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.
Students’ Conceptual Challenges
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:
1. 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.
2. 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. 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?
xy + xy + xy
x + x + x + y + y + y