For y = a(x + p)² + q, the turning point is (−p, q). If a > 0 it is a minimum with least value q. If a < 0 it is a maximum with greatest value q.
This lesson follows completing the square within quadratic structure and discriminants, and it feeds directly into questions on range and optimisation.
How does completed-square form show the turning point?
A squared bracket is never negative. In 2(x − 2)² + 3, the bracket part is 0 when x = 2 and positive everywhere else. So the smallest total is 3, reached at x = 2.
That gives three separate facts: the position x = 2, the value 3 and the point (2, 3). Keeping them apart avoids most answer-format errors.
How do I read each piece?
- Complete the square to get a(x + p)² + q.
- Turning point x-coordinate: make the bracket zero, so x = −p.
- Turning point y-coordinate: this is q.
- Decide maximum or minimum from the sign of a.
- Write the range from q, in the direction the curve opens.
Worked example
The function is f(x) = 2x² − 8x + 11. Find the turning point, the minimum value and the range of f.
Step 1, complete the square: 2(x² − 4x) + 11 = 2[(x − 2)² − 4] + 11 = 2(x − 2)² − 8 + 11 = 2(x − 2)² + 3.
Step 2, turning point: x − 2 = 0 gives x = 2, and the value there is 3. The turning point is (2, 3).
Step 3, type: a = 2 is positive, so it is a minimum with least value 3.
Step 4, range: f(x) ≥ 3.
Check: f(2) = 8 − 16 + 11 = 3. Also f(0) = 11 and 2(0 − 2)² + 3 = 11.
A context version: the height of a ball is h = −5t² + 20t + 1 metres.
Then h = −5(t² − 4t) + 1 = −5[(t − 2)² − 4] + 1 = −5(t − 2)² + 21. So the greatest height is 21 m, at t = 2 s. Check: h(0) = −20 + 21 = 1.
The mistake to watch for
A frequent slip is to copy the sign inside the bracket into the coordinate.
Mistaken answer: for 2(x − 2)² + 3, the turning point is (−2, 3)
The student read “−2” from the bracket instead of the value that makes the bracket zero.
Test it: f(−2) = 8 + 16 + 11 = 35, nowhere near the minimum of 3. The correction is to solve “bracket = 0” every time, so that (x − 2) gives x = 2 and (x + 2) gives x = −2.
Check yourself
1. Find the turning point of y = x² + 10x + 20.
Show answer
x² + 10x + 20 = (x + 5)² − 25 + 20 = (x + 5)² − 5. The turning point is (−5, −5). Check: 25 − 50 + 20 = −5.
2. Find the greatest value of 7 + 6x − x² and the value of x where it occurs.
Show answer
7 + 6x − x² = −(x² − 6x) + 7 = −[(x − 3)² − 9] + 7 = −(x − 3)² + 16. Greatest value 16 at x = 3. Check: 7 + 18 − 9 = 16.
3. For f(x) = 3x² − 12x + 5, state the range of f and find the values of k for which f(x) = k has no real solution.
Show answer
f(x) = 3(x² − 4x) + 5 = 3[(x − 2)² − 4] + 5 = 3(x − 2)² − 7. The range is f(x) ≥ −7. There is no real solution for k < −7, because the curve never goes that low. Check: f(2) = 12 − 24 + 5 = −7.
Where this leads next
The next lesson, solve a parameter condition for repeated roots, links the minimum value to the discriminant. Try it on the quadratic structure explorer, where the turning point appears on the graph, and use the non-calculator working trainer to practise the arithmetic.
If giving the right answer to the wrong question keeps costing marks, our teachers in online one-to-one Additional Mathematics tuition can help you build a checking routine that fits how you work.