Home

Episode 2 Supports

  • Exploring: Keoni and Sasha develop an equation that relates the x-value to the y-value for a general point on a particular parabola.

  • Students might wonder how to interpret more than one symbol in a general equation. Students are often asked to isolate a variable on one side of the equals sign with a number on the other side, e.g., they “solve for x” by finding a numerical value for x. Having it make sense that the symbols are representing a general point in the parabola is challenging.

     

    Keoni protests “but there are two variables” when Sasha writes (y–1)2 + b2  = (y+1)2 [1:11-1:34]. By going back to work through a specific example 62 + b2 =  82, Sasha and Keoni identify the method to solve for b in the general case [1:47-4:50].



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

     

    1. [Pause video at 1:34]. What problem are Sasha and Keoni trying to solve?

     

    2. [Pause video at 2:44]. Sasha and Keoni are working to solve for b. Work ahead from here. Then we will compare our work with theirs.

  • Invite students to engage in revoicing. Place a blank student worksheet under a document camera.

     

    1. Can someone come up to the graph to draw and label the right triangle that Sasha and Keoni used in this problem?

    2. Ask a student to revoice how a fellow student determined the side lengths.

  • Use the Pythagorean theorem to represent the lengths of the right triangles described below.

     

    1. A right triangle where one leg is half the length of the other leg.

     

    2. A right triangle where the two legs are the same length.

     

1

2

3

4

5

 

Mathematics in this Lesson

Lesson Description

 

Keoni and Sasha create a general method for representing the x-value for any point on a particular parabola, given the y-value of that point. By using their previous results, along with the Pythagorean theorem, they are able to determine the equation for the parabola.