Master Guide: Maths Week 1 - Sets, Relations, and Functions

Core Idea: This week introduces the fundamental language of mathematics. We learn how to group objects into sets, define logical relations between them, and establish precise rules called functions for mapping inputs to unique outputs. Mastering this vocabulary is key to understanding everything that follows.

Mental Algorithm: The TAA Framework for Week 1

This is your systematic approach to solving any Week 1 problem.

Triage (10-Second Scan & Categorize): Scan the problem for keywords to instantly classify it.

Keywords / Structure Seen	Immediate Thought / Category
”survey”, “how many like…”, ”% of people”, “at least one/both”	Venn Diagram / Set Theory
`√`, `(√a+√b)`, fractions, `p/q`, “rational or irrational”	Number Systems & Simplification
”domain of f(x)”, “not in the domain”, denominators, `√` in function	Function Domain Analysis
”relation R on set A”, “reflexive”, “symmetric”, “transitive”	Relation Properties
”function f: A → B”, “one-to-one”, “onto”, “bijective”	Function Classification
Set definitions like A = {x…}, R = {(x,y)…}	Set & Relation Cardinality

Abstract (Connect to the Core Tool):

Venn Diagram: Immediately draw circles and think Inclusion-Exclusion Principle: $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣ - ∣ A \cap B ∣$ .
Number Systems: Your goal is simplification. Combine radicals, rationalize denominators. The final form will tell you if it’s rational or not.
Function Domain: List the two main restrictions: 1. Denominator can’t be zero. 2. Inside of a square root can’t be negative.
Relation Properties: Systematically test the three properties:
- Reflexive: Is (a,a) in the relation for every element a?
- Symmetric: If (a,b) is in, is (b,a) also in?
- Transitive: If (a,b) and (b,c) are in, is (a,c) also in?
Function Classification: Apply the two key tests:
- Injective (One-to-One): Can two different inputs give the same output? (Test: Assume $f (x_{1}) = f (x_{2})$ and see if it forces $x_{1} = x_{2}$ ).
- Surjective (Onto): Can you reach every element in the codomain? (Test: Set $f (x) = y$ and try to solve for $x$ in terms of $y$ ).

Act (Execute the Plan): With a clear tool and plan, you can now confidently perform the calculations.

Detailed Patterns, Examples, and Solutions

Pattern 1: Venn Diagram Word Problems

Frequency: High
Difficulty: Medium
Core Skill: Translating percentages and counts into set notation and applying the Inclusion-Exclusion Principle.

Example Problem:

In a Zoo, there are 6 Bengal white tigers (W) and 6 Bengal royal tigers (R). Out of the total 12 tigers, 5 are males (M) and 7 are females (F). We are told that 10 tigers are either Bengal royal tigers or males. Find the number of female Bengal white tigers.

TAA in Action:

Triage: Keywords “either…or”, counts of overlapping groups. This is a Venn Diagram problem.
Abstract: The phrase “either Bengal royal tigers or males” means $∣ R \cup M ∣ = 10$ . The tool is the formula: $∣ R \cup M ∣ = ∣ R ∣ + ∣ M ∣ - ∣ R \cap M ∣$ . My goal is to find the number of tigers that are Female AND White.
Act:
- Step 1: Use the formula to find the overlap. We know: $∣ R ∣ = 6$ , $∣ M ∣ = 5$ , and $∣ R \cup M ∣ = 10$ . $10 = 6 + 5 - ∣ R \cap M ∣$ $10 = 11 - ∣ R \cap M ∣$ $∣ R \cap M ∣ = 1$ . (This means there is 1 tiger that is both Royal and Male).
- Step 2: Find the number of Male White tigers. Total males = (Male Royals) + (Male Whites). $5 = 1 + (Male Whites)$ Number of Male White tigers = 4.
- Step 3: Find the number of Female White tigers. Total White tigers = (Male Whites) + (Female Whites). $6 = 4 + (Female Whites)$ Number of Female White tigers = 2.

Final Answer: 2.

Pattern 2: Function Domain & Cardinality

Frequency: High
Difficulty: Medium
Core Skill: Identifying all conditions that an input variable x must satisfy for the function to be defined.

Example Problem:

Suppose $f : D ⟶ R$ is a function defined by $f (x) = \frac{x ^{2} - 16}{x + 4}$ , where $D \subset Z$ . Let $A$ be the set of integers which are not in the domain of $f$ . Find the cardinality of the set $A$ .

TAA in Action:

Triage: Keywords “domain”, “set of integers not in the domain”. The function has a fraction and a square root. This is a domain restriction problem.
Abstract: I need to list all the rules x must follow, and then find the integers that break those rules.
- Rule 1 (Denominator): $x + 4 \neq = 0$ .
- Rule 2 (Square Root): $x^{2} - 16 \geq 0$ .
Act:
- Step 1: Solve for each restriction.
  - Rule 1: $x \neq = - 4$ .
  - Rule 2: $x^{2} \geq 16$ . This is true for $x \geq 4$ and $x \leq - 4$ .
- Step 2: Identify the excluded integers. The set A contains any integer that violates at least one of these rules. Let’s check the number line:
  - Integers from -3 to 3 violate the square root rule (e.g., $0^{2} - 16$ is negative). This gives us the set ${- 3, - 2, - 1, 0, 1, 2, 3}$ .
  - The integer -4 violates the denominator rule.
- Step 3: Combine the excluded integers. The set $A$ is ${- 4, - 3, - 2, - 1, 0, 1, 2, 3}$ .
- Step 4: Find the cardinality. The number of elements in A is 8.

Final Answer: 8.

Pattern 3: Identifying Relation Properties

Frequency: High
Difficulty: Medium
Core Skill: Systematically testing a relation for reflexivity, symmetry, and transitivity.

Example Problem:

Consider the relation $R_{2} = {(x, y) : x, y \in Z and 5 divides x - y}$ . Is this an equivalence relation?

TAA in Action:

Triage: Keywords “relation”, “divides”, “equivalence relation”. This is a relation properties problem.
Abstract: I must test three properties:
- Reflexive: Is $(x, x) \in R_{2}$ for any integer $x$ ?
- Symmetric: If $(x, y) \in R_{2}$ , is $(y, x) \in R_{2}$ ?
- Transitive: If $(x, y) \in R_{2}$ and $(y, z) \in R_{2}$ , is $(x, z) \in R_{2}$ ?
Act:
- Step 1: Test Reflexivity. Let $a = x, b = x$ . The condition is “5 divides $x - x$ “. Since $x - x = 0$ , and 5 divides 0, this is true for all integers. So, it is reflexive.
- Step 2: Test Symmetry. Assume $(x, y) \in R_{2}$ . This means $x - y$ is a multiple of 5, so $x - y = 5 k$ for some integer $k$ . Now we check the condition for $(y, x)$ : is $y - x$ a multiple of 5? $y - x = - (x - y) = - (5 k) = 5 (- k)$ . Since $- k$ is an integer, $y - x$ is also a multiple of 5. So, it is symmetric.
- Step 3: Test Transitivity. Assume $(x, y) \in R_{2}$ and $(y, z) \in R_{2}$ . This means $x - y = 5 k_{1}$ and $y - z = 5 k_{2}$ . Now we check the condition for $(x, z)$ : is $x - z$ a multiple of 5? Add the two assumptions: $(x - y) + (y - z) = 5 k_{1} + 5 k_{2}$ . $x - z = 5 (k_{1} + k_{2})$ . This is a multiple of 5. So, it is transitive.

Final Answer: Yes, since it is reflexive, symmetric, and transitive, it is an equivalence relation.

Memory Palace: Week 1 Formulas & Concepts

To make these ideas stick, connect them to a story or image.

Inclusion-Exclusion: Imagine two overlapping party guest lists (A and B). To get the total unique guests ( $A \cup B$ ), you add the lists ( $∣ A ∣ + ∣ B ∣$ ), but then you have to subtract the people you counted twice—the ones on both lists ( $∣ A \cap B ∣$ ).
Domain Restrictions: Think of a function as a machine.
- Denominator: The machine has a base it stands on. If the base becomes 0, it falls over (undefined).
- Square Root: The machine has a gear that only works with non-negative numbers. Feed it a negative, and it breaks (not a real number).
Function Types (The Party Analogy):
- Injective (One-to-One): Every person brings a unique dish. No two people bring the same thing.
- Surjective (Onto): The host made a list of desired dishes (the codomain). By the end, every single dish on that list has been brought by someone (the range equals the codomain).
- Bijective: A perfect potluck! Every person brings a unique dish, AND every desired dish is present.

Master Guide: Maths Week 2 - Coordinate Geometry and Straight Lines

Core Idea: This week is about giving pictures to equations. We are translating abstract algebra into a visual world of points, lines, and distances on a plane. Every formula we learn is a tool to measure or describe something in this visual world.

Mental Algorithm: The TAA Framework for Week 2

This framework is your systematic process for decoding any coordinate geometry problem.

Triage (10-Second Scan & Categorize): Quickly scan the problem for keywords to classify it.

Keywords / Structure Seen	Immediate Thought / Category
”equation of the line”, “passes through”, “slope”	Line Equation Construction
”parallel”, “perpendicular”	Slope Relationships
”intersection”, “collision”, “meeting point”	System of Linear Equations
”distance between points”, “length of segment”	Distance Formula
”divides the line in ratio…”, “midpoint”	Section/Midpoint Formula
”shortest distance from a point to a line”	Point-to-Line Distance Formula
”best-fit line”, “data points”, “Sum of Squared Errors”, “SSE”	Linear Fit / SSE Calculation
”reflection”, “ray of light”, “mirror”	Reflection Geometry

Abstract (Connect to the Core Tool):

Line Equation Construction: Ask: What information do I have?
- Point and slope? $\to$ Use Point-Slope Form: $y - y_{1} = m (x - x_{1})$ .
- Two points? $\to$ First, find the slope, then use Point-Slope Form.
- Slope and y-intercept? $\to$ Use Slope-Intercept Form: $y = m x + c$ .
System of Equations: Your goal is to eliminate one variable. The easiest way is to get both equations into the form y = ... and then set the right-hand sides equal to each other.
Distance Formula: Straightforward application of $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .
Section Formula: Identify $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $m$ , and $n$ , and plug them into the formula carefully.
Point-to-Line Distance: The line must be in the form $A x + B y + C = 0$ before you can use the formula $d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}$ .
SSE Calculation: This is a procedural task. Create a table with columns: x | y_actual | y_predicted | Error (Actual - Predicted) | Error². Sum the last column.
Reflection Geometry: The core trick is collinearity. The reflected source point, the point on the mirror, and the destination point all lie on a single straight line.

Act (Execute the Plan): With the right tool chosen, execute the algebraic steps. Double-check your signs and arithmetic.

Detailed Patterns, Examples, and Solutions

Pattern 1: Line Intersection & Collision

Frequency: High
Difficulty: Easy
Core Skill: Solving a system of two linear equations.

Example Problem:

A bird is flying along the straight line $2 y - 6 x = 6$ . An aeroplane’s path has a slope of 2 and passes through the point $(4, 8)$ . Find the collision point $(α, β)$ .

TAA in Action:

Triage: Keywords “flying along”, “path”, “collision point”. This is a line intersection problem.
Abstract: I need to find the equation for both paths and solve the system.
- Path 1: Given as $2 y - 6 x = 6$ . I will convert this to $y = m x + c$ .
- Path 2: Given slope and a point. I will use the Point-Slope Form.
Act:
- Step 1: Find the equation for the bird’s path (Line 1). $2 y = 6 x + 6 ⟹ y = 3 x + 3$ .
- Step 2: Find the equation for the aeroplane’s path (Line 2). Using $y - y_{1} = m (x - x_{1})$ with $m = 2$ and $(4, 8)$ : $y - 8 = 2 (x - 4) ⟹ y - 8 = 2 x - 8 ⟹ y = 2 x$ .
- Step 3: Solve the system. Set the two expressions for y equal: $3 x + 3 = 2 x ⟹ x = - 3$ . Substitute $x = - 3$ into the simpler equation ( $y = 2 x$ ): $y = 2 (- 3) = - 6$ .

Final Answer: The collision point is $(- 3, - 6)$ .

Pattern 2: Geometric Properties & Formulas

Frequency: High
Difficulty: Medium
Core Skill: Correctly applying coordinate geometry formulas (section, midpoint, area, etc.).

Example Problem:

Consider $△ A BC$ with $A (- 3, 3), B (1, 7), C (2, - 2)$ . Point M divides AB in the ratio 1:3. Point O is the midpoint of BC. Find the coordinates of M and O.

TAA in Action:

Triage: Keywords “divides… in ratio”, “midpoint”. This is a direct application of the Section and Midpoint formulas.
Abstract:
- For M: Use Section Formula with $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ , $m = 1, n = 3$ .
- For O: Use Midpoint Formula with $B (x_{1}, y_{1}), C (x_{2}, y_{2})$ .
Act:
- Step 1: Find coordinates of M. $x_{M} = \frac{1 ( 1 ) + 3 ( - 3 )}{1 + 3} = \frac{1 - 9}{4} = - 2$ . $y_{M} = \frac{1 ( 7 ) + 3 ( 3 )}{1 + 3} = \frac{7 + 9}{4} = 4$ . So, $M = (- 2, 4)$ .
- Step 2: Find coordinates of O. $x_{O} = \frac{1 + 2}{2} = 1.5$ . $y_{O} = \frac{7 + ( - 2 )}{2} = 2.5$ . So, $O = (1.5, 2.5)$ .

Final Answer: $M (- 2, 4)$ and $O (1.5, 2.5)$ .

Pattern 3: Sum of Squared Errors (SSE)

Frequency: High
Difficulty: Medium (can be tedious)
Core Skill: A procedural calculation: for each point, find the predicted value, the error, and the squared error, then sum.

Example Problem:

A fitness trainer’s model for weight loss is $W = - 8 t + 98$ . Is this a “good fit” (SSE < 5) for the data points $(5, 57)$ and $(6, 49)$ ? (Assuming other points have zero error).

TAA in Action:

Triage: Keywords “equation”, “data”, “Sum of Squared Errors (SSE)“. This is an SSE calculation.
Abstract: I need to build a table to calculate $\sum (W_{a c t u a l} - W_{p re d i c t e d})^{2}$ .
Act:
- Step 1: Calculate error for the point (t=5, W=57).
  - Predicted Weight: $W_{p re d} = - 8 (5) + 98 = - 40 + 98 = 58$ .
  - Error: $W_{a c t u a l} - W_{p re d} = 57 - 58 = - 1$ .
  - Squared Error: $(- 1)^{2} = 1$ .
- Step 2: Calculate error for the point (t=6, W=49).
  - Predicted Weight: $W_{p re d} = - 8 (6) + 98 = - 48 + 98 = 50$ .
  - Error: $W_{a c t u a l} - W_{p re d} = 49 - 50 = - 1$ .
  - Squared Error: $(- 1)^{2} = 1$ .
- Step 3: Calculate total SSE. Assuming all other data points fit perfectly (error=0), the total SSE is the sum of the squared errors we found. $SSE = 1 + 1 = 2$ .
- Step 4: Compare to the threshold. Since $2 < 5$ , the model is considered a good fit.

Final Answer: Yes, it is a good fit.

Pattern 4: Reflection Geometry

Frequency: Medium
Difficulty: Hard
Core Skill: Understanding that the reflected source, the reflection point, and the destination are collinear.

Example Problem:

A light ray from $A (1, 2)$ reflects off the X-axis at point B and passes through $(5, 3)$ . Find the equation of the line segment AB.

TAA in Action:

Triage: Keywords “ray”, “reflects”, “X-axis”. This is a reflection problem.
Abstract: The key insight is to find the reflection of point A across the X-axis. Let’s call it A’. The line connecting A’ and the destination point (5,3) will contain the reflection point B.
Act:
- Step 1: Find the reflected point A’. Reflection across the x-axis negates the y-coordinate. So, $A^{'} (1, - 2)$ .
- Step 2: Find the equation of the line passing through A’(1,-2) and (5,3).
  - Slope $m = \frac{3 - ( - 2 )}{5 - 1} = \frac{5}{4}$ .
  - This line’s equation is not what we need yet. We first need to find point B.
- Step 3: Find the reflection point B. Point B lies on the X-axis (so its y-coordinate is 0) and on the line we just conceptualized. Let’s find where the line through A’ and (5,3) hits the x-axis. Using point-slope form with A’: $y - (- 2) = \frac{5}{4} (x - 1)$ . Set $y = 0$ : $2 = \frac{5}{4} (x - 1) ⟹ 8 = 5 x - 5 ⟹ 5 x = 13 ⟹ x = 13/5$ . So, point $B = (13/5, 0)$ .
- Step 4: Find the equation of the original line segment AB. We now have two points for the incident ray: $A (1, 2)$ and $B (13/5, 0)$ .
  - Slope of AB: $m_{A B} = \frac{0 - 2}{13/5 - 1} = \frac{- 2}{8/5} = - \frac{10}{8} = - \frac{5}{4}$ .
  - Using point-slope form with $A (1, 2)$ : $y - 2 = - \frac{5}{4} (x - 1)$ .
  - $4 (y - 2) = - 5 (x - 1) ⟹ 4 y - 8 = - 5 x + 5 ⟹ 5 x + 4 y = 13$ .

Final Answer: $5 x + 4 y = 13$ .

Memory Palace: Week 2 Formulas & Concepts

Distance Formula: Imagine a right-angled triangle on the coordinate plane. The distance between two points is the hypotenuse. The legs are the difference in x-coordinates ( $Δ x$ ) and the difference in y-coordinates ( $Δ y$ ). By Pythagoras, $d^{2} = (Δ x)^{2} + (Δ y)^{2}$ .
Slope: m stands for mountain. It’s the rise over the run. rise is the vertical change ( $Δ y$ ), run is the horizontal change ( $Δ x$ ).
Perpendicular Slopes: Think of a plus sign +. One line is horizontal (slope 0), one is vertical (slope undefined). Now rotate it. One slope will be positive, one will be negative. They are “negative reciprocals” of each other.
Point-to-Line Distance: The formula looks complex: $d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}$ .
- Top part: It’s like you’re plugging the point $(x_{0}, y_{0})$ into the line’s equation. The absolute value $∣...∣$ ensures the distance is positive.
- Bottom part: $A^{2} + B^{2}$ is a “normalization factor” related to the Pythagorean theorem for the line’s coefficients. It scales the result correctly.

Of course. Here is the complete, untruncated guide for Week 3: Quadratic Functions, created with the same detailed structure.

Master Guide: Maths Week 3 - Quadratic Functions

Core Idea: This week is a deep dive into second-degree polynomials. Their graphs, called parabolas, are everywhere: from the path of a thrown ball to the shape of satellite dishes. We will learn how to analyze, manipulate, and solve problems involving these crucial functions.

Mental Algorithm: The TAA Framework for Week 3

This is your systematic approach for mastering quadratic function problems.

Triage (10-Second Scan & Categorize): Identify the problem type from its keywords.

Keywords / Structure Seen	Immediate Thought / Category
”maximum value”, “minimum value”, “highest/lowest point”, “vertex”	Vertex Problem
”axis of symmetry”	Vertex Problem (This gives you the x-coordinate of the vertex)
“slope of the parabola at point…”	Derivative/Slope Problem
”roots”, “x-intercepts”, “product of numbers is…”, “area is…”	Quadratic Equation Problem
”intersection of a parabola and a line/parabola”	System of Equations Problem
An equation of the form $y = a x^{2} + b x + c$ is given	General Quadratic Analysis

Abstract (Connect to the Core Tool):

Vertex Problem: The vertex is the key.
- To find its location (the x value where the max/min occurs), use the formula: $x = - \frac{b}{2 a}$ .
- To find the value (the max/min itself), plug that x value back into the function: $f (- \frac{b}{2 a})$ .
Derivative/Slope Problem: The slope is not constant.
- The tool is the derivative formula: Slope at x = $2 a x + b$ .
- This allows you to create linear equations if slopes are given at certain points.
Quadratic Equation Problem: Your goal is to get the equation into the form $a x^{2} + b x + c = 0$ .
- Tool 1: Factoring. Look for two numbers that multiply to c and add to b.
- Tool 2: The Quadratic Formula. The ultimate tool that always works: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ .
System of Equations Problem: To find where two curves meet, their y values must be equal.
- Tool: Set the two equations equal to each other ( $f (x) = g (x)$ ). This will create a new equation to solve.

Act (Execute the Plan): Carefully apply the chosen formula or method. Pay close attention to signs and order of operations.

Detailed Patterns, Examples, and Solutions

Pattern 1: Finding Maximum/Minimum Values (Vertex Problems)

Frequency: High
Difficulty: Easy
Core Skill: Finding and interpreting the vertex of a parabola.

Example Problem:

A water fountain is designed to shoot a stream of water in the shape of a parabolic arc. The equation is given by $h (t) = - 0.5 t^{2} + 4 t + 1$ , where $h (t)$ is the height and $t$ is the time. Determine the time it takes for the water to reach its maximum height.

TAA in Action:

Triage: Keywords “maximum height”. This is a vertex problem. The question asks for the “time”, which is the input variable ( $t$ ), so I need the t-coordinate of the vertex.
Abstract: The formula for the x-coordinate (or t-coordinate) of the vertex is $t = - \frac{b}{2 a}$ .
Act:
- Step 1: Identify coefficients. In $h (t) = - 0.5 t^{2} + 4 t + 1$ , we have $a = - 0.5$ and $b = 4$ . (The negative ‘a’ confirms there is a maximum).
- Step 2: Apply the formula. $t = - \frac{4}{2 ( - 0.5 )} = - \frac{4}{- 1} = 4$ .

Final Answer: It takes 4 seconds to reach the maximum height.

Pattern 2: Finding Coefficients from Slopes

Frequency: High
Difficulty: Medium
Core Skill: Using the derivative formula Slope = 2ax + b to form a system of linear equations.

Example Problem:

If the slope of the parabola $y = a x^{2} + b x + c$ at the point where $x = 3$ is 16, and the slope at the point where $x = 2$ is 12, find the value of $a$ .

TAA in Action:

Triage: Keywords “slope of parabola at points…“. This is a derivative problem.
Abstract: I will use the formula Slope = 2ax + b twice to create two linear equations with variables a and b. Then I will solve the system.
Act:
- Step 1: Set up the equations.
  - At $x = 3$ , slope is 16: $2 a (3) + b = 16 ⟹ 6 a + b = 16$ (Eq. 1).
  - At $x = 2$ , slope is 12: $2 a (2) + b = 12 ⟹ 4 a + b = 12$ (Eq. 2).
- Step 2: Solve the system. Subtracting Eq. 2 from Eq. 1 is the fastest way to eliminate b. $(6 a + b) - (4 a + b) = 16 - 12$ $2 a = 4$ $a = 2$ .

Final Answer: The value of $a$ is 2.

Pattern 3: Real-World Quadratic Modeling

Frequency: High
Difficulty: Easy-Medium
Core Skill: Translating a word problem into a quadratic equation.

Example Problem:

The product of two consecutive odd natural numbers is 143. Find the largest number among them.

TAA in Action:

Triage: Keywords “product”, “consecutive odd natural numbers”. This hints at a quadratic relationship.
Abstract: I need to represent the two numbers algebraically, form an equation, and solve it.
- Let the first odd number be $x$ .
- The next consecutive odd number is $x + 2$ .
- The equation is $x (x + 2) = 143$ .
Act:
- Step 1: Form the quadratic equation. $x^{2} + 2 x = 143 ⟹ x^{2} + 2 x - 143 = 0$ .
- Step 2: Solve the equation. We can try to factor or use the quadratic formula. Let’s try factoring. We need two numbers that multiply to -143 and have a difference of 2. These are 13 and -11. $(x + 13) (x - 11) = 0$ . The solutions are $x = - 13$ and $x = 11$ .
- Step 3: Choose the correct solution. The problem specifies “natural numbers”, so we must discard $x = - 13$ . The first number is 11.
- Step 4: Find the largest number. The second number is $x + 2 = 11 + 2 = 13$ .

Final Answer: The largest number is 13.

Pattern 4: Intersection of Curves

Frequency: Medium
Difficulty: Medium
Core Skill: Setting two function equations equal to each other to find common points.

Example Problem:

Find the points where the parabola $y = x^{2} + 3 x + 2$ and the parabola $y = - x^{2} - 5 x - 4$ intersect.

TAA in Action:

Triage: Keyword “intersect”. This is a system of equations problem.
Abstract: I will set the two equations equal to each other. This will give me a new quadratic equation to solve for the x-coordinates of the intersection points.
Act:
- Step 1: Set the equations equal. $x^{2} + 3 x + 2 = - x^{2} - 5 x - 4$ .
- Step 2: Rearrange into standard quadratic form ( $a x^{2} + b x + c = 0$ ). Move all terms to one side: $2 x^{2} + 8 x + 6 = 0$ .
- Step 3: Simplify and solve. Divide the entire equation by 2: $x^{2} + 4 x + 3 = 0$ . This can be factored easily: $(x + 1) (x + 3) = 0$ . The solutions are $x = - 1$ and $x = - 3$ .
- Step 4: Find the corresponding y-coordinates. Substitute these x-values back into either of the original equations. Let’s use the first one, $y = x^{2} + 3 x + 2$ .
  - For $x = - 1$ : $y = (- 1)^{2} + 3 (- 1) + 2 = 1 - 3 + 2 = 0$ . Point is $(- 1, 0)$ .
  - For $x = - 3$ : $y = (- 3)^{2} + 3 (- 3) + 2 = 9 - 9 + 2 = 2$ . Point is $(- 3, 2)$ .

Final Answer: The intersection points are $(- 1, 0)$ and $(- 3, 2)$ .

Memory Palace: Week 3 Formulas & Concepts

The Vertex King: The vertex is the “king” of the parabola. Its throne is at $x = - b / (2 a)$ . The height of its throne (its power) is the maximum or minimum value, $f (- b /2 a)$ .
The Slope Messenger: To find the slope at any location x, you send a messenger. The message it brings back is 2ax + b. At the king’s throne (the vertex), the messenger reports a slope of 0, because the ground is perfectly flat there.
The Quadratic Formula Story: Imagine a dramatic hero, -b. He’s unsure whether to add or subtract ( $\pm$ ). He enters a dangerous cave under a square root sign. Inside, he fights a beast, b², but is attacked by 4 angry cats (4ac). He barely escapes, all over a 2a-lane bridge.
Vertex Form vs. Standard Form:
- Standard Form ( $a x^{2} + b x + c$ ): Good for finding the y-intercept (it’s just c) and using the quadratic formula.
- Vertex Form ( $a (x - h)^{2} + k$ ): A “cheat code” for graphing. It tells you the vertex is at $(h, k)$ immediately.

Of course. Here is the complete, untruncated guide for Week 4: Polynomial Functions, following the established detailed structure.

Master Guide: Maths Week 4 - Polynomial Functions

Core Idea: This week, we generalize from lines and parabolas to polynomials of any degree. These functions can have multiple “wiggles” (turning points), allowing them to model much more complex data. Our focus is on understanding the graphical behavior of these functions by analyzing their equations.

Mental Algorithm: The TAA Framework for Week 4

This is your systematic approach for analyzing any polynomial function.

Triage (10-Second Scan & Categorize): Identify the problem type from its keywords and structure.

Keywords / Structure Seen	Immediate Thought / Category
”end behavior”, “as x → ∞”, “as x → -∞“	Leading Term Analysis
”turning points”, “local maxima/minima”	Degree-1 Rule & Graph Behavior
A factored form: `(x-a)²(x-b)³`	Roots and Multiplicity Analysis
”x-intercepts”, “find the roots”, “intersects the X-axis”	Finding Roots
”intersection of polynomials”, “connects… districts”	System of Equations
”construct a polynomial”, “equation of the polynomial”	Polynomial Construction

Abstract (Connect to the Core Tool):

Leading Term Analysis:
- Tool: The end behavior is dictated only by the leading term, $a_{n} x^{n}$ .
- Procedure: 1. Find the degree n. 2. Find the sign of the leading coefficient a_n. 3. Apply the end behavior rules.
Degree-1 Rule:
- Tool: A polynomial of degree n has at most n-1 turning points.
Roots and Multiplicity Analysis:
- Tool: The exponent on a factor (x-r)^m is its multiplicity m.
- Procedure:
  - If m is odd (1, 3, 5…), the graph crosses the x-axis at x=r.
  - If m is even (2, 4, 6…), the graph touches the x-axis at x=r (and “bounces” off). This is also a turning point.
Finding Roots:
- Tool: Set the polynomial equation $P (x) = 0$ . If it’s in factored form, the roots are the values of x that make each factor zero.
System of Equations:
- Tool: Set the two polynomial equations equal to each other ( $P (x) = Q (x)$ ) to find their intersection points.

Act (Execute the Plan): Combine these tools. A great strategy is to: 1. Find the roots. 2. Determine the end behavior. 3. Sketch a rough graph connecting the end behaviors through the roots, respecting the “cross” vs. “touch” rule.

Detailed Patterns, Examples, and Solutions

Pattern 1: Graph Behavior from Factored Form

Frequency: High
Difficulty: Medium
Core Skill: Using root multiplicity to predict the graph’s shape and direction in intervals.

Example Problem:

Consider the polynomial $p (x) = - (x^{2} - 16) (x - 3)^{2} (2 - x)^{2} (x + 9)$ . Determine if the graph first increases then decreases in the interval $(2, 3)$ .

TAA in Action:

Triage: Keywords “factored form”, “increases then decreases in the interval”. This is a graph behavior problem.
Abstract: I need to analyze the roots at the endpoints of the interval (2 and 3) and check the sign of the function between them.
Act:
- Step 1: Simplify and Analyze Roots.
  - Factor $x^{2} - 16$ into $(x - 4) (x + 4)$ .
  - Rewrite $(2 - x)^{2}$ as $(x - 2)^{2}$ . The negative sign is squared, so it vanishes.
  - The roots are: 4, -4, 3, 2, -9.
  - At $x = 2$ , the factor is $(x - 2)^{2}$ , so multiplicity is 2 (Even) → Touches.
  - At $x = 3$ , the factor is $(x - 3)^{2}$ , so multiplicity is 2 (Even) → Touches.
- Step 2: Check the Sign in the Interval (2, 3). Pick a test point, e.g., $x = 2.5$ . $p (2.5) = - ((2.5)^{2} - 16) ((2.5) - 3)^{2} (2 - (2.5))^{2} ((2.5) + 9)$ $p (2.5) = - (6.25 - 16) (- 0.5)^{2} (- 0.5)^{2} (11.5)$ $p (2.5) = - (negative) (positive) (positive) (positive)$ $p (2.5) = positive$ .
- Step 3: Sketch the Behavior. The graph comes from the x-axis at $x = 2$ , goes into positive territory, and must come back down to touch the x-axis again at $x = 3$ . This describes an upward arc. Therefore, the function first increases, then decreases in the interval (2, 3).

Final Answer: The statement is correct.

Pattern 2: End Behavior & Turning Points

Frequency: High
Difficulty: Easy
Core Skill: Identifying the leading term to determine end behavior and using the degree to find the maximum number of turning points.

Example Problem:

Consider the polynomial $p (x) = - x^{5} + 5 x^{4} - 7 x - 2$ . Describe its end behavior and state the maximum number of turning points.

TAA in Action:

Triage: Keywords “end behavior”, “turning points”. This is a leading term analysis.
Abstract: I will find the leading term $a_{n} x^{n}$ . Then I’ll use the rules for end behavior and the degree-1 rule for turning points.
Act:
- Step 1: Identify the Leading Term. The term with the highest power is $- x^{5}$ .
- Step 2: Analyze the Leading Term.
  - Degree n = 5 (Odd).
  - Leading Coefficient a_n = -1 (Negative).
- Step 3: Apply the End Behavior Rule. For an Odd/Negative polynomial, the graph goes Up on the left and Down on the right ( $↖ \dots ↘$ ). As $x \to \infty$ , $p (x) \to - \infty$ .
- Step 4: Apply the Turning Point Rule. The degree is $n = 5$ . The maximum number of turning points is $n - 1 = 4$ .

Final Answer: The end behavior is up-left, down-right. It has at most 4 turning points.

Pattern 3: Finding and Using Roots

Frequency: Medium
Difficulty: Easy-Medium
Core Skill: Setting a polynomial to zero and solving for x.

Example Problem:

An ant finds food at the x-intercepts of the function $f (x) = (x^{2} - 23) ((x - 10)^{3} - 1)$ . Find the sum of the x-coordinates of all food locations.

TAA in Action:

Triage: Keywords “x-intercepts”, “sum of the x-coordinates”. This is a root-finding problem.
Abstract: I need to set $f (x) = 0$ and solve for all possible values of $x$ . The product of factors is zero if any of the factors is zero.
Act:
- Step 1: Set each factor to zero. The equation is $f (x) = 0$ , so either:
  - $x^{2} - 23 = 0$
  - OR $(x - 10)^{3} - 1 = 0$
- Step 2: Solve the first equation. $x^{2} = 23 ⟹ x = 23$ and $x = - 23$ .
- Step 3: Solve the second equation. $(x - 10)^{3} = 1$ . Take the cube root of both sides: $x - 10 = 1 ⟹ x = 11$ .
- Step 4: Find the sum of the roots. Sum = $23 + (- 23) + 11 = 0 + 11 = 11$ .

Final Answer: The sum is 11.

Memory Palace: Week 4 Concepts

End Behavior (The Dance): Think of the graph as a person dancing.
- Even Degree (n=2, 4…): “Even Stevens.” They are balanced. Both arms finish in the same direction (both up or both down).
  - If a_n is positive: They finish celebrating, arms up. $(↖ \dots ↗)$
  - If a_n is negative: They finish tired, arms down. $(↙ \dots ↘)$
- Odd Degree (n=1, 3…): “Odd Bods.” They are unbalanced. The arms finish in opposite directions.
  - If a_n is positive (like y=x³): Starts down, finishes up. $(↙ \dots ↗)$
  - If a_n is negative (like y=-x³): Starts up, finishes down. $(↖ \dots ↘)$
Root Multiplicity (The Bouncer): The x-axis is a velvet rope at a club.
- Odd Multiplicity (m=1, 3…): The root is a guest who just crosses right through the rope.
- Even Multiplicity (m=2, 4…): The root is a troublemaker. It comes up to the rope, touches it, and gets “bounced” back the way it came.

🧠 IIT Madras Notes

Explorer

Maths Abstractions

Master Guide: Maths Week 1 - Sets, Relations, and Functions

Mental Algorithm: The TAA Framework for Week 1

Detailed Patterns, Examples, and Solutions

Pattern 1: Venn Diagram Word Problems

Pattern 2: Function Domain & Cardinality

Pattern 3: Identifying Relation Properties

Memory Palace: Week 1 Formulas & Concepts

Master Guide: Maths Week 2 - Coordinate Geometry and Straight Lines

Mental Algorithm: The TAA Framework for Week 2

Detailed Patterns, Examples, and Solutions

Pattern 1: Line Intersection & Collision

Pattern 2: Geometric Properties & Formulas

Pattern 3: Sum of Squared Errors (SSE)

Pattern 4: Reflection Geometry

Memory Palace: Week 2 Formulas & Concepts

Master Guide: Maths Week 3 - Quadratic Functions

Mental Algorithm: The TAA Framework for Week 3

Detailed Patterns, Examples, and Solutions

Pattern 1: Finding Maximum/Minimum Values (Vertex Problems)

Pattern 2: Finding Coefficients from Slopes

Pattern 3: Real-World Quadratic Modeling

Pattern 4: Intersection of Curves

Memory Palace: Week 3 Formulas & Concepts

Master Guide: Maths Week 4 - Polynomial Functions

Mental Algorithm: The TAA Framework for Week 4

Detailed Patterns, Examples, and Solutions

Pattern 1: Graph Behavior from Factored Form

Pattern 2: End Behavior & Turning Points

Pattern 3: Finding and Using Roots

Memory Palace: Week 4 Concepts

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