Polynomial Review - Student Explain Everything movies

On Thursday in class we started to learn about y intercepts on graphs of functions. Here is a graph we looked at:

Equation: y = (x+7) (x+1) (x-6) (x-10)^2

y intercept : -4200

Though we can't see it this picture of the graphed function (the window would have to be huge!!) the y - intercept of this function is -4200.
After graphing this with our calculators in class and playing around with our windows for a while, Lisa asked us to now change the y - intercept to -42. At first I was like, "what?!" but then I realized that a few weeks ago we learned about reflection over the axes, vertical shift, vertical stretch, vertical shrink, horizontal shift, horizontal stretch, and horizontal shrink. Changing the y-intercept of this function from -4200 to -42 is a vertical shrink. 
But how do you change the y-intercept?
Like this: 
y = (x+7) (x+1) (x-6) (x-10)^2 (.01)
By multiplying our old equation by (.01), the function will undergo a vertical shrink.
Now the function looks like this:

Equation: y = (x+7) (x+1) (x-6) (x-10)^2 (.01)

y- intercept: -42

Let's try something else, like a reflection over the x axis keeping our vertical shrink. 
Take the equation of the function:
y = (x+7) (x+1) (x-6) (x-10)^2 (-.01)
By making the (.01) negative, the function is now the reflection of what it used to be. 
Here's what it looks like now:

Equation: y = (x+7) (x+1) (x-6) (x-10)^2 (-.01)

y- intercept: 42

See the difference?

On Friday, we talked about x intercepts, and how do find their exact location on a graph (without even really needing a calculator!) Sometimes x intercepts are easy to see with our eyes:

The function is clearly going through the x axis at (-2,0) (-1,0) and (2,0). Therefore, it is easy to find an equation to this function:

y = (x+2) (x+1) (x-2)

The +2, +1, and -2 are the negative version of the x - intercepts on the graph - 2, - 1, and 2.
However, sometimes it is hard to tell where exactly the function is hitting and/or going through the x axis (I say and/or because remember - if the x is squared then it will be a parabola on the axis instead of going through.)
Let's take this confusing function for example:

Again, we can clearly see that the function goes through the x axis as -3, but WHERE does it go through in the in the other two places??? It goes through somewhere between -1 and -2, and then again somewhere between 1 and 2. But where exactly?

Here's how to find out:

This is the equation of the function: f(x) = 
1. Factor the equation 
2. Factor it completely  y = (x+3) (x + route2) (x - route2)
3. This means that - 3, - route2 and positive route 2 are our exact x intercepts.

Then we learned how to graph functions on our own by picking apart an equation with what we know and how we can use synthetic division to locate points of the function on a graph.
Let's take this equation: (x + 1)^3 (x) (x-1)
1. By looking at this equation, we can see that this is a 5th degree function because there are five xs total. This means that the function will start low, because it is odd.

SIDE NOTE: a function with an odd number of xs starts low, and a function with an even number of xs starts high.

2. Now we need to find the x intercepts in order to graph it without a calculator. x-intercepts: -1, 0, and 1.
3. Always pay attention to x^2, x^3, x^4s and so on when graphing. In this function in particular, we have an x^3, meaning that the function WILL NOT look like this:

The function is not this simple on the graph. The function pictured here is (x+1) (x) (x-1) not 

(x+1)^3 (x) (x-1)

SIDE NOTE: When we have an x^2: 

When we have an x^3:

When we have an x^4:

When we have an x^5:

etc.

So, with this in mind, we would graph our equation (x + 1)^3 (x) (x-1) like this:

Using Synthetic Division to Locate Exact x-intercepts!

Lastly, I would like to go over synthetic division using the homework problem we had for Monday as an example.

Function Equation:

f(x) = x^4 - 6x^3 + 4x^2 + 15x + 4

Step 1. Plug this equation into your calculator. You will get a graph that looks like this:

See how the function goes through the axis 4 times? That is because it is a fourth degree polynomial, its leading exponent is 4. In 2/4 places we can see exactly where these x - intercepts are - one is at (-1,0) and the other one is at (4,0). 

Chose either one of these x-intercepts for your synthetic division.

Step 2. I'm going to use the -1 x-intercept for my factoring. Then, I take all of the coefficients of our equation f(x) = x^4 - 6x^3 + 4x^2 + 15x + 4 and divide them by negative one using synthetic division

Step 3. now I'm going to run these coefficients through synthetic division once again until it is all factored, still dividing by negative one: 

Now, our equation looks like this: y = (x+1)(x-4) (1x^2 -3x -1)

Step 4: To find those two other x-intercepts, we are going to factor the remaining un-factored part of our equation using the quadratic formula to find x. 

Those are your two other x-intercepts!

Polynomials and calculator use

POLYNOMIALS = many terms

A polynomial is a sum of many terms.
It is made up of terms that are added, subtracted, or multiplied. NOT divided.

Example:  5xy^2 + 2x - 3

Polynomials have constants (-3) , variable (x and y), and exponents (^2)

These are polynomials: (positive, integer exponents)
f(x) = x^2+3x-2
y =2
y = 12x^5-11x+3

These are not polynomials: (negative exponents, fractional exponents)
f(x) = 4/x
f(x) = 3^x (this is an exponential function)

EXAMPLE: 4x^7 + 2x^3 - 11x +5

leading term: 4x^7
Leading coefficent: 4
Degree: 7

SKETCHING POLYNOMIALS 

here are some examples: 
here is a video: http://www.youtube.com/watch?v=y8iUngkQFwg (her voice is super obnoxious, sorry)

END BEHAVIOR
- an easy way to simply sketch a polynomial is by using its end behavior
we know that simple even functions tend to look parabola like, and that simple odd functions start high and end low or vice versa.
-look at the coefficient, if it is positive then the graph opens upward. If it is negative, it opens downward. and if the exponent is positive (for simple polynomials) it is a parabola, and for odd exponents it is starting high and ending low.

Finding Zeroes On Your Calculator 



On a calculator:

plug in equation, y=
then click 2nd then calc
click 2:zero then enter
then left bound, move the cursor to the left of the zero you want to find, then click enter
then do the same but on the right side for right bound
it will calculate your zeroes for you and be displayed at the bottom of the screen.



HERE IS A VIDEO WHICH WILL EXPLAIN FINDING ZEROES ON YOUR CALCULATOR
http://www.youtube.com/watch?v=KVy3h3VUBGg

HERE IS AVIDEO WHICH EXPLAINS FINDING MINIMUMS AND MAXIMUMS 
http://www.youtube.com/watch?v=keYBfngzopo

Thursday January 17th:

Today in class, we started by going over the homework we had from Wednesday which was on page 234, numbers 13-29 every other odd. After we finished that, we took notes about graphing with factored polynomials. The first example that was put on the board was as follows:

y=(x+3)(x-7)(x+6) ---------> From this equation, we learned that the x-intercepts can be found by taking the opposite of each number and putting it in point form. In this equation, the x-intercepts would be (-3,0), (7,0), (-6,0). For a lot of the class, we were all up at the board doing examples of problems. 

The next part of class was spent going over long-division and we learned that: 
dividend/divisor= quotient+remainder/diviser
An example of this is:
x^3+2x^2-5x-6/x+1

          x^2+x-6          <---------quotient
         ---------------------
x+1 |x^3+2x^2-5x-6
      -(x^3+1x^2)
      ------------------------
                  x^2-5x
                -(x^2+x)
              -----------------
                        -6x-6
                      -(-6x-6)
                     ------------
                          0       <---------remainder

Here is a video that goes over some basic problems involving long division: http://www.youtube.com/watch?v=l6_ghhd7kwQ

The homework for Friday was on page 245, numbers 9-10.

Friday January 18th:

On Friday, the majority of class was spent learning about synthetic division. After going over the long division homework, we took some notes from the board. We learned that synthetic division is a short cut that works when dividing by an x-constant. [ x+c= x-(-c)] 
Here is an example of synthetic division that we did in class:

3x^4-8x^2-11x+1/ x-2
The coefficients are:     2⏐   3   0   -8   -11   1
                                                     6   12    8   -6
                                       -----------------------------
                                               3    6   4     -3   -5     <------remainder
= 3x^3+6x^2+4x-3-5/x-2

We also learned about the remainder theorem which means that if you plug in the 2 from the example above into the equation, you will get the remainder (-5). 
One thing that is very important to remember is that synthetic division only works in certain circumstances. You can't use synthetic division if the "x" in the denominator is squared. 

This is a video that shows an example of synthetic division: http://www.youtube.com/watch?v=bZoMz1Cy1T4

The homework for Wednesday is on page 245, number 10, 11-17 odd.
                                       

1/10/13 and 1/11/13 (Wasita)

1/10/13 - Thursday

On Thursday we went over homework (p. 136: 70-74 and another worksheet), wherein we were given formulas and had to plot and sketch transformations that matched up with the given transformed functions.

The problems consisted of the following transformations in several combinations:
*Horizontal shift
*Vertical shift
*Reflection over the x-axis
*Reflection over the y-axis
*Vertical stretch
*Vertical shrink
*Horizontal stretch
*Horizontal shrink
Here I will give an example of each transformation by itself:
Horizontal Shift:
f(x-2)
*Since whatever is in the parentheses is the opposite of what it actually is, similar to the vertices of conic sections, the ‘x-2’ means that it is a horizontal shift by 2 units to the right/positive.
*To plot this, find all of the coordinates from your given baseline formula, and then plug in all of your x-values.
*Say you had the point (1,2). Plug 1 into the f(x-2) formula. You will get 3. Since the y is not adjusted nor transformed in the new formula, the y-value, 2, stays the same. As a result, your (1,2) from your baseline formula turns into (3,2) with the transformed formula.
*All of your x-value points undergo this change. Here is a visual example of a horizontal shift by 2 units to the right: In this example, the blue line is the original function y=x^2 and the red line is the transformed function y=(x-2)^2. Vertical Shift:
f(x)-2
*Since whatever is outside of the parentheses contributes to what the actual function looks like, we do not do the opposite operation of what is given, unlike for x-value/horizontal changes.
*That means that the -2 really is a vertical shift by 2 units down/towards negative.
*If you had the point (1,2), the 2, when plugged into the formula [f(x)-2] comes out as a 0. The x-value (1) stays the same, giving you the new point (1,0).
*All of the y-values from your baseline/given formula undergo this change. Here is a visual example of a vertical shift by 2 units down: In this example, the blue line is the original function of y=x^2 and the red line is the transformed function y-2=x^2.
Reflection Over the x-Axis:
-f(x)
*Because there is a negative in front of the f, that means the formula is a reflection over the x-axis.
*This means that you take all of your y-values in your coordinates and multiply them by -1, or make them their opposite integers.
*If you had the point (1,2), after multiplying the y-value, 2, by -1, you get -2. Your new point would then be (1,-2). Here is a visual example of a reflection over the x-axis: In this example, the blue line is the original function: y=|x| and the red line is the transformed function: -y=|x|.
Reflection Over the y-Axis:
f(-x)
*Because there is a negative in front of the x, that means that the x/all of the x-values get flipped. You can also think of it as multiplying all of the x-values by -1.
*Plugging the point (1,2) into the formula would give you (-1,2).
*All of your x-values undergo this change. Here is a visual example of a reflection over the y-axis:
In this example, the blue line is the original function y=.25x^3+.5x^2 and the red line is the transformed function y=.25(-x)^3+.5(-x)^2.

***Another example of a reflection of the y-Axis from our worksheet (one that I had trouble grasping at first, actually) is:
-g(-(x+3))+4
*You do not distribute the negative (the second one/the one in front of the parenthesis that is in front of the x) throughout the formula. What you do do is understand that the negative sign in front of the x means that the x is flipped.
*The +3 after the x, as per usual, means that the x is shifted horizontally by 3 units to the left/towards negative.
*Granted, all of the other transformations act like usual (the -g means that the y-values are flipped and the +4 means that you add 4 to every y-value).
*Plugging in the point (1,2) into the equation would give you (2,2)
Here is a step-by-step explanation of the plugging in of this point:
*Your x-value, 1, is first multiplied by -1 → -1
*+3 is then added to the -1 → 2, your resulting x-value.
*2 is flipped/multiplied by -1 → -2
*+4 is then added to the -2 → 2, your resulting y-value. Vertical Stretch:
2f(x)
*The 2 in front of the f means that all of your y-values get multiplied by the coefficient 2.
*Plugging in the point (1,2) would give you (1,4).
*All of your y-values undergo this change and your x-values remain the same.  Here is a visual example of a vertical stretch: In this example, the blue line is the original function y=(sin x) and the red line is the transformed function y=4(sin x).
Vertical Shrink:
½f(x)
*The ½ in front of the f means that all of your y-values get halved. You can also think of this as multiplying all of your y-values by ½ or .5.
*Plugging in the point (1,2) would give you (1,1) Here is a visual example of a vertical shrink: In this example, the blue line is the original function y=5(sin x) and the red line is the transformed function y=2(sin x).
Horizontal Stretch:
f(1/2x)
*The ½ in front of the x means that all of your x-values get multiplied by 1/2’s reciprocal: 2.
*In other words, all of your x-values get multiplied by 2/get doubled.
*Plugging in the point (1,2) would give you (2,2). Here is a visual example of a horizontal stretch: In this example, the blue line is the original function y=x^3 and the red line is the transformed function 50y=x^3. Horizontal Shrink:
f(2x)
*The 2 in front of the x means that all of your x-values get divided by 2. *Plugging in the point (1,2) would give you (1/2,2). Here is a visual example of a horizontal shrink: In this example, the blue function is the original function 8y=x^3 and the red function is the transformed function y=x^3. We then worked on a worksheet in class, which was then assigned for homework. The worksheet gave us transformations (where more than 1 transformation occurred) and we had to graph/sketch the transformations ourselves. That assignment basically pulled together everything that we've been working on. Homework from the book, which contained similar exercises, was also assigned: p. 137: 86-89, 107, 109.


1/11/13 - Friday On Friday we went over the homework (p. 137: 86-89, 107, 109) assigned from last class. In the homework we were given the function f(x)=x^3-3x^2, which was our baseline function. In each problem, we were then given a graph which contained a function, which was the "transformed function" that derived from the baseline function. We had to identify the transformations and come up with the formulas for each graph. Now, I will do an example of such transformations, that being problem number 86 from the homework: You are given the formula f(x) = x^3 - 3x^2 (pictured left) as your baseline function. *You can see that the points from 86's graph show a vertical shrink by 1/2. (-1,-4) on the baseline function turned into (-1, -2), (2,-4) on the baseline function turned into (2,-2), and so on and so forth. *86's function is therefore 1/2f(x) = 1/2(x^3 - 3x^2) *Because the y-value is what is being halved, 1/2 goes in front of the f on the left side as well as in front of the first parenthesis on the right side. On the worksheet, in each problem we were given a graph of a function [f(x)]. Also in each problem was a transformation that we had to sketch on the same coordinate axes. These problems helped us visually see multiple transformations in one formula and helped us learn how to plot and graph transformations. Now, I will do 8a from the said worksheet: Here, I recreated the graph from 8a. The blue line represents the given function f(x). The transformation formula given was 2f(x+3), meaning we must take all of the blue line's points and adjust them accordingly. The +3 after the x means that there is a horizontal shift by 3 units to the left and the 2 in front of the f means that there is a vertical stretch. Specifically, this means that every x-value must have 3 subtracted from it and every y-value must be multiplied by 2.
Baseline Function's Points Transformation's Points (-6,-2) (-9,-4) (-4, 3) (-7,6) (-1,3) (-4,6)
(2,2) (-1,4) (4,-4) (1,-8) (10,-3) (7,-6) This is what the transformation looks like once plotted and graphed:  As you can see, there is an apparent horizontal shift by 3 units to the left and a vertical stretch, just like the formula indicated. After discussing the homework we started a graded problems worksheet in class. It covers symmetry and transformations and is due this Monday. Anyways, I hope this post helped. -Wasita Links to check out if you are somehow still confused about symmetry and transformations involving functions: http://math.kennesaw.edu/~sellerme/sfehtml/classes/math1113/transformation.pdf http://www.purplemath.com/modules/symmetry3.htm

Symmetry with Functions! 

              The last couple of classes in math we have been learning about symmetry with relations and functions. We learned that if something is symmetric over the X-Axis, the (x.y) coordinate implies that there will be a (x,-y) coordinate as well. In the case of something being symmetric over the Y axis there will be an (x,y) coordinate along with a (-x, y) point. If something is symmetric over the origin there will be an (x,y) point and a (-x,-y) coordinate.

An even function = symmetry over Y axis (to test for even plug in -x)
An odd function = symmetry over origin (to test for odd plug in -y)

A problem that might present it's self with symmetry will look something like this...

5y=2x^2-3

There are two ways to solve this, one being algebraically the other is using a calculator.

We also learned about transformations with functions
These transformations include....
Vertical shift
horizontal shift
reflection over the x or y axis
vertical stretch or shrink
horizontal stretch or shrink

Say you have an set like f(x)=sinX Versus  2f(x)=2(sinX)

Plug these forms into your graphing calculator and graph the functions.
This one turns out to a vertical stretch because of the 2's in the second equation.

This link has quality examples of what an Y-axis, X-axis and origin symmetry looks like
http://library.thinkquest.org/20991/alg2/quad2.html

It's 2013, and we're blogging!!

I'm very excited about the ideas, questions, concepts, debates, theories, etc. you will all post on this site this semester!  For inspiration, take a moment to view some of the best math scribe blog posts in The Scribe Post Hall of Fame!