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Completing the Square for Maximum/Minimum Values

CSEC Additional Mathematics Essential Knowledge: Completing the square is a powerful algebraic technique that transforms a quadratic expression into vertex form. This form reveals the vertex (turning point) of the parabola, allowing us to easily identify maximum or minimum values—a crucial skill for optimization problems in CSEC exams.

Key Concept: For any quadratic function $f(x) = ax^2 + bx + c$ (where $a \neq 0$), we can rewrite it in the form $f(x) = a(x – h)^2 + k$. The point $(h, k)$ is the vertex of the parabola. If $a > 0$, the vertex represents a minimum point. If $a < 0$, the vertex represents a maximum point.

Part 1: Why Complete the Square?

🎯

The Power of Vertex Form

📊

Three Forms of Quadratic Functions

Standard Form

$f(x) = ax^2 + bx + c$

Advantages: Easy to identify y-intercept (c)

Disadvantages: Hard to see vertex

Vertex Form

$f(x) = a(x – h)^2 + k$

Advantages: Vertex (h, k) is visible

Disadvantages: Need to complete square to get it

Factored Form

$f(x) = a(x – r_1)(x – r_2)$

Advantages: Roots are visible

Disadvantages: Doesn’t exist if no real roots

Vertex Form: $f(x) = a(x – h)^2 + k$
Vertex = $(h, k)$ • $a > 0$ = Minimum • $a < 0$ = Maximum

⚡

Why Vertex Form Matters for Max/Min

When a quadratic is in vertex form $f(x) = a(x – h)^2 + k$:

The squared term $(x – h)^2$ is always ≥ 0
If $a > 0$: The minimum occurs when $(x – h)^2 = 0$, giving $f(x) = k$
If $a < 0$: The maximum occurs when $(x - h)^2 = 0$, giving $f(x) = k$
The vertex $(h, k)$ is the turning point of the parabola

∪

$a > 0$
Minimum at Vertex
Opens Upward

∩

$a < 0$
Maximum at Vertex
Opens Downward

Part 2: The Step-by-Step Process

📝

Completing the Square: Method Explained

🔢

Standard Procedure for $x^2 + bx + c$

Step 1: Start with $x^2 + bx + c$ (coefficient of $x^2$ must be 1)

Step 2: Take half of $b$, square it: $\left(\frac{b}{2}\right)^2$

Step 3: Add AND subtract this square: $x^2 + bx + \left(\frac{b}{2}\right)^2 – \left(\frac{b}{2}\right)^2 + c$

Step 4: Group the perfect square: $\left[x^2 + bx + \left(\frac{b}{2}\right)^2\right] – \left(\frac{b}{2}\right)^2 + c$

Step 5: Factor the perfect square: $\left(x + \frac{b}{2}\right)^2 + \left[c – \left(\frac{b}{2}\right)^2\right]$

\[x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 + \left[c – \left(\frac{b}{2}\right)^2\right]\]

📝 Example 1: Basic Completing the Square

Express $x^2 + 6x + 5$ in the form $(x + p)^2 + q$.

Identify $b = 6$, $c = 5$

Half of $b$: $\frac{6}{2} = 3$

Square it: $3^2 = 9$

Add and subtract 9: $x^2 + 6x + 9 – 9 + 5$

Group: $(x^2 + 6x + 9) + (-9 + 5)$

Factor: $(x + 3)^2 – 4$

Answer: $(x + 3)^2 – 4$ where $p = 3$, $q = -4$

Visual Check: The vertex is at $(-3, -4)$. Since $a = 1 > 0$, this is a minimum point with minimum value $-4$ when $x = -3$.

🎯

When $a \neq 1$: $ax^2 + bx + c$

Step 1: Factor $a$ from first two terms: $a\left(x^2 + \frac{b}{a}x\right) + c$

Step 2: Complete square inside brackets: Half of $\frac{b}{a}$ is $\frac{b}{2a}$, square it: $\left(\frac{b}{2a}\right)^2$

Step 3: Add and subtract inside brackets: $a\left[x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 – \left(\frac{b}{2a}\right)^2\right] + c$

Step 4: Factor perfect square: $a\left[\left(x + \frac{b}{2a}\right)^2 – \left(\frac{b}{2a}\right)^2\right] + c$

Step 5: Expand: $a\left(x + \frac{b}{2a}\right)^2 – a\left(\frac{b}{2a}\right)^2 + c$

\[ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c – \frac{b^2}{4a}\right)\]

📝 Example 2: When $a \neq 1$

Express $2x^2 – 8x + 5$ in the form $a(x + p)^2 + q$.

Factor 2 from first two terms: $2(x^2 – 4x) + 5$

Half of $-4$ is $-2$, square it: $4$

Add and subtract 4 inside brackets: $2[x^2 – 4x + 4 – 4] + 5$

Group: $2[(x^2 – 4x + 4) – 4] + 5$

Factor perfect square: $2[(x – 2)^2 – 4] + 5$

Expand: $2(x – 2)^2 – 8 + 5$

Simplify: $2(x – 2)^2 – 3$

Answer: $2(x – 2)^2 – 3$ where $a = 2$, $p = -2$, $q = -3$

Key Insight: Vertex is at $(2, -3)$. Since $a = 2 > 0$, minimum value is $-3$ when $x = 2$.

Part 3: Finding Maximum and Minimum Values

📈

Applying Completed Square to Optimization

🎯

The Max/Min Decision Rule

$f(x) = a(x – h)^2 + k$

Vertex: $(h, k)$

⇒

If $a > 0$:

Minimum value = $k$

Occurs when $x = h$

⇒

If $a < 0$:

Maximum value = $k$

Occurs when $x = h$

Remember: The squared term $(x – h)^2$ is always non-negative (≥ 0). Therefore:

When $a > 0$, $a(x – h)^2 ≥ 0$, so minimum is $k$ (when square = 0)
When $a < 0$, $a(x - h)^2 ≤ 0$, so maximum is $k$ (when square = 0)

📝 Example 3: Finding Minimum Value

Find the minimum value of $f(x) = x^2 – 10x + 28$ and the value of $x$ at which it occurs.

Complete the square: $x^2 – 10x + 28$

Half of $-10$ is $-5$, square it: $25$

Add and subtract 25: $x^2 – 10x + 25 – 25 + 28$

Group: $(x^2 – 10x + 25) + 3$

Factor: $(x – 5)^2 + 3$

Identify: $a = 1 > 0$, so minimum exists

Minimum occurs when $(x – 5)^2 = 0$ ⇒ $x = 5$

Minimum value = $3$

Answer: Minimum value is 3 when $x = 5$

📝 Example 4: Finding Maximum Value

Find the maximum value of $f(x) = -2x^2 + 12x – 7$ and the value of $x$ at which it occurs.

Factor -2 from first two terms: $-2(x^2 – 6x) – 7$

Half of $-6$ is $-3$, square it: $9$

Add and subtract 9 inside: $-2[x^2 – 6x + 9 – 9] – 7$

Group: $-2[(x^2 – 6x + 9) – 9] – 7$

Factor: $-2[(x – 3)^2 – 9] – 7$

Expand: $-2(x – 3)^2 + 18 – 7$

Simplify: $-2(x – 3)^2 + 11$

Identify: $a = -2 < 0$, so maximum exists

Maximum occurs when $(x – 3)^2 = 0$ ⇒ $x = 3$

Maximum value = $11$

Answer: Maximum value is 11 when $x = 3$

Memory Aid: “Positive ‘a’ points Up (minimum), Negative ‘a’ points Down (maximum)”. The vertex (h, k) gives the turning point coordinates.

Part 4: CSEC Past Paper Questions

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Real CSEC Exam Questions

CSEC May/June 2019 Paper 2

Question: The function $f(x) = 2x^2 – 12x + 19$ is defined for all real values of $x$.

(a) Express $f(x)$ in the form $a(x + b)^2 + c$.

(b) State the minimum value of $f(x)$.

(a)

$f(x) = 2x^2 – 12x + 19$

Factor 2: $2(x^2 – 6x) + 19$

Complete square: Half of $-6$ is $-3$, square = 9

$2[x^2 – 6x + 9 – 9] + 19$

$2[(x – 3)^2 – 9] + 19$

$2(x – 3)^2 – 18 + 19$

$2(x – 3)^2 + 1$

Answer: $2(x – 3)^2 + 1$

(b)

Since $a = 2 > 0$, minimum exists

Minimum value = constant term = 1

Answer: Minimum value is 1

(c)

Minimum occurs when squared term = 0

$x – 3 = 0$ ⇒ $x = 3$

Answer: $x = 3$

CSEC January 2018 Paper 2

Question: The function $g(x) = -x^2 + 4x – 3$ is defined for all real values of $x$.

(a) Express $g(x)$ in the form $a(x + h)^2 + k$.

(b) Hence, or otherwise, determine the maximum value of $g(x)$.

(a)

$g(x) = -x^2 + 4x – 3$

Factor -1: $-1(x^2 – 4x) – 3$

Complete square: Half of $-4$ is $-2$, square = 4

$-1[x^2 – 4x + 4 – 4] – 3$

$-1[(x – 2)^2 – 4] – 3$

$-(x – 2)^2 + 4 – 3$

$-(x – 2)^2 + 1$

Answer: $-(x – 2)^2 + 1$

(b)

Since $a = -1 < 0$, maximum exists

Maximum value = constant term = 1

Answer: Maximum value is 1

Exam Insight: CSEC frequently asks: “Express in the form $a(x + p)^2 + q$” followed by “State the minimum/maximum value” and “Write down the value of x at which it occurs.” Always show your completing the square steps to earn method marks.

Part 5: Applications & Real-World Problems

🌍

Optimization in Real Life

📦

Maximizing Area/Volume

Typical Problem: “A farmer has 100m of fencing to create a rectangular pen. What dimensions maximize the area?”

Application Example: Maximum Area

A farmer has 60m of fencing to create a rectangular enclosure against a straight wall (needs only 3 sides fenced). Find the maximum possible area.

Let width = $x$ m, length = $y$ m

Fencing: $2x + y = 60$ ⇒ $y = 60 – 2x$

Area $A = x \times y = x(60 – 2x) = 60x – 2x^2$

Rewrite: $A = -2x^2 + 60x$

Complete square: Factor -2: $-2(x^2 – 30x)$

Half of $-30$ = $-15$, square = 225

$-2[x^2 – 30x + 225 – 225] = -2[(x – 15)^2 – 225]$

$A = -2(x – 15)^2 + 450$

Maximum when $x = 15$ m

Then $y = 60 – 2(15) = 30$ m

Maximum area = $450$ m²

💰

Maximizing Profit

Typical Problem: “A company sells items at price $p$. Revenue = price × quantity sold. Costs include fixed and variable costs. Find price that maximizes profit.”

Common Optimization Problems:

Maximum area given fixed perimeter

Maximum revenue/profit given price-demand relationship

Minimum cost of materials for containers

Maximum height of projectile (physics applications)

Optimum dimensions for packaging

Quiz: Test Your Understanding

Completing the Square & Max/Min Quiz

Question 1: Express $x^2 + 8x + 15$ in the form $(x + p)^2 + q$. Hence state the minimum value and the value of $x$ at which it occurs.

Answer:
Complete square: $x^2 + 8x + 15 = (x^2 + 8x + 16) – 16 + 15 = (x + 4)^2 – 1$
So $p = 4$, $q = -1$
Since $a = 1 > 0$, minimum exists.
Minimum value = $-1$ when $x = -4$.

Question 2: Express $3x^2 – 12x + 14$ in the form $a(x + b)^2 + c$. State the minimum value.

Answer:
Factor 3: $3(x^2 – 4x) + 14$
Complete square: $3[x^2 – 4x + 4 – 4] + 14 = 3[(x – 2)^2 – 4] + 14$
$= 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2$
Since $a = 3 > 0$, minimum exists.
Minimum value = $2$ when $x = 2$.

Question 3: The function $f(x) = -x^2 + 6x – 5$ has a maximum value. Find this maximum value and the corresponding value of $x$.

Answer:
Complete square: $-x^2 + 6x – 5 = -(x^2 – 6x) – 5$
$= -[x^2 – 6x + 9 – 9] – 5 = -[(x – 3)^2 – 9] – 5$
$= -(x – 3)^2 + 9 – 5 = -(x – 3)^2 + 4$
Since $a = -1 < 0$, maximum exists.
Maximum value = $4$ when $x = 3$.

Question 4: A rectangular garden is to be fenced on three sides (one side is against a house). If 40m of fencing is available, find the maximum area that can be enclosed.

Answer:
Let width = $x$ m, length = $y$ m
Fencing: $2x + y = 40$ ⇒ $y = 40 – 2x$
Area $A = x(40 – 2x) = 40x – 2x^2 = -2x^2 + 40x$
Complete square: $-2(x^2 – 20x) = -2[(x – 10)^2 – 100] = -2(x – 10)^2 + 200$
Maximum area = $200$ m² when $x = 10$ m, $y = 20$ m.

Question 5: The profit $P$ in dollars from selling $x$ items is given by $P = -2x^2 + 100x – 800$. Find the number of items that must be sold to maximize profit, and the maximum profit.

Answer:
Complete square: $P = -2x^2 + 100x – 800 = -2(x^2 – 50x) – 800$
$= -2[(x – 25)^2 – 625] – 800 = -2(x – 25)^2 + 1250 – 800$
$= -2(x – 25)^2 + 450$
Since $a = -2 < 0$, maximum exists.
Maximum profit = $450 when $x = 25$ items.

🎯 Key Concepts Summary

Vertex Form: $f(x) = a(x – h)^2 + k$ gives vertex $(h, k)$
Completing Square Steps:
- Ensure coefficient of $x^2$ is 1 (factor if needed)
- Take half of $x$-coefficient, square it
- Add and subtract this square inside
- Factor perfect square trinomial
- Simplify constant terms
Max/Min Determination:
- If $a > 0$: Minimum at vertex, opens upward (∪)
- If $a < 0$: Maximum at vertex, opens downward (∩)
Optimization Strategy:
- Write expression for quantity to optimize
- Express in quadratic form
- Complete the square
- Identify vertex as max/min point
Common CSEC Questions:
- “Express in the form $a(x + p)^2 + q$”
- “State the minimum/maximum value”
- “Write down the value of $x$ at which it occurs”
- Applied problems: area, profit, projectile height
Exam Strategy:
- Always show completing square steps for method marks
- Check sign of $a$ to determine max vs min
- Include units in applied problems
- Verify answer makes sense in context

Common Mistakes to Avoid: 1. Forgetting to factor $a$ when $a \neq 1$
2. Incorrect sign when taking half of $b$
3. Forgetting to multiply by $a$ when expanding the subtracted square
4. Confusing max vs min based on sign of $a$
5. Not stating both the max/min value AND the $x$ value where it occurs
6. In applied problems: not defining variables or writing equations

CSEC Exam Strategy: When answering completing square questions: (1) Show clear step-by-step work, (2) Circle your final vertex form, (3) Clearly state “minimum” or “maximum” based on sign of $a$, (4) Give both the max/min value AND the $x$-value, (5) For applied problems, include units and a sentence answer. Practice makes perfect—master this technique as it appears in almost every CSEC Add Math exam!