🏈 Mathematics 1 - Week 3: Quadratic Functions & Algebra

“Quadratic functions model the world around us - from projectile motion to optimization problems.”

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📋 Week 3 Overview

Week 3 introduces quadratic functions and their applications:

• Quadratic Equations - Solving and word problems
• Parabola Properties - Vertex, symmetry, maximum/minimum
• Calculus Applications - Derivatives and rates of change
• Intersection Problems - Curves meeting at points
• Real-World Models - Projectile motion, optimization
• Polynomial Operations - Building and analyzing quadratics

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🔢 Pattern 1: Quadratic Equations & Word Problems

📖 Concept Explanation

Standard Form: ax² + bx + c = 0 Vertex Form: a(x - h)² + k = 0 Solutions: x = [-b ± √(b² - 4ac)] / (2a)

Word Problems: Translate English to mathematical equations

• Consecutive numbers: n, n+1, n+2
• Area problems: length × width = area
• Age problems: current age relationships

🧠 Mental Algorithm

1. Identify quadratic pattern in word problem
2. Define variable (usually n or x)
3. Set up equation with given conditions
4. Solve using formula or factoring
5. Verify solution makes sense

📝 Pattern-Based Examples

Example 1: Product of Consecutive Numbers

Question Pattern: Product of consecutive odd/even numbers = k

Problem: Product of two consecutive odd natural numbers is 143. Find largest number.

Solution:

$$Letnumbersbe:2n-1,2n+1(consecutiveodd)(2n-1)(2n+1)=1434n^2-1=1434n^2=144n^2=36n=6(sincenaturalnumbers)Numbers:2(6)-1=11,2(6)+1=13Product:11\times 13=143✓Largest:13$$

Answer: 13

Example 2: Arrangement Problems

Question Pattern: Students in rows with specific relationships

Problem: 140 students arranged in rows where number of students per row is one less than thrice the number of rows.

Solution:

$$Letr=numberofrowsStudentsperrow=3r-1Totalstudents:r(3r-1)=1403r^2-r-140=0Usingquadraticformula:r=[1\pm \surd (1+1680)]/6=[1\pm \surd 1681]/6=[1\pm 41]/6Positiveroot:(42)/6=7Studentsperrow:3(7)-1=20$$

Answer: 20

⚠️ Common Pitfalls

• Forgetting consecutive numbers are n, n+1, n+2
• Not checking if solution satisfies original conditions
• Arithmetic errors in quadratic formula calculations

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🏟️ Pattern 2: Parabola Properties & Vertex

📖 Concept Explanation

Vertex: Minimum/maximum point (h,k) Axis of Symmetry: x = h Maximum/Minimum Value: k For ax² + bx + c:

• Vertex x-coordinate: h = -b/(2a)
• Vertex y-coordinate: k = c - b²/(4a)

🧠 Mental Algorithm

1. Find vertex x-coordinate (-b/2a)
2. Substitute to find y-coordinate
3. Determine if maximum or minimum (based on a)
4. Use symmetry for other properties

📝 Pattern-Based Examples

Example 3: Maximum Value Problems

Question Pattern: Find maximum value of quadratic function

Problem: Maximum value of f(x) is -3, axis of symmetry x=2, f(0)=-9. Find coefficient of x².

Solution:

$$Given:vertex(2,-3),f(0)=-9Vertexform:a(x-2{)}^2-3Atx=0:a(0-2{)}^2-3=-94a-3=-94a=-6a=-1.5Verification:a<0confirmsmaximumatvertex$$

Answer: -1.5

Example 4: Derivative Applications

Question Pattern: Slope of parabola at given point

Problem: Slope of y = ax² + bx + c at (3,2) is 32, at (2,3) is 2. Find a.

Solution:

$$Derivative:dy/dx=2ax+bAt(3,2):2a(3)+b=32\Rightarrow 6a+b=\mathrm{32...}(1)At(2,3):2a(2)+b=2\Rightarrow 4a+b=\mathrm{2...}(2)Subtract(2)from(1):2a=30\Rightarrow a=15From(2):4(15)+b=2\Rightarrow 60+b=2\Rightarrow b=-58Verification:At(3,2):2(15)(3)+(-58)=90-58=32✓$$

Answer: 15

⚠️ Common Pitfalls

• Forgetting sign of coefficient a determines max/min
• Incorrect derivative calculation
• Not using given points to verify solution

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🎯 Pattern 3: Intersection of Curves

📖 Concept Explanation

Solving Systems: Set equations equal, solve for intersection points Multiple Solutions: Quadratic with linear may have 0, 1, or 2 solutions Tangent: Exactly one solution (repeated root)

🧠 Mental Algorithm

1. Set equations equal to each other
2. Rearrange into standard quadratic form
3. Solve using factoring or quadratic formula
4. Verify each solution in original equations

📝 Pattern-Based Examples

Example 5: Parabola-Line Intersection

Question Pattern: Find intersection points of quadratic and line

Problem: Curve y = 4x² + x + 6 and line through (1,6) and (4,5)

Solution:

$$Lineequation:Points(1,6),(4,5)Slopem=(5-6)/(4-1)=(-1)/3=-1/3Equation:y-6=(-1/3)(x-1)y=(-1/3)x+1/3+6=(-1/3)x+19/3Setequal:4x^2+x+6=(-1/3)x+19/3Multiplyby3:12x^2+3x+18=-x+1912x^2+4x-1=0Solutions:x=[-4\pm \surd (16+48)]/24=[-4\pm \surd 64]/24=[-4\pm 8]/24x_1=4/24=1/6x_2=-12/24=-1/2y_1=4(1/6{)}^2+(1/6)+6=4/36+1/6+6=1/9+1/6+6y_2=4(-1/2{)}^2+(-1/2)+6=4(1/4)-0.5+6=1-0.5+6=6.5Points:(1/6,113/18),(-1/2,13/2)$$

Answer: Both points

Example 6: Multiple Parabola Intersection

Question Pattern: Two parabolas intersecting, analyze tangent lines

Problem: y = x² + 3x + 2 and y = -x² - 5x - 4 intersect at A and B Line through A with slope of second parabola at A, etc.

Solution Strategy:

1. Find intersection points A and B
2. Find derivatives (slopes) at each point
3. Analyze parallel/tangent conditions

⚠️ Common Pitfalls

• Not finding both intersection points
• Incorrect algebraic manipulation
• Forgetting to verify solutions

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🚀 Pattern 4: Projectile Motion Applications

📖 Concept Explanation

Projectile Motion: h(t) = - (1/2)gt² + v₀t + h₀ Time of Flight: Total time in air Maximum Height: Vertex of parabola Range: Horizontal distance covered

🧠 Mental Algorithm

1. Identify initial conditions (height, velocity)
2. Use quadratic formula for time calculations
3. Find maximum height using vertex formula
4. Solve systems for intersection problems

📝 Pattern-Based Examples

Example 7: Water Fountain

Question Pattern: Time to reach maximum height

Problem: h(t) = -0.5t² + 4t + 1, find time to maximum height

Solution:

$$Standardform:-0.5t^2+4t+1Vertexx-coordinate:t=-b/(2a)=-4/(2\times -0.5)=-4/-1=4Maximumatt=4seconds$$

Answer: 4

Example 8: Ballistic Missile

Question Pattern: Multi-part projectile motion problems

Problem: h(t) = -8t² + 32t + 40 Find: (1) Maximum height, (2) Time to hit ground, (3) Intersection with defense system

Solution:

$$(1)Maximumheight:t=-32/(2\times -8)=-32/-16=2secondsh(2)=-8(4)+32(2)+40=-32+64+40=72meters(2)Hitground:h(t)=0-8t^2+32t+40=0Divideby-8:t^2-4t-5=0t=[4\pm \surd (16+20)]/2=[4\pm \surd 36]/2=[4\pm 6]/2t=5seconds(positiveroot)(3)Defensesystem:h(t)=10tSetequal:10t=-8t^2+32t+408t^2-22t-40=0t=[22\pm \surd (484+1280)]/16=[22\pm \surd 1764]/16=[22\pm 42]/16t_1=64/16=4seconds{t}_2=-20/16=-1.25(discard)Height:10(4)=40meters$$

Answer: (1) 72, (2) 5, (3) 40 meters

⚠️ Common Pitfalls

• Forgetting negative sign in acceleration
• Using wrong time in calculations
• Not discarding negative time values

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📐 Pattern 5: Polynomial Construction & Analysis

📖 Concept Explanation

Building Polynomials: Start with roots, multiply factors Given Conditions: Use points, derivatives, behavior to find coefficients Vertex Problems: Use vertex form then convert

🧠 Mental Algorithm

1. Start with known information (roots, vertex, points)
2. Build polynomial step by step
3. Use conditions to solve for unknowns
4. Verify all conditions satisfied

📝 Pattern-Based Examples

Example 9: Polynomial Through Vertex

Question Pattern: Polynomial passes through vertex of quadratic

Problem: p(x) = a(x-4)(x-6)(x-8)(x-10) passes through vertex of q(x) = -(x-7)² - 9

Solution:

$$Vertexofq(x):x=7,y=-9p(7)=-9a(7-4)(7-6)(7-8)(7-10)=-9a(3)(1)(-1)(-3)=-9a(9)=-9a=-1Verification:p(x)=-(x-4)(x-6)(x-8)(x-10)Atx=7:-(3)(1)(-1)(-3)=-9✓$$

Answer: -1

Example 10: System of Equations

Question Pattern: Multiple conditions determine quadratic

Problem: Lines: ax + by + c = E (equation 1) bx + cy + d² = F (equation 2) E = F = 0, arithmetic mean of a,b is c, geometric mean is d

Solution Strategy:

1. Set up system with given conditions
2. Solve for intersection point
3. Use means to find relationships

⚠️ Common Pitfalls

• Incorrect order of operations in polynomial multiplication
• Forgetting to verify all given conditions
• Sign errors in vertex calculations

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📚 Pattern-Based Exercises

Set 1: Quadratic Word Problems (4 questions)

1. Product of two consecutive even numbers is 168
2. Students in rectangular arrangement with given total
3. Age problem: person’s age 5 years ago and 5 years from now
4. Number theory: sum of squares equals given value

Set 2: Parabola Properties (4 questions)

1. Find vertex of y = 2x² - 8x + 6
2. Maximum value of y = -3x² + 12x - 5
3. Axis of symmetry of y = x² + 6x + 8
4. Range of quadratic function with given vertex

Set 3: Derivatives & Slope (4 questions)

1. Slope of y = 3x² - 6x + 2 at x = 1
2. Find a given slopes at two points
3. Point where derivative equals given value
4. Rate of change interpretation

Set 4: Intersection Problems (3 questions)

1. Line intersects parabola at two points, find them
2. Two parabolas intersect, analyze tangent conditions
3. Circle and parabola intersection

Set 5: Projectile Motion (3 questions)

1. Time to reach maximum height
2. Maximum height reached
3. Time to hit ground from given height

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🎯 Mental Algorithms Summary

Quick Reference for Exam

1. Word Problems: Define variable, set up equation, solve quadratic
2. Properties: Vertex at -b/2a, use for max/min problems
3. Derivatives: 2ax + b, use for slope problems
4. Intersection: Set equal, solve quadratic equation
5. Applications: Identify physics context, use appropriate formulas

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📈 Progress Tracking

Week 3 Completion Checklist:

• Master all 5 question patterns
• Complete 15+ practice exercises
• Achieve 90%+ accuracy on pattern sets
• Review all solution explanations
• Identify personal error patterns

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Next: Week 4 - Polynomial Functions

Remember: Quadratic functions are everywhere in the physical world. Understanding their properties helps solve real optimization problems.