# Completing the Square, Max/Min and Range

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u2-completing-the-square-max-min/

This guide teaches you to rewrite quadratics in vertex form via completing the square or differentiation, find max/min values, and state correct range for CIE IGCSE Add Maths 0606 exam questions.

**Prerequisites:** Expanding and factoring quadratic expressions; Basics of quadratic function graphs

## Learning objectives

- Rewrite $ax^2+bx+c$ in vertex form $a(x+h)^2+k$ using completing the square
- Identify vertex, maximum/minimum value of a quadratic from vertex form
- Calculate the range of a quadratic function for all real x or given domain
- (Optional, introduced in Unit 14) Use differentiation as a cross-check on the vertex coordinates

## Completing the Square for $ax^2 + bx + c$

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

**Worked example:** Rewrite $3x^2 - 12x + 7$ 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 $2x^2 + 5x + 3$ 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 $\left(-\frac{5}{4},\, -\frac{1}{8}\right)$. 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.

## Finding Maximum/Minimum Values from Vertex Form

**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.)

> **tip**
>
> For now, completing the square is the method to reach for — it only needs the algebra from this unit. Differentiation (Unit 14) can later serve as a quick independent check, but it is not required here.

**Worked example:** Find the maximum value of the function $f(x) = -2x^2 + 8x - 3$.

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): $f'(x) = -4x + 8 = 0$ gives $x=2$, and $f(2) = -2(4) + 16 - 3 = 5$ — the same maximum.

## Stating the Range of a Quadratic Function

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 $f(x) ≥ k$ (if $a>0$) or $f(x) ≤k$ (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 $f(x) = x^2 -6x + 4$ for the domain $0 ≤x ≤5$.

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 $-5 ≤ f(x) ≤4$.

> **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.

## Using Vertex Form for Graph Sketching

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 $y = 2x^2 +4x -1$, 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 $2(x+1)^2 -3=0 → (x+1)^2= 3/2 → x = -1 ±√(1.5)$, mark these points on the x-axis.
5. Draw a smooth upward opening parabola passing through all marked points.

## Common pitfalls

- **Wrong:** Forgetting to multiply the subtracted constant by the factored coefficient $a$ when completing the square.
  - Why it fails: 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: 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:** Forgetting the sign flip when reading the vertex from $a(x+h)^2 + k$.
  - Why it fails: 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: 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:** Stating $k$ as the maximum value when $a>0$, or minimum when $a<0$.
  - Why it fails: The orientation of the parabola determines if the vertex is a maximum or minimum, which is controlled by the sign of $a$.
  - Correct: Always check the sign of $a$ first: $a>0$ = upward opening = minimum at vertex; $a<0$ = downward opening = maximum at vertex.
- **Wrong:** Ignoring restricted domain when calculating range, using only the vertex value.
  - Why it fails: If the vertex is outside the given domain, the maximum/minimum will be at the endpoints of the domain.
  - Correct: 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:** Writing range in terms of $x$ instead of $f(x)$ or $y$.
  - Why it fails: Range refers to output values, not input values, so using $x$ leads to incorrect notation.
  - Correct: Always write range using the output variable: e.g. $f(x) ≥ 3$ or $y ≤ 7$, not $x ≥3$.

## Cheatsheet

| Task | Method | Result |
| --- | --- | --- |
| Rewrite $ax^2+bx+c$ to vertex form | 1. 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 $x$ | $a>0$: $f(x) ≥k$; $a<0$: $f(x) ≤k$ | Inequality or interval notation |
| Range for restricted domain | Calculate $f(x)$ at vertex (if inside domain) and endpoints, take min/max | Bounded inequality |

## 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.

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