Keoni and Sasha develop an equation that relates the x-value to the y-value for a general point on a particular parabola.
Students’ Conceptual Challenges
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.