Precal Blog

Howdy. We started off class today by taking a picture outside because it was “color war” day and our class was green. Next we talked about our tests, however we did not get them back. Mr. B did say that they were overall good. After the test discussion was over we got down to business and started off on a new topic, Composition of Functions.

The word “composition” is the act of combining parts or elements to form a whole. So for our purposes, composition of functions basically means combining previous functions to create new, transformed functions. Here is an example that we used in class.

When Damien is hungry, he gets grouchy. When he is not hungry, he stays in a good mood. In other words, mood is a function of hunger.

Damien’s friend Karla usually has positive feelings toward him. When he is grouchy, however, Karla gets irritated with him. In other words, Karla’s feelings are a function of Damien’s mood.

So in this example, the first independent variable is Damien’s hunger. This directly affects his mood, the dependent variable. Next, Damien’s mood affects Karla’s feelings. You are taking one input and inserting it into the first function to get an output, and then using that output as an input in the next function. This process can keep going on forever and ever if there was a function that was dependent on the previous one.

In that case, the next question is how to write out these composition of functions mathematically. This is simply called notation, or the notation of functions. Let’s use the previous example to define these functions.

h: Damien’s hunger level

m: Damien’s mood

k: Karla’s feelings

Because Damien’s mood is a function of his hunger level, we can write this as:

m(h)= h

Next we take this output and use it as the input in the next function, which is to find Karla’s feelings. So we write this as:

k(m(h)): m(h)

That is basically the process of how to write these functions algebraically.

In class today we also learned that composition of functions also transform the graph.

So for an example we had to use this graph and transform the function f into function g.

So first we had to describe the changes that the function undergoes. So for the first thing, the graph needs to be squared because function g is a parabola. Another change was that the slope became negative. The graph was also translated vertically 2 units. So first lets define some variables.

s: squaring

n: negative (causes graph to flip)

t: translation

Next lets use these as a composition of functions to work this out step by step.

1.  x: simply the independent variable

2.  f(x): the blue line that we are transforming

3.  s(f(x)): made the function become a parabola

4.  n(s(f(x))): made the graph “flip”

5.  t(n(s(f(x)))): moved the graph up

Now if we used more accurate variables it would look like this:

f(x)= x

s(f(x))= x²

n(s(f(x)))= -x²

t(n(s(f(x))))= -x²+2

so basically g(x)=-x²+2

But one thing you have to look out for is doing these transformations in the wrong order. Lets say you made the function negative before squaring the x value. Because when a number is squared it is always positive, you would not have the result of the desired function g.
So in review, today we learned that composition of functions is basically the combining of functions and using the output of one function as the input in another. Another fact that we learned was that translations and dilations generally keep their shape, but when involving composition of functions the graph can completely change.

The next scribe will be Pablo.

Today’s class was mostly a review for the test. Mr. B began class with claims and presentations from Section 1-3. After that we worked on a problem from the book (pg 48, C1).

This problem was a perfect way to review Translations and Dilations. g(x) (shown in blue below) is a image function of f(x) ( shown in red below), which is x^2. The problem is to write an equation for g(x)  in terms of f(x) and in terms of x. This problem seems simple but there are both horizontal and vertical dilations, as well as translations.

It’s easiest to start this problem by finding the translations. The easiest way to do this is by looking at the vertex of the parabola. The vertex is (0,0) in the pre-image f(x). In the image, g(x), the vertex is (3,-5). So the graph was translated 3 units to the right and 5 units down.

Next we need to figure out the dilations. In the pre-image the width of the graph was 4, and in the image was 12. This means there was a dilation by a factor of 3 horizontally. We can do the same vertically. The original height was 4 and the height of the image is 8. So the vertical dilation factor was 2.

• Horizontal Dilation: x3
• Vertical Dilation: x2
• Horizontal Translation: +3
• Vertical Translation: -5

The basic formula for translation and dilation is g(x) = a + b(cx-d), where a and b are vertical translations and dilations respectively, and c and d are horizontal dilations and translations. We need to be careful within the parenthesis as the opposite of the expected operation takes place. So a -5 is really +5 on the graph etc. So our two equations are as follows:

• g(x) = -5 + 2 * f(1/3x – 1)
• g(x) = -5 + 2 * ((1/3x – 1)^2)

The reason the equations have -1 instead of -3 for d is due to a weird technicality with the parenthesis, but don’t worry Mr. B said we don’t need to understand that yet.

Good luck on the test everyone. The next scribe will be Taylor.

Scroll down to the “On-line study group” post and click the link.

Nothing in this post should be new to anyone that has been in class. This is going to be a review post on what we’ve learned for the past moth and what will be on our test.

I will start with the definition of a function. A Function is a relationship between two sets of data. For every input there must be an output. Every x-value must have a definite y-value. Think back to the soda machine. Every time you hit the Mellow Yellow button, you will always get a specific soda. It must be consistent. If you hit the Mellow Yellow and get 3 different sodas on 3 different tries, we don’t have a function. One way to check if a function works is to perform the dotted line test. Also remember that the x-values are the domain and the y-values are the range.

In the second section, we learned about the different types of functions:

Polynomial Function

f(x)=a_nxⁿ+a_n-1x^n-1

Power and Quadratic Function

Power: f(x)=axⁿ

Quadratic: f(x)=ax²+bx+c

Linear Function

f(x)=ax+b

Direct Variation Function

f(x)=ax

Exponential Function

f(x)=ab^x

Inverse Variation Function

f(x)=a/xⁿ

Rational Algebraic Function

f(x)=p(x)/q(x)

This list came from Jenny’s post so credit goes to her for the list.

Finally, lately we’ve been going over transformations.

The transformations we’ve been talking about are translations and dilations. A translation moves the graph whereas a dilation streches the graph.

f(x)=a+b(cx+d)

Everything inside the parenthesis affects the graph’s x-values inversly. For example, if c=1/3, you would dilate the graph by three. The same goes for if d=3, you would translate the image -3. Everything outside of the parenthesis affects the y-values as is. I.e- a=4, translate +4, b=4, dilate by 4. When you multiply, you dilate, add/subtract, translate.

That’s basically our overview of the past 3 sections. Make sure to tune into the study session tonight on the website if you need additional help. That’s it for now. The next scribe will be Anchal (Sorry if i spelled the name wrong)

During class on Friday, we spent most of the time going over Exploration 1-3c: Transformations from graphs. It asks us to describe the transformation from the pre-image (dotted line) to the image (solid line) verbally and algebraically in terms of g(x).

1.

For this problem,  we know that it is a vertical translation by -6 because the image is 6 units below the pre-image.  The equation would be g(x)=f(x)-6.

2.

This problem is a horizontal translation because the y values stay the same and the x values of the pre-image are increased by 10.  The equation ends up being g(x)=f(x-10).  Note that it is f(x-10) and not f(x+10).  This is because when you go from image to pre-image it is a loss of 10 units.

3.

This graph shows a vertical dilation by 3.  You can tell this because if you take the height of the image and divide it by the height of the pre-image you get 3.  The equation is g(x)=3f(x).  The 3 is on the outside of the parenthesis because only the y values are changed.

4.

This graph shows a horizontal dilation by 2 because the x values of the image are 2 times as big as those of the pre-image.  The equation is g(x)=f(1/2x).  X is multiplied by 1/2 because horiz0ntal movement is always inverse.

5.

This graph shows an absolute value relationship in the y axis.  You can tell it is in the y axis because all of the negative y values in the pre-image are positive in the image.  The equation is g(x)=abs(f(x)).

6.

This graph is a reflection over the y axis because the pre-image is basically flipped.  The equation is g(x)=f(-1x).  To find the equation you need to look at the table.

The tables show the x values being multiplied by -1.

After going over the worksheet, we worked on the problems for section 1-3.

The next scribe will be………………..Scotty P

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

 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

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

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

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

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

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

Image

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.

We started today’s class out with Mr. B explaining the group work that we started the day before involving transformations in functions.  He did this by giving us the basic equation for describing a function during a transformation, which was f(x)= a + b (cx-d).  With this equation we are able to figure out what type of transformation the image will go through, based on which variable we change.  The 4 types of transformations are vertical translation, vertical dilation/stretch, horizontal dilation/stretch, and horizontal translation.

Change in the a variable

This would make the image move, while staying the same shape/ figure up and down along the y axis.  Which is called a Vertical tranlsation.  Translation meaning moving a shape, without rotating or flipping it. “Sliding”. The shape still looks exactly the same, just in a different place.

For example, g(x) = f(x) + 3, with g(x) being the y values for the new image, the image would move up 3 on the y axis.

Change in the b variable

This would allow the image to be stretched or dilated, meaning it is a similarity transformation in which a figure is enlarged or reduced using a scale factor, without altering the center.  So the image would be dilated on the y axis, meaning the x values are unchanged after the dilation.

For example g(x) = 2f(x), the y values are increased by a scale factor of 2, making the image longer and more stretched out.

Change in the c variable

This would have the same idea as variable b, but the dilation or stretch is on the x axis, so it would make a horizontal dilation/ stretch.  We have to remember that if the values are inside the parentheses, we look at them as an inverse, so 1/2 would really mean an increase of 2.

For example, g(x) = f(1/2x), its an increase on the scale factor of 2 along the horizontal axis, widening the image.

Change in the d variable

This would just mean a horizontal translation, so moving the whole either right or left on the x axis, without changing the shape, as explained in vertical translation. Once again since the d value would be inside the parentheses, we have to think of the inverse, and thats how the image will move, with the inverse doing the opposite of what you would think.

For example g(x)= f(x-3), since its says x-3, you know it really means your x values are increasing by 3, so the image would be moved right on the graph.

So basically variables c and d have the effect on the x values direction and variables a and b have an effect on the f(x) or y values.  Also another way to look at it, inside ( ) means changing x values, and outside ( ) means changing y values.

Another way that Mr. B told us we could write the function describing equation would be 1/b*f(x) – a = (cx-d),  now with this we would always be thinking inverse, when working with this equation.

Now we are finishing up working on worksheet 1-3b, which allows us to find the answers algebraically rather than numerically.

PS – sorry, for each example I wanted to put the graph of the pre image and the new image based on the transforamtion, but i got confused on how to do that, and couldn’t figure it out, even though it’s probably really easy.

The next scribe will be Dushan!!

In todays class we began by finishing going over the Exploration 1-3a: Translations and Dilations, Numerically sheet.  By completing the worksheet and having a discussion about it today, we learned all about different transformations. 

We learned that a translation is a graph thats been moved any number of places left, right, up, or down, but still looks the same as the original graph.  A verticle translation is a graph that’s been moved up or down, while a horizontal translation is a graph that’s been moved left or right.

A dilation makes a graph stretch or shrink vertically or horizontally.  When you put a number in greater than 0 to multiply with f(x) or Y, the graph stretches vertically.  However when you put a number between 1 and 0 in, the graph shrinks.  When you put a number in to multiply with x, the graphy stretches horizontally.  However to make it stretch wider, you have to multiply the x value by a number between 1 and 0.  The opposite is true to make the graph shrinks inward.

Next, “reversearp” showed us an equation on the board of how dilations actually worked.  The equation is…

f(x)=a+b(cx+d)

In this equation the variables on the right side of the equals sign outisde of the paranthises, a and b, represent changes in the equation that deal with Y transformations.  Variable a represents a verticle translation.  Variable b represents a verticle stretch.  The variables inside the paranthises, c and d, represent an X calue transformation.  Variable c represents a horizontal stretch in the graph while variable d stands for a horizontal translation.  Another way to write this equation is to transfer all of the variables that deal with Y, or f(x) to the f(x) side of the equation.  For example the new equation would be written as…

1/b*f(x)-a=(cx-d)

This equation basically means the same thing as the other, but it groups all of the variables that deal with f(x) together, and all of the variables that deal with x together. 

After this quick lesson we finished up worksheet 1-3b and started on worksheet 1-3c.

The next scribe will be Bo-dog.

Click Here

Today at the beginning of class, it was each student’s responsibility to put in their claims for the designated problems.  The problems that were claimable were 9, 27, 30, and 39.  These problems were taken from section 2.  Section 2 was about different kinds of functions.

I will do a walk through of each problem that was presented in class today to help others have a better understanding:

9.  Heather

Power function f(x)=3x^2/3            The domain for the function is 0<=x<=8

This is the equation graphed.  The question asked for the graph to be plotted, the x,y intercepts, and the range.  Since this is a power function, it contains the origin.  Therefore, the x and y intercepts are both zero.  The range for this function would be 0<=y<=12.  To show this on the graph, we would use a boolean variable.  Heather showed us how to type that into our calculator.  It is pictured below:

27.  Jocelyn

In this problem, we were given a graph that looked a little like this:

Jocelyn explained to us how this was an inverse variation function.  The general equation for this is:     f(x)=a/x^n

Also, for this graph there are asymptotes at the x and y axis because the graphed functions will never be zero.

30.  Eric (Me)

For this problem i was to sketch a reasonable graph and explain how the variables are related.  I was also supposed to identify the type of function that it resembled.  The temperature of a cup of coffee and the time since it was poured were my two variables.  My graph came out to be like this:

As shown in my graph, i have the temperature of the coffee decreasing as time elapses.  When the temperature reached room temperature it leaveled off.  It leaveled off because it is impossible for it to decrease below room temperature if the coffee is sitting out.  This concept has created a horizontal asymptote at room temperature.  Also written near my graph is “exponential function” and the general equation that goes with it.

39. Jared

The main concept covered in Jared’s problem was the vertical line test.  This test is just an easy way of explaining the defenitin of a function graphically.  The defenition of a function states that for each x-value (input), there is one unique y-value (output).  Therefore, when a line is drawn vertically on a graph, it should only pass through the graphed function at one unique point.  If the vertical line cuts through two or more points on a graphed line, it is considered a non-function.  Here is an example below:

The line that is graphed in red would not be considered a function because if a vertical line was drawn on the graph, it would cut through 2 different points.

The green line on the other hand would be considered a function because a vertical line would only cut through one point.

That concludes the problem claim and presentation portion of class today.

For the last part of class we began a worksheet that dealt with transformations of functions.  This sheet walked through a couple different ways of translating or dialating a graphed function.  We will continue with this sheet in class tomorrow.

The scribe for tomorrows class will be Ben!

p.s.  I was obviously having troubles with writing equations and inequalities.  So just stay with me and interpret them the best you can.  Thanks and sorry for the inconvenience!