Posts Tagged ‘scribe post’

Piecewise Functions

Amol
Sep 9 |22:11

As we walked into class today, the daily problem on the board was a bit different from the ones we have had in the past. It was different because in the problem, two new operations were created: up arrow(ua) and down arrow(da). The problem was if a ua b means ab and if a da b means b(radical symbol) a, what is the value of [(2 ua 6) da 3] ua 2?

To simplify this equation, one would replace the arrows with the given operation. By replacing the arrows with the operation they represent, the expression reads (3(radical symbol)26)2. Now we simplify everything in the parentheses first. 26=64. 3(radical symbol)64 = 4. This just leaves us with 42 which equals 16. So the final answer to this problem was 16. The class worked well and efficiently and even put the problem up on the board to explain to the students who did not understand the problem.

The addition of two new operations made this problem confusing to most of the class. However, most people understood the problem after it was explained and drawn on the board.

After the practice problem was explained and talked about in a class discussion, Mr. Bieniek passed back our assessments that were taken at the end of yesterday’s class. The assessment was about whether or not a given set of inputs and outputs was a function.  The function’s inputs, a, b, and c all led to the same output, 1. We eventually came to the conclusion that the set of inputs and outputs was, in fact, a function. We referred again to the soda machine example. Mr. Bieniek explained that a soda machine can have multiple buttons(inputs) that will get you the same soda(output) and it will still “function” properly. However, if  a button has two outputs or two possible soda outcomes, the soda machine will not function. A question that came up was that if one input did not have an output, could the system be a function. The answer to this was no. In the soda machine example, it was said that if a button does not work and doesn’t give you an output or soda, the soda machine is “not functioning” properly. It was also noted that this function would create a horizontal line on a graph.

We also discussed what domain and ranges are. A domain is all the possible inputs or x-values. A range is all the possible outputs or y-values.

Next we were introduced to piecewise functions. When Mr. Bieniek put the first piecewise function on the board, much of the class was intimidated by it. The first piecewise function we received was:

f(x)= (x+4)2-2       If x is less than or equal to -2

-x                    If x is greater than -2 and less than or equal to 2

-(x-4)2+2      If x is greater than 2

Mr. Bieniek said that the best method to solve this problem was by making a table and then graphing it. He stressed that this was not the best way necessarily to graph a piecewise function all the time. The table that we made in class can be viewed by clicking on the link below

http://public.iwork.com/document/?a=p1017344002&d=Untitled.numbers

So now all what was left to do was to graph the points here and connect the points depending on the inequalities.

Next, we received another piecewise function. However, this time the class was told by Mr. Bieniek to try and graph this function without making a table. The piecewise function was as shown below.

f(x)=x2 If x is less than 2

f(x)=6               If x=2

f(x)=10-x         If x is greater than 2 and less than or equal t0 6

The class graphed the equation fairly easily. After, we participated in a class wide discussion about the function. It was found that the f(x)=6 part of the function was confusing but after it was discussed, everybody understood it meant to graph the point (2,6). Also, there was controversy on whether to connect the parabola created by f(x)=x2 to the line created by f(x)=10-x. The line would go through the point (2,6). After a discussion, it was determined there should not be line because it would be a vertical line, meaning that two or more inputs would have the same output. This is not possible in a function so it would be illogical to connect the parabola and the line

After this, we received two piecewise function problems for homework and class was dimissed.

For help with piecewise function, see the following sites

http://www.analyzemath.com/Graphing/piecewise_functions.html- A site that gives tutorials for piecewise functions

http://www.youtube.com/watch?v=Kac2I5ao49E- A video about piecewise functions

http://www.prenhall.com/divisions/esm/app/graphing/ti83/Graphing/Special_graphs/piece_wise/piece_wise.html- A site that shows how to graph piecewise functions on your TI calculators
 

Functions

Kate
Sep 7 |16:49

Hey everyone! Today we started off class with this practice problem. Mr. B said it would be good to refer back to these problems to study, so I would make sure you have this in your notebook and understand how to solve it.

We were given: http://prec.alcul.us/?attachment_id=2367

And had to substitute it in the following function: http://prec.alcul.us/?attachment_id=2368

To solve that, we learned that you were to put the (5+h/h) and (5/h) in place of x in the function shown here: http://prec.alcul.us/?attachment_id=2369 | http://prec.alcul.us/?attachment_id=2371

The goal for today was “I can tell the difference between a function and a non-function” We are going to have a different goal everyday, so be sure that by the end of each one, you fully comprehend what’s going on.

A function is simply: A special relationship between two variables. But as I looked online, it gets more specific saying that,  it’s a relation between two sets in which one element of the second set is assigned to each element of the first set, as the expression y  = x 2.  basically stating that, for every ‘y’ value, there has to be 2 ‘x’ values because when you square a number there is a positive and negative answer.

We also sorted through some graphs to see which we thought were functions, and which were not.

I’m sure tomorrow we’ll go into more depth about functions and whatnot, but this is what we have so far. Also, I’m sorry the pictures of the math type equations are links… It wouldnt let me copy and paste them here, or let me phost them as pictures on the blog, but I’m sure you all understood just fine.

-Katy

The scribe for tomorrow is Brittney.

 

Transformations from Graphs

Student 09_10
Sep 26 |11:18

For most of class on Friday we worked on completing worksheet 1-3c, which is on transformations from graphs.

Given each pre-image and its image on a coordinate plane, we were asked to identify the transformation of f(dotted) to g(solid).  We had to figure out the transformation both verbally, and then write the equation using g(x) and f(x).

Question 1

 1

 a) verbally: vertical translation of -6       b) equation: g(x)=f(x)-6

Since the image is 6 units lower than the pre-image and only the y-values changed between the two graphs, the modification to f(x) is going to be outside the parenthesis.

Question 2

 2

a) verbally: horizontal translation of 10       b) equation: g(x)=f(x-10)

On this graph the image is moved 10 units to the right.  Only the x-values were changed, therefore the change must be made to the x.  Also  it must be minus ten because inside the parenthesis you must “think inverse.”  So subtracting 10 in the equation ends up moving the graph at postive 10 units.

Question 3

3

a) verbally: veritcal stretch by a factor of 3      b) equation: g(x)=3f(x)

Only the y-values were changed and they were changed by a multiple of 3.  This causes the 3 to be on the outside of the parenthesis.

Question 4

 4

a) verbally: horizontal stretch by a factor of 2         b) equation: g(x)=f(x/2)

To get the image for this graph only the x-values were altered.  The image is 2 times wider than the pre-image.  Since it is a horizontal transformation the number that affects the graph must go inside the parenthesis and we must “think inverse” so x must be multiplied by 1/2 to stretch the graph by a factor of 2.

Question 5

 5

a) verbally: absolute value transformation of y-values       b) equation: g(x)=|f(x)|

On the two graphs both of the x-values remain the same so any change must be in the y-values.  In the image all of the y-values are positive as compared to the pre-image where some were negative.  The only way this could be achieved is by taking the absolute value of the y-values.  This is why the absolute value signs are outside the whole function f(x).

Question 6

 6

a) verbally: horizontal reflection over y-axis       b) equation: g(x)=f(-x)

I’m pretty sure just about everyone in our class struggled with this problem, suprisingly even me (kidding), until we got some help from Mr. B.  However, in reality, its really fairly simple.  Just look at a table of values for the pre-image and the image and you should come to the answer.

Pre-image

x y
-6 2
-2 0
0 -2
3 0
5 4

Image

x y
-5 4
-3 0
0 -2
2 0
6 2

From these tables you should be able to see that the x-values are simply flipped from negative to positive or vice versa.  Since the x-values are changed the switch must come inside the parenthesis. This particular change causes a refleciton over the y-axis.

Once we finished working on the worksheet, we were supposed to start on the book work for section 1-3, all 20 problems.  Hopefully most of the class got a good start on that.  Overall, it was a very solid friday in math class.

The next scribe may or may not be new girl.  It will, however, be stefan.

 

Introduction to the Pythagorean Property

Student 08_09
Dec 9 |21:27

Today in class we learned about the Pythagorean Property.  This states that cos2x + sin2x = 1. (Note, Graphing calculators read (cos(x)2) as cos2x.)

Graphically, cos2x and sin2x appear as:

(Click Picture to enlarge)