Completing the square rewrites a quadratic as a(x + p)² + q. The turning point is then (−p, q): a minimum if a > 0 and a maximum if a < 0.
This lesson is part of quadratic functions. For the matching ideas on roots, see using the discriminant to classify roots.
Why does the form a(x + p)² + q show the turning point?
A squared term is never negative, and it is exactly zero when x + p = 0. So a(x + p)² + q is smallest when x = −p, with value q, if a is positive.
This one fact gives you the turning point at once, without any calculus or graphing.
Worked example 1: a = 1
Here is an illustrative example. Write f(x) = x² − 6x + 11 in the form (x + p)² + q.
- Take half of the coefficient of x: −6 ÷ 2 = −3.
- Write (x − 3)², which expands to x² − 6x + 9.
- Adjust the constant: 11 − 9 = 2.
- So f(x) = (x − 3)² + 2.
The minimum value is 2, at x = 3. The turning point is (3, 2). Check: f(3) = 9 − 18 + 11 = 2.
Worked example 2: a = 2
Write g(x) = 2x² + 8x + 3 in the form a(x + p)² + q.
- Factor out 2 from the x-terms only: g(x) = 2(x² + 4x) + 3.
- Complete the square inside: x² + 4x = (x + 2)² − 4.
- Substitute: g(x) = 2[(x + 2)² − 4] + 3.
- Expand the outer 2: g(x) = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5.
The minimum value is −5, at x = −2. Check: g(−2) = 2(4) − 16 + 3 = −5.
Worked example 3: a negative a
Write h(x) = −x² + 4x + 1 in the form a(x + p)² + q.
- Factor out −1 from the x-terms: h(x) = −(x² − 4x) + 1.
- Complete the square: x² − 4x = (x − 2)² − 4.
- Substitute: h(x) = −[(x − 2)² − 4] + 1 = −(x − 2)² + 4 + 1.
- So h(x) = −(x − 2)² + 5.
Since a is negative, the graph opens downward and 5 is the maximum value, at x = 2. Check: h(2) = −4 + 8 + 1 = 5.
The mistake that costs marks
The common slip in example 2 is to forget to multiply the −4 by the 2 that was factored out. The working then reads 2(x + 2)² − 4 + 3 = 2(x + 2)² − 1.
| Step | Wrong | Right |
|---|---|---|
| After completing the square | 2[(x + 2)² − 4] + 3 | 2[(x + 2)² − 4] + 3 |
| Expanding the outer 2 | 2(x + 2)² − 4 + 3 | 2(x + 2)² − 8 + 3 |
| Result | 2(x + 2)² − 1 | 2(x + 2)² − 5 |
Substitute the turning point to check. With the wrong form, the minimum would be −1, but g(−2) = −5. The numbers disagree, so the wrong form is found out.
Check yourself
Write f(x) = 3x² − 12x + 5 in the form a(x + p)² + q, and state the minimum value and where it occurs.
Answer
f(x) = 3(x² − 4x) + 5 = 3[(x − 2)² − 4] + 5 = 3(x − 2)² − 12 + 5 = 3(x − 2)² − 7.
The minimum value is −7, at x = 2.
Check: f(2) = 12 − 24 + 5 = −7.
What to study next
Knowing the turning point and the roots lets you describe where a quadratic is positive or negative. Continue with solving quadratic inequalities using intervals, then work through the practice set.
For a teacher to watch your bracket step on new cases, see online one-to-one Additional Mathematics tuition.