Introduction: The Sweet Spot

Hook: The Billion-Dollar Question

Imagine you run a lemonade stand, and you want to set the perfect price. If you charge too little, you don't make much money. If you charge too much, nobody buys it. There is a "sweet spot"—a price that maximizes your profit. How do mathematicians find that perfect peak?

We are going to learn how to find the maximum or minimum point of a curve. This skill is used by engineers, economists, and athletes every day.

Learning Objectives (Tell them what you'll teach)

By the end of this lesson, you will be able to:

1. Identify the standard form of a quadratic equation ($y = ax^2 + bx + c$).
2. Use the vertex formula to calculate the exact maximum or minimum point (the Vertex) of a quadratic function.
3. Apply this skill to solve real-world optimization problems (e.g., maximizing area, height, or profit).

Body: The Power of the Vertex

Phase 1: I Do (Modeling the Concept)

Content Presentation: Introducing the Vertex

A quadratic function graphs as a U-shape called a parabola. If the parabola opens upward, the lowest point is the minimum. If it opens downward, the highest point is the maximum. This crucial turning point is called the Vertex.

The Magic Formula: To find the x-coordinate of the Vertex for any standard quadratic equation ($y = ax^2 + bx + c$), we use the formula:

$$x = -\frac{b}{2a}$$

Success Criteria: We know we have solved for the optimum x-value when we have successfully plugged the coefficients $a$ and $b$ into the formula and calculated the resulting value.

Modeling Example: The Farmer’s Fence

A farmer has 100 meters of fencing and wants to create the largest rectangular field possible. The area (A) can be modeled by the function $A = -x^2 + 50x$, where $x$ is the width of the field.

1. Identify a, b, and c: In $A = -x^2 + 50x + 0$, we have $a = -1$, $b = 50$, $c = 0$. (Since $a$ is negative, we know the parabola opens down, so we are finding a MAXIMUM area.)
2. Calculate the optimal width (x-value):

   $$x = -\frac{50}{2(-1)} = -\frac{50}{-2} = 25$$

   The optimal width is 25 meters.
3. Find the Maximum Area (y-value): Substitute $x=25$ back into the original equation:

   $A = -(25)^2 + 50(25) = -625 + 1250 = 625$

   The maximum possible area is 625 square meters.

Phase 2: We Do (Guided Practice & Discussion)

Activity: Optimizing Revenue (Think-Pair-Share)

A company finds that the profit ($P$) generated by selling a certain product depends on the price ($p$) according to the function: $P(p) = -4p^2 + 160p - 700$.

1. Identify the coefficients (5 minutes): Working individually or in a pair (if in a classroom/group setting, otherwise Jaspen works independently), identify $a$, $b$, and $c$. (Formative Assessment Check: Does $a$ tell us this is a max or min?)
2. Calculate the Optimal Price ($p$): Use the vertex formula to find the price that generates the maximum profit.
3. Calculate the Maximum Profit ($P$): Substitute the optimal price back into the profit equation.

Phase 3: You Do (Independent Application & Creativity)

Project: The Launch Angle Challenge

You are designing a prototype catapult (or optimizing a paper airplane design). The trajectory (height, $H$, in meters) of your projectile over time ($t$, in seconds) is often modeled by a quadratic function. Your goal is to maximize the height before it starts falling.

Scenario 1: The initial launch is modeled by: $H(t) = -5t^2 + 30t + 2$.

1. Find the time (t) it takes to reach maximum height.
2. Calculate the actual maximum height (H).
3. (Optional Extension) Use a graphing tool (Desmos) to graph the function and visually confirm the vertex location.

Scenario 2 (Choice/Autonomy): Jaspen, choose ONE of the following real-world optimization problems to solve completely:

• A: Bridge Arch Design (Minimum): A suspension bridge cable is modeled by $y = 0.05x^2 - 12x + 1000$. Find the minimum height of the cable above the water. (Hint: $a$ is positive, so you are finding a minimum.)
• B: Advertising Return (Maximum): The effectiveness of an ad campaign ($E$) based on money spent ($m$, in thousands) is $E(m) = -0.1m^2 + 4m + 50$. How much money should be spent to achieve maximum effectiveness, and what is that effectiveness rating?

Success Criteria for "You Do"

Your solution is successful if:

1. You correctly identify the coefficients (a, b, c) for the chosen problem.
2. You use the vertex formula ($x = -b/2a$) correctly to find the optimal input value (time, money, etc.).
3. You substitute the optimal input value back into the original function to find the maximum or minimum output value (height, effectiveness, etc.).

Conclusion: Reinforcing the Takeaways

Recap and Review (Tell them what you taught)

We use quadratic functions to model scenarios where an initial increase is followed by a decrease (or vice versa). The Vertex Formula is our mathematical tool to pinpoint the optimal result.

• What does the sign of 'a' tell us about whether we are looking for a maximum or minimum? (Negative 'a' = maximum peak; Positive 'a' = minimum valley).
• What is the first step in solving any optimization problem modeled by a quadratic equation? (Finding the x-coordinate using $x = -b/2a$).

Summative Assessment: Application Check

Explain in one short paragraph how a theme park designer could use the vertex formula to ensure a new roller coaster drop (modeled by a quadratic curve) maximizes the thrill while keeping the minimum height above ground safe.

(Teacher/Mentor Note: Assess based on whether the explanation correctly links the vertex formula to finding the minimum point (the lowest safe point) of the curve.)

Adaptability and Differentiation

Scaffolding (For Struggling Learners)

• Pre-Calculated Steps: Provide pre-filled templates that separate the calculation of the optimal x-value from the calculation of the resulting y-value, ensuring the student focuses on one step at a time.
• Visual Aid: Require the student to sketch a graph of the function (even if rough) to visually confirm if their calculated vertex is indeed the max/min.

Extension (For Advanced Learners)

• Advanced Modeling: Instead of providing the quadratic equation for the fence problem, challenge the learner to derive the Area function ($A(x) = -x^2 + 50x$) themselves, starting only with the perimeter constraint (100m).
• Optimization Constraint: Introduce a non-linear constraint. For example, in the Revenue problem, add a cost constraint that is not quadratic, requiring comparison of the optimal quadratic result with the new constraint boundary.
• Cubic Functions: Research methods (calculus) used to find maximums and minimums of higher-degree polynomial functions (like cubic functions).

Context Adaptation

• Homeschool: Use the Launch Angle Challenge to physically test different launch angles of a paper plane or toy catapult, recording the height data to see how it aligns with the mathematical model.
• Classroom: Run the Launch Angle Challenge as a competitive group lab, where teams must calculate and justify the optimal launch parameters.
• Training: Use the Revenue/Profit problem to discuss business modeling. The focus shifts to interpreting the meaning of the vertex in terms of market analysis (Price vs. Demand).