Complete a square and read a quadratic vertex
Build a square without changing the expression, then use it to explain a turning point and a minimum.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Paper 1 is part of every route. Written practice is self-checked; it does not award official examination marks.
Before you begin
- Expand (x + p)² and collect like terms.
- Substitute a value into a quadratic expression.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which is the expansion of (x − 3)²?
What you will learn
- Rewrite x² + bx + c in completed-square form.
- Read a vertex and minimum from that form, with attention to the domain.
Add a square and compensate
The expansion (x + p)² = x² + 2px + p² shows why the number inside the bracket is half the coefficient of x. For x² + bx + c, choose p = b/2. The square introduces b²/4, so subtract that same amount outside the square. You are rewriting an expression, not solving an equation.
For x² − 8x + 3, (x − 4)² introduces a constant 16. Subtract 16 and add the original 3: (x − 4)² − 13. Expanding back gives x² − 8x + 16 − 13, so the original expression is unchanged.
The smallest square is zero
For real x, (x − h)² ≥ 0, with equality exactly when x = h. Thus y = (x − h)² + k has vertex (h, k), symmetry line x = h and minimum y = k when the domain includes h. Read x from the point at which the bracket becomes zero: (x + 3)² has vertex x = −3, not 3.
A sketch should show the opening direction, vertex and relevant intercepts. For x² − 8x + 3, the y-intercept is 3 and the roots are 4 ± √13. If the domain is restricted and excludes the vertex, consider the permitted interval instead of assuming its endpoint behaviour matches the unrestricted graph.
Pause and explain
What is the vertex of y = (x + 2)² − 7?
Worked example
Complete the square for y = x² − 8x + 3 and give its vertex, symmetry line and minimum over all real x.
Show the worked solution
- Half of −8 is −4, so begin with (x − 4)².
- Its constant is 16; compensate: x² − 8x + 3 = (x − 4)² − 13.
- The square is zero at x = 4, giving the lowest y-value −13.
Answer y = (x − 4)² − 13; vertex (4, −13); symmetry line x = 4; minimum −13.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Complete the square for x² + 6x + 2 and locate the vertex of its graph.
Give me a hint
Half the coefficient of x, then subtract the new square's constant.
Compare my reasoning
- Half of 6 is 3, giving (x + 3)².
- Expand the square's constant: 3² = 9, so subtract 9 and add 2.
- The expression is (x + 3)² − 7; the bracket vanishes at x = −3.
(x + 3)² − 7; vertex (−3, −7).
Look for these in your work
- Compensate for all 9 introduced by the square.
- Give both vertex coordinates with their signs.
Write x² − 5x + 1 in completed-square form using exact fractions.
Give me a hint
Halving an odd coefficient produces a fraction.
Compare my reasoning
- Half of −5 is −5/2, so use (x − 5/2)².
- The square adds 25/4; the outside constant is 1 − 25/4 = −21/4.
- Expand (x − 5/2)² − 21/4 to confirm x² − 5x + 1.
(x − 5/2)² − 21/4.
Look for these in your work
- Keep the half-coefficient and correction exact.
- Check the rewritten expression by expansion.
A supplied model is H(t) = −t² + 8t + 1 metres for 0 ≤ t ≤ 8 seconds. Use a square to find the largest modelled height and its time.
Give me a hint
Factor a minus sign from the quadratic terms before completing the square.
Compare my reasoning
- Write H = −(t² − 8t) + 1.
- Since t² − 8t = (t − 4)² − 16, H = 17 − (t − 4)².
- The square is smallest at t = 4, which belongs to the supplied interval; H = 17.
17 m at t = 4 s.
Look for these in your work
- Explain why subtracting a square gives a maximum.
- Check that the vertex time is in the model's stated domain.
Common mistakes
- Subtract the constant introduced by the square; do not change the expression's value.
- The vertex x-coordinate makes the bracket zero, so its sign is opposite to the written constant.
- A restricted domain may exclude the unrestricted vertex.
What can you explain now?
Rewrite x² + 10x + 7 and give its minimum over all real x.
Compare with the explanation
(x + 5)² − 18; minimum −18 at x = −5.
The square adds 25, so 7 − 25 = −18. Its zero is at x = −5.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Without notes, derive the half-coefficient rule by expanding a square. Then explain why the sign inside a bracket differs from the vertex x-coordinate.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.