it so the turning point can be read off, then state that point.

• −a(x − b)²the form already has the minus outside; \(a\), \(b\), \(c\) are positive
• vertexthe turning point: here a maximum, because the curve is ∩

2 What is it asking for? the target, part by part

• (a)\(a,\ b,\ c\) and the vertex3 marks

3 What do we know? everything the question gives us

• \(-2x^2+8x+11\)the quadratic

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findthe vertex
  • Use\(-a(x-b)^2+c\) has vertex \((b,\ c)\)
    • Firstcomplete the square with \(-2\) taken out
    • Trap\(8x\div(-2) = -4x\): the sign flips

When you take −2 out of 8x, what is left inside?

\(-4x\). Check: \(-2\times(-4x) = 8x\).

Is the vertex a maximum or a minimum?

A maximum: the squared term is multiplied by \(-2\), so it can only pull \(y\) down from 19.

5 Now solve it have a go on paper first, then open the worked solution

1. \(-2x^2+8x+11 = -2(x^2-4x)+11\) M1
2. \(= -2[(x-2)^2-4]+11 = -2(x-2)^2+19\) DM1
3. So \(a=2,\ b=2,\ c=19\); vertex \(\mathbf{(2,\ 19)}\) A1

6 Check it does the answer make sense?

Put \(x=2\) into the original: \(-8+16+11 = 19\) ✓. Expanding \(-2(x-2)^2+19\) gives \(-2x^2+8x-8+19\) ✓.

Example 3 · Completed square, "hence" solve, then sketch (Paper 1, 9 marks) ★★ Medium

1 Read it with me what is actually happening?

A three-part question where each part feeds the next. Part (a) is completing the square with \(a=4\) and an odd \(x\)-coefficient after dividing, so expect fractions. Part (b) says "hence": solve using the form from (a). Part (c) wants the shape and the turning point.

• Henceuse (a): isolate the bracket, then square-root - no quadratic formula
• stationary pointthe vertex - read it from the completed square
• not required to indicate where the curve crosses the axesso don't waste time on intercepts

2 What is it asking for? the target, part by part

• (a)\(a,\ b,\ c\)3 marks
• (b)both values of \(x\)3 marks
• (c)a sketch with the vertex labelled3 marks

3 What do we know? everything the question gives us

• \(y=4x^2+20x+6\)the curve
• \(=45\)for (b)

• \(b = \tfrac52\) is a fraction: keep it as a fraction, not 2.5 rounded later

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)\(a(x+b)^2+c\)
  • Usetake 4 out: \(x^2+5x\), half of 5 is \(\tfrac52\)
• (b)solve = 45
  • Use\(4\left(x+\tfrac52\right)^2-19=45\Rightarrow\left(x+\tfrac52\right)^2=16\)
    • Trap±4: there are two answers
• (c)sketch
  • Vertex\(\left(-\tfrac52,\ -19\right)\) from (a)
  • Shape\(a=4>0\): ∪

Why is the vertex at x = −5/2 and not +5/2?

The bracket \(x+\tfrac52\) is zero when \(x=-\tfrac52\).

Once (x + 5/2)² = 16, what is the next line?

\(x+\tfrac52 = \pm4\). Taking only +4 loses an answer.

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \(4(x^2+5x)+6 = 4\left[\left(x+\tfrac52\right)^2-\tfrac{25}{4}\right]+6 = \mathbf{4\left(x+\tfrac52\right)^2-19}\) B1 B1 B1
2. (b) \(4\left(x+\tfrac52\right)^2-19 = 45\Rightarrow\left(x+\tfrac52\right)^2 = 16\) M1
3. \(x+\tfrac52 = \pm4\) DM1 \(\Rightarrow x = \mathbf{\tfrac32}\) or \(\mathbf{-\tfrac{13}{2}}\) A1
4. (c) ∪-shaped curve, seen on both sides of the vertex B1; minimum point \(\left(-\tfrac52,\ -19\right)\) labelled B1FT B1FT

6 Check it does the answer make sense?

Substitute \(x=\tfrac32\): \(4(2.25)+30+6 = 45\) ✓. Both roots are equally far (4) from \(-\tfrac52\): symmetry ✓.

Example 4 · f(x) > 2 for every x (Paper 1, 4 marks) ★★ Medium

1 Read it with me what is actually happening?

The graph of \(y = x^2-6x+c\) is a ∪ shape. "\(f(x)>2\) for all \(x\)" means the whole curve sits above the line \(y=2\) - in particular its lowest point does.

• for all values of xit must be true at the worst point: the minimum
• set of possible valuesthe answer is an inequality in \(c\)

2 What is it asking for? the target, part by part

• the set of values of \(c\)4 marks

3 What do we know? everything the question gives us

• \(f(x) = x^2-6x+c\)\(c\) unknown
• \(f(x)>2\)for every x

• the least value of \(f(x)\) must be more than 2

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• Findthe set of \(c\)
  • Useleast value of f > 2
    • Needsthe least value: complete the square, \((x-3)^2-9+c\)
  • Trap\(c\ge11\): at \(c=11\) the least value equals 2, which is not \(>2\)

Where is f(x) smallest?

At \(x=3\), where \((x-3)^2=0\); the least value is \(c-9\).

Could you use the discriminant instead?

Yes: \(x^2-6x+c-2>0\) for all \(x\) means no real roots, so \(36-4(c-2)<0\). Same answer, \(c>11\).

5 Now solve it have a go on paper first, then open the worked solution

1. \(x^2-6x+c = (x-3)^2-9+c\), so \((x-3)^2-9+c>2\) M1 A1
2. \((x-3)^2\ge0\), so the condition is \(-9+c>2\) M1
3. \(\mathbf{c>11}\) A1

Alternative 1 (calculus): \(f'(x)=2x-6=0\Rightarrow x=3\), \(f(3)=c-9>2\Rightarrow c>11\). Alternative 2 (discriminant): \(x^2-6x+(c-2)>0\) always, so \(b^2-4ac = 36-4(c-2)<0\Rightarrow c>11\). All three are on the scheme.

6 Check it does the answer make sense?

Try \(c=12\): \(f(x)=(x-3)^2+3\ge3>2\) ✓. Try \(c=11\): \(f(3)=2\), not \(>2\) ✓ so 11 itself is excluded.

Example 5 · A bracket starting 4x, and the one root (Paper 1, 4 marks) ★★★ Hard

Why it is hard: the June 2021 report says most candidates found k and stopped, or never linked (b) to (a).

1 Read it with me what is actually happening?

Part (a) is completing the square in an unusual form. Part (b) is the clever bit: "exactly one root" means the line \(y=k\) just touches the curve - at its vertex.

• (4x + a)²the bracket starts with 4x; do not take 16 out
• exactly one roota repeated root: \(k\) is the least value, and the root is where the square is zero

2 What is it asking for? the target, part by part

• (a)\(a\) and \(b\)2 marks
• (b)the root (x)2 marks

3 What do we know? everything the question gives us

• \(16x^2-24x+10\)
• \(=k\)exactly one root

• the question asks for the root (an \(x\)-value), not for \(k\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

• (a)\((4x+a)^2+b\)
  • Useexpand: \(16x^2+8ax+a^2+b\), compare
• (b)the single root
  • Use\((4x-3)^2+1=k\) has one root only when \((4x-3)^2=0\)
    • So\(k=1\) and \(4x-3=0\)
  • Trapstopping at k = 1: the question wants x

Why does 'exactly one root' happen at the vertex?

The line \(y=k\) cuts a ∪ twice, misses it, or touches it once - at the lowest point.

What if you used the discriminant instead?

\(24^2-4(16)(10-k)=0\) gives \(k=1\); you then still need the root, \(x = -\tfrac{b}{2a} = \tfrac{24}{32} = \tfrac34\).

5 Now solve it have a go on paper first, then open the worked solution

1. (a) \((4x+a)^2+b = 16x^2+8ax+a^2+b\): \(8a=-24\Rightarrow a=-3\); \(9+b=10\Rightarrow b=1\). \(\mathbf{(4x-3)^2+1}\) B1 B1
2. (b) One root when \((4x-3)^2+1=k\) has a repeated root: \(k=1\) B1FT
3. \((4x-3)^2=0\Rightarrow x = \mathbf{\tfrac34}\) B1

The June 2021 report: most candidates "did not see the connection between the two parts" and used the discriminant; "only the strongest responses realised that this value of k needed to be used to find the value of the root."

6 Check it does the answer make sense?

\(16\left(\tfrac34\right)^2-24\left(\tfrac34\right)+10 = 9-18+10 = 1 = k\) ✓.

Summary

• \(x^2+bx = \left(x+\tfrac b2\right)^2-\left(\tfrac b2\right)^2\). If \(a\ne1\), take \(a\) out of the \(x\) terms first; multiply the subtracted number by \(a\) when you put it back.
• \(y=a(x+p)^2+q\): vertex \((-p,\ q)\); minimum if \(a>0\), maximum if \(a<0\); line of symmetry \(x=-p\).
• Match the exact form the question asks for; compare coefficients when the form is unusual.
• A square is never negative: that proves "always positive" / "for all \(x\)" statements.

Can I do this?

Quiz

Concept checks

Answer each one in your head first, then open it.

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.