Ok, so today we had on claims from 1-6. The problems available for claims were 1, 8, 9, and 11. Thank you Sara and Scott for presenting today. Im sorry for my lack of pictures.
  Sara presented problem 1 for us todayFor problem 1a., -f(x), it is a reflection of the pre-image across the x-axis making all negative y values positive and vice versa. For 1b., we were asked to find f(-x). This reflects the  pre-image across the y-axis making all negative x values positive and vice versa. Problem 1c. asks you to find |f(x)|. This makes it so that all the negative y-values from the pre-image become positive but all positive y-values stay the same on the graph. Problem 1d. asks you to find f(|x|). In doing this, all the points of the pre-image in the I and IV Quadrant stay the same, because all of their x-values are positive. In quadrants II and III, you find the absolute value of all the x-values of the pre-image. Then you take the |x| values and plug them into the f(x) equation. After doing this, the image should look somewhat like a parabola. The points in the II and  III quadrant will be a reflection across the y-axis of the points of the pre-image.
Sadly noone claimed problem number 8, so our class moved onto problem 9 which was well explained by Scott, but caused some confusion for some people including myself, so I do not want to try and explain this problem because I am not sure if how I am explaining it is correct.
  Because of lack of time, at the end of class we were told that we would work on the claim for problem number 11 tomorrow.
The next scribe will be………………………Jocelyn.

Alright, so today we started a new section, 1-6. First off we reviewed what happened to a graph for each of the following functions:
f(x)=f(-x): the graph flips over the y-axis
f(x)=-f(x): the graph flips over the x-axis
f(x)=|f(x)|: all negative y values become positive
f(x)=f(|x|): the positive values of x stay the same while all negative values for x are eliminated. Then the positive values for x are reflected over the y-axis
*the exploration 1-6a worksheet gives examples of all of these

After that we started talking about symmetry in a graph.We learned that there are two types of symmetry- odd and even- and then we looked at six graphs to determine if they were symmetrical.

Two of the graphs had even symmetry. So f(-x)=f(x). This means that the graph flips over the y-axis and that there will always be two x values that have the same y value.

Two more graphs demonstrated odd symmetry where f(-x)=-f(x) and the graph was reflected over both the x and y axes.

After we identified the graphs that were symmetrical and the types of symmetry used in the ones that were we learned how to algebraically determine if a graph was symmetrical and if it was, if it had even or odd symmetry.

We can figure out if the graph has even or odd symmetry by applying f(-x) and –f(x) to the f(x) equation and seeing which two are equal. If f(-x) and f(x) are equal then the graph has even symmetry, if f(-x) and –f(x) are equal then the graph has odd symmetry. If none of these functions are equal then the graph has no symmetry.

The first problem we did was this one:

We know that this graph has odd symmetry because f(-x)=-f(x).

The second promblem was

Again, we know that this graph is odd using the same rules from above.

The third example problem we had didn’t have symmetry at all:

When you look at each of the functions you’ll see that none of them are equal to each other meaning that it is neither even or odd symmetry.

The final example problem:

Quite a few people got confused with this one because at first glance none of the function were equal to each other. In order for us to see that there was really even symmetry we had to multiply our simplified version of f(-x) to see that it was the same as the equation for f(x).

At the end of class we were assigned problems and told which ones were available for claims.

The next scribe will be Shannon

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Alrighty so we started off class by talking about new types of transformations. We focused on what happens when f(x) = f(-x), f(x) = -f(x), f(x)= |f(x)| and when f(x) = f(|x|). When f(x) = f(-x), the graph flips over the x-axis. When f(x) = -f(x) the graph flips over the y-axis. When f(x)= |f(x)| all of the y values are made positive. And finally when f(x) = f(|x|) the graph where x is positive is flipped so that when x is negative it equals the same. After that we began discussing symmetry in a graph. We then looked at 6 graphs and determined if they were symmetrical. We also learned that there are two types of symmetry, odd and even. Even symmetry is when (f-x) =f(x) and it is a flip across the y axis and x equals the same. Here are examples that we looked at

As you can see with both of these the y values are the same if x is positive or negative. That is because f(x) = f (-x). So if x is positive or negative y always equals the same. These graphs demonstrate even symmetry.

The graphs that showed odd were

In these f (-x) = f(x) so as you can see the y values or OPPOSITES. These are reflections across the y AND x axes.

It is important to remember that there are plenty of graphs that are not odd or even and that reflections across just the x axis aren’t functions because then you would have multiple y values for a single x value.

Next we looked at how to algebraically determine if graphs are odd or even.

Here’s a picture of the 1st problem we did

We know the graph is odd if f (-x) is -f(x). To do this you simply plug into the equation and simplify and see if you get the same results. In this case you did.

Here’s another example of odd which tripped quite a few people up

The important thing to remember in this is with fractions you are only multiplying into the TOP part (since in essences you are multiplying by ).

Here.s an example when the graph is neither odd nor even

And now for the final problem…

This one is even and caused some problems for people. It is important to remember that you can factor out of your final result. By doing this you actually see that the two equations are equal.

At the end of class we were assigned 1-6 and told the claims would be problems 1, 8,9,11 but that we should NOT just do the claims.

The next scribe will be Adam K

Hi all – sorry I didn’t post this sooner but your precal book is available on-line here: http://www.keymath.com/PreCalc.

You will need a “ClassPass” in order to access the entire contents of the book so you’ll have to send me an email to get it. I carry my ipod touch with me most of the time now so you should get a pretty prompt reply.