Phoebe's Post for 3/7 and 3/8

I will be blogging for Thursday and Friday’s class. We couninted to work on inverse relations.  

The homework for thursday was on page 294 19,21,27 and 31
We went over

21.
f(g(x))=h(x)=x-5x+2
f(x)=x... f(x) is the outside
g(x)=x-5x+2... g(x) is always the inside

27. Find the inverse of this relation
(7,8)   --> (8,7)
(-2,8)       (8,-2)
(3,-4)       (-4,3)
(8,8)        (-8,8)

The coordinates in blue are a function, and the ones in black are not a function.  

We graphed some more functions and their inverses

                 

The red is the inverse of the green function.  

Whenever you have a repeating y value, when flipped it becomes a repeating x value making it not a function.  If you do the vertical line test on the parts drawn in red, it will not pass.  

We learned about the horizontal line test, which is when, if a horizontal line intersects more than once then the inverse will not be a function.  
If a horizontal line intersects just once the original functions is called a “one-to-one” one x and one y.                    

The first graph’s inverse is not a function, but the second graph’s inverse is a fuction.  The second graph is also a one-to-one.
Here is a video that explains the horizontal line test more.  

On Friday, We began to learn about exponential growth.  

We did an activity which involved folding a piece of paper as many times as we could.  Most of us reached 7 or 8 times.  Lisa then had us cut the pieces of paper to eliminate the problem of the creases in the paper.  We found that 2 inches = 500 sheets. to determine the height of one sheet, we did

2inches per 500 sheets=.004 inches

# of folds (cuts)                                                        Thickness

0	.004
1	.008 = .004(2)
2	.016 = .004 (2)2
3	.032 = .004 (2)3
4	.064 = .004 (2)4
5	.128 = .004 (2)5
6	.256 =.004 (2)6
7	.512 = .004 (2)7
30	4,294,967.296 = .004 (2)30

Exponential growth multiplying by the same number makes the number go up very fast.  

Exponential functions
    - have variable in exponent
    - y= abx
    the a is the initial value/ y intercept
    the b is the growth/ decay factor ( B > 0 )
If b > 1 growth
if B < 1 deacy

I am sure that we will be given more examples in class to practice, but this was all we got to in class on friday. I hope everyone enjoyed the long weekend!

Ellie's Blog for 3/4/13 & 3/5/13

Hey Guys! I'm blogging for Monday and Tuesday of this week.

On Monday, we started out by talking about function operations. These operations are addition, subtraction, multiplication, division and composition. Here's the notation for each of these operations:

Addition: f(x) + g(x) or (f+g)(x)

Subtraction: f(x) - g(x) or (f-g)(x)

Multiplication: f(x) * g(x) or (f*g)(x)

Division: f(x)/g(x) or (f/g)(x)

Composition: f(g(x)) or (f ◌ g)(x)

Next, we looked at these tables and evaluated some expressions:

For example, we can determine that:

f(3) = 5

g(4) = 3

(f-g)(4) = -11 because f(4)-g(4) = -8 - 3 which equals -11

(f/g)(2) = undefined because there is no f(2) in the table

Once we had a handle on that, we moved on to composition. 

First of all, what is composition?

Composing two functions means plugging one function into the other function or plugging the output of the "inside" function into the "outside" function. For example, if you're looking at f(x) composed with g(x) which is written f(g(x)) or (f ◌ g)(x), the "inside" function is g(x) and the "outside" function is f(x).

Now for some examples:

We looked at the functions f(x) = 6x+1 and g(x) = √x

In order to evaluate f(g(0)), first you look at the "inside" function which is g(0). When you plug 0 into the function g(0) =√x, the output is 0 so you plug this output in for g(0) so now you're evaluating f(0) which ends up equalling 1

Let's look at another example: 

f(g(25)) = 31 because g(25)=5 and f(5)=31

These are some specific examples, but to put this in more general terms:

f(g(x)) = f(√x) = 6√x+1

g(f(x)) = g(6x+1) =√6x+1

This means that if you're asked to find f(g(x)), you first plug in √x for g(x) and then you plug in √x for x in the other function as shown. The opposite is true for g(f(x)).

Then we moved on to doing some problems on the board:

This same substitution idea can be seen in the problem above

Here's another similar one:

Then we tried doing these problems in reverse order. Here's an example:

This problem was fairly straightforward because when you look at (x+1)³ you can see that (x+1) already looks like it's on the "inside" and can be plugged in for g(x) as shown above.

 And if g(x) = (x+1), then f(x) has to equal x³ in order for f(g(x)) to equal (x+1)³

Our homework was to do p. 294: 1,3,5,7,15,17

These problems are like what we did on the board and are good practice of these concepts.

On Tuesday, we talked about inverse relations. We started out by graphing 3x-7 and (x+7)/3 which are inverses of each other. The graph looked like this:

Then we talked about what makes two functions inverses.

- If two functions are inverses, then f(g(x)) = g(f(x)) = x

- If point (a,b) is on f(x), then (b,a) is on its inverse which is f ^-1(x)

We also discussed how f(x) and f ^-1(x) are symmetric over y=x, as illustrated in the graph below:

Then we had to determine the inverse of y = 1/2x - 5

In order to do this we had to first switch the x's and y's so then we had: x = 1/2y - 5

Then we solved for y and got that y = 2x + 10

Then we checked our work:

f(x) = 1/2x - 5

f ^-1(x) = 2x + 10

f(f ^-1(x)) = f(2x+10) = 1/2(2x+10)-5 = x

f ^-1(f(x)) = f ^-1(1/2x-5) = 2(1/2x-5)+10 = x

We know this is correct because if two functions are inverses, then f(g(x)) = g(f(x)) = x, as I said above. 

Alright, I think that covers what we've talked about for the past two days. If anyone has any questions or if I missed something, please let me know :)

And now for some more good math humor....

Rational Functions Continued

By Timmy Bollinger

Alright guys I am just going to pick up right where Jessie left off... In the last couple of days we have learned a couple of new things that have to do with rational equations: a short cut to finding the horizontal asymptote, and what a "hole" is and how to find its location and account for it on the graph.

The Horizontal Asymptote (H.A.) Short Cut:

               

                              Divide the leading coefficients in a rational function equation

      *This only works when the degree of the numerator is the same as the degree of the denominator*

   Ex.'s:

5x^2+3x-2

2x^2-7

For this example (^) the H.A.= 5/2 or 2.5

__x^2-9__

x^2-x-20

For this example (^) the H.A.= 1

3x^2-5

2x+1

For this example you need to put a place holder in to make the denominator's degree (x or x^1) match that of the numerator's (x^2). Adding the place holder makes the equation look like this:

__3x^2-5__

0x^2+2x+1

For this example (^) their is no H.A. because 3 / 0 does not work

However if the place holder (meaning the coefficient of 0) is in the numerator, then the H.A always = 0 because 0 divided by any other number is still 0

Identifying and Graphing "Holes" on Rational Functions:

A hole occurs when an identical factor appear on both the numerator and denominator of a graph

y= _2(x+2)(x-5)_

      (x+2)(x-3)(x-6)

As you can see the identical factor in this example is: (x+2)

This creates a hole on a graph because the equation places an intercept on an asymptote. When only the asymptotes and the intercepts are marked on the x and y axis, It looks like this:

In order to graph this equation, (x+2) must be cancelled from both the numerator and denominator.

The equation then becomes:

y= _2(x-5)_

      (x-3)(x-6)

*Even though your equation now works, you still have to mark the hole where the graph crosses a vertical line running through -2 on the x axis.*

In order to find out what the new y value of the hole is on your graph (it is no longer 0 because the x intercept: (-2,0) went away when you cancelled the factor (x+2)) you have to plug -2 in for x and solve for y. This looks like:

 y= _2([-2]-5)_

      ([-2]-3)([-2]-6)

y=-7/20

meaning the hole must be marked at (-2, -7/20)

Here is the function graphed with the hole marked correctly:

More information on Rational Functions can be found on pg 254 in the text books

-for specifically H.A. Short Cut: pg 262

-for specifically "holes": pg 267

Rational Functions - 2/5/13

We learned about rational functions. To understand what this phrase meant, we remembered that a rational number was simply a fraction with integer values in both the numerator and denominator.

Example of a rational number : 3 / 5

Rational fractions are very similar to this ration above, except the integers are replaced with polynomials in the numerator and denominator.

Example of a rational function: y =  x  /  (x+1)

If you take

y =  x  /  (x+1)

and graph it on your calculator, it will look like this (with zoom standard):



The graph shows two curves that are separated by a gap at roughly (-1 , 0). That means it’s a discontinuous function.

But we don’t always need the graph to find most of the characteristics of the function, which include:  x-intercept, y-intercept, horizontal asymptote, vertical asymptote.

I’ll use y= x / (x+1) to show how to find each. (this section can be found in our book starting on page 254)

X-intercept: Plug in zero for y. Then look at the numerator and see how you could get zero by plugging in a number for x. In this case, because the term in the numerator is only x, we plug in 0. So the x-intercept is at (0 , 0)


Y-intercept: Plug in zero for all the x terms. It would look like this : 

y = 0 / (0 + 1)

And that becomes y = 0, so the y-intercept is also at (0 , 0). By looking at the graph on your calculator, you can estimate the intercepts fairly easily, but with other function equations it’s more difficult to do so.


Horizontal Asymptote: Examine the “end behavior” of the function as it curves along the x-axis towards infinity (∞) and negative infinity. What y value on the coordinate plane is it seemingly approaching, yet never touching? If you look at the graph of y= x / (x+1), the function seems to move towards y = 1 in the negative infinity direction, and also rises towards (yet never reaches) y = 1 in the positive infinity direction. Therefore, the horizontal asymptote is :

y = 1

Another way to check this mathematically is to plug in a large number for all x-values in the function. Use a number like 1,000,000:

y = 1,000,000 / (1,000,000 + 1)    =    0.999999

This shows that the function rises towards 1 in the x direction for infinity.

Vertical Asymptote: The vertical asymptote occurs when the denominator equals zero. Plug in a number for x in the denominator to get zero:

y= x / (x+1)    plug in -1 for the x in the denominator

y = x / (-1 + 1)   

y = x / 0

This means that the vertical asymptote exists at x =  - 1

Here is a picture from pg. 264 in our book explaining the occurrence of asymptotes:

---

Homework Problems
Our homework on Tuesday was pg. 268 #1-6, and 7-10. We tried some of these problems on Monday.

9).  f(x) =   -2  /  (x - 5)

X-intercept: 

0 =   -2  /  (x-5)

*because there is no x variable in the numerator, this function has no x-intercept

Y-intercept:

Plug in zero for all x-values

y = -2 / (0 - 5)

y = -2 / -5

So the y-intercept is at (0 , 2/5)

Horizontal Asymptote:

The function appears to be heading towards positive/negative infinity in the x-direction at y = 0

Plug in 1,000,000 for all x-values

y = -2 / (1,000,000 - 5)

y = -0.000002

That is very close to y = 0

Vertical Asymptote:

Try plugging in a number for x in the denominator so that it will equal zero:

f(x)  = -2 / (5 - 5)

f(x) = -2 / 0

Because 5 works to make the denominator 0, the vertical asymptote is at x = 5

Alright, I hope this post helped you understand rational functions.

 -Jesse

Polynomial Review - Student Explain Everythings

Gleaning X-Intercepts of Functions Using Synthetic Division to Factor

On Thursday and Friday we experimented with using long and synthetic division to find x-intercepts of functions. We used problems from the handout which we were given in class on Tuesday I believe, and I will be referring to problems from it. I will also include it subsequently in this post.

To jog your memories of the remainder theorem, a shortcut which we learned early in the week, I'll go to #29 on the worksheet, the example we used in class.

Since the divisor(bottom term) is (x+3), and -3 is the zero to be derived from that, we plug in -3 as x into the dividend(top term). Don't plug it into the whole equation, just the top of the division. The number you should get(in this case -1) is your remainder.

And since I know that an ineradicable elevated speed of learning is a formidable combatant, I have embedded a video which should explain my feelings on modern organized education's interplay on our thoroughly exacerbated sentiments. If you look closely, you should notice that it is a cat.

Moving on from that, I would like to re-edify the concept of precisely calculating the occasional uncongenial x-intercepts of functions, as well as augment a sense of efficacy that we should all carry with us in our spoken rhetoric when we find ourselves referring to mathematics, as well as assist you in your quest for knowledge concerning the relationships of factoring functions and x-intercepts. To complete both of these tasks at once, assuming you all have read this far and not simply scrolled to the various cat videos which I would neither blame you for nor put past you, I will be using problem 33 from page 154 in our book.

The most intelligible and simple way to find the x-intercepts is by plugging the equation into your graphing calculator, hopefully everyone has theirs fully charged, because we are going to be doing some hardcore mathing.

With a standard window, your graph of x³ + 3x² - 2x - 6 should look like this:

Right away, we could guess that -3 could be an x-intercept, and by simply pressing -3, then hitting "enter", it will confirm our suspicions. Now, we could use our calculators to calculate the other zeros of this function, but it would only give them as decimals, and the ones that remain are not particularly copacetic round numbers. To combat this dilemma, we will divide the original cubic function by the only factor which we now know exists: (x + 3). Using synthetic division on this is the easiest way, so I will gu hold your horses I just found another cat video.

Once you complete the synthetic division of the cubic you should get (x² - 2) as the quotient. The best thing to do with a quadratic, as it turns out, it set it equal to zero. This one is rather easy; no use of the quadratic formula is required. Your answer for x is the other two x-intercepts of the original cubic function.

In addition to using synthetic division to precisely calculate zeros, we also learned what multiplicity and roots of multiplicity are. For an example, in the function f(x) = (x + 5)2(x - 4), a root of multiplicity is -5, one of the x-intercepts. For the multiplicity of -5, look to the exponent of the factor in which it resides. The multiplicity of -5 is 2. The multiplicity of 4 is 1 in this function, but when the multiplicity is 1, we usually don't say it.

To take these concepts further, lets look at problem 64 on the back of the worksheet. To determine an equation for this degree 3 function, look at its x-intercepts. even though it is degree 3, it only has two, -1 and 2. Right from there you can start building your equation. f(x) = (x + 1)(x - 2). But wait, this is supposed to be a cubic, so we need a way to fit one more x in the equation. If you look at the graph, the section where the function touches the x-axis at (2,0) looks nearly parabolic. If you try squaring the (x - 2) part of the equation, you will find that that is the missing piece to this cubic. Your final equation is f(x) = (x + 1)(x - 2)2.

Now I realize that I am currently overdue for another cat video, and knowing that I simply cannot exceed the strangeness posed by Colby "drop it like its hot" Harvey's demonstration video on synthetic division, I will embed another, featuring an old guy someone in the comments named "Gandalf on vacation". For a real treat, go to the statistics of the video, and check out the top demographics.

I think I will let that conclude my blog post, its been a pleasure helping you understa I found another cat video. Dat face.

~Austin

Synthetic & Long Division &other stuff

ayyy guys r u ready 4 sum math bloggin??!!

On Thursday we were introduced to Graphing With Factored Polynomials

y= (x-2)(x+7)(x-5) --> Degree 3 because there are 3 x's, also meaning that it will cross the x axis 3 times.
so then we found that in this problem it crossed the x axis at (-7, 0), (2, 0), (5, 0) and we were all like "daanngg its so easy to see the x-intercepts when the problem is factored!!!!!"
But just to make sure everyone remembers why we thought it was so easy, lets do it out.

(x-2)  - then we ask ourselves "hmm..what do i plug in for x to make this come out 0? OH A 2!!! "and thats how you got (2, 0).

(x+7) "hmmmm...how do i get to 0???" OH a -7!! --> (-7, 0)

(x-5) "Oh I get it, I get it, a 5 would make it zero.Wow, this is so much easier when factored!"--> (5, 0)

My x-intercepts are (2, 0), (-7, 0), and (5, 0) because when I plug them in I get 0.

Next in class we went up to the board and drew some examples:

#1

#2

#3

Then we were introduced to Long Division.

1st we recalled doing long division with numbers.

This is just a general idea of what your answer should look like. The quotient is what you get after you do the long division

And if you're having a hard time remembering how to do long division with numbers here's a video/song that will help you understand how to do it in a catchy way.

http://www.youtube.com/watch?v=R_cqrdZNmr0

Then we tied long division in with polynomials.

Here's an example:

1. x times what gives you x^2 (put that answer on top of the dividing bar.)

2. multiple the answer you got with the leading term on the outside of the dividing bar. (x time x) **You should always get the same as the first term on the inside of the bar, so they cancel out**

3. 1 times x^2 = x bring that answer down next to the x^2

4. subtract

5. repeat with other terms of the dividend

Another example:

That was the end of Thursday's class.

Friday, we reviewed a lot of long division but then we were introduced to Synthetic Division.

**** Synthetic Division only works when you are dividing by x-c [x+c = x-(-c)] ****

two key things to remember during synthetic division are:

1.write down coefficients.

2. flip the sign of the divider.

http://www.youtube.com/watch?v=fdUQuQ-AYM4

Hopefully that video helped.

Here's one last example:

ok thxxx c u thursday!! (cuz we gon get a snowday 2morro) pce out gurlz 'n boiz