Find quadratic turning points and ranges
Complete the square to locate a quadratic's maximum or minimum and use it to state the range.
Section 2 · Quadratic functions
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does the form a(x − h)² + k reveal?
What you will learn
- Complete the square for a quadratic function
- Identify its maximum or minimum value
- Use a turning point and a stated domain to find the range
Completing the square exposes the vertex
Rewrite ax² + bx + c as a(x − h)² + k. The turning point is (h, k). If a is positive, the square term is non-negative and k is the minimum; if a is negative, k is the maximum.
For q(x) = x² − 6x + 5, q(x) = (x − 3)² − 4. Its minimum is −4 at x = 3. On the unrestricted real domain, the range is q(x) ≥ −4.
A restricted domain can change the extrema
If the turning point lies outside the allowed interval, compare endpoint values. For a continuous quadratic on a closed interval, the least and greatest outputs occur at a turning point inside the interval or at an endpoint.
Use interval notation or inequalities that clearly include or exclude endpoints as the question requires.
Pause and explain
If q(x) = 2(x − 1)² + 3 for all real x, what is its range?
Worked example
Find the minimum and range of q(x) = x² − 6x + 5 for 0 ≤ x ≤ 2.
Show the worked solution
- Complete the square: q(x) = (x − 3)² − 4.
- The turning point x = 3 is outside [0, 2], and the parabola decreases across this interval.
- Compare endpoints: q(0) = 5 and q(2) = −3.
Answer Minimum −3; range −3 ≤ q(x) ≤ 5.
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 and find the minimum of x² + 8x + 13.
Give me a hint
Take half the x-coefficient, square it, and balance the constant.
Compare my reasoning
- x² + 8x + 13 = (x + 4)² − 16 + 13.
- Simplify to (x + 4)² − 3.
- The square is zero when x = −4.
Minimum −3 at x = −4.
Look for these in your work
- Square completed
- Constant balanced
- Minimum located
Find the range of r(x) = −2(x − 1)² + 5 for −1 ≤ x ≤ 2.
Give me a hint
The maximum is at the vertex; compare the endpoints for the minimum.
Compare my reasoning
- The vertex is (1, 5), so the maximum is 5.
- r(−1) = −3 and r(2) = 3.
- The least endpoint value is −3.
−3 ≤ r(x) ≤ 5.
Look for these in your work
- Maximum uses negative leading coefficient
- Both endpoint values checked
- Range includes endpoints
A rectangular garden has area A(x) = 20x − x² square metres for 0 ≤ x ≤ 20. Find the width x that gives the greatest area and state that area.
Give me a hint
Complete the square, then check the vertex is in the physical domain.
Compare my reasoning
- A(x) = −(x² − 20x) = −(x − 10)² + 100.
- The coefficient is negative, so the vertex gives a maximum.
- x = 10 lies in [0,20].
A width of 10 m gives a maximum area of 100 m².
Look for these in your work
- Model transformed
- Maximum identified
- Domain checked and units stated
Common mistakes
- Changing the sign of the completed-square constant
- Using a turning point outside the stated domain
- Giving a maximum or minimum without a range when one is requested
What can you explain now?
What is the minimum of (x + 2)² − 7?
Compare with the explanation
−7 at x = −2
A square is at least zero, so the least value occurs when x + 2 = 0.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Complete the square and explain how the leading coefficient determines a maximum or minimum.
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.