Warm Up for Lesson 3.6 Identify the vertex of each parabola: (1). y = -(x + 3) 2 – 5 (2). y = 2x (3). y = 3(x – 1) (4). y = -5(x + 4) 2 (5). Solve: 0 = x 2 – 8x – 9

Warm Up Answers for Lesson 3.6 Identify the vertex of each parabola: (1). y = -(x + 3) 2 – 5 V = (-3, -5) (2). y = 2x 2 + 7V = (0, 7) (3). y = 3(x – 1) V = (1, 4) (4). y = -5(x + 4) 2 V = (-4, 0) (5). Solve: 0 = x 2 – 8x – 9 x = 9, -1

Graphing Quadratic Functions Standard Form and Completing the Square Section 3.6 Standard: MM2A3 bc Essential Question: How do I analyze and graph quadratic functions in standard form?

Let’s begin by squaring a few binomials: 1. (x + 3) 2 = _______________ 2. (x – 4) 2 = _______________ 3. (x + 6) 2 = _______________ 4. (x – 1) 2 = _______________ Is there any kind of a pattern that we notice within each problem?

Let’s test our theory by factoring a few perfect square trinomials: 1. x x + 25 = _______________ 2. x 2 – 6x + 9 = _______________ 3. x x + 49 = _______________

Finally, let’s look at some examples where we must determine the missing value so that the polynomial is a perfect square trinomial. 1. x 2 + 4x + ___ 2. x 2 – 8x + ___ 3. x x + ___

Using the information from these few examples, we are going to examine a method called “completing the square”. This method will appear in other areas of mathematics, but in this unit it will be used to change a quadratic from general form to vertex form. Start with: y = x 2 + 6x + 8 and change to y = (x + 3) So how does it work?

Example 1: y = x 2 + 6x + 8 Rewrite your equation so that there is a space between The 6x and 8, like this: y = (x 2 + 6x + ____ ) + 8 Now we have to think: What value can I add in the space I have so that the first three terms create a perfect square trinomial? I can determine this quickly and easily by halving the middle coefficient and then squaring it.

Example 1: y = x 2 + 6x + 8 y = (x 2 + 6x + ____) + 8 Half of 6 is 3, and 3 2 = 9, so I will add 9 in the space. In algebra, we know that we cannot simply add 9 to an equation without doing something to balance it out. In this case, since we added 9 on the right hand side, let’s also subtract 9 on the right hand side. This is like adding zero to the right side of the equation, which does not change its value.

Example 1: y = x 2 + 6x + 8 y = (x 2 + 6x + __9__) We’re almost finished! I can factor the polynomial now that it is a perfect square trinomial and I can combine like terms outside of the parentheses. Doing this results in: y = (x + 3) 2 – 1 Now we can easily see the vertex for this quadratic is (-3, -1) and we could graph it more easily.

Example 2: y = x x + 7 Example 3: y = x 2 - 8x - 7

It can be tricky if our leading coefficient isn’t 1, but we can always make our leading coefficient 1 by factoring. Example 4: y = -x 2 + 6x + 5 Let’s leave a space between the last 2 terms as we have been doing: y = (-x 2 + 6x ___) + 5 And let’s factor a -1 out of our polynomial, like this: y = -1(x 2 – 6x _____) + 5

From our previous examples, we know that we need to add 9 inside of the parentheses to complete the square. But what have we really added? What? Why? If we distribute the -1 to the term we added (9), we get -9. So since we really added a -9 inside the parentheses, we need to add 9 outside of the parentheses. Simplifying, this gives us y = -(x – 3)

(1). Graph: y = x 2 – 4x + 8

(1). Graph: y = x 2 – 4x + 8 xy

Identify each characteristic of: y = x 2 – 4x + 8 (a). Domain: (b). Range: (c). Vertex: (d). Axis of symmetry: (e). Opens: (f). Max or Min: (g). x-intercept: (h). y-intercept: (i). Extrema: (j). Increasing: (k). Decreasing: (l). Rate of change (-2 ≤ x ≤ 1):

Estimated Rate of change over the interval (-2 ≤ x ≤ 1): y = x 2 – 4x + 8

(2). Graph: y = x 2 – 6x + 5

(2). Graph: y = x 2 – 6x + 5 xy

Identify each characteristic of: y = x 2 – 6x + 5 (a). Domain: (b). Range: (c). Vertex: (d). Axis of symmetry: (e). Opens: (f). Max or Min: (g). x-intercept: (h). y-intercept: (i). Extrema: (j). Increasing: (k). Decreasing: (l). Rate of change (-6 ≤ x ≤ -3):