Quadratic functions

1. Overview

Quadratic functions are polynomial functions of degree 2, of the form $f(x) = ax^2 + bx + c$. Mastering quadratic functions is crucial for IGCSE Additional Mathematics (0606). You'll use them to:

• Analyze parabolas and their properties (vertex, axis of symmetry, intercepts).
• Solve optimization problems using differentiation (finding maximum or minimum values).
• Determine the nature of roots using the discriminant.
• Solve quadratic inequalities.
• Apply these concepts in coordinate geometry problems involving intersections of lines and curves.

This topic is examined in both Paper 1 (without a calculator) and Paper 2. Paper 1 emphasizes algebraic manipulation and exact answers (surds, fractions). Paper 2 allows calculators for numerical solutions, but you must still show your working.

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Key Definitions

• Quadratic Function: A function of the form $f(x) = f(x) = ax^2 + bx + c$ where $a, b, c$ are constants and $a \neq 0$.
• Roots (or Zeros): The values of $x$ for which $f(x) = 0$; the points where the graph intersects the $x$-axis.
• Discriminant ($\Delta$): The value $b^2 - 4ac$ that determines the nature of the roots.
• Vertex (Turning Point): The maximum or minimum point of the parabola.
• Completing the Square: The process of converting $ax^2 + bx + c$ into the form $a(x-h)^2 + k$.
• Tangent: A line that touches a curve at exactly one point (where the discriminant of the resulting quadratic is zero).

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Core Content

A. Solving Quadratic Equations

Equations can be solved for real roots using three primary methods. For Paper 1, you must be proficient in algebraic manipulation without a calculator.

1. Factorisation: The fastest method if the quadratic has integer roots.
2. The Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Use this when factorisation is not obvious.
3. Completing the Square: Essential for finding the vertex and for "Show that" questions.

Worked example 1 — Solving by Completing the Square (Exact Form)

Solve $3x^2 + 5x - 2 = 0$ by completing the square. Give your answer in exact form.

Step 1: Factor out the coefficient of $x^2$ from the $x^2$ and $x$ terms. $3(x^2 + \frac{5}{3}x) - 2 = 0$ (Factor out 3)

Step 2: Complete the square inside the bracket. Add and subtract $(\frac{1}{2} \cdot \frac{5}{3})^2 = (\frac{5}{6})^2 = \frac{25}{36}$ inside the bracket. $3[(x + \frac{5}{6})^2 - \frac{25}{36}] - 2 = 0$ (Complete the square)

Step 3: Expand the outer bracket and simplify. $3(x + \frac{5}{6})^2 - \frac{25}{12} - 2 = 0$ (Expand) $3(x + \frac{5}{6})^2 - \frac{25}{12} - \frac{24}{12} = 0$ (Common denominator) $3(x + \frac{5}{6})^2 - \frac{49}{12} = 0$ (Simplify)

Step 4: Solve for $x$. $3(x + \frac{5}{6})^2 = \frac{49}{12}$ (Isolate the squared term) $(x + \frac{5}{6})^2 = \frac{49}{36}$ (Divide by 3) $x + \frac{5}{6} = \pm \sqrt{\frac{49}{36}}$ (Take the square root) $x + \frac{5}{6} = \pm \frac{7}{6}$ (Simplify the square root) $x = -\frac{5}{6} \pm \frac{7}{6}$ (Isolate $x$)

Step 5: Find the two solutions. $x = -\frac{5}{6} + \frac{7}{6} = \frac{2}{6} = \frac{1}{3}$ $x = -\frac{5}{6} - \frac{7}{6} = -\frac{12}{6} = -2$

Solution: $x = \frac{1}{3}$ or $x = -2$

Worked example 2 — Solving using the Quadratic Formula (Paper 2)

Solve $5x^2 - 9x + 2 = 0$ using the quadratic formula. Give your answer to two decimal places.

Step 1: Identify $a$, $b$, and $c$. $a = 5$, $b = -9$, $c = 2$

Step 2: Apply the quadratic formula. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ (Quadratic formula) $x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(5)(2)}}{2(5)}$ (Substitute values) $x = \frac{9 \pm \sqrt{81 - 40}}{10}$ (Simplify) $x = \frac{9 \pm \sqrt{41}}{10}$ (Simplify)

Step 3: Calculate the two solutions. $x = \frac{9 + \sqrt{41}}{10} \approx 1.54$ $x = \frac{9 - \sqrt{41}}{10} \approx 0.26$

Solution: $x \approx 1.54$ or $x \approx 0.26$

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B. Maximum and Minimum Values

The turning point $(h, k)$ represents the maximum (if $a < 0$) or minimum (if $a > 0$) value.

• Method 1: Completing the Square: If $f(x) = a(x - h)^2 + k$, the vertex is $(h, k)$.
• Method 2: Differentiation:
  1. Find $f'(x) = 2ax + b$.
  2. Set $f'(x) = 0$ to find the $x$-coordinate of the stationary point.
  3. Substitute $x$ back into $f(x)$ to find the $y$-value (the max/min value).

Worked example 3 — Finding Minimum Value and Range

Find the minimum value of the function $f(x) = 2x^2 + 8x - 3$ and determine its range.

Step 1: Complete the square. $f(x) = 2(x^2 + 4x) - 3$ (Factor out 2) $f(x) = 2[(x + 2)^2 - 4] - 3$ (Complete the square) $f(x) = 2(x + 2)^2 - 8 - 3$ (Expand) $f(x) = 2(x + 2)^2 - 11$ (Simplify)

Step 2: Identify the vertex. The vertex is $(-2, -11)$.

Step 3: Determine the minimum value. Since $a = 2 > 0$, the parabola opens upwards, and the vertex represents the minimum point. The minimum value is $f(x) = -11$.

Step 4: Determine the range. Since the parabola opens upwards and the minimum value is -11, the range is $f(x) \ge -11$.

Solution: Minimum value = $-11$, Range: $f(x) \ge -11$

Worked Example 4: Range for a Given Domain

Find the range of $f(x) = x^2 - 4x + 7$ for the domain $0 \le x \le 5$.

Step 1: Complete the square to find the vertex. $f(x) = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$. The vertex is $(2, 3)$. Since $a > 0$, the minimum value is $3$.

Step 2: Check the boundaries of the domain. $f(0) = (0)^2 - 4(0) + 7 = 7$ $f(5) = (5)^2 - 4(5) + 7 = 25 - 20 + 7 = 12$

Step 3: Determine the range. The lowest $y$-value is the vertex (3) and the highest $y$-value is at the boundary $x=5$ (12). Range: $3 \le f(x) \le 12$.

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C. The Discriminant and Intersections

The nature of the roots for $ax^2 + bx + c = 0$ is determined by $\Delta = b^2 - 4ac$.

Condition	Nature of Roots	Intersection of line and curve
$b^2 - 4ac > 0$	Two distinct real roots	Line cuts curve at two points
$b^2 - 4ac = 0$	Two equal real roots	Line is a tangent to the curve
$b^2 - 4ac < 0$	No real roots	Line does not intersect curve

Worked Example 5: Intersection with a Line

Find the values of $k$ for which the line $y = kx - 5$ is a tangent to the curve $y = x^2 + 4x - 1$.

Step 1: Set the equations equal to each other. $x^2 + 4x - 1 = kx - 5$ (Equate $y$ values)

Step 2: Rearrange into standard quadratic form $ax^2 + bx + c = 0$. $x^2 + (4 - k)x + 4 = 0$ (Rearrange)

Step 3: For a tangent, set the discriminant to zero. $a = 1, b = (4 - k), c = 4$ $(4 - k)^2 - 4(1)(4) = 0$ (Apply $b^2 - 4ac = 0$)

Step 4: Solve for $k$. $16 - 8k + k^2 - 16 = 0$ (Expand) $k^2 - 8k = 0$ (Simplify) $k(k - 8) = 0$ (Factorise) $k = 0$ or $k = 8$ (Solve)

Solution: $k = 0$ or $k = 8$.

Worked Example 6: No Real Roots

Determine the range of values for $m$ such that the equation $x^2 + (m+2)x + 4 = 0$ has no real roots.

Step 1: Identify $a$, $b$, and $c$. $a = 1$, $b = m+2$, $c = 4$

Step 2: Apply the discriminant condition for no real roots. $b^2 - 4ac < 0$ (Condition for no real roots) $(m+2)^2 - 4(1)(4) < 0$ (Substitute values)

Step 3: Simplify and solve the inequality. $m^2 + 4m + 4 - 16 < 0$ (Expand) $m^2 + 4m - 12 < 0$ (Simplify) $(m+6)(m-2) < 0$ (Factorise)

Step 4: Find the critical values. $m = -6$ or $m = 2$

Step 5: Determine the range of values for $m$. Sketching the parabola $y = (m+6)(m-2)$ shows that the region where $y < 0$ is between the roots. Therefore, $-6 < m < 2$.

Solution: $-6 < m < 2$

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D. Quadratic Inequalities

To solve $ax^2 + bx + c > 0$ or $< 0$:

1. Find the critical values (roots) by solving $ax^2 + bx + c = 0$.
2. Sketch the graph.
3. Identify the region:
   • If $> 0$, take the regions above the $x$-axis.
   • If $< 0$, take the region below the $x$-axis.

Worked Example 7: Solving a Quadratic Inequality

Solve the inequality $x^2 - 5x + 6 > 0$.

Step 1: Find the critical values (roots). $x^2 - 5x + 6 = 0$ (Set to zero) $(x - 2)(x - 3) = 0$ (Factorise) $x = 2$ or $x = 3$ (Solve)

Step 2: Sketch the graph. The parabola $y = x^2 - 5x + 6$ opens upwards (since $a = 1 > 0$) and intersects the $x$-axis at $x = 2$ and $x = 3$.

Step 3: Identify the region where $y > 0$. The region where $y > 0$ is when $x < 2$ or $x > 3$.

Solution: $x < 2$ or $x > 3$

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Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

• Standard Form: $\mathbf{f(x) = ax^2 + bx + c}$
• Vertex Form: $\mathbf{f(x) = a(x-h)^2 + k}$
• The Quadratic Formula: $\mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$ (Given in formula sheet)
• The Discriminant: $\mathbf{\Delta = b^2 - 4ac}$ (Not given in formula sheet - MEMORIZE)
• Axis of Symmetry: $\mathbf{x = -\frac{b}{2a}}$

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Common Mistakes to Avoid

• ❌ Wrong: Forgetting to distribute the coefficient $a$ when completing the square, e.g., $3(x-2)^2 - 4$ instead of $3[(x-2)^2 - 4] = 3(x-2)^2 - 12$. ✓ Right: Always use square brackets to isolate the $(x-h)^2$ expansion before multiplying by $a$.
• ❌ Wrong: Using the discriminant on the original curve equation when looking for intersections. ✓ Right: You must first equate the line and curve ($y_1 = y_2$) and find the new $a, b,$ and $c$ values.
• ❌ Wrong: Writing the solution for $k^2 > 9$ as $k > \pm 3$. ✓ Right: Use a sketch to see that the solution is two separate intervals: $k > 3$ or $k < -3$.
• ❌ Wrong: Omitting $k=0$ as a solution in discriminant problems involving $k^2$ terms. Always check $k=0$ separately.
• ❌ Wrong: Giving decimal approximations when the question requires an exact answer (surd form). ✓ Right: Leave your answer in surd form (e.g., $2 + \sqrt{3}$) unless the question explicitly asks for a decimal approximation.
• ❌ Wrong: Making sign errors when expanding brackets or rearranging equations. ✓ Right: Double-check each step, especially when dealing with negative signs. Use brackets carefully.

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Exam Tips

• Command Word "Hence": If part (a) asks you to complete the square and part (b) asks for the maximum value or roots, you must use the result from part (a). Do not start from scratch using differentiation.
• Paper 1 Exactness: Never use decimals for $\sqrt{3}$ or $\frac{2}{3}$ unless specified. If you use a calculator to find roots on Paper 2, write down the formula and the substitution first to secure method marks.
• Domain Restrictions: Always check if the vertex $(h, k)$ falls within the domain given in the question before stating the range.
• "Show that": Every algebraic step must be shown. If you are showing that a line is a tangent, you must explicitly state: "Since $b^2 - 4ac = 0$, the line is a tangent."
• Calculator Shortcut (Paper 2): Most modern scientific calculators have a polynomial solver. Use this to check your answers for $x$, but ensure all manual working is shown on the paper.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

The function $f(x) = 2x^2 + 8x + 5$ is defined for all real values of $x$.

(a) Express $f(x)$ in the form $a(x+b)^2 + c$, where $a$, $b$, and $c$ are integers. [4 marks]

(b) Write down the coordinates of the vertex of the graph of $y = f(x)$. [2 marks]

(c) Find the range of $f$. [2 marks]

Worked Solution:

(a)

1. Factor out the coefficient of $x^2$ from the first two terms: $f(x) = 2(x^2 + 4x) + 5$ [Isolate the x terms to complete the square]

2. Complete the square inside the parentheses: $f(x) = 2[(x+2)^2 - 4] + 5$ [Add and subtract $(4/2)^2=4$ inside the parentheses]

3. Expand and simplify: $f(x) = 2(x+2)^2 - 8 + 5$ $f(x) = 2(x+2)^2 - 3$ [Simplify to the required form]

$\boxed{f(x) = 2(x+2)^2 - 3}$

How to earn full marks: Show each step of completing the square, especially factoring out the leading coefficient and distributing it back in.

(b)

1. The vertex of the parabola $y = a(x+b)^2 + c$ is $(-b, c)$. [Recognize the vertex form of a quadratic]

2. From part (a), the vertex is $(-2, -3)$. [Identify b and c]

$\boxed{(-2, -3)}$

How to earn full marks: Remember the vertex form of the equation and make sure you get the signs right for the x-coordinate.

(c)

1. Since $a > 0$, the parabola opens upwards. The minimum value of $f(x)$ is the $y$-coordinate of the vertex. [Recognize the parabola shape and minimum point]

2. The range of $f$ is $f(x) \ge -3$. [Write the range]

$\boxed{f(x) \ge -3}$

How to earn full marks: State the range as an inequality, and make sure it's a "greater than or equal to" since the parabola has a minimum.

Common Pitfall: When completing the square, double-check your arithmetic on the constants after you complete the square, as errors with fractions and minus signs here are common. Also, remember that the range is a y value, so it should be $f(x) \ge$ or $y \ge$, not $x \ge$.

#### Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [7 marks]

**Question:**

(a) Solve the equation $x^2 + 4x - 8 = 0$, giving your answers in the form $a + b\sqrt{c}$, where $a$, $b$, and $c$ are integers. [5 marks]

(b) Hence, without using a calculator, find the exact value of $\frac{1}{ \alpha} + \frac{1}{\beta}$, where $\alpha$ and $\beta$ are the roots of the equation $x^2 + 4x - 8 = 0$. [2 marks]

**Worked Solution:**

**(a)**
1. Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
   *[Recall the quadratic formula]*

2. Substitute $a=1$, $b=4$, $c=-8$ into the formula:
   $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-8)}}{2(1)}$
   *[Substitute the coefficients]*

3. Simplify:
   $x = \frac{-4 \pm \sqrt{16 + 32}}{2}$
   $x = \frac{-4 \pm \sqrt{48}}{2}$
   *[Simplify the discriminant]*

4. Simplify the surd: $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
   *[Simplify the surd]*

5. Final solutions:
   $x = \frac{-4 \pm 4\sqrt{3}}{2} = -2 \pm 2\sqrt{3}$
   *[Divide by 2]*

$\boxed{x = -2 + 2\sqrt{3}, x = -2 - 2\sqrt{3}}$

**How to earn full marks:** Show all steps of simplification, especially when dealing with the square root and fractions.

**(b)**
1. Recall that for a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$.
   *[Use Vieta's formulas]*

2. $\alpha + \beta = -\frac{4}{1} = -4$ and $\alpha\beta = \frac{-8}{1} = -8$.
   *[Find the sum and product of the roots]*

3. $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-4}{-8} = \frac{1}{2}$
   *[Express the required expression in terms of sum and product]*

$\boxed{\frac{1}{2}}$

**How to earn full marks:** Clearly state the sum and product of the roots before using them to find the final answer.

**Common Pitfall:** On non-calculator questions with surds, don't skip steps: show every expansion and rationalization. Also, remember that if a question says 'Hence' after you've solved something in part (a), you MUST use the result from part (a) in part (b).

```markdown
#### Exam-Style Question 3 — Paper 2 (Calculator Allowed) [8 marks]

**Question:**

The line $y = kx - 3$ is a tangent to the curve $y = x^2 - 5x + 1$.

(a) Find the two possible values of $k$. [5 marks]

(b) For each value of $k$ found in part (a), find the coordinates of the point where the line is tangent to the curve. Give your answers to 2 decimal places. [3 marks]

**Worked Solution:**

**(a)**
1. Since the line is tangent to the curve, they intersect at one point. Equate the two equations:
   $kx - 3 = x^2 - 5x + 1$
   *[Equate the equations]*

2. Rearrange into a quadratic equation:
   $x^2 - (5+k)x + 4 = 0$
   *[Rearrange to standard quadratic form]*

3. For tangency, the discriminant must be equal to zero: $b^2 - 4ac = 0$.
   *[Recall condition for tangency]*

4. Substitute $a=1$, $b=-(5+k)$, and $c=4$ into the discriminant:
   $[-(5+k)]^2 - 4(1)(4) = 0$
   $(5+k)^2 - 16 = 0$
   *[Substitute into the discriminant formula]*

5. Solve for $k$:
   $25 + 10k + k^2 - 16 = 0$
   $k^2 + 10k + 9 = 0$
   $(k+1)(k+9) = 0$
   $k = -1$ or $k = -9$
   *[Solve the quadratic equation]*

$\boxed{k = -1, k = -9}$

**How to earn full marks:** Remember to set the discriminant to zero and show all the steps in solving the resulting quadratic equation for *k*.

**(b)**
1. When $k = -1$, the equation $x^2 - (5+k)x + 4 = 0$ becomes $x^2 - 4x + 4 = 0$, which factors to $(x-2)^2 = 0$. Thus, $x = 2$.
   *[Solve for x when k=-1]*

2. Substitute $x = 2$ into $y = kx - 3 = -1(2) - 3 = -5$.
   *[Find y]*

3. When $k = -9$, the equation $x^2 - (5+k)x + 4 = 0$ becomes $x^2 + 4x + 4 = 0$, which factors to $(x+2)^2 = 0$. Thus, $x = -2$.
   *[Solve for x when k=-9]*

4. Substitute $x = -2$ into $y = kx - 3 = -9(-2) - 3 = 15$.
   *[Find y]*

$\boxed{(2, -5), (-2, 15)}$

**How to earn full marks:** For each *k* value, solve the quadratic to find *x*, then substitute *x* back into either equation to find *y*.

**Common Pitfall:** When solving an inequality with a discriminant, some forget to include 'k' inside the discriminant calculation itself. Also, remember to substitute your *k* values back into the original equations to find the points of tangency.

```markdown
#### Exam-Style Question 4 — Paper 2 (Calculator Allowed) [9 marks]

**Question:**

The quadratic function $f(x) = ax^2 + bx + c$ has a minimum value of $-4$ at $x = 1$. The graph of $y = f(x)$ intersects the $y$-axis at the point $(0, -1)$.

(a) Find the values of $a$, $b$, and $c$. [6 marks]

(b) Solve the inequality $f(x) \le 0$. Give your answer in terms of inequalities. [3 marks]

**Worked Solution:**

**(a)**
1. Since the minimum value is $-4$ at $x = 1$, we know that $f(1) = -4$.
   $a(1)^2 + b(1) + c = -4$
   $a + b + c = -4$  --- (1)
   *[Use the minimum value to create an equation]*

2. Since the graph intersects the $y$-axis at $(0, -1)$, we know that $f(0) = -1$, so $c = -1$.
   *[Use the y-intercept to find c]*

3. Substitute $c = -1$ into equation (1): $a + b - 1 = -4$, so $a + b = -3$ --- (2)
   *[Substitute c into the first equation]*

4. Since the minimum occurs at $x=1$, the line of symmetry is $x = 1$. Using $x = -\frac{b}{2a}$, we get:
   $1 = -\frac{b}{2a}$, so $2a = -b$ or $b = -2a$ --- (3)
   *[Use the x-coordinate of the vertex to find a relationship between a and b]*

5. Substitute $b = -2a$ into equation (2): $a - 2a = -3$, so $-a = -3$, and $a = 3$.
   *[Solve for a]*

6. Substitute $a = 3$ into equation (3): $b = -2(3) = -6$.
   *[Solve for b]*

$\boxed{a = 3, b = -6, c = -1}$

**How to earn full marks:** Form three independent equations using the given information and solve them simultaneously to find *a*, *b*, and *c*.

**(b)**
1. With $a=3$, $b=-6$, and $c=-1$, the quadratic function is $f(x) = 3x^2 - 6x - 1$.
   *[Write the full quadratic]*

2. To find the roots, solve $3x^2 - 6x - 1 = 0$ using the quadratic formula:
   $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(-1)}}{2(3)} = \frac{6 \pm \sqrt{36 + 12}}{6} = \frac{6 \pm \sqrt{48}}{6} = \frac{6 \pm 4\sqrt{3}}{6} = 1 \pm \frac{2\sqrt{3}}{3}$
   *[Find the roots of the quadratic]*

3. $x_1 = 1 - \frac{2\sqrt{3}}{3} \approx -0.155$ and $x_2 = 1 + \frac{2\sqrt{3}}{3} \approx 2.155$. Since $a > 0$, the parabola opens upwards. The function $f(x) \le 0$ between the roots.
   *[Find the approximate decimal values of the roots and determine the region between the roots]*

$\boxed{1 - \frac{2\sqrt{3}}{3} \le x \le 1 + \frac{2\sqrt{3}}{3}}$ or $\boxed{-0.155 \le x \le 2.155}$ (3 dp)

**How to earn full marks:** Find the roots accurately using the quadratic formula, and express the solution as a "between" inequality since the parabola opens upwards.

**Common Pitfall:** When solving quadratic inequalities, sketch a quick graph to visualise whether you want the region *between* the roots or *outside* the roots. Also, remember to include the "equal to" part in your inequality if the question asks for $f(x) \le 0$ rather than $f(x) < 0$.