Lesson Plan: Applications of Quadratic Functions

Lesson Summary

In this 50-minute lesson, students will explore real-world applications of quadratic functions, focusing on modeling problems involving area and projectile motion. Multimedia resources from Media4Math.com are integrated throughout the lesson, providing rich visual and interactive content. Students will engage in warm-up activities, detailed teaching examples, and collaborative problem-solving exercises. The lesson concludes with a 10-question quiz and an answer key.

Lesson Objectives

• Model real-world problems using quadratic functions.
• Solve quadratic equations to determine maximum or minimum values.
• Analyze projectile motion and other applications of quadratic functions.

Common Core Standards

• HSA.CED.1: Create equations and inequalities in one variable and use them to solve problems.
• HSF.IF.4: For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.

Prerequisite Skills

• Problem-solving with algebraic expressions.
• Graphing linear and quadratic functions.
• Understanding the structure of quadratic equations.

Key Vocabulary

• Maximum: The highest point of a parabola when the graph opens downward. 
  • Multimedia Resource: https://www.media4math.com/library/74589/asset-preview
• Minimum: The lowest point of a parabola when the graph opens upward. 
  • Multimedia Resource: https://www.media4math.com/library/74588/asset-preview
• Projectile motion: The curved path an object follows when thrown or launched, influenced by gravity. 

Warm-up Activities

Choose from one or more of these activities.

Activity 1: Desmos

Students will use the Desmos graphing calculator (https://www.desmos.com/calculator) to explore the maximum and minimum points of quadratic functions.

• Open Desmos and input the quadratic equations: y = -2x² + 4x + 6 and y = x² - 4x + 3.
• Identify the vertex for each parabola and classify it as a maximum or minimum.
• Discuss how the coefficient of x² determines the direction of the parabola.

Activity 2: Drag-and-Drop Game

Students complete a drag-and-drop activity to identify the discriminant of various quadratic equations and relate it to the nature of the graph.

Interactive Game: https://www.media4math.com/library/4833/asset-preview

Activity 3: Slide Show

Introduce students to the concept of parabolas by observing their appearance in real-world contexts, sparking curiosity and discussion about their properties.

Materials Needed

• Slide deck with suggested images (or use your own):
  • Space ship launch
  • Path of a football in mid-air
  • Satellite dish
  • Fountains with water arcs
  • Fireworks in the sky

Activity Steps

1. Slide Show Presentation:

   Begin the lesson with a short, visually engaging slideshow. Each slide features one of the real-world examples listed above, showing clear parabolic shapes. Use this slide show or create one of your own.

   Slide Show

2. Guided Observation:

   For each slide, guide students through the following discussion points:

   • What do you see? (Encourage descriptive observations of the image.)
   • Where is the curve? (Have students identify the parabolic shape in the image.)
   • Why might this curve be important in this context? (E.g., trajectory of the spaceship, focus of the satellite dish.)

3. Class Discussion:

   After the slideshow, ask students:

   • What do all these examples have in common? (Elicit "curves" or "parabolas" as the unifying feature.)
   • What questions do you have about these shapes?
   • Where else might you see this kind of shape? (Prompt students to think beyond the examples provided.)

4. Quick Activity: Matching Parabolas:

   Distribute a short handout with simplified diagrams of parabolas. Ask students to match the diagrams to the real-world examples from the slideshow. This reinforces recognition of parabolic shapes.

Transition to the Lesson

Conclude the activity by explaining:

“Parabolas are not just beautiful shapes; they have mathematical properties that make them essential in science, engineering, and everyday life. Today, we’ll explore how to describe and analyze these curves using algebra.”

Teach (25 minutes)

Summary

Mathematical modeling is a powerful tool that allows us to represent real-world situations using mathematical concepts and structures. By creating these models, we can simplify complex phenomena, analyze patterns, make predictions, and solve problems. For example, models are used in fields like physics, engineering, finance, and even sports to represent everything from the motion of objects to the trajectory of projectiles and the growth of investments.

Quadratic equations and functions, in particular, play a significant role in mathematical modeling. A quadratic function, characterized by its parabolic graph, is often used to model situations where there is a peak or a minimum value, such as the height of a ball thrown in the air, the shape of a suspension bridge, or the profit of a business as it depends on production levels. These models allow us to predict key points like maximum height, time to hit the ground, or optimal outcomes.

Through this lesson, students will explore how quadratic functions are applied in real-world contexts and learn how to interpret and create quadratic models. This will deepen their understanding of quadratic equations while equipping them with practical problem-solving skills applicable to their everyday lives and future careers.

A ball is thrown upward with a velocity of 20 m/s from a height of 2 meters. The height (h) in meters after (t) seconds is given by the equation:

h(t) = -5t² + 20t + 2

• Identify the maximum height by finding the vertex of the parabola.
• The vertex formula is t = -b/(2a) or t = -20/(2*(-5)) = 2 second.
• Substitute t = 2 into h(t): h(2) = -5(2)² + 20(2) + 2 = 22 meters.

Summary: The ball reaches a maximum height of 22 meters after 2 seconds.

A farmer wants to build a rectangular pen with 120 meters of fencing. One side is against a barn, so only three sides require fencing. What dimensions maximize the area?

• Let w be the width, then the length is 

L = (120 - 2w)

• The area is:

A = w(120 - 2w)

• Rewrite as a quadratic equation in standard form:

 A = -2w² + 120w

• Find the maximum by calculating the x coordinate of the vertex:

x = -b/(2a)
 x = -120/(2*(-2)) 
x = 30 meters

• This is the width for the maximum area. Calculate length:

L = 120 - 2(30) = 60

• Maximum area is 30 times 60 = 1800 square meters.

Summary: The maximum area of the pen is 1800 square meters when the width is 30 meters and the length is 60 meters.

A company finds that revenue R(x) in thousands of dollars is modeled by R(x) = -2x² + 12x + 15, where x is the price per unit.

• Find the price that maximizes revenue by calculating the x-coordinate of the vertex: 

x = -b/(2a)
x = -12/(2(-2)) = 3

• Substitute x = 3: 

(R(3) = -2(3)² + 12(3) + 15 = 39

• The maximum revenue is \$39,000 at a price of \$3 per unit.

Summary: The company maximizes its revenue at \$39,000 when the price is set at \$3 per unit.

Additional Multimedia Resources

• Applications of quadratic equations: https://www.media4math.com/MathClipArtCollection--ApplicationsLinearQuadraticFunctionsSpeedAcceleration

Review (10 minutes)

Summarize the lesson by revisiting how quadratic functions model real-world problems. Reinforce the importance of the vertex in optimization problems.

Additional Examples

Example 1: Throwing a Ball

A ball is thrown upward with an initial velocity of 20 m/s from a height of 5 m. The height h(t) of the ball after t seconds is modeled by the equation:

h(t) = -4.9t2 + 20t + 5

Find the time when the ball hits the ground.

Solution:

1. Set the height to zero: h(t) = 0, resulting in -4.9t2 + 20t + 5 = 0.
2. Use the quadratic formula: t = (-b ± √(b² - 4ac)) / 2a with a = -4.9, b = 20, c = 5.
3. Calculate the discriminant: Δ = 20² - 4(-4.9)(5) = 498.
4. Substitute into the quadratic formula:
   • t = (-20 ± √498) / -9.8
   • √498 ≈ 22.32
5. Find the roots:
   • t₁ = (-20 + 22.32) / -9.8 ≈ -0.24 (not valid).
   • t₂ = (-20 - 22.32) / -9.8 ≈ 4.32.

Conclusion: The ball hits the ground after approximately 4.32 seconds.

Example 2: Profit Maximization

A company produces and sells x units of a product. The profit P(x) (in dollars) is given by:

P(x) = -2x2 + 40x - 150

Determine the number of units x the company must produce to maximize its profit.

Solution:

1. Recognize the quadratic form: P(x) = -2x2 + 40x - 150 is a downward-opening parabola.
2. Find the vertex using x = -b / 2a:
   • x = -40 / 2(-2) = 10
3. Substitute x = 10 into P(x):
   • P(10) = -2(10)² + 40(10) - 150
   • P(10) = -200 + 400 - 150 = 50

Conclusion: The company maximizes its profit when it produces 10 units, with a maximum profit of $50.

Multimedia Resources

Videos

These videos provide real-world examples of quadratic functions.

Slide Show

This slide show also includes applications of quadratic functions:  https://www.media4math.com/library/slideshow/illustrated-math-dictionary-quadratic-functions-math-application

Quiz

Directions: Solve each problem and show your work.

1. Find the vertex of y = -x² + 6x - 5.
   
    
2. Determine the maximum value of h(t) = -4.9t² + 19.6t + 1.5.
   
    
3. Solve x² - 4x - 5 = 0.
   
    
4. Write the equation of a parabola with vertex at (3, -2).
   
    
5. Find the discriminant of x² + 6x + 9 = 0.
   
    
6. A ball is thrown upward, and its height is modeled by h(t) = -5t² + 20t + 3. Find the time it takes to reach the maximum height.
   
    
7. A rectangular garden has a perimeter of 100 meters. One side is against a wall. Write an equation to model the area as a function of the garden's width.
   
    
8. A company’s profit is modeled by P(x) = -3x² + 18x - 5. Find the price (x) that maximizes the profit.
   
    
9. A projectile is launched, and its path is modeled by y = -4x² + 8x + 3. What is the maximum height?
   
    
10. Write a quadratic equation in standard form that has roots x = 2 and x = -3.

Answer Key

1. Vertex: (3, 4).
2. Maximum height: 20.4 meters.
3. Roots: x = 5, x = -1.
4. Equation: y = (x - 3)² - 2.
5. Discriminant: 0.
6. Time to maximum height: 2 seconds.
7. Area equation: A = w(50 - 2w).
8. Price for maximum profit: x = 3.
9. Maximum height: 7 units.
10. Equation: y = x² + x - 6.