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Episode 4 Supports

  • Episode Description

    Making Sense: Keoni and Sasha examine an equation of a parabola in a different form, y = x2 – 4x + 4. When they look for geometric information, the p-value and vertex are not apparent. They start by rewriting the equation.

     

  • Students’ Conceptual Challenges

    Sasha tries to create something that looks like the parenthesis (x – h) from the vertex form by factoring out an x [1:08–1:24].

     

    • When asked if that form is helpful, she quickly realizes this doesn’t satisfy the need for a squared term, i.e., like (x – h)2 [1:26-1:34].

  • Focus Questions

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

     

    1. [Pause the video at 1:23]. Is y = x2 – 4x + 4 equivalent to y = x(x – 4) + 4? How do you know? Does the re-expressing the equation this way support the goal of finding geometric information?

     

    2. [Pause the video at 2:52]. Explain why the p-value is ¼.

  • Supporting Dialogue

    Support the opportunity to revoice the mathematical ideas of others. Ask students to answer the focus questions with their partner. Ask them to prepare their answers to share with the class.

     

  • Math Extensions

    1. Graph y = x2 – 4x + 4. List and label all of the geometric information.

    2. Find and label the “special points.” How did you find them?

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Mathematics in this Lesson

Lesson Description

Targeted Understanding

CC Math Standards

CC Math Practices

Lesson Description

 

Given the equation of a parabola in any form, Sasha and Keoni find geometric information (such as the focus, directrix, p-value, and vertex) about the parabola.

Targeted Understandings

 

This lesson can help students:

 

  • Understand that when the equation representing a parabola is written in vertex form [y = (x–h)2/(4p) + k], then the vertex can be easily determined as (h ,k) and the p-value can located in the denominator and can be used, along with the vertex, to determine the focus and directrix of the parabola.

  • Understand that geometric information (e.g., the vertex, p-value, focus, and directrix) can also be located from a parabola that is given in a different form, by first re-expressing the equation in an equivalent vertex form.

Common Core Math Standards

 

• CCSS.M.HSF.IF.C.8: Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.

In this lesson, Sasha and Keoni re-express quadratic functions like y = 2(x – 3)2 + 1 and y = x2 – 4x + 4 in vertex form [y = (x–h)2/(4p) + k] in order to locate geometric information about the graphs (such as the vertex, p-value, focus and directrix). To rewrite the function y = 2(x – 3)2 + 1 in vertex form, they think of 2 as 1 divided by ½ [see 2:20 – 2:31 in Episode 3] and ½ as 4 multiplied by 1/8 [3:21 – 3:43, Episode 3]. To rewrite the function y = x2 – 4x + 4 in vertex form, Sasha and Keoni figure out that they need to factor, but that not every way of factoring is helpful. For example, factoring out the x from x2 – 4x yields y = x(x – 4) + 4 [1:06 – 1:30, Episode 4], which doesn’t help them. However, factoring the entire trinomial yields y = (x – 2)2, which is very close to being in vertex form [3:14 – 2:52, Episode 4]

 

Common Core Math Practices

 

CCSS.Math.Practice.MP1: Make sense of problems and persevere in solving them.

An important aspect of Math Practice 1 is not giving up, even when several attempts have not been fruitful. In Episode 5, Sasha and Keoni face a challenging task of rewriting y = x2 – 4x + 5 in vertex form. They begin with a false start, as Keoni incorrectly factors x2 – 4x + 5 as (x – 5)(x + 1) [1:08 – 1:44, Episode 5]. Instead of getting discouraged, Keoni tells Sasha, “Let’s not give up” [2:19 – 2:26]. And they don’t! Sasha tries factoring out the x [2:53 – 3:07], which doesn’t end up helping. However, when they explain the correspondences between y = x2 – 4x + 5 and the function they worked with in Episode 4 [y = x2 – 4x + 4], they are able to successfully rewrite the given equation as y = (x – 2)2 + 1 [4:25 – 4:46].