Aprenda FUNÇÃO DO SEGUNDO GRAU de uma vez por todas (aula super didática)

Main ideas / lessons

• Quadratic functions model “trade-offs”: when you change a decision variable (price), you typically affect both:

  • quantity demanded (number of customers),
  • and unit profit/revenue, leading to a nonlinear outcome where revenue/profit can increase then decrease.

• Revenue vs. profit

  • Revenue = (number of buyers) × (selling price).
  • Profit = revenue − (fixed cost per item).

• From a real-world story to a quadratic function

  1. Define how many buyers depend on price: buyers often follow a linear rule (e.g., (500-x)).
  2. Define profit per item as “selling price minus cost”: (x-100).
  3. Total profit multiplies the two expressions (buyers × profit per buyer), producing a quadratic: [ P(x) = (500-x)(x-100) ] Expands to: [ P(x) = -x^2 + 600x - 50{,}000 ]

• Zeros/roots have meaning

  • Solve (P(x)=0) to find prices where total profit is zero (break-even).
  • The roots found are (x=100) and (x=500).
  • Graphically, these are the points where the parabola intersects the x-axis.

• Vertex gives the maximum profit

  • The vertex is the maximum point (since the quadratic opens downward: leading coefficient is negative).
  • Using quadratic-vertex formulas yields:
    • Vertex y-value (maximum profit) = 40,000
    • Vertex x-value (optimal price) = 300

• Symmetry property

  • For a quadratic, the vertex is centered between the two roots.
  • With roots 100 and 500: [ \text{Axis symmetry} = \frac{100+500}{2}=300 ]

• Method emphasis (conceptual workflow)

  • Don’t only memorize formulas: understand that the quadratic comes from multiplying two linear functions (buyers function × profit-per-item function).

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Detailed methodology / instructions presented (step-by-step)

A) Build revenue/profit from an economic scenario

• Let:

  • (x) = selling price of the product (skateboard/course).

• Find quantity of buyers as a function of price

  • Given by market research: [ \text{buyers}(x)=500-x ]

• Find profit per unit

  • Cost to produce per skateboard is fixed: 100
  • Profit per unit when selling for (x): [ \text{profit-per-unit}(x)=x-100 ]

• Compute total profit

  • Multiply: [ P(x) = \text{buyers}(x)\cdot \text{profit-per-unit}(x)=(500-x)(x-100) ]

• Expand to standard quadratic form

  • Multiply binomials: [ P(x) = -x^2 + 600x - 50{,}000 ]

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B) Evaluate profit at specific prices (using the profit function)

• To test a price (x=a):

  • Substitute into: [ P(a) = -a^2 + 600a - 50{,}000 ]

• Examples referenced in the lesson:

  • (P(100)=0) (break-even)
  • (P(200)=30{,}000)
  • (P(300)=40{,}000) (maximum)
  • (P(400)=30{,}000)
  • (P(450)=17{,}500)
  • (P(500)=0) (break-even)
  • (P(0)=-50{,}000) (all cost, no selling price)

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C) Find the roots using the quadratic formula

• Start with: [ P(x)= -x^2+600x-50{,}000 ]

• Identify coefficients in (ax^2+bx+c):

  • (a=-1), (b=600), (c=-50{,}000)

• Compute discriminant: [ \Delta = b^2 - 4ac ]

• Use: [ x=\frac{-b\pm\sqrt{\Delta}}{2a} ]

• Results claimed:

  • (\boxed{x=100}) and (\boxed{x=500})

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D) Find the vertex (maximum profit) using vertex formulas

• Vertex formulas given:

  • [ y_{\text{vertex}}=\frac{-\Delta}{4a} ]

  • [ x_{\text{vertex}}=\frac{-B}{2A} ]

• With values from the lesson ((\Delta=160{,}000), (a=-1), (b=600)):

  • Maximum profit: [ y_{\text{vertex}}=40{,}000 ]

  • Optimal price: [ x_{\text{vertex}}=300 ]

• Interpret:

  • Price 300 gives maximum profit 40,000.

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E) Alternative root reasoning based on product = 0

• Since: [ P(x)=(500-x)(x-100) ] and profit is zero when the product is zero:

• Either:

  • (500-x=0) or (x-100=0)

• Solve:

  • (500-x=0 \Rightarrow x=500)
  • (x-100=0 \Rightarrow x=100)

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Speakers / sources

• Primary speaker/instructor: the narrator/teacher (mentions “Pedro” as an example character; Pedro is addressed but is not presented as another distinct real speaker).
• No other distinct named speakers or external sources are clearly featured in the subtitles.