Summaries

*Molly checked! Session Summaries Please try and follow the guidelines on writing your summaries. Avoid writing a linear "Here's what we did first, then second, and so on". The intent is that you begin to think more broadly about the ideas and find more connections; beginning to think like a teacher! So try it out.
 * Here is a sample format I'd like you to refer to when writing the summary: **


 * The following mathematical ideas were the focus of today's class:
 * • idea 1
 * • idea 2
 * • idea ......
 * The way we developed idea 1 (2, 3, etc) was ...
 * An important thing to remember about idea 1(2, 3, etc) is ...
 * Idea 3 (idea developed late in the class session) is something that we'll revisit in later class periods but we got a start on this idea by discussing ...

Session 23: Dec 1 We strated class off by talking about multiplying 2 linear functions together. the product produces a parabola.(there was a good idea about dividing the graph up into sections, but i missed how to divide the last 2 sections up if anyone wants to add it.) the problems we were looking at are y=(x+2)3x. make y=0 so you would know that (x+2) x=-2 and 3x, x=0. those are the x coordinates for the parabola. then you find the x coordinate in between -2 and 0 to find the coordinate of x for the vertex. you would plug -1 (x coordinate for the vertex) into the original equation. so y=(-1+2)3(-1). the y coordinate is -3, and the vertex is (-1,-3). The big idea here is finding the vertix /axis of symmetry for functions. we had one more problem, but this time we had to factor it out to produce the answer. so, y=5x-2x^2, if you factor it out it becomes y=x(5-2x) so you set y to 0 and must find what x needs to be. 0=x(5-2x), x outside of the parenthesis=0, x inside the parenthesis=2.5 find the x coordinate in between 0 and 2.5 which is 1.25. plug 1.25 into x(5-2x)=y, 1.25(5-2(1.25))= 3.125 the vertex is (1.25, 3.125). next we wanted to find the differences in values for any given equation. we used the equations in the worksheet on page 169. while in L1 on your calculator go to stat, ops, #7. type in x,x,-5,5,1. your values range from -5 to 5 and are increasing by 1. then go to the header in L2... type 3(L1)(L1+2). then go to L3 and go to stat, ops, delta (triangle) List. enter L2. this produces your differences in values. so we found the difference for 4 different equations and decided that from the form ax+bx^2+c, 2a=delta2. or a =delta2/2. the big idea here is if you double the coefficient of x^2, you will get the differences in values. we then talked about f(1)-f(0) for 3x^2+6x=y. for x=1, the value is 9, x=0, the value is 0. so 9-0=9. we also tried x^2+6x+9=y. for x=1, the value is 16, x=0, the value is 9. 16-9=7. we found f(1)-f(0)=a+b if you look back at the first equation we tried, 3x^2+6x=y, 3+6=9. also, 1+6=7 for the second equation. if you have the equation, how do you find a and b? this is the generic proof: y=ax^2+bx+c f(0)=c f(1)=a+b+c f(1)-f(0)= a+b that was the coolest part! Homework for next class: pg 495. investigation 2, problems 1-7 and 8.3 white worksheet we need to finish.

Margie Checked

Session 22: November 29, 2010
 * Review of Check Your Understanding, pg. 472: The main idea behind the homework was the idea of symmetry when it comes to parabolas. We discussed the idea that there are two y's for one x, and they are on opposite sides of the parabola.
 * We also reviewed the effects of a,b, and c on the graphs of quadratic equations by doing problems from the Check Your Understanding on page 479.
 * A long-time mathematics teacher raised a concern with terminology: His concern was with how teachers were allowing students to focus only on visual cues when exploring connections between graphs of quadratic functions and their equations. Next we explored the idea that some guy had (don't remember his name) and he did not like that people teachers were okay with having their students say "wider" and "narrower" to describe the shape of the graph of a quadratic, when that graph is changed by the coefficient "a". His problem was that "wider" and "narrower" are not mathematically correct, were not mathematically describing what was happening to the function but this is what is appears to happen visually. What is happening mathematically is called "vertical stretch/shrink". We looked at pg. 479 Rule 3 which was y = -x^2 + 2 and Rule 4 which was y = -0.5x^2 +2 to observe this idea. After much discussion in class, this is what we came up with (or at least what I got out of it): rule 4's y changes at a slower rate than rule 3's y, which makes the graph of rule 4 look wider, or it is shrinking vertically. So basically: vertical stretch is when you have a larger coefficient, which makes the graph look narrower. And, vertical shrink is when you have a smaller coefficient, and it makes the graph look wider. (In comprehending this idea, it really helps me to pretend that I have a rubber band around both my hands and that when my "a" value increases, the rubber band is stretching more vertically and therefore appears to look thinner. However, when my "a' value is smaller (example when it is 1/2 is instead of 1) it vertical shrinks and therefore it starts to look wider than it would it would than if my "a'' was 1 because it is not being stretched out as much. When the a is large, for a given change in x, the related change in y values are larger as compared to the parent function y=x^2, thus "stretching" the graph of the function, y=ax^2 as compared to y = x^2. This mathematical difference between "stretching" and "looking narrower" perhaps isn't such a problem for us now. HOWEVER, when your students continue in more advanced mathematics, they can struggle with advanced ideas because the ground work laid by a middle school teacher is faulty or incomplete. Hence this mathematics teacher's concern. He has to help his students "unlearn" some faulty thinking just like middle school teachers sometimes have to help their students unlearn notions such as "always take the smaller number from the larger" or "multiplying by a fraction always makes the answer smaller".
 * Problem #5, pg. 476: This was a practice problem we did at the end of class. This problem was mainly looking at factored and expanded forms of equations. And it also looked at using graphs and equations to find special points on the graph including x and y intercepts, and maximum points.
 * The y-intercept: when x = 0, what does y equal? Put in 0 for x
 * The x-intercepts: where the two points are on the x-axis so when y = 0
 * So what did we find out about finding intercepts and the forms of equations?? Big connections here I hope!
 * So what did we find out about finding intercepts and the forms of equations?? Big connections here I hope!

No homework problems today, but...
 * Don't forget that we have a small quiz on Wednesday!!!**

And fellow group members, I am probably missing a lot on the explanation of that wider/narrower thing, so feel free to add a lot if necessary!! You had essential points. I think the hassle was "why is this important to think about?" I hope I've added some info to make that a bit more clear.

This sounds good to me Sara, Hailey

Session 21 November 22,2010 Today we reviewed some homework problems from last week and came up with some 'big ideas' from them. We specifically looked at problems 4.1 and 4.2 in the CMP handout. In problem 4.1 we talked about what we would want/expect from our students when we ask them to describe the graphs they see in the problems. For example, is it enough for the student to simply state that the graph forms a parabola, or should we expect them to go deeper? In class we decided that these kind of judgement calls need to be made by the teacher. We would want them to be able to state what kind of graph is formed and describe patterns in a table, but not necessarily why this is. As college students we are expected to 'dig deeper'. Another major idea that came from 4.1 was how to answer questions using the table or equation when the answer was not given directly in the table. These questions are designed to push the students thinking, and required us to use our calculator to find the answer. By plugging the equation into the y= function, we could use the table to find the answer to the question. Problem 4.2 was similar, in that the basic steps to solving the problems were the same as in 4.1. The main ideas we focused on in 4.2 was how to compare multiple quadratic equations at once, how to find the maximum height of the jump, and how long it took for one jump. In order to find the maximum height of the jump, we did the "scroll and search" method. We put the equation into our calculators and used the table to find where the y value was the 'largest'. This just required us to find the y value that was the largest, and in this case, gave us the answer to the maximum height of the jump. In order to find out how long each jump lasted we also used the tbale feature on the calculator. In this scenario we used the starting height of the animal/person, and scrolled until we found where that same y value occured again in the table. When this happened, we simply needed to look over to the x-value in the table to see how long the jump lasted. In the case of the flea, we needed to start with 0 because the flea did not have a heoght. The basketball player had an original height of 6.5, and the frog had an original height of 0.2. One last thing we noted was that we need to be careful of what our window is set to when we graph these equations. If the window is too small we can only see part of the parabola, and this can be decieving. In our group we found that if the window is too small, it can look like a straight line, when we know in fact, that what we are seeing is part of the parabola. We also discussed how different parts of the quadratic equation can affect the graph. Here are the big ideas we came up with: y=ax^2+bx+c (original equation) y=ax^2 is the parent function, and defines the quadratic When looking solely at y=ax^2: If a>0, the parabola opens up (concave up) If a<0, concave down as absolute value of a increases, the graph gets skinnier/steeper or the concavity is getting more narrow as absolute value of a decreases, the graph gets wider/less steep or concavity gets wider the vertex is at the origin and the line of symmetry is x=0 When looking solely at y=ax^2+c: c>0, ax^2 moves up from the origin c<0, ax^2 moves down from the origin vertex is at (0, c) line of symmetry is x=0 We looked at how a and c impact the grpah in order to undertsand how all of the parts work together to make the quadratic 'work'. We also started discussing how the product of two linear factors produces a graph of a parabola. Homework: pg. 472 check your understanding Complete the chart on the bottom of page 169 on the handout we recieved at the end of class (entitled 8.3 Graphs, Tables, and Equations of Quadratic Functions). Hope everyone has a good Thanksgiving :) ~Ashley F Great Job Ashley!! Very informative!! ~Kristin J

Session 20 November 17, 2010 Big ideas P. 466 #6 P. 480 #4 ACE #5
 * V0 = initial velocity
 * We can use same variables in equations but when it has different sub scripts, it means these are not the same velocities
 * h vs. h0
 * h = height of object (can change over time)
 * h0 = the initial height of the object (does not change over time)
 * h = h0 + v0t – 16t2
 * **does not determine distance, just height**
 * h0 = height of the shooter
 * a) h = 24 when t = 0
 * plug in zero for t in above equation
 * h = h0 when t = 0
 * IN CONTEXT: h0 is always 24 no matter how much time has passed
 * b) How to figure out v0
 * h = h0 + v0t-16t2
 * 0 = 24 + v06 – 16(62)
 * 552 = v06
 * 92 = v0
 * Using equation and then looking at the table, find when h = 0 to figure out how long an object was in the air for
 * to find distance of the object, use the t from the first part ( context: t = 3.48 sec) into the given equation of d (d=70t)
 * Has a fixed perimeter, not Area
 * lengths given are only for ONE side of the rectangle
 * P = 2L + 2W
 * A = LW
 * d) given area, fixed perimeter, no side measurement, but we know length of rope.
 * A = 11.5 m2, also A = XY
 * To figure out L & W:
 * Length can equal X, while width can equal (32 – 2X)/2 or 16 – X
 * plug in y = X(16-X)
 * use table and find where y = 1.5 to figure out what X (or the length of the rectangle) is
 * Once you know X, figure out Y -> 16 – X

We need to go over P.466 # 7 on Monday No new homework, just work on unfinished homework for 11/15/10

Session 19 November 15th, 2010 BIG IDEAS! parabola y= ax(power)2(we looked at this function in class first) y= ax(power)2 +bx + c
 * The graph of a quadratic function is a curve called a
 * Quadratic functions include: y=x(power)2
 * Quadratic functions include: y=x(power)2
 * All parabolas are symmetric.
 * Time related to height can help determine the distance of an object.
 * The function for gravity(found by Galileo) is d = 16t(power)2
 * The graph of a quadratic function of (time, height) is not the actual path of an object when in flight

Mathematical Ideas: To help us develop our ideas about quadratic functions, we first used the CBR to create a quadratic function. (Definition of a Function: For every x-value there will be only one y-value. Vertical Line Test.) Go to CBL/CBR Go to Ranger Go to setup/sample Change smoothing to medium Than push start now To create a quadratic function on our calculator, you would walk forward at a quick steady pace(to keep the line smooth) gradually slowing to a stop and then, you would walk backwards slowly at first then up to the same quick steady pace. This created a quadratic function. (A quadratic function has a line of symmetry, meaning that if you put a line through the middle both sides will be mirror reflections of each other). The equation for a quadratic function is: y=ax^2 + bx + c. In investigation #1 we looked at how (time, height) are related to each other to find how far the pumpkin will travel. Not the distance traveled but the height of the pumpkin from the ground as it travels. I think you will look at distance traveled later in Investigation 1. We also looked at the graph on page 463 to determine which graph represents the pattern relating the pumpkins height to time in flight. Graph 2 represents the pumpkin shooting up at a fast rate. Graph 3 represents the pumpkin shooting up at a constant rate. Graph 4 represents the pumpkin shooting up at a constant rate, but then stops in mid-air and falls straight to the ground (Not so much a graph of height vs time but more a sketch of the path of the pumpkin). (this cannot be a function due to our rule that any x-value should only have one y-value, in this case the x-value has multiple y-values). Graph 1 represents a quadratic function because of the steady rate of the line going up than dropping back down. Remember not to confuse the path of the graph to that of the actual path of the pumpkin, they are not the same. >
 * Quadratic Functions
 * CBR Directions: Go to your Applications(Apps)
 * Punkin Chunkin Video(helped to develop our idea on this function by relating it to real life situations).
 * Page 463(Unit 7) Investigation #1(looking more at the quadratic function)
 * Page 464(number 1) We had to create a table to show the distance that a pumpkin has fallen compared to the height that it was dropped at. This introduced us to the affect of gravity on the object falling, the function we used(created by Galileo) was d=16t(to the power)2 ( this is time in seconds and distance in feet).
 * For this problem we had to find the distance of the pumpkin at a certain place in time when dropped at a height of 100ft. To do this we take the time and plug it into our function d=16t(power)2. This helped us to find the distance of the pumpkin from when it was dropped. To find how far it has before it hits the ground, we took the height at which we started, 100ft, and subtracted that from the distance at which the pumpkin has fallen so far.
 * Page 465(#2) To answer these questions we had to use the data from problem one, relating the height and time.
 * a. How is the pumpkins height related to time: Function: 100 - 16t(power)2 (starting height is 100)
 * b. We want to know the time when the pumpkin is at 10ft from the ground. Function: 90(the height of 100ft minus the 10ft) = 16t(power)2
 * c. Finding the time when the pumpkin hits the ground. Function: 100 = 16t(power)2
 * d. What happens when the height of the drop point is changed from 100ft to 75ft. For question A: the function would than be 75 - 16t(power)2; Question B: the function would than be 75 - 10 = 65, so instead of 90ft it would be 65ft= 16t(power)2; Question C: the function would be 75 = 16t(power)2.
 * Page 465(#3) The big idea for this question is, what if there was no gravity and how would it affect the pumpkins height as time changes. Also, what would be the function for this situation. For the first part we found that without gravity the function d=16t(power)2 would not be used because it represents the presence of gravity. So, as time passes the height of the pumpkin would increase and never fall(due to no gravity). This than represents a linear function: d = 90t + h(90 represents the constant rate and h will represent the starting point). For question c: you would change the rate from 90 to 120(d=120t + 20). For question d: you would change the starting point from 20ft to 15ft(d=90t + 15).
 * Page 466(#4) The big idea for this question is there are three factors that affect the flight of an object: 1. the height at which the object is released; 2. the initial upward velocity produced by the device that is launching the pumpkin; 3. the affect of gravity on an object. With these three factors you can determine the height, time and distance of an object being launched. This question relates back to question three when there was no gravity acting upon the pumpkin. Now they want to know how the function in question three changes when gravity is present. The function would change from d=90t +h TO d=90t + h - 16t(power)2. We do this because this is the function for gravity and since gravity is in affect here, we need to subtract it to find the actual distance of the pumpkin when launched.
 * HOMEWORK: Finish investigation #1 and 2( just #1) on pages 466-470; ON the pink handout do 1.1 through 1.3, and #5 and #8(ACE), than pick any other questions that you may need to work on. In Unit 7 pick any of these problems you need work on: 2,4,7,10,12,28,30 and 3

Session 18 November 8th, 2010 Hey dudes, -we started out by looking at problem 35 on page 351. we looked at a set of numbers and tried to find the relationships of the expressions b^ 1/n and b ^ m/n. we decided that the out comes would always have to result in more than zero. As long as b > 0. -next we looked at problem 16 on page 344. we had to find equivalent equations for a few problems with negative exponents. G. (4ax)^-2= 1/ ( 4ax^2 ) don't forget parentheses C.(2/5) ^ -1 = (5/2)^1 a. 4.5^-2= (1/4.5)^2 h. 5/t^3 = 5(1/t3) While this is true, I think you were to use negative exponents? Yes we swtiched 5/t^3 to 5t^ -3 since the exponent is only associated with the t variable. -Margie -after we did all these we had about 20 (I think a bit more) minutes left in class and we had to take a look (I know, such torture, we had to look at it!) at problem 1 on page 355. Remember, the other half of the class examined another problem! the question asked how many 2-digit codes can be made using digits 0-9. because you can choice 10 numbers for the first digit and ten numbers for the second digit. the answer is 10 x 10= 100. 3 digits= 10 x 10 x 10 = 1000 4 digits= 10 x 10 x 10 x 10 =10,000 -question 5 on page 358 asks if the value of 3^2 = 9 and 3^3 = 27,what is the approximate value 3^2.4 and 3^2.7 it tells you the values are between 9 and 27. Anything else we can say? Yes once you have identified that 3^2.4 and 3.^2.7 are between the square roots of 9 (3^2=9) and 27 (3^3=27), you can then draw out a number line and estimate what the answers will be. -Margie b. find the numbers to make the expressions true. I think you had to find two ways for each? square of 48 = 1 square root of 48 square of 9 x square of 4 = square of 36

please add if you need more information...... i know I'm prettttty reliable but i may have missed some stuff :) Thanks Craig!!!! ﻿We decided that a negative or positive exponent doesn't affect the negative or positiveness of the answer. ex: (9/2)^-2 = (2/9)^2  We also cleared up when to take the root of a number: if the exponent is a fraction then you are taking the root, so 27^2/3 = the 3rd root of 27 to the 2nd power  Ashleigh Checked :)


 * Margie Checked **

Session 17 November 3, 2010 We covered many important ideas that will be helpful when solving exponential equations and understanding exponents. We looked at the following table and talked about the relationship between how we move forward and backward within the table. To move forward in the table, you multiply by another 2, and to move backwards in the table, you divide by 1/2. 2 (or you multiply by 1/2). 2^-2 1/4

2^-1 1/2 2^0 1 2^1 2x1 2^2 2x2 2^3 2x2x2

We also began talking about negative exponents. If the exponent is even, the solution will be a positive number. If the negative exponent is odd, the solution will be negative. This is not quite right. The negative exponent indicates you want a reciprocal of the base; it doesn't directly suggest something about the positiveness or negativeness of the final value. Perhaps we are really talking about negative //bases// here. If so, would the statement be true then?

Yes, this is true if we are talking about negative bases this is true. If the base is negative and the exponent is odd, the answer will be odd. If the base is negative and the exponent is even, the answer will be even. -2 x -2 x -2 = -8 (3 times, odd number of times/exponent) -2 x -2 x -2 x -2 = 16 (4 times, even number of times/exponent)

Rena :)

We discussed the affects of B on a graph as well. B is what shows growth or decay and the steepness of the line. -If B is greater than one, the graph will show exponential growth, and the larger that B is, the steeper the line will be and the closer to the Y axis the line will get.

-If B is less than 1 but greater than 0, it will show exponential decay. The smaller B is, the closer to the Y axis the line will be and the steeper the line will be.

-If B is negative, the graph will be oscillating, depending upon the exponent.

Y=a(b^x)+k: We looked at graphs on our calculator to compare what K does to the graph. We found that K will create an identical line curve to a(b^x), however the curve will simply shift upwards or downwards depending on K. For example, if the equation is a(b^4), then you add +10, the curve will be identical, however shifted upwards 10 units. If you were subtract 10, the curve would shift downwards.

We also discovered that fractional exponents are roots. So therefore, if you have 5^1/2, that is the same of taking the square root of 5.

Homework: For monday, pg. 343 #12-17 (As many as you need), 20, 21, 22, Read 29, 35, 42, 44, 45 if you need the practice.

Don't forget...Quiz on Wednesday! Looks great! -Rena Christyne Checked!

= Session 16 11-1-10 = Today we went back and looked at problems we didn’t get to last week. We started on page 311 of On Your Own with problem 12.


 * We determined that you have to be careful with saying that the value of “a” is the starting value. Remember; it is where we are starting (the initial value in a domain), not where the function "starts". For problem 12, our “a” value is 700,000 which was the price Walter O’Malley paid to buy the Dodgers, so at year 0.
 * To find the value of “b” we used ExpReg. (among other techniques) First put the points we have in L1 and L2. For problem 12, we put 0 and 40 in L1 and 700,000 and 350,000,000 in L2. Then go to the homepage, click 2nd List, go left to go right to CALC, click 7 (ExpReg), and press enter. This will give you the values of a and b. We determined that b=1.168. This gives us the exponential equation V=700,000(1.168^t) (It doesn't hurt to think about using the homescreen and an explicit view of exponential growth by using the constant function or the ANS function; it takes a little more time but may make connections on what ExpReg is finding).


 * An important thing to remember: annual percent growth rate= 17% not 117%. 17% is how much you are increasing each year. It doesn’t tell you the total amount that you have. Therefore the worth after 1 year= 700,000(1.168^1). This can be written out as 700,000 x (1 + .168). The 1 represents the 100% of what you started with (in this case the 700,000) and the .168 represents the increase each year.

Next we worked on problem 23c,d where we looked at determining how many years it would take for $5,000 to double if it earns interest that is compounded annually at a rate of 2%, 4%, 6%, and 8%.

Then go to the table and look for when y=10,000
 * An important thing to remember: Rule of 72- The product of the interest and years should be close to 72.**
 * -This only "works" when doubling your initial investment. Don't you wonder if there is a rule for tripling the initial investment?? **
 * -For the problem: (2,36) 2x36=72 **
 * (4,18) 4x18=72 **
 * (8,9) 8x9=72 **
 * (interest rate, # of years) **


 * Lastly we looked at Exponential Properties and discovered:


 * (a^m)a^n=a^m+n If you have the same bases, you add exponents **
 * (a^m)b^m=ab^m If you have the same exponent, you multiply the bases **
 * (a^m)^n= a^mn If you have a base risen to a specific power, which is raised to another power, you multiply the two powers **
 * a^m/a^n= a^m-n If you divide the bases, you subtract the exponents **

We ended with looking back at our M & M data, trying to create a rule to model it. We will work on these again on Wednesday. Our next biggie quiz has been moved to Wednesday the 10th. Dr B added stuff
 * Great job summarizing Kristin :)**
 * I think that you covered everything really well and hit the main points that we determined were important. Looking back at this now, it really helps me to review what we learned/went over. Great job! ~Ashley F **

Session 15: 10/27/10 pg.298 Check your understanding, check point to make sure that groups were ok What was the main idea behind the "check your understanding"? Someone please write that here! we discussed what makes sense in the context of the problem? Should we say only whole days, or have partial days included?

from the graph, we think that it looks exponential because of the curve





>>

1. Reminders: You can discuss your quiz with Dr. Browning at her office hours. Inequalities: When you multiply/divide by a negative, it causes the sign to flip. We still haven't proven that. but we'll trust it and use it.

13d: pink packet p.27 - first discussed A and C. We cannot find exact for C because we do not have enough info. For part D we're trying to find the growth factor using the graph. To do this we need to take points (1,20) and (2,40) and realize that the lizards are doubling, therefore there is a growth factor of 2. Then to create the equation we use the y-intercept to find the initial population (10). NOW-NEXT form for exponential functions Some questions that came up: Are they the same equation? What is meant by starting value? Is the y-intercept the starting value? Domain: the independent values that make sense in the problem example: you would not use partial generations Larger growth factor means getting steeper faster

Hailey Checked :) > Session 13 10/20/2010

We started off our discussion with problem 22 (page 13) of the first packet of pink sheets on exponential growth. We decided every half hour the amoebas split into two new ones. 1/2 hour = 2 ambeos. 1 hour = splits into 4 ambeos. Quadruplets every hour. NOW NEXT RULE: Next = Now * 4. starting at 1 amobea. In the calculator: 1 *4. Then press *4 and keep pressing enter until you get to 8 hours. Equation is: A=4^t Does this fit to the equations A(B^x)?

In the calculator go to the table set function. TblStart = 8. DeltaTbl= 1 10 hours in the table shows as 1.05x10^6

For question A we will plug 0 in for d because we haven't started the experiment yet. This will give us the answer of 50mm.

EX: 3^-3 = 1/27

Next= Now*2 Starting at 25 A changes shift the graph, B changes will change the overall graph A is the start value

Great Job Chelsea! You have got a lot of valuable information and it is easy to follow!! ~Kristin Chelsea this is great! I agree with Kristin that it is very easy to follow, and was helpful to refer back to after class :) ~ Ashley

Inverse proportions (y=r/x) Deductive Reasoning Now/Next Formulas for ax + b, how to interpret a and b in a context of a problem Greater than, Less than, Greater than or equal too, Less than or Equal too

We used a number line to help describe this. Using a positive and negative number line, the farther you go away from zero on the postiive side the bigger the number gets, and the farther you go away from zero on the negative side, the smaller the number gets. so the distance away from zero is getting bigger/smaller on the right/left side (respecitively). we then performed a few tests on our theory with b-a is greater than 0. AHE: we were asked to validate it for all cases.



put y=96 + 2.1x into the calculator tablestart 0 xmin = -5 make sure the plots are off or you are tracing the function you want. press the up or down key to get the correct function if you are tracing your plot. change the table interval to 1 and start at y given. CMP Page 58

Page 58d Using the table solve both for y - because it makes them equivalent y = 5/2 - x and y = -2 + x/2

Great Job Jess! Very informative! -Courtney Awesome job! Love the organization! -SaraThanks Jessica! This was super helpful & I appreciate the time you spend being so clear! - Hailey

= =

We started out the class with an announcement that we have an Exam (Biggie Quiz) over Linear Functions on Monday October 18. We will have just over 1 hour to complete the test, from 4:30-5:40pm. In addition, to figure out our grade so far you need to do what points you have earned so far/total points possible. However, by the end of our class the grades will be weighted based on the percentages stated in the class rubric. = =

-When you have a Linear Regression problem (ax +b), how are you going to interpret the a and the b in the context of this problem? (*Could be a possible TEST question) a = for every 1000 feet added, the temperature drops 3.61 degrees Fahrenheit b = at the altitude of 0 feet, the temperature is about 63.41 degrees Fahrenheit (remember to put the ABOUT and not exact)

What inequality did you have for this problem?



600 > 1.25s + .75c (note: it is really less than or equal to) -.75c -.75c 600 - .75c > 1.25s 1.25 1.25 (divided by 1.25 on both sides)

480 - 0.6c > s (note: is really less than or equal to) My goal is to isolate s by itself in order to find what it will be. So I started by subtracting .75c from both sides, so the sides will remain balanced. Since I subtracted .75c-.75c, it becomes an additive inverses which means their difference will equal 0. Then I divided 1.25 from both sides so the sides will remain balanced. Since when I divide 1.25s by 1.25, it becomes a multiplicative inverse which means their difference will also equal 0. Therefore I found that s < 480-.6x

Then in using our calculators we put in- Y = 480 - .6x We went to the / before the Y= and pressed enter one time until we got the less than section blinking (you can press enter two times in order to get the greater than sign) Once that was complete, we adjust our window and the graphed our inequality.

= = We did a calculator poll on what each person answered for questions #1, 3, 5, and 7 and came up the correct answers as 1. B, 2. C, 3. E, 4. F To check which way your sign of greater than or less than should be facing: = = = = 1. No because if we plugged in (2,12) to our equation of y < 3x +6, we would see that- 1. 12 < 3(2) + 6 2. 12 < 6 + 6  3. 12 < 12, this does not work because 12 is not less than 12, it is equal to it but we do not have a less than or equal to sign, we just have a less than sign and therefore these coordinates would not satisfy our inequality. = = = = How to graph the inequality of 0.75c + 1.25s < 400 and s < 3c

Y2 = 1/3X = = We know that our variable t = time as in number of years We put your equations in Y= then [2nd] [Table}, go to [2nd] [TBLSET], change Tbl=10, then back to [2nd] [Table]. Go to Y1 and press [ENTER] and it will show you what y will become with your equation in Y= = =

We began by spending time on the homework problem #3 of Looking Back and Looking Ahead. We discussed whether or not the relationship was linear by relating it to the formula Distance = Rate x Time. We substituted the information from the problem to get 5000 meters = Speed x Time. Since we were looking for an equation relating speed to time we came up with Speed = 5000/Time. This equation we compared to the generic inverse proportion equation y=r/x and concluded that the line was an inverse proportion. We briefly reviewed the graph, table, and generic equation of direct proportions. It was noted that, should we be asked to explain how we know a linear table is linear, we should indicate that the rate of change for consecutive inputs and outputs remains the same. The remainder of the class was spent working with the data on page 165 of the Core Plus packet (Unit 3, Linear Functions). We began by creating lists named ALT (altitude) and TEMP (temperature) with the data on page 165.



To name a list: Select the header of an unnamed list, press [2nd] [TEXT] and enter the name. We then plotted the data (Remember to change your Xlist and Ylist to ALT and TEMP.) and manually fitted a line.

To manually fit a line to data: Press [2nd] [STAT], scroll to CALC, select 3: Manual-Fit. Next, create a list of predicted temperatures (PRED) using the equation of your manually fitted line. For example, in the header of this list you would enter something similar to: PRED= -3.61ALT + 64.05 and press [ENTER]. Now you have predicted temperatures based on your manually fitted line. Once we had predicted temperatures we calculated the absolute error between the actual and predicted temperatures. To do this we created another list (ERROR). For this list, in the heading should be entered: ERROR= abs(TEMP - PRED).

To find absolute value: Press [MATH], scroll to NUM, and select 1:abs(. The final step in determining the accuracy of our manually fitted line is to determine the mean absolute error. This can be done on the main screen of the calculator. Press [2nd] [STAT], scroll to MATH, and select 3:mean(. Next, press [2nd] [STAT], and select your error list. The closer this number is to zero, the more accurate the line is. We then determined the most accurate line within our groups and then shared them with the class. We spent the majority of the class working on this exercise and at the end of class we learned an alternative method to find a line much quicker. To have the calculator fit a line to your data, go to the main screen, press [2nd] [STAT], scroll to CALC, select 5:LinReg(ax+b). The linear regression is automatically set as L1 for X and L2 for Y. If you wish to use other lists you must indicate so by listing them in X, Y order before pressing enter. For example, to do a linear regression for this data, the full sequence would be: [2nd] [STAT], CALC, 5:LinReg(ax+b), [2nd] [STAT], ALT, [, ], [2nd] [STAT], TEMP, [ENTER].Linear regression finds the best fit line for a set of data points. Essentially, what the calculator does is the entire process we just went through multiple times until it determines the closest fitting line.

To plot this line over your data: Press [Y=], [2ND] [VARS], scroll to EQ, select 1:RegEQ.

HOMEWORK (due Wednesday):

Read through Investigation # 5 (p. 69) of the CMP handout (has an airplane on the front). Make sure you can answer every question asked in that investigation.



I think the homework changed. I have written down your number 1, the investigation 5 from CMP, but from the core plus packet I have written down read p.186, then on p.188 do Investigation 1 #1-3 because we ran out of time during class. EB

Homework: None! Just review what we already have had. Ashley Checked!

Session 6 9-27-10 we started by reviewing proportional relationships:





Next we looked at Look Back/Ahead #2  a. we made the graph on our calculator by plugging the numbers into LIST and creating a scatter plot. You could see on the graph how a linear line could be made so we don't want to connect the dots but instead draw a line through and see where it hits the y axis or the x axis.  b. we needed to find a linear equation that fits for y=mx+b with this problem. So all we had to do was pick two points that seemed closest to the linear line we drew (points from the table). (x1, y1) (x2,y2) We picked (45, 3.4) & (55, 3). We have to take y2 from y1 then x2 from x1 so:



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>> >>

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Margie checked, looks great Molly! Ashleigh Checked :)  Session 5 9-22-10  During this class period we discussed Investigations 1- 1e and 2d.





Christyne Jarosz-checked Session 4 9/20

//__Summary of Direct, Inverse and Non-proportional Relationships Chart.__//

//__Task 1__// **from the Candle worksheet as an example of a** //__Direct Proportion__//**. With a Direct Proportion it's important to remember that the equation will be set up like so...**//__y = rx__//

== and when making a chart the r and the x will both increase or decrease at constant rates. The graph for this type of equation will be linear and in class we discussed how the straight line must pass through the origin of the graph. ==

//__Task 3__// **to explain the** //__Inverse Proportion__//**, which could be represented by** //__y = r/x__//**. This task could be classified as a** //__'missing value'__//

problem. We also talked about how the graph will be curved instead of a straight line.
//non-proportion////__y = mx + b__//

__//Task 5//__ **on the Candle worksheet. With this problem we graphed two linear lines, both of which need to go through the origin of the graph. Each candle is burning at a different rate, so the lines are not parallel. We also discovered that finding the** //__unit rate__//

The homework assignment for Wednesday is
Courtney checked :) jessica checked. Sara checked too! Session 3 9/15