**Factor the following. Notice any pattern?**

**x² + 2x + 1**

**x² - 4x + 4**

**x² - 6x + 9**

**x² - 8x + 16**

**x² - 12x + 36**

**When the above equations are factored out, you should notice a pattern starting to appear. The following pattern should look like this:**

x² + 2x + 1

= (x - 1)(x + 1)

**= (x+1)²**

x² - 4x + 4

= (x - 2)(x - 2)

**= (x - 2)²**

x² - 6x + 9

= (x - 3)(x - 3)

**= (x - 3)²**

x² - 8x + 16

= (x - 4)(x - 4)

**= (x - 4)²**

x² - 12x + 36

= (x - 6)(x - 6)

Not only are all these equations trinomials, but because what we see as the result of them being factored appears to be a pattern, we call these equations, '

**= (x - 6)²**Not only are all these equations trinomials, but because what we see as the result of them being factored appears to be a pattern, we call these equations, '

**perfect square trinomials'.****A few days ago, we worked and discussed about the following equation of a parabola:**

When the above general form is factored, we get:

= (x + 4 )(x - 2)

Using the factored form above, the roots and the y-intercept of the parabola can now be easily identified:

= (x - 4)(x - 2)

(Looking at the general form, we know immediately that -8 is the y-intercept)

Now that we know the roots, we can determine the vertex of the parabola, by simply adding the two roots together, and dividing by 2. After all, the vertex

= -2/2

= -1

Using the vertex, and roots, the points on the graph would appear to be:

*f*(x) = x² + 2x - 8When the above general form is factored, we get:

= (x + 4 )(x - 2)

Using the factored form above, the roots and the y-intercept of the parabola can now be easily identified:

= (x - 4)(x - 2)

*Roots:*-4, 2(Looking at the general form, we know immediately that -8 is the y-intercept)

Now that we know the roots, we can determine the vertex of the parabola, by simply adding the two roots together, and dividing by 2. After all, the vertex

*is*the middle point of the parabola.*Vertex*= -4 + 2= -2/2

= -1

Using the vertex, and roots, the points on the graph would appear to be:

**The black points located on -4 and 2 are the roots.**

The blue point located on -1 is the vertex, which is also the location for the axis of symmetry.

The red point located on -8 of the y-axis is the y-intercept.

----------------------------------

Are you still with me? I'm hoping you are! If you noticed above, the equation

The blue point located on -1 is the vertex, which is also the location for the axis of symmetry.

The red point located on -8 of the y-axis is the y-intercept.

----------------------------------

*was written in its gerenal form*

**f(x) = x² + 2x - 8***and its factored form. Yet there's one form we still haven't written it as, which is its*

**standard form**. The standard form for this equation would look like this

*:*

**f(x) = (x + 1)² - 9**

**You're probably all wondering how we came up with -9, and what happened to the -8.**

**This is what we did:**

f(x) = x² + 2x + 1 -9

= f(x) = x² + 2x - 8

**There you have it! This beautfiully drawn**(sarcasm)**parabola was created with the equation**

**f(x) = (x + 1)² - 9**

**...**And also appears to be a perfect square.**-------------------------------**You can do many things to a parabola. Push it up or down, shift it to the left or right, squeeze it, widen it, etc.

**We discussed in class what you can do to push the parabola up 1 unit. We did the following:**

**i.e**

**g(x) = x² + 2x - 7**

**= (x + 1)² -8**Where -8 is parameter k. With parameter k, you can position the parabola either up or down by changing the number.

**Let's look at another example of an equation and turn it into a perfect square:**

**i.e**

**g(x) = x² +2x + 5**

(general to standard)

**To turn that into a perfect square, you must add zero.**

**g(x) = x² + 2x + 1 - 1 - 5**

**This is legit, because +1 and -1 equal zero, therefore we get:**

**g(x) = (x + 1)² - 6**

**The process you have just seen and we discussed in Friday's class is called,**

**'completing the square'.**

**-------------------------------------------------------------------****How do you know if a trinomial is a perfect square?**

- Last digit and first terms are always perfect squares.
- Constant terms
**always**positive. - Half of
*b***(***f(x)*= ax² +**b**x + c**)**is the square root of the last digit.

**Here's an example of an equation of a parabola written in perfect square:**

**y = x² - 8x - 11**

(as we look back, x² - 8x + 16 is a perfect square trinomial, so to turn this into a perfect square, we must turn the '11' into '16')

**So ..**

**y = x² - 8x + 16 - 16 - 11**

**= (x - 4)² - 27**

**Where the vertex would be the opposite of -4, which will be +4.**

**And there you have it! A lesson in**

**perfect square trinomials and**

**completing the squares.**

*---------------------------------------------*

**REMINDER: The "Quiz 2 - Quadratic Functions"**

**worksheet we did in class, finish it, and review it!**

*SIMILAR QUIZ ON MONDAY, SEPTEMBER 18!*---------------------------------------------

**ANNNNNDDDDDD, FINALLY.**

**The next scribe will be ..**

**S - A - N - D -Y !!!!!!!!!!**

Yes,

**Sandy**, you're the next scribe (:

**Seeee ya tomorrow.**

Hi Cherrieeeeee,

ReplyDeleteCongratulations on your Hall of Fame Scribe Post! I felt I was in class with a master teacher as I read! Using questions to keep your reader focused is a great technique.

Best,

Lani