CIE AMath Summary

Chapter 1:Sets and Venn Diagrams

Introduction

Set theory is a fundamental branch of mathematics that studies sets, which are collections of objects. It serves as a foundational system for nearly all mathematical concepts.

Basic Definitions and Notation

A set is a well-defined collection of distinct objects. The objects are called elements or members.

  • $x \in A$: $x$ is an element of set $A$.
  • $x \notin A$: $x$ is not an element of set $A$.
  • $\emptyset$ or $\{\}$: the empty set (no elements).

Examples:

  1. $A = \{1, 2, 3, 4\}$ (set of first four natural numbers)
  2. $B = \{x \mid x \text{ is an even prime number}\} = \{2\}$
  3. $\mathbb{N} = \{1, 2, 3, \dots\}$ (natural numbers)
  4. $\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$ (integers)

Set Operations

Let $A$ and $B$ be sets. The following operations are defined:

Operation Symbol Definition
Union $A \cup B$ $\{x \mid x \in A \text{ or } x \in B\}$
Intersection $A \cap B$ $\{x \mid x \in A \text{ and } x \in B\}$
Difference $A \setminus B$ $\{x \mid x \in A \text{ and } x \notin B\}$
Complement $A^c$ $\{x \in U \mid x \notin A\}$ (relative to universal set $U$)

Example

Let $U = \{1,2,3,4,5,6\}$, $A = \{1,2,3\}$, $B = \{3,4,5\}$.

$A \cup B = \{1,2,3,4,5\}$
$A \cap B = \{3\}$
$A \setminus B = \{1,2\}$
$A^c = \{4,5,6\}$

Venn Diagrams (Figures)

Venn diagrams visually represent sets and their relationships.

Figure 1: Union $A \cup B$ (conceptual diagram)

Figure 2: Intersection $A \cap B$ (conceptual diagram)

Figure 3: Universal set $U$ with sets $A$ and $B$ (conceptual Venn diagram)

Subsets and Power Sets

  • $A \subseteq B$ means every element of $A$ is also an element of $B$.
  • $A \subset B$ means $A \subseteq B$ and $A \neq B$ (proper subset).
  • The power set of $A$, denoted $\mathcal{P}(A)$, is the set of all subsets of $A$.

Example:

If $A = \{1, 2\}$, then
$\mathcal{P}(A) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$

Cardinality: $2^{|A|} = 2^2 = 4$

Cardinality

The cardinality of a set $A$, denoted $|A|$, is the number of elements in $A$.

  • $|\{a,b,c\}| = 3$
  • $|\emptyset| = 0$
  • $|\mathbb{N}| = \infty$ (countably infinite)

Set Identities (Laws)

Law Expression
Commutative $A \cup B = B \cup A$, $A \cap B = B \cap A$
Associative $(A \cup B) \cup C = A \cup (B \cup C)$
Distributive $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
De Morgan's $(A \cup B)^c = A^c \cap B^c$, $(A \cap B)^c = A^c \cup B^c$
Identity $A \cup \emptyset = A$, $A \cap U = A$
Complement $A \cup A^c = U$, $A \cap A^c = \emptyset$

Cartesian Product

The Cartesian product of $A$ and $B$ is:

$A \times B = \{(a,b) \mid a \in A, b \in B\}$

Example:

$A = \{1,2\}$, $B = \{x,y\}$:
$A \times B = \{(1,x), (1,y), (2,x), (2,y)\}$

Conclusion

Set theory provides the language and basic constructs for nearly all areas of mathematics, including relations, functions, and number systems. Understanding sets, operations, Venn diagrams, and cardinality is essential for advanced mathematical study.

Chapter 2: Functions

Introduction

Functions are fundamental to calculus, analysis, and applied mathematics. This summary covers relations, function notation, domain and range, the modulus function, composite functions, sign diagrams, and inverse functions.

A. Relations and Functions

Relation

A relation from set $A$ to set $B$ is any subset of the Cartesian product $A \times B$. It pairs elements of $A$ with elements of $B$.

Function

A function $f: A \to B$ is a relation such that every element $x \in A$ is paired with exactly one element $y \in B$. We write $y = f(x)$.

Example of a relation that is NOT a function

Let $A = \{1,2\}$, $B = \{a,b\}$. The relation $R = \{(1,a), (1,b)\}$ is not a function because 1 maps to two different outputs.

Vertical Line Test

A graph represents a function if and only if no vertical line intersects the graph more than once.

B. Function Notation

Functions are denoted by letters such as $f$, $g$, $h$. The notation $f(x)$ means "the value of $f$ at $x$".

Examples

  1. $f(x) = x^2 + 3x - 5$. Then $f(2) = 2^2 + 3(2) - 5 = 4 + 6 - 5 = 5$.
  2. $g(t) = \sqrt{t-1}$. Then $g(5) = \sqrt{5-1} = \sqrt{4} = 2$.
  3. $h(x) = \frac{1}{x-2}$. Then $h(4) = \frac{1}{2}$.

Piecewise Functions

A function defined by different formulas on different intervals.

$$ f(x) = \begin{cases} x^2, & x \lt 0 \\ x+1, & x \geq 0 \end{cases} $$

Then $f(-2)=4$, $f(0)=1$, $f(3)=4$.

C. Domain and Range

  • Domain: The set of all possible input values ($x$) for which the function is defined.
  • Range: The set of all possible output values ($y = f(x)$) as $x$ varies over the domain.

How to find domain (common cases)

  1. For polynomial: domain = $\mathbb{R}$.
  2. For rational function: denominator $\neq 0$.
  3. For square root: radicand $\geq 0$.
  4. For $\log(x)$: $x \gt 0$.

Examples

  1. $f(x) = \sqrt{4 - x^2}$

    Domain: $4 - x^2 \geq 0 \implies x^2 \leq 4 \implies -2 \leq x \leq 2$

    Range: $y \in [0,2]$.
  2. $g(x) = \frac{1}{x-3}$

    Domain: $x \neq 3$, i.e., $\mathbb{R} \setminus \{3\}$

    Range: $\mathbb{R} \setminus \{0\}$.

D. The Modulus Function

The modulus (absolute value) function is defined as:

$$ |x| = \begin{cases} x, & x \geq 0 \\ -x, & x \lt 0 \end{cases} $$

Properties:

  • $|x| \geq 0$ for all $x \in \mathbb{R}$.
  • $|x| = |-x|$.
  • $|ab| = |a||b|$.
  • Triangle inequality: $|a+b| \leq |a| + |b|$.

Examples

  1. $|-5| = 5$, $|3| = 3$.
  2. Solve $|x-2| = 5$:

    Either $x-2 = 5 \implies x = 7$, or $x-2 = -5 \implies x = -3$.
  3. Solve $|2x+1| \lt 3$:

    $-3 \lt 2x+1 \lt 3 \implies -4 \lt 2x \lt 2 \implies -2 \lt x \lt 1$.

Graph of $y = |x|$

V-shaped, vertex at $(0,0)$, slope $1$ for $x \gt 0$, slope $-1$ for $x \lt 0$.

E. Composite Functions

Given $f: A \to B$ and $g: B \to C$, the composite function $g \circ f$ (read "g of f") is defined by:

$$ (g \circ f)(x) = g(f(x)) $$

Domain of $g \circ f$ = $\{x \in \text{dom}(f) \mid f(x) \in \text{dom}(g)\}$.

Examples

  1. $f(x) = x+1$, $g(x) = x^2$.

    $(g \circ f)(x) = g(f(x)) = (x+1)^2$

    $(f \circ g)(x) = f(g(x)) = x^2 + 1$

    Note: In general, $g \circ f \neq f \circ g$.
  2. $f(x) = \sqrt{x}$, $g(x) = 2x-3$.

    $(g \circ f)(x) = 2\sqrt{x} - 3$, domain $x \geq 0$.

F. Sign Diagrams

A sign diagram (or sign chart) shows where a function is positive, negative, or zero.

Steps to construct a sign diagram

  1. Find zeros of $f(x)$ (where $f(x)=0$).
  2. Find points where $f$ is undefined (vertical asymptotes).
  3. Mark these points on a number line, dividing it into intervals.
  4. Test a point in each interval to determine sign ($+$ or $-$).

Example: $f(x) = \frac{(x-2)(x+1)}{x-3}$

  • Zeros: $x = 2$, $x = -1$.
  • Undefined: $x = 3$.
  • Intervals: $(-\infty, -1)$, $(-1, 2)$, $(2, 3)$, $(3, \infty)$.
  • Test points:

    $x=-2$: $\frac{(-)(-)}{-} = \frac{+}{-} = -$ (negative)

    $x=0$: $\frac{(-)(+)}{-} = \frac{-}{-} = +$ (positive)

    $x=2.5$: $\frac{(+)(+)}{-} = \frac{+}{-} = -$ (negative)

    $x=4$: $\frac{(+)(+)}{+} = +$ (positive)

Sign diagram: $- \quad | \quad + \quad | \quad - \quad | \quad +$

with zeros at $-1$ and $2$, undefined at $3$.

G. Inverse Functions

A function $f$ has an inverse $f^{-1}$ if and only if $f$ is one-to-one (injective): different inputs give different outputs.

Properties

  • $f^{-1}(f(x)) = x$ for all $x$ in domain of $f$.
  • $f(f^{-1}(y)) = y$ for all $y$ in domain of $f^{-1}$.
  • Graph of $f^{-1}$ is the reflection of graph of $f$ across line $y=x$.

How to find inverse

  1. Replace $f(x)$ with $y$.
  2. Swap $x$ and $y$.
  3. Solve for $y$.
  4. Replace $y$ with $f^{-1}(x)$.

Examples

  1. $f(x) = 3x + 2$

    $y = 3x + 2$, swap: $x = 3y + 2 \implies 3y = x-2 \implies y = \frac{x-2}{3}$

    So $f^{-1}(x) = \frac{x-2}{3}$.
  2. $f(x) = \sqrt{x-1}$, domain $x \geq 1$, range $y \geq 0$

    $y = \sqrt{x-1}$, swap: $x = \sqrt{y-1} \implies x^2 = y-1 \implies y = x^2 + 1$

    Domain of $f^{-1}$ is $x \geq 0$ (range of $f$), so $f^{-1}(x) = x^2 + 1$, $x \geq 0$.

Horizontal Line Test

A function has an inverse if and only if every horizontal line intersects its graph at most once.

Summary Table

Concept Key Idea
Relation Any pairing from $A$ to $B$
Function Each input has exactly one output
Domain All valid $x$ values
Range All possible $y$ values
Modulus $|x| = \max(x, -x)$
Composite $(g \circ f)(x) = g(f(x))$
Sign diagram Shows intervals of positivity/negativity
Inverse Swaps domain and range, requires one-to-one

Chapter 3: Quadratic function

Introduction

A quadratic function is a polynomial function of degree 2. It has the general form:

$f(x) = ax^2 + bx + c \quad \text{where } a, b, c \in \mathbb{R} \text{ and } a \neq 0.$

Quadratic functions appear in physics (projectile motion), economics (profit maximization), and engineering (optimization).

A. Quadratic Equations

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$ with $a \neq 0$.

Methods of Solution

  1. Factoring: Write as $(x - r_1)(x - r_2) = 0$.
  2. Completing the square: Rewrite as $(x + p)^2 = q$.
  3. Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Examples

  1. $x^2 - 5x + 6 = 0$
    Factor: $(x-2)(x-3) = 0 \implies x = 2$ or $x = 3$.
  2. $x^2 + 6x + 5 = 0$
    Complete square: $(x+3)^2 - 9 + 5 = 0 \implies (x+3)^2 = 4 \implies x+3 = \pm 2 \implies x = -1$ or $x = -5$.
  3. $2x^2 - 4x - 3 = 0$
    Quadratic formula: $x = \dfrac{4 \pm \sqrt{16 - 4(2)(-3)}}{4} = \dfrac{4 \pm \sqrt{40}}{4} = \dfrac{4 \pm 2\sqrt{10}}{4} = 1 \pm \dfrac{\sqrt{10}}{2}$.

B. Quadratic Inequalities

A quadratic inequality involves $\leq$, $\geq$, $\lt$, or $\gt$ rather than $=$.

Method: Sign Diagram Approach

  1. Rearrange so one side is $0$: $ax^2 + bx + c \; (\text{sign}) \; 0$.
  2. Find roots of $ax^2 + bx + c = 0$.
  3. Draw a number line with roots as boundary points.
  4. Test each interval to determine sign.
  5. Write solution set based on inequality direction.

Examples

  1. $x^2 - 4x + 3 \gt 0$
    Roots: $(x-1)(x-3)=0 \implies x=1, x=3$
    Intervals: $(-\infty,1)$, $(1,3)$, $(3,\infty)$
    Test $x=0$: $0-0+3=3\gt0$ (positive)
    Test $x=2$: $4-8+3=-1\lt0$ (negative)
    Test $x=4$: $16-16+3=3\gt0$ (positive)
    Solution: $(-\infty,1) \cup (3,\infty)$.
  2. $x^2 + x - 2 \leq 0$
    Roots: $(x+2)(x-1)=0 \implies x=-2, x=1$
    Test $x=-3$: $9-3-2=4\gt0$ (+)
    Test $x=0$: $0+0-2=-2\lt0$ (-)
    Test $x=2$: $4+2-2=4\gt0$ (+)
    Solution: $[-2,1]$.

C. The Discriminant of a Quadratic

For $ax^2 + bx + c = 0$, the discriminant is:

$\Delta = b^2 - 4ac$

Interpretation

Discriminant Meaning
$\Delta \gt 0$ Two distinct real roots
$\Delta = 0$ One repeated real root (tangent to x-axis)
$\Delta \lt 0$ No real roots (complex conjugates)

Examples

  1. $x^2 - 5x + 6 = 0$: $\Delta = 25 - 24 = 1 \gt 0$ → two real roots.
  2. $x^2 - 4x + 4 = 0$: $\Delta = 16 - 16 = 0$ → repeated root $x=2$.
  3. $x^2 + x + 1 = 0$: $\Delta = 1 - 4 = -3 \lt 0$ → no real roots.

D. Quadratic Functions

The graph of $f(x) = ax^2 + bx + c$ is a parabola.

Key Features

  • Vertex: $x_v = -\dfrac{b}{2a}$, $y_v = f(x_v)$.
  • Axis of symmetry: $x = -\dfrac{b}{2a}$.
  • Concavity: $a \gt 0$ opens upward (minimum), $a \lt 0$ opens downward (maximum).
  • y-intercept: $(0, c)$.
  • x-intercepts: Roots of $ax^2+bx+c=0$ (if $\Delta \geq 0$).

Vertex Form (Completed Square Form)

$f(x) = a(x - h)^2 + k$

where $(h, k)$ is the vertex, $h = -\frac{b}{2a}$, $k = f(h)$.

Example

For $f(x) = 2x^2 - 8x + 5$:

$\text{Vertex: } x_v = -\frac{-8}{2(2)} = \frac{8}{4} = 2$

$y_v = 2(4) - 8(2) + 5 = 8 - 16 + 5 = -3$

$\text{Vertex } (2, -3), \quad a=2\gt0 \text{ opens upward.}$

E. Finding a Quadratic from Its Graph

Given three points on a parabola, or the vertex and one other point, we can determine $a, b, c$.

Method 1: Using Vertex Form

If vertex $(h,k)$ and another point $(x_1, y_1)$ are known:

$f(x) = a(x - h)^2 + k$

Substitute $(x_1, y_1)$ to solve for $a$.

Method 2: Using Standard Form

Given three points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$:

$ \begin{cases} ax_1^2 + bx_1 + c = y_1 \\ ax_2^2 + bx_2 + c = y_2 \\ ax_3^2 + bx_3 + c = y_3 \end{cases} $

Solve the linear system in $a,b,c$.

Method 3: Using x-intercepts (if known)

If roots $r_1, r_2$ and one other point $(x_0, y_0)$:

$f(x) = a(x - r_1)(x - r_2)$

Substitute to find $a$.

Example

A parabola has vertex $(2, -3)$ and passes through $(0, 5)$:

$f(x) = a(x-2)^2 - 3$

$5 = a(0-2)^2 - 3 \implies 5 = 4a - 3 \implies 8 = 4a \implies a = 2$

Thus $f(x) = 2(x-2)^2 - 3 = 2x^2 - 8x + 5$.

F. Where Functions Meet

To find intersection points of two functions $y = f(x)$ and $y = g(x)$:

  1. Set $f(x) = g(x)$.
  2. Solve for $x$.
  3. Substitute back to find $y$.

Examples

  1. Find intersection of $f(x)=x^2-2x+1$ and $g(x)=2x-2$:
    $x^2-2x+1 = 2x-2 \implies x^2 - 4x + 3 = 0$
    $(x-1)(x-3)=0 \implies x=1$ or $x=3$
    $x=1 \implies y=0$, $x=3 \implies y=4$
    Intersection points: $(1,0)$ and $(3,4)$.
  2. Line and parabola intersection:
    For $y = x^2$ and $y = x + 2$:
    $x^2 = x+2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1)=0$
    Points: $(2,4)$ and $(-1,1)$.

G. Problem Solving with Quadratics

Many real-world problems reduce to quadratics.

Example: Projectile Motion

The height of a projectile is $h(t) = -16t^2 + 64t + 5$ (feet, $t$ seconds).

  1. Time to reach maximum height: $t = -\frac{b}{2a} = -\frac{64}{2(-16)} = 2$ seconds.
  2. Maximum height: $h(2) = -16(4) + 128 + 5 = -64 + 133 = 69$ feet.
  3. Time to hit ground ($h=0$):
    $-16t^2+64t+5=0$
    $t = \dfrac{-64 \pm \sqrt{4096 - 4(-16)(5)}}{-32} = \dfrac{-64 \pm \sqrt{4096+320}}{-32} = \dfrac{-64 \pm \sqrt{4416}}{-32}$
    $\sqrt{4416} \approx 66.45$, $t \approx 4.08$ seconds (positive root).

Example: Number Problems

The product of two consecutive integers is 306. Find them.

Let $n$ and $n+1$: $n(n+1)=306 \implies n^2+n-306=0$

$\Delta = 1 + 1224 = 1225 = 35^2$

$n = \frac{-1 \pm 35}{2} \implies n=17$ or $n=-18$

Numbers: $17, 18$ or $-18, -17$.

H. Quadratic Optimisation

Since a quadratic has a single vertex, it attains a global maximum or minimum there.

Procedure

  1. Identify the quantity to optimize (area, profit, etc.) as a quadratic function.
  2. Find the vertex $x_v = -\frac{b}{2a}$.
  3. If $a \gt 0$, vertex gives minimum; if $a \lt 0$, vertex gives maximum.
  4. Compute the optimum value $f(x_v)$.
  5. Check domain restrictions (e.g., positive lengths).

Example: Maximizing Area

A farmer has 200 meters of fencing to enclose a rectangular field against a barn (so only 3 sides need fencing). Maximize area.

Let width = $x$, length = $y$ (parallel to barn).

Fencing: $2x + y = 200 \implies y = 200 - 2x$.

Area: $A(x) = x \cdot y = x(200 - 2x) = 200x - 2x^2$.

This is a downward parabola ($a = -2 \lt 0$), so vertex gives max.

$x_v = -\frac{b}{2a} = -\frac{200}{2(-2)} = \frac{200}{4} = 50$ meters.

$y = 200 - 2(50) = 100$ meters.

Maximum area $A = 50 \times 100 = 5000$ m$^2$.

Example: Minimizing Cost

A company finds its daily cost is $C(x) = x^2 - 60x + 1200$ where $x$ is units produced.

$C$ is a upward parabola ($a=1\gt0$), minimum at vertex.

$x_v = -\frac{-60}{2(1)} = 30$ units.

Minimum cost $C(30) = 900 - 1800 + 1200 = 300$ dollars.

Summary Table

Concept Key Formula/Method
Quadratic equation $ax^2+bx+c=0$, solve by factoring/quadratic formula
Quadratic inequality Sign diagram using roots
Discriminant $\Delta = b^2-4ac$ (nature of roots)
Quadratic function $f(x)=a(x-h)^2+k$, vertex $(h,k)$
Finding from graph Use vertex form or intercept form
Intersection Set $f(x)=g(x)$, solve quadratic
Problem solving Model real situations with quadratics
Optimisation Vertex gives max (if $a\lt0$) or min (if $a\gt0$)

Chapter 4: Surds, indices, and exponentials

Introduction

Surds, indices, and exponentials form the foundation of algebraic manipulation and calculus. This summary covers irrational roots, exponent rules, rational powers, exponential equations, and both general and natural exponential functions.

A. Surds

A surd is an irrational root of a rational number, typically expressed in radical form $\sqrt[n]{a}$ where $a$ is not a perfect $n$th power.

Laws of Surds

For $a, b \gt 0$ and $n \in \mathbb{N}$:

$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}$

$\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}}$

$\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}$

$(\sqrt[n]{a})^m = \sqrt[n]{a^m}$

Simplifying Surds

Factor out perfect squares (for square roots):

$\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$

Rationalising the Denominator

  • Single term: $\dfrac{1}{\sqrt{a}} = \dfrac{\sqrt{a}}{a}$
  • Binomial: $\dfrac{1}{a + \sqrt{b}} \times \dfrac{a - \sqrt{b}}{a - \sqrt{b}} = \dfrac{a - \sqrt{b}}{a^2 - b}$

Examples

  1. $\sqrt{50} + 3\sqrt{2} - \sqrt{18} = 5\sqrt{2} + 3\sqrt{2} - 3\sqrt{2} = 5\sqrt{2}$
  2. $\dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$
  3. $\dfrac{5}{\sqrt{5} - 1} = \dfrac{5(\sqrt{5}+1)}{5-1} = \dfrac{5\sqrt{5}+5}{4}$

B. Indices

An index (plural: indices) or exponent indicates repeated multiplication.

Basic Definition

For $n \in \mathbb{N}$, $a^n = a \times a \times \cdots \times a$ ($n$ factors).

Special Cases

  • $a^0 = 1$ (for $a \neq 0$)
  • $a^{-n} = \dfrac{1}{a^n}$ (for $a \neq 0$)

C. Index Laws

For $a, b \gt 0$ and $m, n \in \mathbb{R}$ (where defined):

Law Example
$a^m \times a^n = a^{m+n}$ $2^3 \times 2^4 = 2^7 = 128$
$a^m \div a^n = a^{m-n}$ $3^5 \div 3^2 = 3^3 = 27$
$(a^m)^n = a^{mn}$ $(5^2)^3 = 5^6 = 15625$
$(ab)^n = a^n b^n$ $(2x)^3 = 8x^3$
$\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$ $\left(\dfrac{2}{3}\right)^4 = \dfrac{16}{81}$

Examples

  1. Simplify $\dfrac{2^5 \times 2^3}{2^4} = 2^{5+3-4} = 2^4 = 16$
  2. Simplify $(3x^2 y^3)^4 = 3^4 x^{8} y^{12} = 81x^8 y^{12}$
  3. Simplify $\dfrac{6a^3 b^2}{3a b^5} = 2a^{2} b^{-3} = \dfrac{2a^2}{b^3}$

D. Rational Indices

Rational exponents connect indices to surds.

Definition

For $a \gt 0$ and $m, n \in \mathbb{Z}$ with $n \gt 0$:

$a^{\frac{1}{n}} = \sqrt[n]{a} \quad \text{and} \quad a^{\frac{m}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$

Examples

  1. $27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$
  2. $16^{\frac{3}{4}} = (\sqrt[4]{16})^3 = 2^3 = 8$
  3. $8^{-\frac{2}{3}} = \dfrac{1}{8^{\frac{2}{3}}} = \dfrac{1}{(\sqrt[3]{8})^2} = \dfrac{1}{2^2} = \dfrac{1}{4}$
  4. $(\frac{4}{9})^{-\frac{1}{2}} = (\frac{9}{4})^{\frac{1}{2}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$

E. Algebraic Expansion and Factorisation

Use index laws to expand and factor expressions containing powers.

Expansion

  1. $x^2(x^3 + x) = x^5 + x^3$
  2. $(x^2 + y^2)(x^2 - y^2) = x^4 - y^4$
  3. $(x^{\frac{1}{2}} + 2)(x^{\frac{1}{2}} - 3) = x - x^{\frac{1}{2}} - 6$

Factorisation

  1. $x^5 - x^3 = x^3(x^2 - 1) = x^3(x-1)(x+1)$
  2. $2^{2x} - 2^{x+1} = 2^x(2^x - 2)$
  3. $e^{2x} - 3e^x + 2 = (e^x - 1)(e^x - 2)$
  4. $x^{\frac{2}{3}} - 4x^{\frac{1}{3}} = x^{\frac{1}{3}}(x^{\frac{1}{3}} - 4)$

F. Exponential Equations

Equations where the unknown appears in the exponent.

Methods

  1. Same base method: If $a^{f(x)} = a^{g(x)}$, then $f(x) = g(x)$.
  2. Taking logarithms: Apply $\ln$ or $\log_{10}$ to both sides.
  3. Substitution: Let $u = a^x$ to form a quadratic.

Examples

  1. $2^{x+1} = 16$
    $2^{x+1} = 2^4 \implies x+1 = 4 \implies x = 3$
  2. $3^{2x-1} = 27^{x}$
    $3^{2x-1} = (3^3)^x = 3^{3x} \implies 2x-1 = 3x \implies -1 = x \implies x = -1$
  3. $4^x - 6 \cdot 2^x + 8 = 0$
    Let $u = 2^x$, then $u^2 - 6u + 8 = 0$
    $(u-2)(u-4)=0 \implies u=2$ or $u=4$
    $2^x = 2 \implies x=1$, $2^x = 4 \implies x=2$
  4. $5^{2x} = 3$
    $\ln(5^{2x}) = \ln 3 \implies 2x \ln 5 = \ln 3 \implies x = \dfrac{\ln 3}{2\ln 5}$

G. Exponential Functions

An exponential function has the form $f(x) = a^x$ where $a \gt 0$, $a \neq 1$.

Properties

  • Domain: $\mathbb{R}$
  • Range: $(0, \infty)$
  • y-intercept: $(0,1)$
  • Horizontal asymptote: $y=0$ (as $x \to -\infty$ for $a\gt1$, as $x \to \infty$ for $0\lt a \lt1$)
  • One-to-one (passes horizontal line test)

Graph Characteristics

Base Growth/Decay Behaviour
$a \gt 1$ Exponential growth Increases, concave up
$0 \lt a \lt 1$ Exponential decay Decreases, concave up

Transformations

For $f(x) = k \cdot a^{x-h} + c$:

  • $k$: vertical stretch/compression
  • $h$: horizontal shift
  • $c$: vertical shift (asymptote becomes $y=c$)

Examples

  1. $f(x) = 2^x$: growth, passes through $(0,1)$, $(1,2)$, $(-1, \frac{1}{2})$
  2. $g(x) = (\frac{1}{2})^x = 2^{-x}$: decay, reflection of $2^x$ across y-axis
  3. $h(x) = 3 \cdot 2^{x-1} + 4$: asymptote $y=4$, y-intercept $h(0)=3\cdot 2^{-1}+4=5.5$

H. The Natural Exponential

The natural exponential function is $f(x) = e^x$, where $e$ is Euler's number ($e \approx 2.71828\ldots$).

Definition

$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = \sum_{n=0}^{\infty} \frac{1}{n!}$

Properties

  • Derivative: $\frac{d}{dx} e^x = e^x$ (unique property)
  • Integral: $\int e^x \, dx = e^x + C$
  • Inverse function: $\ln x$ (natural logarithm)

Key Relationships

$\ln(e^x) = x \quad \text{for all } x \in \mathbb{R}, \qquad e^{\ln x} = x \quad \text{for } x \gt 0$

Examples

  1. Simplify $e^{3x} \cdot e^{2x} = e^{5x}$
  2. Solve $e^{2x-1} = 5$:
    $\ln(e^{2x-1}) = \ln 5 \implies 2x-1 = \ln 5 \implies x = \frac{1+\ln 5}{2}$
  3. Solve $e^{2x} - 3e^x + 2 = 0$:
    Let $u = e^x$, then $u^2 - 3u + 2 = 0 \implies (u-1)(u-2)=0$
    $e^x = 1 \implies x=0$, $e^x = 2 \implies x = \ln 2$
  4. Simplify $\frac{e^{2x} - e^{-2x}}{e^x - e^{-x}}$:
    Factor numerator as $(e^x - e^{-x})(e^x + e^{-x})$, cancel to get $e^x + e^{-x}$

Applications

The natural exponential models continuous growth/decay:

$P(t) = P_0 e^{kt}$

where $k \gt 0$ for growth, $k \lt 0$ for decay (e.g., population, radioactive decay, compound interest).

Example: Continuous Compounding

If \$1000 is invested at 5% annual interest compounded continuously:

$A(t) = 1000 e^{0.05t}$

After 10 years: $A(10) = 1000 e^{0.5} \approx 1000 \times 1.64872 = \$1648.72$

Summary Table

Concept Key Points
Surds $\sqrt[n]{a}$, rationalise denominator, simplify perfect squares
Indices $a^n$, $a^0=1$, $a^{-n}=1/a^n$
Index laws Multiplication, division, power of power, product/quotient rules
Rational indices $a^{m/n} = \sqrt[n]{a^m}$
Expansion/factorisation Use index laws, common factor, substitution
Exponential equations Same base or logarithms
Exponential functions $f(x)=a^x$, domain $\mathbb{R}$, range $(0,\infty)$
Natural exponential $f(x)=e^x$, derivative is itself, inverse is $\ln x$

Chapter 5: Logarithms

Introduction

Logarithms are the inverse operations of exponentials. They are essential for solving exponential equations, modelling phenomena across science and finance, and simplifying multiplicative relationships into additive ones.

Logarithms in Base $a$

For $a > 0$, $a \neq 1$, the logarithm base $a$ of $x$ is defined as:

$$ \log_a x = y \iff a^y = x $$

where $x > 0$ (logarithm of a non-positive number is undefined).

Definition

$\log_a x$ answers the question: “To what power must $a$ be raised to obtain $x$?”

Examples

  1. $\log_2 8 = 3$ because $2^3 = 8$
  2. $\log_3 81 = 4$ because $3^4 = 81$
  3. $\log_5 5 = 1$ because $5^1 = 5$
  4. $\log_7 1 = 0$ because $7^0 = 1$
  5. $\log_{10} 0.01 = -2$ because $10^{-2} = 0.01$
  6. $\log_2 \frac{1}{4} = -2$ because $2^{-2} = \frac{1}{4}$

Special Values

  • $\log_a 1 = 0$ (since $a^0 = 1$)
  • $\log_a a = 1$ (since $a^1 = a$)
  • $\log_a (a^x) = x$ (inverse property)
  • $a^{\log_a x} = x$ for $x > 0$

Logarithms in Base $10$

Base 10 logarithms are called common logarithms. They are often written without the base:

$$ \log x \text{ means } \log_{10} x $$

Examples

  1. $\log 100 = 2$ because $10^2 = 100$
  2. $\log 1000 = 3$ because $10^3 = 1000$
  3. $\log 0.1 = -1$ because $10^{-1} = 0.1$
  4. $\log 1 = 0$

Laws of Logarithms

For $a > 0$, $a \neq 1$, and $M, N > 0$:

Law Formula
Product Law $\log_a (MN) = \log_a M + \log_a N$
Quotient Law $\log_a \left(\frac{M}{N}\right) = \log_a M - \log_a N$
Power Law $\log_a (M^k) = k \log_a M$
Change of Base $\log_a M = \dfrac{\log_b M}{\log_b a}$

Examples

  1. $\log_2 (8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5$
    Check: $\log_2 32 = 5$ Correct
  2. $\log_3 \left(\frac{81}{9}\right) = \log_3 81 - \log_3 9 = 4 - 2 = 2$
    Check: $\log_3 9 = 2$ Correct
  3. $\log_5 (125^2) = 2 \log_5 125 = 2 \times 3 = 6$
  4. Simplify $\log_a (x^2 y^3)$:
    $\log_a (x^2 y^3) = \log_a x^2 + \log_a y^3 = 2\log_a x + 3\log_a y$
  5. Simplify $\log_a \left(\frac{x^4}{y^2}\right) = 4\log_a x - 2\log_a y$

Logarithmic Equations

Equations where the unknown appears inside a logarithm.

Methods

  1. Use logarithm laws to combine terms.
  2. Rewrite in exponential form: $\log_a (f(x)) = c \implies f(x) = a^c$.
  3. If $\log_a f(x) = \log_a g(x)$, then $f(x) = g(x)$ (must check domain: $f(x)>0$, $g(x)>0$).
  4. For equations with multiple logs, isolate then exponentiate.

Examples

  1. $\log_3 (x+2) = 2$
    Exponential form: $x+2 = 3^2 = 9 \implies x = 7$
    Check: $7+2=9>0$ Correct
  2. $\log_2 x + \log_2 (x-3) = 2$
    Combine: $\log_2 [x(x-3)] = 2$
    $x(x-3) = 2^2 = 4 \implies x^2 - 3x - 4 = 0$
    $(x-4)(x+1)=0 \implies x=4$ or $x=-1$
    $x=-1$ invalid (domain $x>0$ and $x-3>0$), so $x=4$ only.
  3. $\ln(x+1) - \ln(x-1) = \ln 3$
    $\ln\left(\frac{x+1}{x-1}\right) = \ln 3 \implies \frac{x+1}{x-1} = 3$
    $x+1 = 3x - 3 \implies 4 = 2x \implies x = 2$
    Check: $x+1=3>0$, $x-1=1>0$ Correct
  4. $\log_5 (x-2) = \log_5 (x+6) - \log_5 2$
    $\log_5 (x-2) = \log_5 \left(\frac{x+6}{2}\right)$
    $x-2 = \frac{x+6}{2} \implies 2x-4 = x+6 \implies x = 10$
    Check: $10-2=8>0$, $(10+6)/2=8>0$ Correct

Natural Logarithms

The natural logarithm $\ln x$ is a logarithm with base $e$, where $e \approx 2.71828\ldots$

$$ \ln x = \log_e x $$

Properties

  • $\ln e = 1$
  • $\ln 1 = 0$
  • $\ln(e^x) = x$ for all $x \in \mathbb{R}$
  • $e^{\ln x} = x$ for $x > 0$
  • $\ln(ab) = \ln a + \ln b$
  • $\ln\left(\frac{a}{b}\right) = \ln a - \ln b$
  • $\ln(a^k) = k \ln a$

Examples

  1. $\ln e^5 = 5$
  2. $e^{\ln 7} = 7$
  3. Simplify $\ln(9x^4) = \ln 9 + 4\ln x$
  4. Simplify $\ln\left(\frac{e^3}{x^2}\right) = 3 - 2\ln x$
  5. Solve $\ln(x-2) = 3$: $x-2 = e^3 \implies x = e^3 + 2$

Solving Exponential Equations Using Logarithms

When bases cannot be made the same, take logarithms of both sides.

Method

  1. Isolate the exponential term if necessary.
  2. Take $\ln$ or $\log$ of both sides.
  3. Use the power law: $\ln(a^x) = x \ln a$.
  4. Solve for the variable.

Examples

  1. $2^x = 7$
    $\ln(2^x) = \ln 7 \implies x \ln 2 = \ln 7 \implies x = \dfrac{\ln 7}{\ln 2} \approx 2.807$
  2. $3^{2x-1} = 5^{x+2}$
    $\ln(3^{2x-1}) = \ln(5^{x+2})$
    $(2x-1)\ln 3 = (x+2)\ln 5$
    $2x\ln 3 - \ln 3 = x\ln 5 + 2\ln 5$
    $2x\ln 3 - x\ln 5 = 2\ln 5 + \ln 3$
    $x(2\ln 3 - \ln 5) = 2\ln 5 + \ln 3$
    $x = \dfrac{2\ln 5 + \ln 3} {2\ln 3 - \ln 5}$
  3. $e^{2x} - 4e^x + 3 = 0$
    Let $u = e^x$, then $u^2 - 4u + 3 = 0$
    $(u-1)(u-3)=0 \implies u=1$ or $u=3$
    $e^x = 1 \implies x=0$
    $e^x = 3 \implies x = \ln 3$
  4. $5 \cdot 10^{2x} = 500$
    $10^{2x} = 100$
    $\log(10^{2x}) = \log 100 \implies 2x \log 10 = 2$
    $2x = 2 \implies x = 1$

The Change of Base Rule

For any positive $a, b, x$ with $a \neq 1$, $b \neq 1$:

$$ \log_a x = \frac{\log_b x}{\log_b a} $$

Common choices: $b = 10$ (common log) or $b = e$ (natural log)

$$ \log_a x = \frac{\log x}{\log a} = \frac{\ln x}{\ln a} $$

Uses

  1. Evaluate logarithms with unusual bases on calculators.
  2. Simplify expressions and solve equations.

Examples

  1. $\log_3 8 = \frac{\ln 8}{\ln 3} \approx \frac{2.07944}{1.09861} \approx 1.893$
  2. $\log_5 100 = \frac{\log 100}{\log 5} = \frac{2}{\log 5} \approx \frac{2}{0.69897} \approx 2.861$
  3. Simplify $\log_a b \cdot \log_b a$:
    $\log_a b \cdot \log_b a = \frac{\ln b}{\ln a} \cdot \frac{\ln a}{\ln b} = 1$
  4. Solve $\log_2 x + \log_4 x = 3$
    Convert to base 2: $\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}$
    Let $y = \log_2 x$:
    $y + \frac{y}{2} = 3 \implies \frac{3y}{2} = 3 \implies y = 2$
    $\log_2 x = 2 \implies x = 4$

Graphs of Logarithmic Functions

The graph of $f(x) = \log_a x$ is the reflection of $g(x) = a^x$ across the line $y = x$.

Properties of $f(x) = \log_a x$

  • Domain: $(0, \infty)$
  • Range: $\mathbb{R}$
  • x-intercept: $(1, 0)$
  • Vertical asymptote: $x = 0$ (y-axis)
  • For $a > 1$: increasing function
  • For $0 < a < 1$: decreasing function
  • One-to-one (passes horizontal line test)

Graph Characteristics

Base Behaviour Shape
$a > 1$ Increasing Concave down, slow growth
$0 < a < 1$ Decreasing Concave up, negative slope

Transformations

For $f(x) = k \cdot \log_a (x - h) + c$:

  • $k$: vertical stretch/compression and reflection
  • $h$: horizontal shift (vertical asymptote at $x = h$)
  • $c$: vertical shift

Examples

  1. $f(x) = \log_2 x$: passes through $(1,0)$, $(2,1)$, $(4,2)$, $\left(\frac{1}{2},-1\right)$
  2. $g(x) = \log_{\frac{1}{2}} x = -\log_2 x$: reflection of $\log_2 x$ across x-axis
  3. $h(x) = \ln(x-2) + 1$: shift right 2, up 1; asymptote at $x=2$; passes through $(3,1)$ since $\ln(1)=0$
  4. $p(x) = -2\log_3 (x+1)$: shift left 1, vertical stretch by 2, reflection across x-axis; asymptote at $x=-1$

Key Points on Logarithmic Graphs

For $y = \log_a x$ (with $a>1$):

  • As $x \to 0^+$, $y \to -\infty$ (approaches asymptote)
  • As $x \to \infty$, $y \to \infty$ (slowly)
  • The graph crosses x-axis at $(1,0)$
  • The graph passes through $(a,1)$ and $\left(\frac{1}{a},-1\right)$

Summary Table

Concept Key Points
Definition $\log_a x = y \iff a^y = x$, $x>0$, $a>0$, $a\neq 1$
Common log $\log x = \log_{10} x$
Natural log $\ln x = \log_e x$, $e \approx 2.71828$
Product law $\log_a(MN) = \log_a M + \log_a N$
Quotient law $\log_a(M/N) = \log_a M - \log_a N$
Power law $\log_a(M^k) = k \log_a M$
Change of base $\log_a x = \frac{\ln x}{\ln a} = \frac{\log x}{\log a}$
Domain of $\log_a x$ $x > 0$
Asymptote $x = 0$ (vertical)

Quick Reference: Exponential $\leftrightarrow$ Logarithmic Form

Exponential Logarithmic
$a^y = x$ $y = \log_a x$
$10^3 = 1000$ $\log 1000 = 3$
$e^2 \approx 7.389$ $\ln 7.389 \approx 2$
$2^{-3} = \frac{1}{8}$ $\log_2 \frac{1}{8} = -3$

Chapter 6: Summary of Polynomials

Introduction

Polynomials are fundamental algebraic expressions that appear throughout mathematics. Understanding their structure, roots, and factorisation is essential for calculus, algebra, and mathematical modelling. This summary covers real polynomials, zeros and factors, the Remainder Theorem, and the Factor Theorem.

Real Polynomials

Definition

A real polynomial in $x$ is an expression of the form:

$$ P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $$

where $n$ is a non-negative integer, and the coefficients $a_n, a_{n-1}, \dots, a_0$ are real numbers with $a_n \neq 0$.

Terminology

  • Degree ($\deg P$): The highest power $n$ with non-zero coefficient.
  • Leading term: $a_n x^n$
  • Leading coefficient: $a_n$
  • Constant term: $a_0$
  • Monic polynomial: Leading coefficient $a_n = 1$

Examples

  1. $P(x) = 3x^4 - 2x^3 + x - 5$
    Degree: 4, Leading coefficient: 3, Constant term: $-5$
  2. $Q(x) = x^3 + 2x^2 - 7$ (monic, degree 3)
  3. $R(x) = 7$ (constant polynomial, degree 0)
  4. $S(x) = 0$ (zero polynomial, degree undefined)
  5. $T(x) = (x-1)(x+2)(x-3) = x^3 - 2x^2 - 5x + 6$

Properties of Real Polynomials

  • The sum, difference, and product of polynomials are polynomials.
  • Polynomials are continuous and differentiable for all real $x$.
  • A polynomial of degree $n$ has at most $n$ real zeros.
  • As $x \to \pm\infty$, the leading term dominates the behaviour.

End Behaviour

For $P(x) = a_n x^n + \cdots + a_0$:

Degree $n$ $a_n > 0$ $a_n \lt 0$
Even $x \to \pm\infty$, $P(x) \to +\infty$ $x \to \pm\infty$, $P(x) \to -\infty$
Odd $x \to -\infty$, $P(x) \to -\infty$;
$x \to \infty$, $P(x) \to \infty$ 		$x \to -\infty$, $P(x) \to \infty$;
$x \to \infty$, $P(x) \to -\infty$

Zeros, Roots, and Factors

Definitions

  • A zero of a polynomial $P(x)$ is a number $r$ such that $P(r) = 0$.
  • A root of the polynomial equation $P(x) = 0$ is the same as a zero.
  • A factor of $P(x)$ is a polynomial $Q(x)$ such that $P(x) = Q(x) \cdot R(x)$ for some polynomial $R(x)$.

The Factor Theorem (Preview)

If $r$ is a zero of $P(x)$, then $(x - r)$ is a factor of $P(x)$. (Formally stated in Section D.)

Multiplicity

If $(x - r)^k$ is a factor of $P(x)$ but $(x - r)^{k+1}$ is not, then $r$ is a zero of multiplicity $k$.

Behaviour at Zeros

  • Multiplicity odd: Graph crosses the x-axis.
  • Multiplicity even: Graph touches but does not cross the x-axis.

Examples

  1. $P(x) = (x-2)(x+3) = x^2 + x - 6$
    Zeros: $x = 2$ and $x = -3$ (each multiplicity 1, graph crosses)
  2. $P(x) = (x-1)^2 (x+2) = x^3 - 3x + 2$
    Zeros: $x = 1$ (multiplicity 2, touches), $x = -2$ (multiplicity 1, crosses)
  3. $P(x) = (x-4)^3$
    Zero: $x = 4$ (multiplicity 3, crosses as odd multiplicity)
  4. $P(x) = x^2 + 1$
    No real zeros (zeros are $x = \pm i$, complex)

Fundamental Theorem of Algebra

Every non-constant polynomial with complex coefficients has at least one complex zero. Consequently, a polynomial of degree $n$ has exactly $n$ complex zeros (counting multiplicities).

The Remainder Theorem

Theorem (Remainder Theorem)

When a polynomial $P(x)$ is divided by a linear divisor $(x - c)$, the remainder is $P(c)$.

$$ P(x) = (x - c) \cdot Q(x) + P(c) $$

where $Q(x)$ is the quotient polynomial (degree one less than $P$).

Proof

From polynomial long division: $P(x) = (x - c)Q(x) + R$, where $R$ is constant (since divisor is linear). Substitute $x = c$:

$$ P(c) = (c - c)Q(c) + R = 0 \cdot Q(c) + R = R $$

Thus $R = P(c)$. $\square$

Applications

  1. Evaluate polynomials quickly without substitution.
  2. Check if $(x - c)$ is a factor (remainder $= 0$).
  3. Find remainders without performing division.

Examples

  1. Find the remainder when $P(x) = x^3 - 2x^2 + 3x - 5$ is divided by $(x - 2)$.
    $P(2) = 8 - 8 + 6 - 5 = 1$
    Remainder = 1.
  2. Find the remainder when $P(x) = 2x^4 - 3x^3 + x - 7$ is divided by $(x + 1)$.
    Note: $x + 1 = x - (-1)$, so $c = -1$
    $P(-1) = 2(1) - 3(-1) + (-1) - 7 = 2 + 3 - 1 - 7 = -3$
    Remainder = $-3$.
  3. For $P(x) = x^3 - 5x^2 + 8x - 4$, find $P(1)$ without substitution:
    Divide by $(x-1)$: synthetic division gives remainder $P(1) = 0$.
  4. If $P(x) = x^4 - 3x^3 + kx^2 + 2x - 1$ has remainder 5 when divided by $(x-2)$, find $k$.
    $P(2) = 16 - 24 + 4k + 4 - 1 = (16 - 24 + 4 - 1) + 4k = (-5) + 4k = 5$
    $4k = 10 \implies k = 2.5$

Synthetic Division (Horner's Method)

A quick algorithm for dividing a polynomial by $(x - c)$:

For $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$:

  1. Write coefficients: $a_n, a_{n-1}, \dots, a_0$
  2. Bring down $a_n$
  3. Multiply by $c$, add to next coefficient
  4. Repeat
  5. Last number is remainder $P(c)$

Example: Synthetic Division

Divide $P(x) = 2x^3 - 5x^2 + 3x - 4$ by $(x - 3)$:

$$ \begin{array}{c|cccc} 3 & 2 & -5 & 3 & -4 \\ & & 6 & 3 & 18 \\ \hline & 2 & 1 & 6 & 14 \\ \end{array} $$

Quotient: $2x^2 + x + 6$, Remainder: $14 = P(3)$.

The Factor Theorem

Theorem (Factor Theorem)

A polynomial $P(x)$ has a factor $(x - c)$ if and only if $P(c) = 0$. That is, $c$ is a zero of $P(x)$.

Proof

($\Rightarrow$) If $(x - c)$ is a factor, then $P(x) = (x - c)Q(x)$. Then $P(c) = (c - c)Q(c) = 0$.

($\Leftarrow$) If $P(c) = 0$, by the Remainder Theorem, $P(x) = (x - c)Q(x) + P(c) = (x - c)Q(x) + 0 = (x - c)Q(x)$. Thus $(x - c)$ is a factor. $\square$

Consequences

  • Every zero corresponds to a linear factor.
  • A polynomial of degree $n$ can be factored as $P(x) = a_n (x - r_1)(x - r_2)\cdots(x - r_n)$ over the complex numbers.
  • Over the reals, non-real zeros occur in conjugate pairs.

Examples

  1. Show that $(x-2)$ is a factor of $P(x) = x^3 - 4x^2 + x + 6$.
    $P(2) = 8 - 16 + 2 + 6 = 0 \implies (x-2)$ is a factor.
    Factor: $P(x) = (x-2)(x^2 - 2x - 3) = (x-2)(x-3)(x+1)$
  2. Factor $P(x) = x^3 - 3x^2 - 4x + 12$ completely.
    Test $x=2$: $P(2) = 8 - 12 - 8 + 12 = 0$, so $(x-2)$ is a factor.
    Divide: $P(x) = (x-2)(x^2 - x - 6) = (x-2)(x-3)(x+2)$
  3. Find $k$ such that $(x+2)$ is a factor of $P(x) = x^3 + kx^2 - 3x + 2$.
    $P(-2) = (-8) + k(4) + 6 + 2 = 4k + 0 = 0 \implies 4k = 0 \implies k = 0$
  4. Factor $P(x) = x^4 - 5x^2 + 4$ using the Factor Theorem.
    Let $u = x^2$: $u^2 - 5u + 4 = (u-1)(u-4)$
    So $P(x) = (x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2)$

Finding Zeros of Higher-Degree Polynomials

Strategy using the Factor Theorem:

  1. Test possible rational zeros using the Rational Root Theorem.
  2. When a zero $c$ is found, factor out $(x - c)$.
  3. Repeat on the quotient until quadratic remains.
  4. Solve quadratic (factor or quadratic formula).

Rational Root Theorem

If $P(x) = a_n x^n + \cdots + a_0$ has integer coefficients, then any rational zero $\frac{p}{q}$ in lowest terms satisfies:

  • $p$ divides the constant term $a_0$
  • $q$ divides the leading coefficient $a_n$

Example: Rational Root Theorem

Factor $P(x) = 2x^3 - 3x^2 - 8x - 3$.

Possible rational zeros: $\pm1, \pm3, \pm\frac{1}{2}, \pm\frac{3}{2}$

Test $x=3$: $P(3) = 54 - 27 - 24 - 3 = 0$, so $(x-3)$ is a factor.

Divide: $P(x) = (x-3)(2x^2 + 3x + 1) = (x-3)(2x+1)(x+1)$

Zeros: $x = 3, -\frac{1}{2}, -1$.

Relationship Between Zeros and Coefficients

For $P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0$ with zeros $r_1, \dots, r_n$:

$$ \begin{aligned} r_1 + r_2 + \cdots + r_n &= -a_{n-1} \\ r_1 r_2 + r_1 r_3 + \cdots + r_{n-1} r_n &= a_{n-2} \\ &\vdots \\ r_1 r_2 \cdots r_n &= (-1)^n a_0 \end{aligned} $$

Example: Sum and Product of Zeros

For $P(x) = x^3 - 5x^2 + 2x - 8$:

  • Sum of zeros $= 5$
  • Product of zeros $= -(-8) = 8$ (since $n=3$, $(-1)^3 = -1$)

Summary Table

Concept Key Points
Real polynomial $P(x) = a_n x^n + \cdots + a_0$, $a_i \in \mathbb{R}$
Degree Highest power with non-zero coefficient
Zero/Root $r$ such that $P(r) = 0$
Multiplicity Number of times $(x-r)$ appears as a factor
Remainder Theorem $P(x) \div (x-c)$ leaves remainder $P(c)$
Factor Theorem $(x-c)$ is a factor $\iff P(c) = 0$
Rational Root Theorem Possible zeros $\frac{p}{q}$ where $p|a_0$, $q|a_n$
Synthetic division Efficient algorithm for division by $(x-c)$

Quick Reference: Remainder vs Factor Theorem

Remainder Theorem Factor Theorem
$P(x) = (x-c)Q(x) + R$ Special case when $R = 0$
$R = P(c)$ $(x-c)$ divides $P(x)$ exactly
Always true True only when $P(c) = 0$
Finds remainder without division Confirms a factor

Chapter 7:Straight line graph

%Straightline graph

Introduction

Straight line graphs are fundamental to algebra and calculus. They model linear relationships between variables and serve as the foundation for understanding more complex functions. This summary covers equations of lines, intersections, transforming relationships to linear form, and extracting relationships from data.

Equations of Straight Lines

General Form

The most general form of a linear equation is:

$$Ax + By + C = 0$$

where $A$, $B$, $C$ are constants, and $A$ and $B$ are not both zero.

Standard Forms

Form Equation Key Information
Gradient-intercept $y = mx + c$ Gradient $m$, $y$-intercept $c$
Point-gradient $y - y_1 = m(x - x_1)$ Gradient $m$, point $(x_1, y_1)$
Two-point $\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}$ Points $(x_1, y_1)$ and $(x_2, y_2)$
Intercept form $\dfrac{x}{a} + \dfrac{y}{b} = 1$ $x$-intercept $a$, $y$-intercept $b$

Gradient (Slope)

The gradient measures the steepness of a line:

$$m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$$

Special Lines

  • Horizontal line: $y = c$ (gradient $m = 0$)
  • Vertical line: $x = k$ (gradient undefined)
  • Lines through the origin: $y = mx$

Parallel and Perpendicular Lines

  • Parallel lines: $m_1 = m_2$
  • Perpendicular lines: $m_1 \cdot m_2 = -1$ (provided neither is vertical)

Examples

  1. Find the equation of the line through $(2,3)$ with gradient $4$.
    $y - 3 = 4(x - 2) \implies y - 3 = 4x - 8 \implies y = 4x - 5$
  2. Find the equation of the line through $(1,2)$ and $(3,8)$.
    $m = \frac{8-2}{3-1} = \frac{6}{2} = 3$
    $y - 2 = 3(x - 1) \implies y = 3x - 1$
  3. Find the equation of the line perpendicular to $y = 2x + 1$ through $(0,3)$.
    Perpendicular gradient $m = -\frac{1}{2}$
    $y - 3 = -\frac{1}{2}(x - 0) \implies y = -\frac{1}{2}x + 3$
  4. Find the $x$ and $y$ intercepts of $2x + 3y = 12$.
    $x$-intercept ($y=0$): $2x = 12 \implies x = 6$
    $y$-intercept ($x=0$): $3y = 12 \implies y = 4$

Intersection of Straight Lines

The intersection point of two lines satisfies both equations simultaneously.

Methods of Solution

  1. Substitution: Solve one equation for one variable, substitute into the other.
  2. Elimination: Add or subtract equations to eliminate one variable.
  3. Graphical: Plot both lines and read intersection.

Possible Outcomes

  • Unique solution: Lines intersect at one point (different gradients).
  • No solution: Lines are parallel and distinct (same gradient, different intercepts).
  • Infinite solutions: Lines are coincident (same line).

Examples

  1. Solve: $\begin{cases} y = 2x + 1 \\ y = -x + 4 \end{cases}$
    Substitute: $2x + 1 = -x + 4 \implies 3x = 3 \implies x = 1$
    $y = 2(1) + 1 = 3$
    Intersection: $(1,3)$
  2. Solve: $\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}$
    Add equations: $3x = 9 \implies x = 3$
    Substitute: $3 - y = 2 \implies y = 1$
    Intersection: $(3,1)$
  3. Solve: $\begin{cases} 3x - 2y = 7 \\ 6x - 4y = 5 \end{cases}$
    Multiply first equation by 2: $6x - 4y = 14$
    Compare with second: $6x - 4y = 5$
    Contradiction $\implies$ no solution (parallel lines).

Intersection of a Straight Line and a Curve

To find intersection points, solve the equations simultaneously: substitute the linear equation into the curve equation.

General Method

Given: Line $y = mx + c$ and curve $y = f(x)$.

$$mx + c = f(x)$$

Solve the resulting equation for $x$, then find corresponding $y$ values.

Quadratic Case (Line meets Parabola)

For a line and a quadratic $y = ax^2 + bx + c$:

  • Two intersections: Line cuts curve at two points (discriminant $\Delta > 0$)
  • One intersection (tangent): Line touches curve (discriminant $\Delta = 0$)
  • No intersection: Line misses curve (discriminant $\Delta \lt 0$)

Examples

  1. Find intersection of $y = x + 2$ and $y = x^2$.
    $x + 2 = x^2 \implies x^2 - x - 2 = 0$
    $(x-2)(x+1)=0 \implies x = 2$ or $x = -1$
    $x=2 \implies y=4$, $x=-1 \implies y=1$
    Points: $(2,4)$ and $(-1,1)$.
  2. Show that $y = 2x + 1$ is tangent to $y = x^2 + 1$.
    $x^2 + 1 = 2x + 1 \implies x^2 - 2x = 0 \implies x(x-2)=0$
    $x = 0$ or $x = 2$ → two intersections? Wait, check carefully:
    $x^2 - 2x = 0 \implies x(x-2)=0$ gives two solutions $x=0$ and $x=2$.
    Not tangent. Let's find a tangent example:
  3. Find $k$ such that $y = x + k$ is tangent to $y = x^2$.
    $x^2 = x + k \implies x^2 - x - k = 0$
    For tangency: $\Delta = (-1)^2 - 4(1)(-k) = 1 + 4k = 0$
    $4k = -1 \implies k = -\frac{1}{4}$
    Intersection point: $x = \frac{1}{2}$ (double root).
  4. Find intersection of $y = 2x + 3$ and $y = \frac{4}{x}$ for $x \neq 0$.
    $2x + 3 = \frac{4}{x} \implies 2x^2 + 3x = 4$
    $2x^2 + 3x - 4 = 0$
    $x = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4}$
    Two intersection points.

Transforming Relationships to Straight Line Form

Many non-linear relationships can be transformed into linear form by applying appropriate transformations to variables.

Common Transformations

Original Equation Transformation Linear Form
$y = a x^n$ Take $\log$: $\ln y = \ln a + n \ln x$ $Y = \ln y$, $X = \ln x$
$y = a b^x$ Take $\log$: $\ln y = \ln a + x \ln b$ $Y = \ln y$, $X = x$
$y = a \cdot e^{kx}$ $\ln y = \ln a + kx$ $Y = \ln y$, $X = x$
$y = \frac{a}{x} + b$ $Y = y$, $X = \frac{1}{x}$ $Y = aX + b$
$y = \frac{ax}{b+x}$ Reciprocal: $\frac{1}{y} = \frac{1}{a} + \frac{b}{a} \cdot \frac{1}{x}$ $Y = \frac{1}{y}$, $X = \frac{1}{x}$

Examples

  1. Transform $y = 3x^4$ to linear form.
    $\ln y = \ln 3 + 4 \ln x$
    Let $Y = \ln y$, $X = \ln x$:
    $Y = 4X + \ln 3$ (gradient $4$, intercept $\ln 3$)
  2. Transform $y = 5 \cdot 2^x$ to linear form.
    $\ln y = \ln 5 + x \ln 2$
    Let $Y = \ln y$, $X = x$:
    $Y = (\ln 2)X + \ln 5$
  3. Express $y = \frac{3x}{2x+5}$ in linear form.
    $\frac{1}{y} = \frac{2x+5}{3x} = \frac{2}{3} + \frac{5}{3} \cdot \frac{1}{x}$
    Let $Y = \frac{1}{y}$, $X = \frac{1}{x}$:
    $Y = \frac{5}{3}X + \frac{2}{3}$
  4. Transform $y = 4e^{2x}$ to straight line form.
    $\ln y = \ln 4 + 2x$
    Let $Y = \ln y$, $X = x$:
    $Y = 2X + \ln 4$ (gradient $2$)

Finding Relationships from Data

Given experimental data points $(x_i, y_i)$, we can determine the underlying relationship by:

  1. Plotting the data.
  2. Guessing the type of relationship (linear, power, exponential, etc.).
  3. Transforming variables to linearize.
  4. Calculating the line of best fit (linear regression).
  5. Converting back to the original relationship.

Procedure for Power Law $y = ax^n$

  1. Compute $X = \ln x$, $Y = \ln y$.
  2. Plot $Y$ against $X$; expect a straight line.
  3. Find gradient $m$ and intercept $c$ of best-fit line.
  4. Then $n = m$ and $a = e^c$.

Procedure for Exponential Law $y = a b^x$

  1. Compute $Y = \ln y$, keep $X = x$.
  2. Plot $Y$ against $X$; expect a straight line.
  3. Find gradient $m$ and intercept $c$.
  4. Then $\ln b = m \implies b = e^m$, and $a = e^c$.

Example 1: Finding a Power Law

Given data: $(1, 2)$, $(2, 8)$, $(3, 18)$, $(4, 32)$

$x$ $y$ $X = \ln x$ $Y = \ln y$
1 2 0 0.693
2 8 0.693 2.079
3 18 1.099 2.890
4 32 1.386 3.466

Gradient $m = \frac{3.466 - 0.693}{1.386 - 0} \approx \frac{2.773}{1.386} \approx 2.00$

Intercept $c \approx 0.693 = \ln 2$

Thus $Y = 2X + \ln 2 \implies \ln y = 2\ln x + \ln 2 = \ln(x^2) + \ln 2 = \ln(2x^2)$

Hence $y = 2x^2$.

Example 2: Finding an Exponential Law

Given data: $(0, 3)$, $(1, 6)$, $(2, 12)$, $(3, 24)$

$x$ $y$ $Y = \ln y$
0 3 1.099
1 6 1.792
2 12 2.485
3 24 3.178

Gradient $m = \frac{3.178 - 1.099}{3 - 0} = \frac{2.079}{3} \approx 0.693 = \ln 2$

Intercept $c \approx 1.099 = \ln 3$

Thus $Y = (\ln 2)x + \ln 3 \implies \ln y = x\ln 2 + \ln 3 = \ln(2^x) + \ln 3 = \ln(3 \cdot 2^x)$

Hence $y = 3 \cdot 2^x$.

Example 3: Finding $y = a/x + b$ Relationship

Given data: $(1, 5)$, $(2, 4)$, $(3, 3.5)$, $(4, 3.25)$

$x$ $y$ $X = 1/x$
1 5 1
2 4 0.5
3 3.5 0.333
4 3.25 0.25

Plot $y$ against $1/x$: expect straight line.

Using points $(1,5)$ and $(0.5,4)$:
Gradient $m = \frac{4-5}{0.5-1} = \frac{-1}{-0.5} = 2$

Equation: $y - 5 = 2(X - 1) \implies y = 2X + 3 = 2(1/x) + 3$

Thus $y = \frac{2}{x} + 3$.

Method of Least Squares (Linear Regression)

When data has errors, the best-fit line $Y = mX + c$ minimises the sum of squared residuals:

$$ m = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2}, \quad c = \frac{\sum Y - m\sum X}{n} $$

Example 4: Linear Regression

Given transformed data:

$X$ $Y$ $XY$ $X^2$
0 0.693 0 0
0.693 2.079 1.441 0.480
1.099 2.890 3.176 1.208
1.386 3.466 4.804 1.921
$\sum X = 3.178$ $\sum Y = 9.128$ $\sum XY = 9.421$ $\sum X^2 = 3.609$

$n = 4$

$$ m = \frac{4(9.421) - (3.178)(9.128)} {4(3.609) - (3.178)^2} = \frac{37.684 - 29.012} {14.436 - 10.100} = \frac{8.672}{4.336} = 2.00 $$

$$ c = \frac{9.128 - 2(3.178)}{4} = \frac{9.128 - 6.356}{4} = \frac{2.772}{4} = 0.693 $$

Thus $Y = 2X + 0.693$, confirming $y = 2x^2$.

Summary Table

Concept Key Points
Gradient $m = \frac{y_2 - y_1}{x_2 - x_1}$
Line equations $y = mx + c$, $y - y_1 = m(x - x_1)$
Parallel lines $m_1 = m_2$
Perpendicular lines $m_1 \cdot m_2 = -1$
Intersection Solve simultaneously
Line-curve intersection Substitute line into curve, solve
Transformation to linear $\ln y = \ln a + n \ln x$, $\ln y = \ln a + kx$, etc.
Power law $y = ax^n \implies \ln y = \ln a + n \ln x$
Exponential law $y = ab^x \implies \ln y = \ln a + x \ln b$

Quick Reference: Transformation Summary

Original Linear Form Variables
$y = a x^n$ $Y = nX + \ln a$ $Y = \ln y$, $X = \ln x$
$y = a b^x$ $Y = (\ln b)X + \ln a$ $Y = \ln y$, $X = x$
$y = a e^{kx}$ $Y = kX + \ln a$ $Y = \ln y$, $X = x$
$y = \frac{a}{x} + b$ $Y = aX + b$ $Y = y$, $X = \frac{1}{x}$
$\frac{1}{y} = a + \frac{b}{x}$ $Y = bX + a$ $Y = \frac{1}{y}$, $X = \frac{1}{x}$

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