# Algebra and Functions (Edexcel IAL P3)

> Edexcel International A-Level Mathematics · IAL P3 2018 Specification
> Source: https://www.owlsprep.com/study/edexcel-ial-math-p3-algebra-and-functions/

This guide covers all Edexcel IAL P3 Algebra and Functions content, including simplifying rational expressions, functions/domain/range, modulus equations/inequalities, and combined graph transformations, aligned to the 2018 WMA13 specification.

**Prerequisites:** [Algebraic factorisation and basic function notation (Edexcel IAL P1)](https://www.owlsprep.com/study/edexcel-ial-math-p1-algebra-basics/); [Single graph transformations (Edexcel IAL P1)](https://www.owlsprep.com/study/edexcel-ial-math-p1-graph-transformations/)

## Learning objectives

- Simplify rational expressions via factorisation and algebraic division (linear/quadratic denominators only)
- Define functions, calculate domain/range, distinguish one-one/many-one mappings, compute composite and inverse functions
- Sketch modulus function graphs, solve modulus equations and inequalities
- Apply combinations of vertical/horizontal stretches, reflections and translations to graphs

## Simplifying Rational Expressions (§1.1)

P3 rational expressions have linear or quadratic denominators only, and are simplified via factorisation of the numerator and denominator, followed by cancelling common non-zero factors. For expressions where the numerator is of higher degree than the denominator, use algebraic long division first before factorising. Partial fractions are not assessed in this unit.

**Rational Expression** — A fraction where both the numerator and denominator are algebraic polynomials, with linear or quadratic denominators for P3 assessment.

*Example:* \(\frac{x^3 + 1}{x^2 - 1}\)

**Worked example:** Simplify \(\frac{x^3 + 1}{x^2 - 1}\) fully, and state excluded values of x.

1. Factorise numerator and denominator using standard polynomial identities:
2. $$x^3 + 1 = (x + 1)(x^2 - x + 1)$$
3. $$x^2 - 1 = (x + 1)(x - 1)$$
4. Cancel the common non-zero factor \((x + 1)\), noting excluded values that make the original denominator zero: \(x \neq -1\) and \(x \neq 1\):
5. $$\frac{(x + 1)(x^2 - x + 1)}{(x + 1)(x - 1)} = \frac{x^2 - x + 1}{x - 1}$$
6. The expression is fully simplified as the quadratic numerator has no common factors with the linear denominator.

> **Exam tip:** Always state excluded values of x (values that make the original denominator zero) for full marks, even if not explicitly asked.

*Calculator:* allowed

## Functions, Composite and Inverse Functions (§1.2)

A function is a mapping from a set of inputs (domain) to a set of outputs (range), where each input maps to exactly one output. One-one functions have unique outputs for every input, while many-one functions have multiple inputs mapping to the same output. Composite functions are written fg(x), meaning apply g first, then f. Inverse functions reverse the action of f, and only exist for one-one functions; their graph is a reflection of f(x) in the line \(y=x\).

**Composite Function** — A function formed by applying one function to the output of another, denoted fg(x) meaning apply g first, then f.

*Example:* If \(f(x) = 2x + 3\) and \(g(x) = x^2\), then \(fg(x) = 2x^2 + 3\)

**Worked example:** Given \(f(x) = 2x + 3\) for \(x \geq 0\) and \(g(x) = x^2\) for all real x, find (a) \(fg(x)\), (b) the range of \(f(x)\), (c) \(f^{-1}(x)\) and state its domain.

1. (a) For \(fg(x)\), substitute the output of g(x) into f(x):
2. $$fg(x) = 2(g(x)) + 3 = 2x^2 + 3$$
3. (b) \(f(x)\) is a strictly increasing linear function, so substitute the boundary of its domain to find the minimum output:
4. $$f(0) = 2(0) + 3 = 3$$
5. Range of \(f(x)\) is \(f(x) \geq 3\).
6. (c) To find the inverse, let \(y = f(x)\), rearrange to make x the subject, then swap x and y:
7. $$y = 2x + 3 \implies x = \frac{y - 3}{2} \implies f^{-1}(x) = \frac{x - 3}{2}$$
8. Domain of \(f^{-1}(x)\) = range of \(f(x)\), so domain is \(x \geq 3\)

> **Exam tip:** Remember the domain of an inverse function is always equal to the range of the original function, not its domain. This is one of the most frequent mark-losing mistakes for this topic.

*Calculator:* allowed

## Modulus Functions: Graphs, Equations & Inequalities (§1.3)

The modulus function returns the non-negative value of its input, so its graph is never below the x-axis. To sketch \(y = |ax + b|\), draw the line \(y = ax + b\) and reflect any part below the x-axis above it, creating a V-shape with vertex at \(x = -\frac{b}{a}\). For \(y = |f(x)|\), reflect negative parts of \(f(x)\) above the x-axis; for \(y = f(|x|)\), reflect the right half of \(f(x)\) across the y-axis.

**Modulus Function** — The function \(|x|\) that returns the non-negative value of its input, defined as \(|x| = x\) for \(x \geq 0\) and \(|x| = -x\) for \(x < 0\).

**Worked example:** Solve the inequality \(|2x - 1| > x + 5\).

1. First solve the corresponding equality \(|2x - 1| = x + 5\) to find critical points, splitting into cases based on the sign of the expression inside the modulus:
2. Case 1: \(2x - 1 \geq 0\) (i.e. \(x \geq 0.5\)): \(2x - 1 = x + 5 \implies x = 6\)
3. Case 2: \(2x - 1 < 0\) (i.e. \(x < 0.5\)): \(-(2x - 1) = x + 5 \implies -3x = 4 \implies x = -\frac{4}{3}\)
4. Test intervals to find which satisfy the original inequality:
5. - For \(x < -\frac{4}{3}\): test \(x = -2\), \(|-5| = 5 > -2 + 5 = 3\) → valid
- For \(-\frac{4}{3} < x < 6\): test \(x = 0\), \(|-1| = 1 > 0 + 5 = 5\) → invalid
- For \(x > 6\): test \(x = 7\), \(|13| = 13 > 7 + 5 = 12\) → valid
6. Final solution: \(x < -\frac{4}{3}\) or \(x > 6\)

> **Exam tip:** Always test intervals for modulus inequalities, never just rearrange without considering the modulus sign, as you can miss valid solutions or include extraneous ones.

*Calculator:* allowed

## Combined Graph Transformations (§1.4)

P3 assesses combinations of four standard transformations: vertical stretch/reflection \(y = af(x)\), vertical translation \(y = f(x) + a\), horizontal translation \(y = f(x + a)\), and horizontal stretch/reflection \(y = f(ax)\). Always apply stretches and reflections first, then translations, to avoid incorrect coordinate mapping.

**Worked example:** Describe the sequence of transformations that maps the graph of \(y = \cos x\) to \(y = -\cos(x + \frac{\pi}{4}) + 3\).

1. 1. Apply a reflection in the x-axis: this transforms \(y = \cos x\) to \(y = -\cos x\).
2. 2. Apply a horizontal translation of \(\frac{\pi}{4}\) units to the left: this transforms \(y = -\cos x\) to \(y = -\cos(x + \frac{\pi}{4})\).
3. 3. Apply a vertical translation of 3 units upwards: this gives the final graph \(y = -\cos(x + \frac{\pi}{4}) + 3\).

> **Exam tip:** List each transformation separately for full marks; do not combine horizontal or vertical adjustments into a single step. Note \(y = f(ax + b)\) is explicitly excluded from P3 assessment.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Cancelling factors in rational expressions without stating excluded x values
  - Why it fails: Cancelling a factor that equals zero for some x leads to undefined expressions, costing 1-2 marks per question
  - Correct: Always list values of x that make the original denominator zero, even if not explicitly asked
- **Wrong:** Calculating composite function fg(x) as apply f first then g
  - Why it fails: Notation fg(x) means apply the rightmost function first, so g is applied before f, not the reverse
  - Correct: Remember composite function order is right-to-left, substitute the right function into the left function
- **Wrong:** Using the domain of the original function as the domain of its inverse
  - Why it fails: The inverse function takes outputs of the original function as inputs, so its domain equals the original function's range
  - Correct: Calculate the range of the original function first to get the domain of the inverse
- **Wrong:** Squaring both sides of modulus equations without testing solutions
  - Why it fails: Squaring introduces extraneous solutions that do not satisfy the original modulus equation
  - Correct: Split into cases based on the sign of the expression inside the modulus, or test all solutions against the original equation
- **Wrong:** Applying translations before stretches/reflections for combined transformations
  - Why it fails: Doing translations first shifts the graph before stretching, leading to incorrect transformed coordinates
  - Correct: Apply all stretches and reflections first, then apply horizontal and vertical translations

## Cheatsheet

| Concept | Key Rule | Exam Reminder |
| --- | --- | --- |
| Simplify rational expressions | Factorise numerator/denominator, cancel common factors | State excluded x values where denominator = 0 |
| Composite functions fg(x) | Apply g first, then f, substitute g(x) into f | Order is right-to-left, not left-to-right |
| Inverse functions | Only exist for one-one functions, graph reflection in y=x | Domain of f⁻¹ = range of original f |
| Modulus equations/inequalities | Split into cases where modulus expression ≥0 and <0 | Test all solutions/intervals for validity |
| Combined transformations | Apply stretches/reflections first, then translations | y=f(ax+b) is not assessed in P3 |

## What's next

Now that you have mastered P3 Algebra and Functions, you are ready to progress to the next core P3 topics, starting with trigonometric identities and equations, which build heavily on the function transformation skills you learned here. Your rational expression simplification skills will also be critical for upcoming P3 calculus topics, including differentiation and integration of more complex functions. This topic is foundational for P4 content like partial fractions and further calculus, so ensure you are confident solving modulus inequalities and finding inverse functions before moving on. Practice past paper questions regularly to avoid common pitfalls and build speed for exam conditions.

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