Graphing |f(x)|

My apologies for the very late scribe post.

So, on Thursday, we have been given a surprise. We had a substitute which was Ms. Nickelson, BUT, that was not the surprise. She had been given us a POP QUIZ on Transformations, using NO calculators. It all turned out very well than we thought. The quiz was quite easy once after we finished.

We then, were given a new lesson to be learned; Graphing f(x).

Graphing Absolute Value Functions




We were given y = x to start off. On the graph above it shows a purple line which indicates y = x. The question then was added by Absolute Values; y = x. The blue line on the graph above indicates y = x. How did i get that V shape looking graph you may ask? Well because of those absolute values in front and after the x, you make the x values that are negative, positive. once you've done that, you'll end up having a V shape graph, and there you have it, you've graphed an absolute value function.


Example #1

we first graphed y = 3x - 4, which is the purple line that is on the graph above.

To find out where the y intercept is, we made x equal zero. there fore, the y intercept is at (0, -4). after we plotted the dot, we would easily find the rest of the dots and connect them by finding out what the slope is. y = mx + b. the m indicates the slope. rise over run. so starting from the point (0, -4) we rise 3 then run 1. once after that is all graphed, it should look like the purple line above.

we were given a second function to graph which is y = 3x - 4, which is the blue line above. what we did to get our original line to that function, was making all the negative values become positive values. then after we factor out the three;

y = 3 x-4/3

we shift 4/3 to the right, and the 3 indicates a horizontal compression. the y intercept is at 0 and the x intercept is at 4/3. once that all is done, it should look like the blue graph i have up above. the domain and range for this absolute graph is D: XER or (-00, 00), R: [0, 00).

Example #2

the blue line indicates y = -x +4. i got that line by finding out that the x and y intercepts were which was (o, 4) and (-4, 0).

we then were given to graph the absolute value of that line. y = -x +4;

first off, we make all negatives become positives and factor out the negative.

y = -1 x+4

it would still look the same because the absolute negative 1 does not affect the graph since it is an absolute value, which can't be negative.

D: (-00, 00)

R: [0, 00)

Zeroes at; X = 4



Example #3

The black absolute function above on the graph is the original function of y = x. we were given to graph 4 x + 2 - 5.

4 = compress

finding where Zero's are;

0 = 4 x+2 - 5

5/4 = 4 (x+2/4) --> (the 4's reduce)

5/4 = x+2

5/4 = x+2 OR -5/4 = x+2

5/4 - 8/4 = x OR -5/4 - 8/4 = x

-3/4 = x OR -13/4 = x

Therefore, Zeros; X = -3/4 , and -13/4.

D: (-00, 00)

R: [-5, 00)

Example #4

we where given to graph y = x^2 - 16 and y = x^2 - 16.
the function y = x^2 - 16 is shown above in black. we already know what shape the function is going to be before graphing it, given by the hint of the x^2 which is obviously a parabola. to find where the zero's are on the x axis, you factor out the y = x^2 - 16. which will become the factors of ;
(x-4)(x+4)
Zero's; X = -4, and 4.

but when graphing y = x^2 - 16, we make all the negative values become positive. and then you should have an upside down parabola just like the red one is shown up above.

Assignment in class was to do Exercise 11 questions # 1 - 20. unfinished is for homework

"drums please................ Friday's scribe... will be..... Haiyan"

Reciprocal Graphs

So lets get this over and done with already...

First of Mr. K gave us some fractions and decimals and we needed to get the Reciprocal. Easy enough eh?

Next he told us that this was going to be the hardest part of this unit. So I hope you all were paying attention to his teachings. Anyways I'll stop typing useless nonsense and start the scribe.

You remember the fractions and decimals he gave us and that he needed to find the reciprocal? Well here's some "notes" that I hope will help you if needed.

As the sequence gets bigger the reciprocal gets smaller.
As the sequence gets smaller the reciprocal gets bigger.
If negative then it gets bigger negatively.
1 is special - reciprocal does not change, Invariant. Invariant means it does not change.

All that was just to help with reciprocals.
Now here are some "notes" on graphing the reciprocal functions. I don’t really get that chart thing he showed us so this is what I can recall during class. Now this is what we really did in class.

Graph the outputs reciprocal, the Y coordinates, not the X coordinates.
Everywhere the outputs = -1 and 1 on the original graph it will be the same on the reciprocal graph.

Steps to graphing... Example: Graph 1/(2-x)

Well first just go and change the equation if that helps 1/(2-x) => 1/(-x+2)

1/(-x+2) => (-x+2)/1 or (-x+2) <====you know how that happened right? Just flip it to get it like that. Now that you have that done, graph this one so it will make things much easier. 1.Graph the original one first.... the reciprocal of the function they give you or the one that isn't in fraction state.


2. After you have the graph done like the one above, get the points that will stay the same. You remember which ones they were right? Where the outputs (y- coords) equal 1 and -1 will always be the same on both graphs.

3. The graph shows you the points in red. Now that you have that done lets get the Asymptote. Of coarse you can just double click it to find out what it is but since im doing this... Asymptote - it is a line on the graph that the graph will never actually touch, but the graph will continue to move towards it.

In our case the asymptote is the root(s). Why? Because when you input the x coord. you get a zero. You can't have a reciprocal of zero... anything over zero is undefined and it just can’t happen.

4. So now you have the graph above. Your two pints and your asymptote should be on there. Now you want to actually plot the graph. On the graph the red dots show you your two points that are always the same. Lets start with the right most point.

You see the point is near the asymptote. Between it and the root on the asymptote you can determine how the graph will look. The graph is going downwards from the asymptote. The x-coords are getting bigger so the reciprocal is it getting smaller. The y-coords are getting bigger negatively so the reciprocal of that is getting smaller negatively.

How do you do that? The x coords need to get smaller so it needs to go downward to get smaller, just along the asymptote. Not along the x axis as the x coords won't be getting smaller. The y coords are getting bigger negatively so it needs to get smaller negatively. This leads it to go along almost touching the x-axis. Not along the asymptote as the y coods would just get bigger negativly again. I know it’s a bit confusing. :S

As you can see it works out fine. The x coords are getting smaller and the y coords are getting bigger negativly

Now finish it off with the other half of the graph. You should be able to do it now right? I don’t know if all these things are correct but if you want to, go ahead and correct me. Don’t quote me on any of this xD

Some things that you need to know…
-Never have your graph curve back
-Don't have the graph touch the Asymptote.

This is basically what you do when you do a problem like this. I hope this scribe helped you guys out at least a little.

Go ahead and correct me on anything you see wrong.

Okay for homework we have...Exercies 10 but Mr. K didn't stop us there. He also gave us the job of finding the Reciprocal graph of the inverse trig functions. So that means we need to get the reciprocal of the graph SEC, CSC, and COT.

Oh yeah the next scibe is wendy. >_> Could you update the list Mr. K?

Today's Slides: October 3

Here they are ...





To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Inverses

This is Mary Ann, your scribe for today!

Inverse functions can be solved in more than one way. Like the block of wood that Mr. K has, you can look at each of it's different sides but it's all the same block. The same goes with a function. You can figure out the inverse of a function by solving it numerically, algebraically, or graphically, and you'll end up with the same answer for all of them.

The whole idea of inverting a function is simple. All you really need is an input and an output. When the function is inverted the input and output switch. OR the output undoes what the input did. Mr. K told us a story as an example of how this inversion works. The original function would be Baby Play. clean room would be the input and the messy room would be the output. But once the parent cleans up the room, the room goes from a messy room to a clean room. This is the inverted function. Notice how they switched? Here's a diagram from the slide for better explanation.

Numerically

To figure out the inverse of an ordered pair ( x , y ) what happens is that the domain becomes the range, and the range because the domain. Domain being x and Range being y.

Algebraically

In this slide we solved the equations' inverse, algebraically. The y and the x switched. But an equation must be written as y = _____ . So we isolated the y. This same equation can be solved by using the table of values, but when doing so, you must undo the last thing you did to get the inverse.

*note: some functions cannot be solved with the table of values.

Graphically

When sketching a function on a graph, it's inverse can be found by switching obvious points. On the graph below, for example, It's points are (-6,-3) , (1,5) , (9,6). To find the inverse, the x's become the y's. y's become the x's.





The function also flipped over the line y = x . Any points that lay on this line will not change because even when those points are inverted, it'll still be the same.



NEXT SCRIBE IS PAULO

Today's Slides: October 2

Here they are ...





To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Transformations

Hello everybody this is Alanna with the scribe post for today.

A.M Class:
This morning Mr. K gave us some time to work on a few questions to start off the period.

For part A the last two questions are the ones the I'll be concentrating on because those are the ones that the class somewhat had trouble on.

The coordinates of point, A, on the graph y= f(x) are (-2, -3). What are the coordinates of its image on each of the following graphs?
1. y= 3f(x)
First off we were given the points for A in the beginning of the question: (-2, -3). Since the 3 is outside of the brackets this will cause Only the output to be tripled. So the x values with remain the same.
(-3)(3)= -9
Leaving us with the coordinates of (-2, -9)


2. y= f(1/2x)
We take the x value of coordinate A, which is (-2) and multiply it by the reciprocal in the equation which is (2/1) or just the number 2.
(-2)(2)= 4
Leaving us with the coordinates of (-4, -3)
*note- Mr.K told us that since we are in a higher class of math we don't divide anymore, we multiply everything.


For part B of the questions we were given the same thing but except this time we have to work backwards.
e.x We are given coordinates (5, -4) and equation y= f(x - 4)-5, now we have to find the original coordinates of B.
Take the x coordinate 5 and add -4, which will give you 1. For the y coordinate take -4 and subtract -5 and it will give you 1. This will give you the coordinates (1, 1).

After doing these questions he gave one more long question and I'm pretty sure everyone will agree that the last part of the question was complicated.

Not until one of the students pointed out an easier way to solve the question and pretty much everybody liked his way better. Not sure of the guys name but good job!
The questions are on slide 2 of todays lesson and "that guys" method is on slide 3. What he did was just fill in the green equation on slide 2 then just worked on it from there by isolating the x variable. That gives you the original coordinates of B that was asked for.


P.M Class


To begin the afternoon class Mr. K rambled on about the calculator and how they can have technical limitations.


After that we applied our knowledge of transformations onto graph. Thank fully we had that excerise because it made things even more clearer.


Later on we learned about odd and even functions.

Even Functions:
To tell if its an even function what you get is kinda like a mirror image on the Y axis.

Even Only IF f(-x) = (x)
f(x)= x^2
f(-x)= (-x)^2
f(-x) = x^2
This is equal because the end result is the same thing as the beginning equation.

g(x)= x^2 + 2x
g(-x)= (-x)^2 + 2(-x)
g(-x)= x^2 - 2x
Not equal because in the end you end up with x^2 - 2x which is not the same as the original.

Odd Functions:
To tell if its an odd function by a glance is if you are able to get the same image if you flipped it over.

e.x

A function is odd only IF f(-x) = -f(x)
f(x) = x^3 - x

f(-x) = (-x)^3 - (-x)

f(x) = -x^3 + x
not even

-f(x) = -(x^3 - x)

-f(x) = -x^3 + x
since f(-x) = -f(x) this is an odd function.

The bell rang for class change so that was the end of the class. Mr. K assigned exercise 9 today.

Tomorrows scribe will be Mary Ann.

Today's Slides: October 1

Here they are ...





To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.