Friday, May 31, 2013

Weekend Homework!

Create a rule for an arithmetic sequence
Find the first 4 numbers in your sequence
Write the formula using your rule and use it to double check your four numbers in the sequence
Write a short description of a real life situation that might fit with your formula / rule

Do the same for a geometric sequence 

Thursday, May 30, 2013

Sequences - Geometric and Arithmetic

Zombies.  Yes, zombies.  Below are the two data sets for the zombie virus cases in the cities that are reporting sequential data.

London Tokyo
Day Incidence Report Day Incidence Report
1 3 1 4
2 4.5 2 8
3 6.75 3 12
4 10.125 4 16
5 15.1875 5 20
6 22.78125 6 24
7 34.171875 7 28
8 51.2578125 8 32
9 76.88671875 9 36
10 115.3300781 10 40
11 172.9951172 11 44
12 259.4926758 12 48
13 389.2390137 13 52
14 583.8585205 14 56


Do I even have a sequence?
When we're working with sequences, we first need to determine if it is actually a sequence.  This step is covered if you can find a pattern in the data.  In other words, what is your rule?  If you can't find a rule to get from one data point to the next, then you don't have a sequence.

Arithmetic Sequences
Next you need to determine if the sequence is arithmetic or geometric.  Arithmetic sequences have a rule that deals with addition (think basic "arithmetic" is adding...).  In our zombie case, we have the data set from Tokyo that has a rule with addition.  Specifically, our rule is to "+4", meaning we add 4 to a data point to get the next data point.
We can also plot the data on a graph.  If your graph shows a linear function, and you can find that rule, you're going to have an arithmetic sequence.  Below is a picture of the graph of our Tokyo data.



We can use our rule to find other terms in the sequence, but if we wanted to know how many people would be infected with the zombie virus after 30 days, that would be a lot of computation.  Instead, we can use the general formula for arithmetic sequences and tweak it to fit our data.

General Formula for Arithmetic Sequences:         y = a + (x - 1)d
     a = initial Amount
     x = the value of the independent variable we want to measure (for example, if we're looking at the Tokyo data and want to find how many will be infected after 30 days, our x = 30)
     d = the common difference (this is your rule - what are you adding to get to the next number?)

In the case of poor Tokyo, there were initially 4 people infected.  This makes our a = 4.  We also know that our rule is to "+4".  This makes our d = 4.  Guess what?  We have everything we need to make the general equation fit our data.
y = a + (x - 1)d
y = 4 + (x - 1)4
But does this equation actually work?  Below is the graph of our Tokyo data with that exact equation superimposed over it.  


Pretty decent fit, wouldn't you say?

Geometric Sequences
Geometric sequences have a rule that deals with multiplication (something we do a lot of in geometry...).  In our zombie case, we have the data set from London that has a rule with multiplication.  Specifically, our rule is to "times 1.5", meaning we take a data point and multiply its value by 1.5 to get the next data point.  

We can also plot the data on a graph.  If your graph shows a curved function, and you can find that rule, you're going to have an geometric sequence.  Below is a picture of the graph of our London data.  




We can use our rule to find other terms in the sequence, but if we wanted to know how many people would be infected with the zombie virus after 30 days, that would be a lot of computation.  Instead, we can use the general formula for geometric sequences and tweak it to fit our data.  

General Formula for Geometric Sequences:         y = a * r^(x - 1)
     a = initial Amount
     x = the value of the independent variable we want to measure (for example, if we're looking at the Tokyo data and want to find how many will be infected after 30 days, our x = 30)
     r = the common ratio (this is your rule - what are you multiplying by to get to the next number?)

For London, there were initially 3 people infected.  This makes our a = 3.  We also know that our rule is to "times 1.5".  This makes our r = 1.5.  Guess what?  We have everything we need to make the general equation fit our data.
y = a * r^(x - 1)
y = 3 * 1.5^(x - 1)
But does this equation actually work?  Check out the graph of our London data with that exact equation superimposed over it:

That's a pretty good fit.  


I'm still confused....:
Here are a couple of links that walk through some examples of sequence problems!


Wednesday, May 29, 2013

the TI-nspire

After some searching, I found a tutorial for using the TI-nspire to plot our zombie virus data.  Tomorrow we'll go over everything discussed today, and I'll show you the data plotted under the document camera so you have a chance to see them.
Sorry about the mix up today - hopefully everything will be cleared up tomorrow!

http://mathbits.com/MathBits/TINSection/Statistics1/ScatterPlot.html

Also, on Friday we're going to be going to the library to use the computers.  We'll talk a little bit about the program tomorrow.

Thursday, May 23, 2013

Our general exponential growth and decay functions:

Growth:  y = a(1 + r)^t

Decay:  y = a(1 - r)^t

Remember, a is our initial Amount, r is our Rate of change, and t is for Time, Toss, or whatever parameter is changing.

Example:

nA microbiologist discovered a new strain of bacteria that can double its population every half hour.  There are 100 bacteria in the colony to begin with.


Write a formula to model this situation
     This is exponential growth.  Our initial amount is 100 bacteria, and we know that every half an hour our population is doubling.  This means it is increasing by 100%.
               y = 100(1 + 1.0)^t

Find the number of bacterial cells after 0.5 hour, 1 hour, and 1.5 hours.
   Our time 0 (t = 0) is before the first set of doubling occurs.  Since we're working with half hours, our t = 1 value will mean that 30 minutes have gone by.  t = 2 indicates an hour.
               y = 100(1 + 1.0)^1 = 200 bacteria
               y = 100(1 + 1.0)^2 = 400 bacteria
               y = 100(1 + 1.0)^3 = 800 bacteria

Double check - do these numbers for bacterial count make sense?

How many bacterial cells were there after 45 minutes?
    For each t value, half an hour passes (30 minutes).  In order to find the population of bacteria at 45 minutes, we have to decide what our t value will be.  45 minutes is exactly half way between 30 and 60 minutes, or t = 1 and t = 2, so we need to evaluate the expression when t = 1.5
                   y = 100(1 + 1.0)^1.5 = 282.84 bacteria 
Since we're not counting partial bacteria cells, we would round to 282 bacteria cells.  


Don't forget the word problem page and the exponential functions page.  We'll be going over them on Tuesday.
Have an awesome 4 day weekend!

Links:
Video Example
Extra Problems

Wednesday, May 22, 2013

Exponential Growth and Decay Notice

Great news!  
At this level of Algebra, you won't need to understand how to simplify equations using the natural log (ln) function.  Eventually you will be able to use logarithms to solve problems where you need to find the exponent value, but for now we can guess and check to get approximate answers.

In class tomorrow (Thursday), we'll go over what you need to be able to do for exponential growth and decay functions, and then we'll do an introduction to general exponential functions and their graphs and I'll update this site with the notes for that section.  Whew!

In the meantime, if you're interested in the natural log, and the number e (even though it WON'T be on the EOC), here are a couple of sites to explore....

Natural logs and exponents

What is the natural log?


Thursday, May 16, 2013

Trinomials: ax^2 + bx + c

This is the second form we come across with trinomials:

                    ax2 + bx + c

Multiply your a value with your c value
Use a t-chart to help you find the factor pairs of that number.
Find the factor pair that then adds up to your b term

Group like terms together and pull out the GCFs.  Then rearrange!

Example:    2x2 + 3x - 9

2*9 = 18




  Our middle term is +3, so we're going to pick the factor pair (-3, +6).

  
Now we write out the middle number so that it is expressed by the factor pair:
      2x2 + 6x - 3x - 9
Then we group the first and last two terms together:
   (2x2 + 6x)(-3x - 9)
And pull out the GCFs
2x(x + 3) -3 (x +3) -->  (x + 3)(2x - 3)

Voila.  Don't worry - there's a video attached.

Links:

Video - factoring ax^2 + bx + c
Extra Practice Problems

Wednesday, May 15, 2013

EOC REVIEW!!!!!

Alright!  As requested, I have included some links that will help you study for the up and coming EOC.  We know how nerve wracking these tests can be, so we made sure that we have time at the end of the quarter to review the material you learned this year.  If you want to get a head start on studying (an excellent idea, hint hint), the following two links are good places to start.  I'll make a review sheet to pass out to everyone and add that to this site too after it's finalized.  

Washington State Standards 
   This link includes the Washington state standards for Algebra.

Washington EOC Practice Test
   This link goes to a Washington EOC practice test.  There are other practice tests out there, but different states have different standards.

Monday, May 13, 2013

Trinomials: x^2 + bx + c

When we're looking at a trinomial, it will show up in one of two general forms:

x2 + bx + c
OR
ax2 + bx + c

When we have the first form (there's an invisible 1 in front of the x2) remember to look for the 'signals':
    * If there's a negative in front of the c value, we know that the factor pieces are going to have DIFFERENT signs - if c is positive, then the factor pieces are going to have the SAME sign
    *  Look at b -->  if b is positive, then the larger factor will be positive - if b is negative, then the larger factor will be negative.
    * Make your X-Factor.  The product goes on top, and the sum goes on the bottom.  The sides are used to find the factors of the top that add up to the bottom (don't forget to keep those signals in mind!)
Ex:  x2 + 6x + 8
X-Factor
We know that the factor pairs of 8 are (1, 8) and (2, 4), and that the factors have to add up to 6.  Which one does that?  (2, 4)

From here, we combine our signals and our factor pair to find the following:
          (x + 2)(x +4)   <--- FACTORED FORM
    To double check, FOIL:   x2 + 4x +2x + 8
  and combine like terms:  x2 + 6x + 8  <-- is this the polynomial we started with?  Yup!  Our factored form is right!


Extra Practice with x2 + bx + c
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Factoring%201.pdf

Sunday, May 12, 2013

Difference of Squares



In General, 
        when finding the factored form of a Difference of Squares problem...

Expression:  a2 – b2

Factored Expression:  (a + b)(a – b)


Links for Extra Help:


Pulling Out the GCF


Steps for Pulling Out the GCF:


1.  Find the GCF

2. Find What’s Left After You “Pull Out”  the GCF (divide the GCF from each term in the polynomial)

3.  Combine the GCF with the leftovers in parentheses

Links for Extra Help: