# Parabolas Lesson 10 (Teachers)

### Getting and Using Geometric Information

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.

##### Episode 1: Making Sense

Keoni and Sasha begin to find geometric information from the equation, y = (x–2.4)2/6, in vertex form. They find the p-value and determine the vertex of the parabola.

##### Episode 2: Exploring

Sasha and Keoni graph the equation y = (x–2.4)2/6. They determine the coordinates of the focus and the equation of the directrix from the geometric information in the equation.

##### Episode 3: Repeating Your Reasoning

Keoni and Sasha look for geometric information of a parabola represented by the equation y = 2(x – 3)2 + 1. They start by finding the vertex and the p-value.

##### Episode 4: 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.

##### Episode 5: Exploring

Sasha and Keoni examine yet another equation of a parabola that is not in vertex form, y = x2 – 4x + 5. They start out by seeking a method to re-express the equation in vertex form.

### Mathematics in this Lesson

Targeted Understandings

This lesson can help students:

1. 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.

2. 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].