Completing the Square, Max/Min and Range

Q: Do I have to use completing the square, or can I differentiate?

For Unit 2, completing the square is the expected method — it uses only the algebra covered here. Differentiation is introduced later in Unit 14; once you know it, it gives a quick way to check the vertex, but any Unit 2 question can be answered by completing the square.

Q: What range notation is accepted in the exam?

Both inequality notation (e.g. ) and interval notation (e.g. ) are marked correct, as long as they match the function's behavior.

CIE IGCSE Additional Mathematics· 12 min read

1. Completing the Square for $ax^2 + bx + c$★★☆☆☆⏱ 3 min

📘 Definition

Completing the Square

Algebraic technique to rearrange a quadratic function into vertex form by constructing a perfect square trinomial from the quadratic and linear terms.

Example:

For $2 x^{2} + 8 x + 3$ , we rewrite the quadratic and linear terms as $2 (x^{2} + 4 x) + 3$ , then complete the square inside the bracket: $2 [(x + 2)^{2} - 4] + 3 = 2 (x + 2)^{2} - 5$ .

📐 Worked Example

Rewrite $3 x^{2} - 12 x + 7$ in the form $a (x + h)^{2} + k$ .

1
Factor out the coefficient of $x^{2}$ from the first two terms:
$3(x^2 -4x) +7$
2
Complete the square for the expression inside the bracket: half of -4 is -2, square is 4, so add and subtract 4:
$3[(x-2)^2 -4] +7$
3
Expand the bracket and simplify the constant terms:
$3(x-2)^2 -12 +7 = 3(x-2)^2 -5$
4
Final form is $3 (x - 2)^{2} - 5$ . Comparing with $a (x + h)^{2} + k$ , we read $h = - 2$ and $k = - 5$ , so the vertex is $(- h, k) = (2, - 5)$ .

📐 Worked Example

Express $2 x^{2} + 5 x + 3$ in the form $a (x + h)^{2} + k$ , giving $a$ , $h$ and $k$ as exact (rational) values.

1
Factor $2$ from the first two terms (the constant $+ 3$ stays outside):
$2\left(x^2 + \frac{5}{2}x\right) + 3$
2
Half of $\frac{5}{2}$ is $\frac{5}{4}$ ; squaring gives $\frac{25}{16}$ . Add and subtract it inside the bracket:
$2\left[\left(x + \frac{5}{4}\right)^2 - \frac{25}{16}\right] + 3$
3
Multiply the outer $2$ through the $- \frac{25}{16}$ (giving $- \frac{25}{8}$ ) and combine with $+ 3 = \frac{24}{8}$ :
$2\left(x + \frac{5}{4}\right)^2 - \frac{25}{8} + 3 = 2\left(x + \frac{5}{4}\right)^2 - \frac{1}{8}$
4
So $a = 2$ , $h = \frac{5}{4}$ , $k = - \frac{1}{8}$ , and the vertex is $(- \frac{5}{4}, - \frac{1}{8})$ . Fractional constants are normal in these questions — leave them as exact fractions, never round to decimals.

Exam tip:

Always check your answer by expanding the vertex form back to standard form to confirm it matches the original quadratic, especially in non-calculator papers.

2. Finding Maximum/Minimum Values from Vertex Form★★☆☆☆⏱ 2 min

📘 Definition

Vertex of a Quadratic

The turning point of the parabola: if $a > 0$ , it is the minimum point; if $a < 0$ , it is the maximum point. Its coordinates are $(- h, k)$ when the function is written as $a (x + h)^{2} + k$ .

The value of $k$ is the minimum or maximum output value of the function. For $a > 0$ the parabola opens upwards, so $k$ is the minimum value; for $a < 0$ it opens downwards, so $k$ is the maximum value. (Later, in Unit 14, you will meet differentiation, which gives another way to locate the vertex — but for this unit, completing the square is all you need.)

📐 Worked Example

Find the maximum value of the function $f (x) = - 2 x^{2} + 8 x - 3$ .

1
Complete the square. Factor out $- 2$ from the first two terms:
$f(x) = -2(x^2 - 4x) - 3$
2
Complete the square inside the bracket:
$f(x) = -2\left[(x-2)^2 - 4\right] - 3 = -2(x-2)^2 + 8 - 3 = -2(x-2)^2 + 5$
3
Since $a = - 2 < 0$ , the parabola opens downwards, so the maximum value is $k = 5$ , at $x = 2$ .
4
Optional check (differentiation, which you meet in Unit 14): $f^{'} (x) = - 4 x + 8 = 0$ gives $x = 2$ , and $f (2) = - 2 (4) + 16 - 3 = 5$ — the same maximum.

3. Stating the Range of a Quadratic Function★★★☆☆⏱ 3 min

The range of a quadratic function depends on its orientation (sign of $a$ ) and its vertex ( $k$ value). For all real values of $x$ , the range is either $f (x) \geq k$ (if $a > 0$ ) or $f (x) \leq k$ (if $a < 0$ ). If a restricted domain is given, you must also check the output values at the endpoints of the domain to find the full range.

📐 Worked Example

Find the range of the function $f (x) = x^{2} - 6 x + 4$ for the domain $0 \leq x \leq 5$ .

1
First rewrite in vertex form by completing the square:
$f(x) = (x^2 -6x) +4 = (x-3)^2 -9 +4 = (x-3)^2 -5$
2
$a = 1 > 0$ , so vertex at $(3, - 5)$ is the minimum point, and $x = 3$ is within the given domain.
3
Calculate $f (x)$ at the endpoints of the domain: $x = 0$ and $x = 5$ :
$f(0) = 0 -0 +4 =4, f(5)=25 -30 +4 = -1$
4
The maximum output in this domain is 4 at $x = 0$ , minimum is -5 at $x = 3$ . So the range is $- 5 \leq f (x) \leq 4$ .

Exam tip:

Always confirm if the domain is all real x or restricted, as this changes the range entirely. Highlight domain restrictions in the question to avoid missing this.

4. Using Vertex Form for Graph Sketching★★☆☆☆⏱ 2 min

For CIE 0606 exam questions asking you to sketch a quadratic graph, you only need to mark three key features: the vertex (turning point), the y-intercept, and any x-intercepts (roots). You do not need advanced curve sketching techniques beyond these points.

📐 Worked Example

Sketch the graph of $y = 2 x^{2} + 4 x - 1$ , marking all key points.

1
Rewrite in vertex form:
$y = 2(x^2 +2x) -1 = 2(x+1)^2 -2 -1 = 2(x+1)^2 -3$
2
Vertex is at $(- 1, - 3)$ , $a = 2 > 0$ so it opens upwards.
3
Y-intercept is at $x = 0$ : $y = - 1$ , so mark $(0, - 1)$ .
4
X-intercepts when $y = 0$ : solve $2 (x + 1)^{2} - 3 = 0 \to (x + 1)^{2} = 3/2 \to x = - 1 \pm \sqrt (1.5)$ , mark these points on the x-axis.
5
Draw a smooth upward opening parabola passing through all marked points.

5. Common Pitfalls

Wrong move:

Forgetting to multiply the subtracted constant by the factored coefficient $a$ when completing the square.

Why:

When you factor $a$ out of the first two terms, any adjustments inside the bracket are scaled by $a$ , leading to incorrect $k$ value if you skip this step.

Correct move:

After completing the square inside the factored bracket, expand the outer coefficient $a$ to both the perfect square and the subtracted constant before combining terms.

Wrong move:

Forgetting the sign flip when reading the vertex from $a (x + h)^{2} + k$ .

Why:

In $a (x + h)^{2} + k$ the vertex is $(- h, k)$ , so its $x$ -coordinate is $- h$ , not $h$ . Reading $(x + 2)^{2}$ as vertex $x = + 2$ instead of $- 2$ loses the mark.

Correct move:

After writing the completed-square form, negate the number inside the bracket to get the vertex $x$ -coordinate: $(x + 2)^{2} \Rightarrow$ vertex $x = - 2$ ; $(x - 2)^{2} \Rightarrow$ vertex $x = + 2$ .

Wrong move:

Stating $k$ as the maximum value when $a > 0$ , or minimum when $a < 0$ .

Why:

The orientation of the parabola determines if the vertex is a maximum or minimum, which is controlled by the sign of $a$ .

Correct move:

Always check the sign of $a$ first: $a > 0$ = upward opening = minimum at vertex; $a < 0$ = downward opening = maximum at vertex.

Wrong move:

Ignoring restricted domain when calculating range, using only the vertex value.

Why:

If the vertex is outside the given domain, the maximum/minimum will be at the endpoints of the domain.

Correct move:

For restricted domains, calculate the function value at the vertex (if it lies inside the domain) and at both endpoints, then take the smallest and largest values for the range bounds.

Wrong move:

Writing range in terms of $x$ instead of $f (x)$ or $y$ .

Why:

Range refers to output values, not input values, so using $x$ leads to incorrect notation.

Correct move:

Always write range using the output variable: e.g. $f (x) \geq 3$ or $y \leq 7$ , not $x \geq 3$ .

6. Quick Reference Cheatsheet

Task	Method	Result
Rewrite $a x^{2} + b x + c$ to vertex form	Factor $a$ from first two terms 2. Complete square inside bracket 3. Expand and simplify	$a (x + h)^{2} + k$
Find vertex coordinates	Read straight from the completed-square form; differentiation ( $f^{'} (x) = 0$ , Unit 14) is an optional check	$(- h, k)$
Find max/min value	Check sign of $a$ : $a > 0$ → $k$ is min, $a < 0$ → $k$ is max	$k$
Range for all real $x$	$a > 0$ : $f (x) \geq k$ ; $a < 0$ : $f (x) \leq k$	Inequality or interval notation
Range for restricted domain	Calculate $f (x)$ at vertex (if inside domain) and endpoints, take min/max	Bounded inequality

7. Frequently Asked

Do I have to use completing the square, or can I differentiate?

For Unit 2, completing the square is the expected method — it uses only the algebra covered here. Differentiation is introduced later in Unit 14; once you know it, it gives a quick way to check the vertex, but any Unit 2 question can be answered by completing the square.

What range notation is accepted in the exam?

Both inequality notation (e.g. $f (x) \geq 3$ ) and interval notation (e.g. $[3, \infty)$ ) are marked correct, as long as they match the function's behavior.

Going deeper

syllabusCIE 0606 2025-2027 SyllabusRefer to sections 2.1 and 2.2

What's Next

Now that you have mastered completing the square, max/min values and range, you can apply these skills to solve quadratic inequality problems, which are frequently tested alongside this topic in CIE IGCSE Additional Mathematics 0606 papers. You can also use your knowledge of vertex form to simplify solving quadratic equations and analyzing quadratic word problems, such as optimization questions involving projectile motion or area maximization. Ensure you practice both calculator and non-calculator questions, as this topic appears on both Paper 1 and Paper 2 of the exam.