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Episode 2 Supports

  • Exploring: Keoni and Sasha use the definition of a parabola and the Pythagorean theorem to solve for the y-value of a point on the parabola when the x-value is 5.

  • Sasha and Keoni are unsure how to represent distances involving the unknown value of y [see 0:46 – 1:14].

     

    • Keoni returns with more confidence to his idea from Episode 1 of representing one of the distances as y – 1 [1:15]

  • For use in a classroom, pause the video and ask these questions:

     

    1. [Pause video at 1:54]. Keoni has written down an equation. Why is that equation true?

     

    2. [Pause video at 2:28]. Examine the expressions on both sides of the equals sign. How are they the same? How are they different?

  • Facilitate the action of providing justification by asking students to:

     

    • Explain why the distance from the point (5, y) to the directrix is y + 1.  Support your argument by labeling y, 1, and y + 1 on the graph.

    • Explain why the length of the vertical side of the right triangle is y – 1. Support your argument by labeling y, 1, and y – 1 on the graph.

  • 1.1. Multiply the binomials and simplify by combining like terms. What pattern do you see emerging?

     

    (4 + x)(x + 4)

    (x2 + x)(x2 + x)

    (x + y)(x + y)

    (5x – 2y)(5x – 2y)

    (4 – x)(4 – x)

     

    2. Justify by showing your work and explaining your reasoning. Are any of the following expressions are equivalent to 3xy?

     

    3x3y

    xy + xy + xy

    xy3

    x + x + x + y + y + y?

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Mathematics in this Lesson

Lesson Description

 

Sasha and Keoni use the definition of a parabola, the Pythagorean theorem, and their methods from previous lessons to represent the y-value for any point on a particular parabola, given the x-value of that point.  In contrast to Lesson 3 in which y-values were given, in this episode x-values will be given.