Our goal for today’s class is to be able to use the correct notation to express a transformation.

Before we started working on our goal, Mr. B. had us do a warm up problem: What is the sum of all the digits of all the possible integers that are less than 100?

Now where to start with this. Well we know that 1-9 is used 20 times when counting up to 100.   Since we are adding each digit, ten is not included;  1+0 is still one, and we don’t need to repeat numbers already used, so we don’t include 10 in the scale.  Now to get to the solving part, we add all the digits from 1-9, and get 45.  As I mentioned earlier, 1-9 is used 20 times, so we multiply the sum of 1-9 (45) times 20.  In other words,  you are adding 1-9 ,20 times.  After doing the math, we come up with an answer of 900; 20(45)=900.

Next, we went over a worksheet that we started working on the other day.  Except today we looked at the back of the worksheet; talking about how to verbally say what happened to the image, and finding an equation to represent the transformation.  For the first graph, the transformation is moved 6 units straight down from the pre-image.  That makes it a vertical translation by -6, and the equation is g(x)= f(x)-6.  In the next graph, the image is moved to the right ten from the pre-image.  Making it a horizontal translation, and the equation is f(x-10).  It is minus 10 and not plus 10 because if you were to take the graph and make -10 the new zero, one of the pre-image’s points would be at (10, -2).  So when you go back to the original graph (before you moved the 0) you would plot that point at (10, -2) and the rest of the points in the same relationship as the pre-image, just now for the f(x-10) image; all the new points are ten units away from the other image.  Mr. B. used the analogy that is you pulled a rug(graph) to the right, and someone is standing on it, they(image) will fall to the left.  The third graph is dilated, shrunk; it’s a vertical dilation by 3, and the equation is 3f(x).  A vertical dilation is the only time that y=0 is a point for both images.  Our last graph is dilated/stretched, it is a horizontal dilation that’s represented by f(.5x).  The reason .5 x dilates the image to be bigger is because it makes a new graph where what used to stand for one unit now stands for two (each tick mark on graph goes up by 2), so when you go back to the original graph it is going to be bigger than the pre-image; and the opposite goes for a whole number with an x (2x).

For the last part of class, we did more practice with noting what the equation for the transformation is.  For every graph we had, we used the same pre-image, just changing our inputs or outputs.  First we found the equation of that pre-image, using the quadratic formula of f(x)= a(x-h)^2 +k.  We then used the information from the graph to replace the variables; getting us -4= a(5-1)^2+4, a=-1/2.  You could also look at the images points:

normal f(x)                                       1/2

X Y X Y

1          1

2         4                                        2              2

3         9

4        16                                      4               8

We then took this knowledge and labeled the new images; one was f(x+6), another was f(x)+3.  We found all of these just by looking at the graphs.  I hope you can now understand transformations, and how to label/classify them.

The next scribe will be from Lauren E.

Pardon the use of our blog, precalc students. Here are the videos I would like the Algebra II students to view today and take notes on.

http://www.khanacademy.org/video/level-1-exponents
http://www.khanacademy.org/video/negative-exponent-intuition
http://www.khanacademy.org/video/level-2-exponents

http://www.brightstorm.com/math/algebra/exponents/introduction-to-exponents
http://www.brightstorm.com/math/algebra/exponents/multiplication-and-division-properties-of-exponents
http://www.brightstorm.com/math/algebra/exponents/zero-and-negative-exponents
http://www.brightstorm.com/math/algebra/exponents/simplifying-expressions-with-exponents

http://www.khanacademy.org/video/simplifying-expressions-with-exponents
http://www.khanacademy.org/video/simplifying-expressions-with-exponents-2
http://www.khanacademy.org/video/simplifying-expressions-with-exponents-3

Today we continued our goal of learning how to transform a function by creating graphs and reflecting on how they originated. However, we started with our warm up first. Today the challenge was to find the smallest integer, greater than 1000, that is divisible by 5 and 13, but not by 4. Several of us tried to make an equation, with no luck. Finally a few of us got it and showed the class the secret. The obvious first step is to multiply 13 and 5. However, this results in a number less than 1000 (65) and therefore cannot be the solution. To get an answer that would be over 1000 and a multiple of 65, we simply divided 1000 by 65.

This tells us that the answer has to be around 65(16).

And that shows us our answer is 1105!

After our warm up we got  right to our activity for the day. Mr.Bieniek passed out sharpies, transparencies and scissors. We had no idea what was in store for us; or that it would be extremely helpful in understanding how to transform a function. The first graph we made was about 8 by 8 with a scale of 1 on both the x and y axes on graph paper. We laid the transparency on top of the graph and all plotted a function that had ups and downs but stayed close to the origin on the transparency:

Next, we began to explore how changing the inputs of a graph affect it. We made another graph which we called 2x. This symbolizes what the graph would look like when it’s equation is f(2x). Keeping the transparency the same, we made a different graph on the graph paper with a scale of 2 on the x axis and a scale of 1 still on the y axis.
**Remember, when you change the input you CHANGE THE AXIS, not the function being graphed. Mr.Bieniek worded it like this: “I’m stretching the x axis, plotting points, and letting the axis back to where it was, like a rubber-band being stretched with writing on it”
The function looked like this on the f(2x) graph– note that it looks the same as the first graph, just with different numbers on the x axis.

When you take the points from that graph f(2x) and put it on the x graph with the original axis you can see the change:
We then made another graph with the function f(1.5x). You should be able to guess what happened– instead of being a “squished” version of the original, it was a “stretched” version:

Now that we understood how multiplying the outputs by whole and fractional numbers affected the graphs, we investigated how adding and subtracting numbers from the outputs change the function. The most common misconception in adding to (x) in a function is that you just add to the numbers on the x axis similar to the way you multiply them in the first half of our lesson. However, you need to think of it as pulling the whole grid to the positive or negative side. For example, we first graphed f(x+2). Because you’re adding, think of a giant hand pulling the x axis UNDERNEATH the function to the positive side (so the function stays the same while the x axis is pulled 2 spaces to the RIGHT):

On the original axes, the function f(x+2) appears to have been shifted 2 spaces to the left, which is confusing UNLESS you keep in mind how we pulled the axis!

Keeping in mind the process in which we made the f(x+2) graph, we moved on to plot f(x-3). You PULL the x axis 3 spaces to the left, and therefore the function appears to shift 3 spaces to the right:

We concluded the hour with Mr.Bieniek telling us that we should experiment with other kinds of transformations we could make to a function… bet that’s what the next blog will be about

The next scribe will be Leah.

Today Mr. B had us start out with a problem that really got us thinking. 

- What is the smallest integer that is divisible by all the integeres 1 through 10?    To solve this problem i first thought of what could possibly be divisible by all of those numbers.  But then i realized it was a lot easier than that.  If you notice, numbers 6, 7, 8, 9, and 10 are all divisible by at least one of the numbers 1 – 5. Then to make it easier we found out the factors of all of the numbers.  10 is divisible by 2 and 5, 9 by 3 and 3, 8 by 2, 2  and 2 but you only need two 2′s because there is already a two in 1-10 to make 8.  7 is by its self and 6 is divisible by 3 and 2 which we already have.  the rest of the numbers are by themselves since they are too small to be divisible by anything but 1.  If we multiply by only the numbers we need to make each number then the number they make will be divisible by everything.  So we take 2^3 * 3^2 * 5 * 7 * 1.  All of these numbers are divisible by 1-10 so we get 2520 as an answer and it works.  It is also the smallest because we took the least number of numbers we needed to find it. 

Next in class we expanded on our learning of transforming functions from yestereday.  We learned how horozontal transformations effect only the inputs(x values) and don’t touch the outputs.  As you can see for f(x-3), the -3 is only effecting the x value and not anything else.  You plug in what ever number x equals, subtract 3, and then evaluate it again to see what that number is for f(x).  For example you plug in 3 for x and 3-3 is 0.  then you look to see what f(x) is when its 0 and you get 1.  The same thing for vertical transformations, only it effects the outputs(y values) instead of the inputs.  An example of this is F(x) + 3.  It’s only effecting what comes out of the problem. Here’s something Mr. B showed us to help us better understand:

x:               -2             -1            0                1                 2                   3   

F(x):          2               3             1              -2                0                   1          

F(x-3):   Undef. Undef. Undef.           2                 3                   1                

Its like that for the first three because its not in the range of what f(x) can be, and as you can see the inputs have moved over 3 to the right. 

Another thing we learned today is the concept of multiplying the input by a certain number.  For example, F(2x).  When you graph out this problem, the graph shows that it squishes what the original graph used to be, but the original stays the same.  The original graph stays the same, you just make the other one smaller, so the original is 2 times as big as the new one, but its still the same graph.  A person in our class put it as taking the reciprocal of what number is inside the parenthesis for f(x). Or, you could say that F(2x) and F(x) are the same graph, only F(2x) has been shrunk by a factor of 2 towards the origin(y axis), or that the x axis has stretched by a factor of 2.  We learned that the same exact reasoning can go along with F(1/2*x) except that the graph stretches wider than the original one, and  F(1/2*x) has been stretched by a factor of 2 away from the origin(y axis), or the x axis has shrunk by a factor of 2.

The next scribe will be Caitlin.

Today in honors precalculus, our goal was to learn how to transform a function numerically/graphically.

We started out with a varsity math meet warm up problem:  Find the units digit for the sum 13^25 +4^81 +5^411

Since our calculators are not capable of giving us the answer to 13^25, we decided to try and look for a pattern. We started out by looking for a pattern when you take 13 to the power of 1, 2, 3,ect. Our results showed that when you take 13 to a power with the last digit 1, the last digit in the answer will be 3. When you take 13 to a power with the last digit 2, the last digit in the answer will be 9. We continued to find similar patterns in the other part of the problem as well. Once we were able to find the patterns in the problem, it was easy to find the answer.

13^1 = 13                                 4^1 = 4                                    13^25 ends in a 3

13^2 = 169                               4^2 = 16                                   4^81 ends in a 4

13^3 = 2197                             4^3 = 64                         +   5^411 ends in a 5

13^4 = 28561                           4^4 = 256                                           Answer: 2

13^5 = 371293                                                                    

Next we talked about how we are going to learn to transform functions to fit specific situations. We were given several problems to work on with our group. The problems each involved discovering how transforming a function numerically will change its appearance graphically.

The first problem demonstrated vertical translation. We learned that vertical translation causes the range values to increase/decrease at an equal rate. It will appear to move all the points on the graph either up or down.

The next problem demonstrated horizontal translation, which causes the domain values to increase/decrease the same amount. It will cause the graph to appear to move left/ right.    

For homework we were to finish solving the problems on the worksheet.

The next scribe will be Trista