Wednesday, Oct 31, 2007

Hey! Today, we didnt do Mental Math. Kind of dissapointed, its always so fun. He let us do some mental math, and while we do that, he is supposed to catch up on his marking. A couple people have troubles with some questions that came up on their Accelerated math. Simple as that, we get lead into a discussion..

e = 2.71828 2nd + LN - those are the buttons you need to press to get to e.

• 
  Here is an example that Mr max introduced:

During 1991,
140000 people visiting the Amusement Park
1995,
373 000 people visiting the Amusment Park

a) F(t)* t is x * = 14 0000 e 0.245t
b) F(t) = 140000 e 0.255t
c) F(t) = 233000 e 0.245t
d) F(t) = 233000 e 0.255t

That A-value is the y-intercept, so we know we have to get rid of C and D.

If you put in 140000 times E to the power of .245x, into the y= ..
Nothing will happen till you do something smart with your window settings.

________________________________________________

There was a type of questions that Chev brought up, here is what I caught from the discussion we had.

Endangered Species type of Question:

There is a list of information on numbers you will get with the question.. with that info, you do the following:
Remember:

Press :Stat + Enter, then you see the Lists. Put the info in!.
You load, the year into L1, population into L2, and Log P into L3.

There will be two parts of it that it'll tell you to do, a) and b).

a) Find the linear equation of the year&LogP
If you press, Stat + Calc + LinReg (ax+b) *number 4* + 2nd + L1*number 1* + comma + L3 .... enter.
b) Exponential equation of year&LogP
If you press, Stat + Calc + ExpReg* number 0* + 2nd + L1 + comma + L3... enter.
a.. LinReg L1,L3 (just to help you).
b.. ExpReg L1,L2

__________________________________________________

Just a couple reminders:

• 25 Accelerated Math Objectives due tomorrow!
• Math test on Tuesday
• Moga Madness today!
• Happy Halloween!

Lesson: Exponential Functions (non-linear functions)

"Hole puncher, where are you!? It usually comes to me when I call it."
- Mr. Max :)

We started with Mental Math. #17

Things from mental math:

x - linear (no curve)

x^2 - quadratic (one curve)

x^3 - cubic (two curves)

x^4 - quartic (three curves)

x^5 - quintic (four curves)

Pascals Triangle

- All Hallow's Day/All Saints Day (Nov 1) is after All Hallow's Eve (Oct 31)

- 7 months have 31 days

Homework Check for pg A-7 to A-9

25 objectives due thursday at 3:40

Exponential Functions

This slide is the summary notes for the lesson (exponential functions will always be in the form y=ab^x), and how exponential functions should generally look (rabbits multiply quickly over a certain period of time).

This slide is of your first example (hope you can read the example). We did two examples today. This one is about the intensity of light at a certain distance below the surface of a pond.

"You are a little frog sitting on a lily pad basking in the August sun. Something happens that the lily pad is pulled out of underneath you, and you find yourself a meter under the surface of the water."

The point of that was to 'illustrate' how the intensity of the light is less as you go deeper into the pond. If you were pulled a meter under the water you still see a lot of light around you because only 3% of the light was lost. We caught on quickly to the obvious - the intensity of light at the surface of the water is at 100%.

With this information we came up with our equation based on the question (get my drift?). We came up with I(d) = 100(0.97)^d. We chose this equation (out of the possible two/four) because it is a.
Where the 0.97 is the percentage of sunlight you see one meter below the surface of the water.
With that equation we entered it into our calculator/graphed it (substituting I(d) for y and d for x), and got something very similar to what Max has on the slide. We determined that a was the correct answer after observing the graph (at 0 meters below the surface we received 100% of the sunlight, at one meter 97% of the sunlight was showing through, at two meters we were able to see 94.09% of sun rays).

This slide is the second example. I'll 'retype' the question since you can hardly see it in the picture.
"Early in the 1900s, an airplane manufacturer was able to increase the time its airplane could stay aloft by constantly refining its techniques. Using a graphing calculator, determine which exponential equation best models the data."
With this question we are given a table with 'years after 1900' and 'time aloft'.
With the green on the slide Max isolated exponential equation and wrote "do regression!", so that's what we are going to do (with our calculators... in point form because this post is beginning to grow like the bunnies).

- Determine which piece of information/variable is dependent (y) and independent (x). In this case 'time aloft' would be dependent so it will be labeled L2. (likewise the independent variable would be L1).
- On you calculator enter your information in the stat plot (If you don't know how/forgot go to 'the old math blog' or :chevy:'s post... apparently. This is also the top left clip on the slide). You might want to turn on that stat plot also.
- After you have your plot you could use zoom 9 or the window settings that are on another clip on the slide above. There you have the plot (right above the window settings).
- To draw the exponential regression (trying hard not to call it linear regression), go to STAT, CALC, 0:ExpReg.
- You end up back at your home screen. Press L1, 'comma', L2, 'comma'. To get the Y1 (as illustrated on the slide) go to VARS, Y-VARS, 1: Function, 1: Y1, Enter.
- ENTER, and you have your a and b variables which should tell you your answer.
- To see the regression simply go back to your graph and 'see' it.

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

All info. came from inside the classroom and out of my keyboard (accept for pascals triangle which is a link in itself).
A point of interest... you can never have zero light. There will never be perfect and complete darkness.
People seemed to think my cartoon on my last post was pretty good, I'll give yea another one (which will likely end up as my signature with a smiley face just below it).

Which variable is dependent and which is independent? Where is the title for that matter?

:)

Reminder: Word Problem Stratigies (mixtures/investments)

Ok, I can see loads of people are having trouble with some of the word problems. I was having problems with the fertilizer questions until I went back in my notes and realized the antifreeze example/question we went through in class is exactly the same. Then I experimented with the other questions in the same fashion and It wasn't too hard. I'll go through the antifreeze question (that we all should have in your notes).

Ex) Antifreeze A is 18% alcohol. Antifreeze B is 10% alcohol. How many liters of each should be mixed to get 20L of a mixture of 15% alcohol?

Step 1: Create Let Statements (be specific!)
Let a = the number of liters required of the 18% solution
Let b = the number of liters required of the 10% solution

Ask yourself... "what am I solving for?"

Step 2: Create two equations with two variables in each
L1: a + b = 20
L2: 0.18a + 0.10b = 0.15(20)

The second statement is basically (based on the let statements):
type of the first solution times liters of that solution + type of the second solution times liters of that solution = type of mixture you want times liters of the solution you want.

Maybe thinking of it like that will help y'all (I'm sure that's how Mr. Max has been saying it), instead of... point eighteen times 'a' plus point ten times 'b' equals point fifteen times twenty.

Step 3: Solve
The easiest way would be to throw it in graphmatica, but I didn't have graphmatica handy so the next easiest way was within my calculator (let's just go down the line from easiest to hardest).

I'm going to isolate b (the number of liters required of the 10% solution).
L1: b = 20 - a
L2: b = (3 - 0.18a)/0.10

I got L2 this way:
0.18a + 0.10b = 0.15(20)
0.18a + 0.10b = 3
0.18a + 0.10b - 0.18a = 3 - 0.18a
0.10b/0.10 = (3 - 0.18a)/0.10
b = (3 - 0.18a)/0.10

Remember ;)
2nd, CALC, 5: (intercept), ENTER * 3

Step 4: Create a statment to summarize your answer.
In order to create 20L of 15% alcohol antifreeze, mix 12.5L of 18% with 7.5L of 10% antifreeze.

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

All raw information came from our lesson in class and out of my keyboard


:)

Thursday, October 25th

So Mr.Max started off the class by telling us that, outside of math class and teaching, he has probably used applied math to solve problems a total of five times or less. And of those five times, he has used the stuff on maximum volume, so for all of you that were feeling depressed about our little chat last week on how we will probably never use anything we learn in math class outside of school, don't lose hope. :)


We started out by looking at the problem above ▲. Our "building material" was to be 22cm by 28cm, just like a piece of paper. So to make it into a box, we had to cut a square in each corner so the paper would fold nicely into a box. [origami anyone?] To make things easy we started with one cm off each side, but when we made this into a box our class found that the box was quite "shallow". So we thought, what would happen if we made the squares bigger?! While after trying this a couple times we found that the volume of the box seemed to be increasing. But then we kept going with this idea and figured out that, eventually, the volume stops increasing, starts decreasing, and the box begins to get very tall and skinny.



Then Mr.Max showed us that we can make an equation for maximum volume and put it into a spread sheet to generate answers or use our calculators. So to get the equation we started with the formula for volume:

V=(l)(w)(h)

Then we worked out that the formula so that we could plug it into our calculators:

y=(28-2x)(22-2x)(x)

So this basically means that x is equal to the height of the box, and the height of the box is the same as the amount we are taking off each side of the paper, or building material.

First of all Mr.Max showed us how to put it into our calculators. Start by going to Y= and type in the formula. Then hit graph and you should see a cubic function. (you will probably have to play with your window settings a bit) Now to get the maximum volume of this box, go to 2nd Trace, then hit 4. Now it asks for left bound, so move left of the vertex and hit enter. Now do the same for right bound, and then press enter again when it says "guess?".
****x will tell us what the height of the box should be, y will tell us the maximum volume****

If you did it right it will look something like that▼

Then we learned how to do it by spreadsheets. We put all the information into a spreadsheet by making a columns called length, width, height, and volume. So you start off by putting in the dimensions for the first couple boxes, then you can just drag the information down because Excel lets us do that. Then using your spread sheet skills work it out so that in the volume column Excel calculates "length x width x height". So now to find the dimensions of the box with the maximum volume, just scroll down until you come across the highest number in the volume column.

Both methods shown above will produce the same answers for calculating maximum volume. This is proven below:

Friday October 26th

Today Mr.Max was away, so we had Mr.Richer as a sub. Today we worked on accelerated math and on the assignment that is due on Monday.

HOMEWORK: Student Exercise Questions: Page A-8, Exercise 2, 1-3, AND Page A-9, Exercise 3, 1a-f. ALL DUE MONDAY.

How to get to the Word Problems folder:
1) go to my computer
2) then open Rsfiles
3) open the coursework folder
4) go to the math folder and open it as well
5) then go to our class which at the moment is the second folder (Applied Math 30SA-1)
6) then you should see a folder which is labelled "word problems". Open it.

How to find the Vertex of a Parabola:
1) graph the parabola on your calculator.
2) press 2nd, then Trace.
3) then choose either minimum(cup-shaped) or maximum(cap-shaped) and press enter.
4) your screen should now show you graph again. at the bottom it will be asking you to go to the left side of where you think the vertex is. Do so with your arrow keys and then slecting enter when you are on the left.
5) now it will ask you to go right of where you think the vertex is. Do this the same as step 4 except you go to the right.
6) after you've hit enter to finalize your point in step 5, you then move your cursor to where you think the vertex is and press enter.
7) the correct vertex should then appear at the bottom of the screen. It will also state whether you used maximum or minimum.

Hopefully this helps.

*Remember that you should have 25 mastered accelerated math objectives by Wednesday October 31.*

The scribe for Monday is Shure.

Wednesday, October 24

Today Mr. Maks wasnt here so the sub told us to work on Accelerated Math and Student Exercise Questions A-7 #10-13 for the long block. Ws should have 25 mastered objectives to be working on for next Wednesday (Halloween!!!!!) The next scriber is "Patzer."

A fun fact that i would like to share is: "You cannot evenly fold a piece of paper more than 7 times."

~Today is day two which means its a shorter class of applied math. We started the day off by doing our mental math quiz number 16.

*We then did our accelerated math while Mr.Maxs corrected our tests that we took October 2nd. It took him awhile but at least hes finally getting them done.

~We were assigned no homework. We just have to work on having our 25 ojectives mastered by Halloween!

Tomarrow the scriber is Minish.

Have a great week everyone! :D

Monday, October 22nd, 2007

-Today is Day one, which means we have a long block of applied math. We started the class by finishing up the video that Mr. Max showed us on Friday on iTunes called TED Talks by Ken Robinson. The videos topic was about "Do schools kill creativity?" The video took 20 minutes of class.

-Next we dicussed about what is happening about accelerated math. We dicided that any 25 objectives will be due October 31st, 2007.

-After we discussed accelerated math, the remainding of the class was a work period.

-The scriber for Tuesday is now Duchesne.

Friday, October 19th, 2007

Sorry, I forgot to scribe right away on Friday so I did it today. This blog is for Friday, October 19, 2007.

Yesterday was block 2 and we planned on going through the questions on page A-6, #9 on Jane's Garden. But first we talked about Chev's blog from our previous class.

- We started doing the questions on Jane's Garden, but Duchesne had asked Mr. Max a question. "Why is Pre-Calc considered the better class if you learn the Applied way and the Pre-Calc way of solving problems using the technology in the Applied course?" Little did we know that it would take up the whole period to have this discussion.

- Mr. Max gave us his opinion on why he thinks Pre-Calculus is considered the better course. He told us that he thinks it is because Pre-Calculus is the harder type of math and that nobody can skim through it with no problems; there are a few people but not very many in the world. Mr. Max said that choosing between the two math courses just seperates the math students from eachother. It seperates them into two groups-- the ones who want it bad enough they are willing to do up to two hours of homework everynight and the ones who know what they are doing later in life and if it doesn't involve pre-calculus they won't take it.

- He later on told us that he thinks that each of us are capable of taking pre-calculus and that anyone can take it. They just need to want it bad enough to complete the math course successfully.

- Carlee added in on the conversation. "I think that applied math is better to take because you learn more important stuff in the course compared to pre-calculus. Like in pre-calculus you learn how to figure out things that you don't need later on in life, but in applied math you learn things you can use later on in life."

- Mr. Max laughed and said, "no you don't. You are never going to go to a hardware store and buy a bunch of wood and say Oh I think I can use the Quadratic function to figure out how to make my garden." He told us his honest opinion. Mr. Max told us that if you are to choose a math course that would benefit you, it would be consumer math because you learn more valuable stuff that you would use later on in life as opposed to the applied of pre-calculus math courses. For example in consumer math you learn about banking and mortgages.

- We also had a conversation about how fortunate our school was compared to other schools. We have 600+ students and we have so many oppurtunities and options in our school that no other schools have. We are lucky that we have many options for electives and all the technology to help us with our courses such as: the computer labs, resource centre, our own laptops, etc.

-Mr. Max then showed us a video on iTunes called TEDTalks by Ken Robinson. We watched what we could for the remainder of the class.

**Homework!-- There was no assigned homework. We were told just to try and finish the questions on Jane's Garden.

The scriber for Monday is now Delaurier.

Thursday, October 18th, 2007

To start off the day, was mental math quiz number 15. Than, our assignment for the day is Jane's Garden on page A-6, all of number 9.

-- Mr. Max first drew us a diagram of the garden, which from there was put into a table and than into a spread sheet.
-- He than taught us the "pre-calc" way of thinking. " Now since we are an applied class we can do pre-calc ways, but pre-calc is not allowed to do applied ways. Pre-calc kids will never use a spreadsheet"

** Remember regression!! We learned about Regression in grade 10.
press these buttons, once you have your dots onto your graph window.

STAT than over to CALC MENU, than option number FIVE, than ENTER 2nd 1 than 2nd 2, commas seperating them.. press VARS, go over to Y VARS, function number 1 then ENTER.

.. now practise everything we learned!
HOMEWORK: answer questions to part B, but most important one is questions C on page A-6..

scriber for tomorrow is... delaurier, but since she's not here, it's Schure

And good luck to both soccer teams in provincials this weekend! hooray.

Tues Oct 16th Math Class

Starting class today, we did mental math. When we finished, we corrected our homework from the weekend.. page A-4 questions 3,4,5. During the correction, we noticed that any quadratic function has infinite domain. This means the parabola will keep getting wider.
The answer key is found on the Rsfiles (J:) drive, Coursework, Provincial Math Booklets, S3 Applied.



Tomorrow, we are working on the Jane's Garden spreadsheet problem on page A-6. So, it is necessary to create the same spreadsheet with the appropriate columns on Excel.
For homework, make sure you're done or ready to scan accelerated math objectives 21-31 by the end of the day! If you're having difficulties with the word problems, you can recieve examples on the Rsfiles (J:) drive, Coursework, Math, Applied Math 30SA, Word Problems. Then, click on the subject that you need help on.
Wednesday's scriber is Jason.

Oct 15th Math Class

In class today, we did our mental math, and went over the blog. We were given more ideas as to how to make our posts. The first idea being, "growing our posts". This means that we make our posts so that we can continue adding to it at later dates.
The second idea is that adding labels is a good idea as well.
The third idea was that it would be a good idea to put in the due dates for our assignments, and also to remind the person who has to scribe next.
After the scribing ideas talk, we worked on Accelerated Math. As a reminder, we need 10 objectives by the end of Wednesday (3:40). The objectives are 21-30.
Bettesworth, you're the scribe for tomorrow.

Scribe for Oct.12

Today in class we started off with mental math as usual. Then we had a summary of what we learned yesterday. (I wasn’t left any screenshots of anything so I will try to summarize what was written.)
We talked about the “a” value, and how the bigger the number the skinnier the opening in the parabola,
Ex: y-k=a(x-h)^2
Y-1=10(x-(-2))^2
And the smaller the number the wider the opening is.
Ex: Y-1=-.15(x-(-2))^2

We also found a way to find the vertex of a parabola on a calculator.
By putting in an equation
Ex: Y=x^2+6+4
You enter “Y=6x+x^2+4” into you Y= screen the graph the equation.
It should look something like this:
By pressing the keys 2nd, TRACE, 3. You can find the minimum. It is the minimum because the vertex is below the x-axis. The go to the left side of the vertex when it asks for the “Left Bound?” then press ENTER. Then it will ask “Right Bound?” so you go to the right of the vertex and press ENTER. *(NOTE) It doesn’t matter how far left or right you go as long as you are on the side it is asking for*. The it will say “Guess?” you press ENTER and you will get you answer which in this case is (-3,-5).
The we worked on our Accelerated math (which is due soon) till the end of class.

Graphmatica Download

Everything in red may be unique to my computer, so you might be able to ignore it.

I'm using Firefox, so if your using some other browser it might be a little different.

Go to download3000.com

A window should automatically pop up asking if you just want to open it and with what or if you want to save it to your disc (if this doesn't happen there are links on the page called Download Link 1 and Download Link 2, If that doesn't work you/I can find you another website to download it from). If you want to keep graphmatica on your computer you should save it to disk. I will save it to disk.

A zip folder will appear on your desktop. Open it (double click).

There are two files in the folder, 1) README and 2) setup. Double click setup. Since I'm not the administrator on my computer a window has popped up asking me if I want to download the program as the administrator. I don't have the password, so I'll just continue and hope nothing goes wrong.

Now it's asking you/us to affirm where you want to save the files and how much space you need and have. I've got more than enough space (you need 501 K, I have 12740572 K). Most home computers (that I've seen) have a folder called program files in the C: drive. This is the default folder where most if not all programs are stored when you download them. I don't think I'll change it.

Start... and I can't. I guess since I'm not the administrator I can't modify the files of some of the C: drive... or something like that, so I'll save it in My Documents.

Start... and just like that, it's done

The next window has view readme file and run installed application automatically selected. If your not doing math this instant you probably don't need to open graphmatica (unless you want to play with it) and if you got this far in the setup process you don't need the readme file (thought it is nicely organized, compared to other readme files). You can delete the zip folder on your desktop now.

and on the side...
I believe (since I got excited about how easy that was and wanted to see if it really worked) a window comes up with the actual files. Know, if you put it on the desktop you are not creating a copy, that is the application. If you delete it you delete the application and you will have to uninstall the files that are on you computer and redo the setup process all over again (which is not all that hard or time consuming).

:)

thursdays class Scribe, Oct. 11th

Oral Excercise Identifying on the graph;

1)linear function(it must pass the vertical line test)

2)Quadraic functions( opens up opens down , not side ways)

3)Cubic functions(it turns twice on the graph)

4)Absolute function (looks like a check mark)

5) Sideways perabola (not a functionnopens sideways , has 2 x's)

Quadratic Functions

A quadratic function is polynomial function that operates off of the equation:

f(x)=ax²+bx+c

In this equation the terms a, b, and c are all real numbers with a ≠ zero. When you graph a quadratic function the result is a “U-shaped” curve, curving either up or down, not sideways. This curve is also known as a parabola.

The standard form of the quadratic equation is:

y=ax²+bx+c

A quadratic function uses the same equations as linear functions, except when you are working with quadratic functions, one of the terms you are using must be squared. The squared term is what creates the parabola (“u-shaped” curve).


Links:
http://en.wikipedia.org/wiki/Quadratic_function
http://www.analyzemath.com/quadraticg/quadraticg.htm
http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/P&r/quadratic/quadratic.html
http://hotmath.com/images/gt/lessons/genericalg1/parabola_width.gif

Domain and range

What is domain and range?
- The domain of a function is the set of all the stuff you can
plug into the function. The domain of a function is the set of all real numbers for which the function is defined.To find the domain, ask yourself the following questions: What can x not be? or Where does this function have an illegal operation?

- The range of a function is the set of all the stuff you can
get out of the function.The range of the function is the set of all values assumed by the dependent variable(y).To find the range, ask yourself the following question: What are all possible values of y?

http://www.math.ksu.edu/~mxw0000/final/domran.html

Let's do an example: f(x) = x^2 (that's x squared)

What's the domain? Well, you can plug any old real number you want
into this function: I can square 4, or -7, 1.01738, or whatever, and
the world doesn't blow up.

What's the range? Well, let's think about it. If I plug any number
into this function, am I ever going to be able to get a negative
number out of it? Nope! (Unless you're dealing with imaginary numbers,
and I bet you're not!) So it looks like the range of this function is
the set of all non-negative numbers (the positive numbers plus zero).
And in fact, that's the right answer.

Date: 08/25/97 at 13:36:18From: Doctor KenSubject: Re: Pre-Cal


The Domain of a Functions is: The set of possible x-values (independent variable)
The Range of a Functions is: The set of possible y-values (dependent variable)
e.g. F = { (1, 3), (2, 5), (3, 10), (4, 17)}
The Domain of the function F is {1,2,3,4}
The Range of the function F is {3,5,10,17}

http://math.clackamas.cc.or.us/kyser/math111/ch2/22stuff/sol2.htm

Definition of the Domain of a Function
For a function f defined by an expression with variable x, the implied domain of f is the set of all real numbers variable x can take such that the expression defining the function is real. The domain can also be given explicitly.
Definition of the Range of a Function
The range of f is the set of all values that the function takes when x takes values in the domain.

Domain: The domain of a function is the set of all possible input values (usually x), which allows the function formula to work.

Most often a function's domain is all real numbers. Consider a simple linear equation like the graph shown below. What values are valid inputs? Every number! It's range is all real numbers because there is nothing that won't work. The graph extends forever in the x directions.
What kind of functions don't have a domain of all real numbers? The kinds of functions that aren't valid for particular input values.

Range: The range is the set of all possible output values (usually y), which result from using the function formula.

The range of a simple linear function is almost always going to be all real numbers. A graph of a line, such as the one shown below on the left, will extend forever in either y direction. There's one notable exception: y=constant. When you have a function where y equals a constant (like y=3), your graph is a horizontal line. In that case, the range is just that one value. Otherwise, the range is all real numbers.
Many other functions have limited ranges. While only a few types have limited domains, you will frequenty see functions with unusual ranges.

Domain and Range

"Domain" and "range" are just two different words for "how far something extends"; specifically, a king's domain is the territory he controls, and an animal's range is the region it wanders through. So it makes some sense that the set of numbers a function "controls" would be called its domain, and the set through which its value can wander is called its range.1

1.http://mathforum.org/library/drmath/view/62497.htm

Domain represents all the x values, and Range represents all the y values. List the x and y values without duplication.

Ex: State the Domain and Range of the following relation.

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

Domain: {2, 3, 4, 6}
Range: {–3, –1, 3, 6}

http://www.purplemath.com/modules/fcns2.htm

What is a function?
If you can solve for “y”, then it is a function or if you can enter it into your graphing calculator. You can also tell if it’s a function by using the vertical line test. If it crosses through the graph twice, it is not a function.

1. This one is a function because there is no vertical line that will cross this graph twice.

2. This one is not a function because any number of vertical lines will intersect this oval twice. For instance, the y-axis intersects twice.
http://www.purplemath.com/modules/fcns.htm

Finding the Domain

The domain of a function is the set of numbers that you can put into the function and get out something that makes sense.

Ex: f(x) = (4+x)/(x^2-9)

(1) If we put in 4, we get f(4) = (4+4)/(4^2 - 9) = 8/7.

(2) If we put in 0, we get f(0) = (4+0)/(0^2 - 9) = 4/-9 = -(4/9)

(3) If we put in 3, we get f(3) = (4+3)/(9-9) = 7/0

You can't divide by 0, so 7/0 is an answer that doesn't make sense, and that f(x) is not defined at x = 3 and the same for x = -3. You will always get a fraction out of this function. (If you get a whole number n, you can think of it as the fraction n/1.) The only fractions that are undefined are those with 0 in their denominators.

This means that the function is undefined only at these two numbers, so its domain is all real numbers but 3 and -3.

http://mathforum.org/library/drmath/view/54554.html

Cubic Function

Cubic Function

• In mathematics, a cubic function is a polynomial equation in which the highest occurring power in the unknown is the third power.
  


• The standard form of a cubic equation is: ax3 + bx2 + cx +d=0.
  


• The cubic function equation is: f(x) = ax3 + bx2 + cx + d.
  


• The factored form of the cubic equation is: g(x)=a(x-p)(x-q)(x-r)
  


• Example:
  

f(x)= x3+2 (Green Line)

f(x)= x3+5 (Blue Line)

f(x)= x3+0 (Pink Line)

f(x)= x3-1 (Grey Line)

f(x)= x3-3 (Purple Line)

• A cubic equation has no exponent larger than 3.

REFERENCES:
http://jwilson.coe.uga.edu/EMT668/EMT668.Folders.F97/Wynne/Cubic/Cubic%20Functions.html

http://orion.math.iastate.edu/algebra/sp/xcurrent/applets/cubicfunction.html

http://www.answers.com/topic/cubic-equation?cat=technology

Definition of Function

Here are some definitions of FUNCTION:

(THESE ARE NOT MY WORDS...I GOT THESE DEFINITIONS FROM THE WEBSITES IN THE LINKS BELOW THE DEFINITIONS OR PICTURES)

• A rule for creating a set of new values from an existing set; for example, the function f(x) = 2x creates a set of even numbers (if x is a whole number).
  From : http://www.pcmag.com/encyclopedia_term/0,2542,t=mathematical+function&i=46629,00.asp
  
• A function associates to each element in the set X (an input) exactly one
  element in the set Y (the output). Different elements in X can have the same
  output, and not every element in Y has to be an output.



• 
  http://en.wikipedia.org/wiki/Function_(mathematics)
  
  
  
  

Domain and Range by Semba04

Domain is associated with 'x' and
Range is associated with 'y'
always remember that!
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A relation is a set of ordered pairs (if you have a line or points on a graph you can have ordered pairs).

ex. (x, y), (3, 5), (4, -6), (-3, -1)

Domain is all the x values of a relation.

D = {x, 3, 4, -3}

Range is all the y values of a relation.

R: {y, 5, -6, -1}

Curly brackets are used to name the ordered pairs. I believe an = and : can be used interchangeably.


A function is when the x and y variables have something common and can be found using an equation.


Vertical Line Test
If a vertical line can be drawn to intersect more than exactly one point on a relation then that relation cannot be a function because it fails the vertical line test.

This states that if a line intersects its self on the (lets say) x axis it fails the vertical line test, so it is not a function.
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The relation of this diagram is:
(1, 5), (2, 6), (3, 5), (4, 6)
D: {1, 2, 3, 4}
R: {5, 6, 5, 6}
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This is a function
D: {'infinite'}
R: {'infinite'}
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This is not a function
D: {0, >5}
The domain is larger than zero
R: {0, >6}
The range is less than six
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This is a function
D: {<-5.5,>5.5}
R: {<0,>-5.5}
________________________________

This is not a function
D: {'infinite'}
R: {'infinite'}

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This is a function (notice how the circle on the bottom is not filled and the one on the top is. If the circle on the bottom was filled this would not be a function because those two points would overlap)
D: {<-6, >6}
R: {<-6, >6}
________________________________

Links for your knowledge (or entertainment): :)
Math Demos
Free Math Help
Bellingham High School Math
Purple Math

Everyting on this post was typed by me, by myself accept the vertical line test which I took from my notes from last year.

October 4 and 5

What to do....according to my in-class discussion:

Domain and Range

The domain of a function is all the possible x values which will make the function "work" by outputting real values.

When finding the domain, remember:
denominator (bottom) of a fraction cannot be zero
the values under a square root must be positive

The range of a function is the possible y values of a function resulting when we substitute all the possible x-values.

When finding the range, remember:
substitute different x-values into the expression for y to see what is happening
make sure you look for minimum and maximum values of y

You may even want to draw a sketch of the graph.

http://www.intmath.com/Functions-and-graphs/2a_Domain-and-range.php

Domain= D=(x)
Range=R=(y)

Example: (2,4),(2,7),(5,6),(3,8)

D=(x) so the domain must be; {2,2,5,3}

R=(y) so the range must be;{4,7,6,8}



Take the graph: y=2-x



The x-intercept is at (2,0) and the y-intercept is at (0,-2)






Domain- x1,x2, all real numbers

Range-y1,y2, all real numbers

So from the graph above the Domain and range would be:

D= {2,0}

R= {0,-2}

http://block5applied.blogspot.com/

Some Links that you may find helpful are:

http://www.purplemath.com/modules/fcns2.htm

http://www.freemathhelp.com/domain-range.html

http://mathdemos.gcsu.edu/mathdemos/domainrange/domainrange.html

http://mathforum.org/library/drmath/sets/select/dm_domain_range.html

"Why are there rules for croquet" punchline

After solving for these questions and using the letters in the appropriate places, the answer to "Why are there rules for croquet?" is "SO THAT WE CAN HAVE LAW N' ORDER."

Chapter 1 Summary---

There are two techniques--
· Solve by Elimination, which is the adding or subtracting technique to eliminate the variables.
Example:  -3x+5y=11
      7x-5y=14
Multiply each equation to get the same.
     -21x+30y=77
     -25x+30y=70
Add/ Subtract to get x or y.
     -21x+30y=77
    - -25x+30y=70
     4x/(4) =7 /(4)
       x= 7/4
Substitute to get the y value.
And Since -->   3(1.75)+5y=11
        5.25-5.25 +5y=16.25-5.25
        5y/(5)=16.25 /(5)
So y now equals  y=3.25
So -->      (7/4, 13/4)

·Solve by Substitution, which is substituting the variables to get the answer in the end.
Step 1: Choose a variable in either equation to be solved for.
Step 2: Solve for that variable.
Step 3: Substitute so you can create a single equation in 1 variable.
Step 4: Use that known value to solve for the unknown by substituting the variable with the answer from before.
You will then get the coordinates for both the equations combined.
_______________________________________________________

Types of Solutions
Ordered Pair solution - Consistent system (lines intersect). The slopes are different; the intercepts are different, unless there is an intersection with the two lines.
No Solution - Inconsistent system (lines are parallel). The slopes are the same, the intercepts are different.
Infinitely many ordered solutions - Dependant system (lines are superimposed). The slopes are the same, the intercepts are different.

That was an example of what I knew how to do. I understood the elimination and substitution methods without any problems. I had a problem with the 4- Step Word Problem solving. I understood the Let a or b equal part. I just didn’t understand how to solve after that. I still have problems with these types of questions, but other than that.. It's all good:)

Linear Equations -- Review.

There isn't very many parts to this unit.. but it was still a lot to learn and take in.
The following is all what I have learned in the last couple weeks:
________________________________________________________

• SYSTEM - Two or more equations that exist in the same plane
• Here is what you can do with two lines in the same plane, there is 3 ways.

1.Parrellel

2. Intersecting

3. Superimposed

I couldnt find a picture that represented this one very well, but; it is when the lines are on top of eachother.

___________________________________________________________________

There are 2 solving methods;

• Substitution
• Elimination

Substitution - -

Step 1. Choose a variable in either equation to be solved for.

Step 2. Solve for it.

Step 3. Substitute so that you create a single equation in one variable.

Step4. Solve for that variable we get.

Step5. Use that known value to solve for the unknown value. By substituting into either original equation.

I was looking around the internet and I found this, so i copied it and pasted it onto here. Take a look, its a simple question but it shows how to do it.

________________________________________________________________

Elimination Method

Step1. Line up variables and coefficients.

Step 2. Pick a variable, depends on each case. (Pick which ever is easier)

Step 3. Use algebra to multiply and divide an entire equation by a valve to create SAME OR OPPOSITE coefficients in either variable.

Step 4. ADD OR SUBTRACT depending on wether you've created SAMES(subtract) OR OPPOSITES(add)

Step 5. SOLVE for the remaining variable, use substitution to find first variable, (once again in either question.)

Example:)

LOOK AT IT. VERY SIMPLE.

http://rachel5nj.tripod.com/NOTC/ssoewog2.html

___________________________________________________________________

Types of Solutions

1. Ordered Pair solution - Consistent system(lines intersect); Slopes, different; Intercepts, Diff unless intersection is on that intersect.
2. No Solution - Inconsistent system(lines are parallel); Slope,SAME; Intercepts, DIFF;
3. Infinitely many ordered solutions - Dependant system; (lines are superimposed); Slope,SAME; Intercept, DIFF;.

Example.

Line 1. 5x-3y=4

Line 2. 7x - 3y=4

How many solutions? ONE.

As soon as the slopes are different, there is only one solution.

________________________________________________________________________________

Out of everything we learned, I noticed that the Elimination method was easiest. Substitution took a lot of time to get used to, It was hard in the beginning, but now that I make myself go throughly through the steps, its a lot easier.

The word problems tended to create a lot of problems. I got to the Let statements, but hardly ever passed that. The joke sheet did help, but not sure how well I know how to do them yet. This test should help a lot and show me what I could and should try harder on.

Those were my notes on what I learned, hope they helped at least a little bit. The others who have posted, They helped me alot, even creating my cheat sheet thing, it helped.

Chapter 1 Math Review

The past few weeks we have been working on linear equations. The things we were taught in this unit were:
1. Substitution Method
-First you pick a variable to solve for
-Then Solve for it
-Next you substitute to create a single equation in 1 variable
-Solve for that variable
-Then you use that known value to find the unknown value

2. Elimination Method
-Line up variables and coefficients
-Pick a variable to solve for
-Use algebra to divide or multiply the whole equation by a value to create the same or opposite coefficients in either variable
-Next you add or subtract dependent on if they are sames or opposites
-Solve for the remaining variable
-Use substitution to get the variable you eliminated in either variable

I am not bad at doing either of these but i could always use more practice at them.

Then we started word problem...I am really really bad at these! I never know how to set up the equations.

First you start by doing a let statement ( Let y =__________),(Let x =__________)
Then you make your equations using numbers from the question.
Next you solve it using 1 of the 3 methods
To finish it up you make a statement summarizing your answer.

Different types of solutions:
1. Ordered Pair Solution(constant system)-The lines intersect, slopes are different, and the intersects are different.
2.No solution:(lines are paralle)-The slopes are the same, and the intersects are different.
3.Indefinitely many ordered pair solutions(Lines are superimposed)-The slopes are the same and the intersects are the same.

Overall this unit wasn't too terribly hard.

Systems Of Linear Equations

In this unit it was explained that 2 or more equations that are on the same plane are PARALLEL, INTERSECTING, OR SUPERIMPOSED.
Also the use of 3 solving methods was explained. The 3 being GRAPHING, SUBSTITUTION, AND ELIMINATION.
The graphing method uses the "y=mx+b", "Ax+By+C=0", And the "y-y1=m(x-x1) formulas, and even the use of the graphing calculator.
The substitution method has 3 steps that help you answer the equations.
Step 1 tells you to put line 1and line 2 into equations.
Step 2 tells you to solve.
Step 3 tells you to put you solution into sentance form.

ex: L1: X+Y=-1
L2:X-2Y=8

Step 1: L1: X=(-1+Y)
L2: (-1+Y)+2Y=8
Step 2: (-1+Y)+2Y=8
-1+3Y=8
3Y=9
Y=3
Step 3: Since Y=3, and since X+2Y=8
Than X+2(3)=8
X+6(-6)=8(-6)
X=2
The solution is therefore (2,3).

The 3rd method, elimination, has 6 to follow.
Step 1 tells you to line up the variables, and the coefficients.
Step 2 tells you to choose a variable and solve for it.
Step 3 says to use algebra to multiply or divide the equation by a value to create Sames or Opposites.
Step 4 explains how to either add or subtract the lines.(Sames are subtracted. Opposites are added.)
Step 5 tells you to solve for the remaining variable.
Step 6 tells you to put you solution into sentance form.

ex: L1:5x+y=16(x3)
15x+3y=48
L2: 2x-3y=3
(add)
(17x)/17=51/17
x=3

And since 5x+y=16
5(3)+y=16
15x(-15)+y=16(-15)
y=1
The solution is (3,1).

This unit also explained how to solve word problems using 4 steps, and any method you choose.
Step 1 tells you to make 2 "Let" statements.
Step 2 tells you to make 2 equations using the 2 variables you created in the first step.
Step 3 tells you to do the math by how ever you find easy.
Step 4 tells you to make a statement to summarize your answer.

Today I finished up the adobe illustrator thingy.

Unit One, Test Review

The things that I have learned in this unit are:

* 1. Systems of Linear Equations

* 2. Substitution Solving
-Step 1: Choose a variable in either equation to be “solved for”.
-Step 2: Solve for x.
-Step 3: Substitute so that you create a single equation in on one variable.
-Step 4: Solve for that variable.
-Step 5: Use that known value to solve for the unknown value by substituting into either original equation

* 3. Algebraic Solving Method (Elimination Method)
-Step 1: Line up variables and coefficients.
-Step 2: Pick a variable.
-Step 3: Use algebra to multiply or divide an entire equation by a value to create same or opposite coefficents in either variable.
-Step 4: Add or subtract depending on whether you've created "sames" or "opposites".
-Step 5: Slove for the remaining variable.
-Step 6: Use substitution to find the first variable.

* 4. Types of Solutions to Linear Systems.
-Ordered Pairs Solution
-No Solution
-Infinitely Many Ordered Pair Solutions

* 5. Word Problems
-The two rules to solving problems are:
a. Name and define two variables
b. Create
-Then solve as desired:
a. Elimination Method
b. Substitution Method
c. Technology such as graghmatica, graghing calculator, or excel.
---------------------------------
The things that I do not have the stronges skills in this unit that I need to improve are:-Understanding how to set up my equations in word problem solving.

The things that I have a strong ability to do is the substitution method and elimination method

What we've learned so far in this unit
How to solve linear equations :
subtraction method

• 
  algebraic method and elimination method
  
• graphing using technology such as, graphmatica, calculator, excel
  
• system solution methods such as ordered pair solutions, no solution or Infinite solutions to keep in mind
  
• We've learned how to solve Word Problems by using the 2 rules
  1.Name and define 2 variables

2.create 2 equations
t then solve as desired ; elimination , subtraction, technologically

• When doing word problems you must fallow the steps
  1.do let statements
  2.Create equations in 2 variables each
  3.Solve (do the math)
  4.Create a statement
  

In this unit I was comfortable with the elimination method. I’m having trouble figuring out when to use the proper solution method. In the word problems I can find the let statements but I get stuck on second and third steps. I think the main thing I should do is to sit down and analyze the questions before I do them. I just need to take my time to thourally understand the steps to all the methods.

*Robyn*

Liner Equations and Systems Unit Summarization

During this unit we learned the following methods in solving linear equations: substitution, elimintaion, and technology using a calculator, graphmatica, or excel. Personally, I am confidnet using the Substitution method over the other two methods. I am least confident using the Graphing method just because I like being able to see my work. I do not have a problem using the Elimination method; but in the case of trying to solve fractions, the substitution method may be more suitable.

Here are some types of solutions to Linear Systems:

1) To solve these equations, use the formula Ax+By=C. So use m=(-A)/B to find the slope of a line and use b=(-C)/B to find the y-intercept.

a) 2x+3y=5
3x-2y=7
- there is one solution for these equations because they have different slopes (m=(-2)/3 and m= 3/2) and the two lines cross.

b) 6x+12y=21
2x+4y= 7
- there is lots or infinitly many solutions to these two lines because they are super improved lines which means they are the same line and have the same slope and intercept.

c) 5x-2y= 12
-10x+4y=6
-there is no solution to these lines because they have the same slope, different intercepts, and parallel lines.

I sometimes have a hard time with these types of questions so this is something I have to work on. But i find, that instead of thinking of solutions, i like to think of them as coordinates. It enables me to understand the concept better.

Word Problems(not so fun)

I think that word problems are one of the toughest concepts to grasp or figure out. But, i find that if you follow 4 steps, you should be able to gather a basic idea and solution for the question.

Step 1: Name and define 2 variable
- remember to use let statements and i find (in some cases) that it helps to draw a pictures

Step 2: Create two equations using variables

Step 3: Solve using the elimination, substitution, or technolgy.

Step 4: Create a statement to summarize your answer

Each step is worth one mark. Even if you use the wrong answer but still follow the same steps, you will get part marks for showing your work.

Test Review

Definition of Linear Equations: Two or more equations that exist in the same plane constitute a system... All equations are linear. This is basically what we've been working on for the past few weeks.
Just a quick recall from last year:
slope and y-intercept...... y=mx+b
standard form...... Ax+By+C=0
y-y1=m(x-x1)
----------------------------------------------------------------------------------------------
Than from there, we were taught the Substitution Method, in a variety of different steps
Step 1) choose a variable in either question to be "solved for"
Step 2) solve for the variable
Step 3) substitute so that you can create a single equation in one variable
Step 4) solve for that variable
Step 5) use that known value to solve for the unknown value by substituting into either original equation

Algebraic Solving Method #2: Elimination Method/ Addition and Subtraction Method/ Linear Combination Method.
Step 1) line up variables and coefficients
Step 2) pick a variable to solve for
Step 3) use algebra to multiply/ divide an entire equation by a value to create the SAME OR OPPOSITE COEFFICIENTS in either variable
Step 4) Add or subtract depending on whether you created sames or opposites... for same you subract, for opposite you add
Step 5) solve for remaining variable
Step 6) use substitution

I'm pretty comfortable using both the elimination method and the substitution method. More so the elimination method, because when sometimes when I do the substitution method, I confuse myself in what variable I should solve for first.

Now for what I find most challenging..... Word Problem Strategies.
Step 1) LET STATEMENTS..
Step 2) the two equations in two variables each
Step 3) Do the math! either elimination method, graphing calculator, graphmatika, etc..
Step 4) create a sentence summarizing your answer

Word Problems to me, are terrible. When we do them as a class, they are easy, but when i take them home and trying to find the let statements is very difficult. These will be something that i'll study tonight.

The different types of solutions to Linear Equations:
1) Ordered Pairs Solution..
Consistent System (lines intersect)
Slopes: different
Intersect: different

2) No solution
Inconsistent System (lines parallel)
Slopes: same
Intersect: all different

3) Infinetly many ordered pair solutions
Dependant System (lines are super imposed)
slope: Same
intersect: Same

Anyway.... So these are a quick summary of all the notes that I've taken in class. In my opinion, this unit wasn't too difficult, but decently challenging. But I need to work on how to understand and set up the equations in the word problems.