Pre-Cal 30S (Fall 2008)

Today's Slides: January 23

noreply@blogger.com (dkuropatwa) — Fri, 23 Jan 2009 14:12:15 PST

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Pre-Cal 30S January 23, 2009

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A hole in the graph

noreply@blogger.com (echizenR18) — Fri, 30 Jan 2009 13:00:01 PST

Hell0 :)

Today in class, we continued doing the unfinished example from yesterday's class. The first thing we did was Mr. k showed us the video of a real "Plug Hole" because it is related to our topic today. Here's what he showed us "Plug Hole (short version)" "Plug Hole (long version)" Feel free to watch it..

Anyways here are the examples we did...

next scribe is Ashley

Today's Slides: January 22

noreply@blogger.com (dkuropatwa) — Thu, 22 Jan 2009 09:52:51 PST

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Pre-Cal 30S January 22, 2009

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More about Rational Functions

noreply@blogger.com (larlyn014) — Wed, 21 Jan 2009 22:13:56 PST

Hi! everyone..

Sorry for the delay I almost forgot that I have to scribe for today's class.

Today, Mr.K reviewed about the Graphing Function that we did for yesterday's class.

Remember: The seven steps of doing the Graphing Function;

1st step: find the y-intercept

2nd step: Factor (if it can be factor)

3rd step: Find the Roots

4th step: Find the V.A (vertical analysis)

5th step: Find the H.A (Horizontal analysis)
6th step: Sign Analysis

7th step: Sketching the graph

Mr. K ask to do this question:

If the graph didn't touch the x-axis then it doesn't have roots

Mr. K give an example of irrational number that can be factor

don't forget to do the exercise review^_^

next scribe will be EchizenR18

Today's Slides: January 21

noreply@blogger.com (dkuropatwa) — Wed, 21 Jan 2009 11:56:23 PST

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Pre-Cal 30S January 21, 2009

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Function mix with Niwatori-san XD

noreply@blogger.com (Niwatori-san) — Thu, 22 Jan 2009 20:56:26 PST

Morning fellow Math ELITES!!

Itz function time last unit for this curriculum but not the last for your school year!

T_T

Oh stop crying about it letz go on with some learning!

Well following with the Synthetic division as well as the RATIONAL ROOTS THEOREM . . .
we can get sketching a graph well worth the time to look at. =p

Please follow these steps as followed to sketch the graph!

Step 1: Find your Y-intercept by letting variable X become zero like this
e.g . . . . . (x=0)
Good job you're smart!

Step 2: Find all rational roots possible while reminded to wiping out the copied numbers
(Use the rational roots theorem if needed)
Okay next step!

Step 3: Determine the sign of the function by looking at the different ways the X intercepts are hitting the X-axis such as -2 hitting at X and another hitting +2 see different spots!

Final Step! *Was going to do a drum beat but I let the drum just roll away* Step 4: Sketch the Graph! XP (what did you expect Oprah Winfrey forget it!)

Of course this is done after you get show the work so these following diagrams will show you how:

1st use Synthetic division mind yourself this is to find Polynomial Functions I will touch with Rational Functions in the next one!

Ok, you got the roots time to graph it with the roots acquired!

Use the equation and get the Y-intercept by letting the X's in the equation F(x)=4X^3+3X^2+4X+5
so...
Y = 4(0)^3+3(0)^2+4(0)+5
= 5

After doing this you can make a number line including all the Roots/X-intercepts depending on the context used in this case it's roots for the number line

Sketching Rational Functions is just a bit of an extra step! All that is different is that you need to look for vertical asymptotes.

Asymptotes are lines on a graph not meant to be there but they are there to show that certain part of the Cartesian plain is not allowed to have a(n) point to touch the line. If the point does touch then there is something wrong with the graph.

*Be careful to not mistake the Asymptote as part of the graph itself!*

Oh yes, quick reminder you have to find both horizontal and vertical Asymptotes not just the vertical.

Ok... that was to get the horizontal asymptote but to get the vertical asymptote you just need the root of the denominator!

*****Please notice!!! that not all the time will there be an asymptote!!! So no need to worry if there isn't one present!!!****

Niwatori-san signing off I'll see you all in the next year. Have fun next Pre-cal class and hopes to always to feel happy as well obliged to use your own voice to help out your peers.

"A little voice from you could change a lot from the many in silent"
-By: Niwatori-san- GAMBETE FOLKS!

> . < So sorry people I forgot to scribe the past two times but yeah I had work and even though that be a bad excuse to skip scribing I'm still sorry!

Graphing Rational Functions

noreply@blogger.com (Jeamille) — Tue, 20 Jan 2009 16:14:37 PST

This morning we continued talking about Graphing Rational Functions. We went back to the example from yesterday's class and finish graphing the rational function.

Example:

Step 1: Find the y-intercept by letting x=0.

Step 2: Factor everything, which in this case it's already factored.

Step 3: Find the roots of the function by finding the roots of the numerator.
This one has NO ROOTS.

Step 4: Find the vertical asymptotes by finding the roots of the denominator.
V.A. → x=2

Step 5: Find the horizontal asymptotes by dividing each term in the function by highest power of x, and take the limit as x goes infinity.

H.A. → y=0

Step 6: Determine the sign of the function over the intervals defined by the roots (step 3) and vertical asymptotes (step 4).

Step 7: Sketch the graph.

This is not the the exact graph, but it gives us idea on how the graph would look like.

We did another example on graphing rational functions, but it's pretty much the exact same thing on how we solve it. The only thing that changed was the graph (we can see this on the slides that Mr. K post everyday).

Homework: Go back to the Exercise that was assigned yesterday and and do the graphing rational functions part of it. Also, don't forget to start on doing the last exercise in the book that covers all the units that we did in Pre-Cal.

Next scribe is Larlyn

- Jeamille

Today's Slides: January 20

noreply@blogger.com (dkuropatwa) — Tue, 20 Jan 2009 08:43:46 PST

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Pre-Cal 30S January 20, 2009

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Today's Slides: January 19

noreply@blogger.com (dkuropatwa) — Mon, 19 Jan 2009 09:27:28 PST

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Pre-Cal 30S January 19, 2009

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Today's Slides: January 16

noreply@blogger.com (dkuropatwa) — Fri, 16 Jan 2009 13:47:55 PST

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Pre-Cal 30S January 16, 2009

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Today's Slides: January 15

noreply@blogger.com (dkuropatwa) — Thu, 15 Jan 2009 13:45:20 PST

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Pre-Cal 30S January 15, 2009

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Remainder Theorem and The Rational Roots Theorem

noreply@blogger.com (EricT) — Wed, 14 Jan 2009 22:54:25 PST

Today in class, we went over one more time how to do synthetic division and how to find the roots of a function using the Remainder Theorem.

After that, we learned how to find the missing coefficient in a polynomial if we already know what the remainder is. Here's how you would do this:

Let's say you have the question:

The first thing you want to do is write it out in a way that is easy to understand:

Then you plug the root of the denominator into the function:

Since we already know the remainder we can rewrite it this way:

Now all we do is isolate K with a little algebra, and solve it:

And there you have it!

Near the end of the class we managed to quickly learn the Rational Roots Theorem. This theorem allows us to find any rational roots of a polynomial function. Here's an example:

So you're given the equation:

The first thing to do is the find all the possible positive and negative factors of the constant term:

Now we find all the positive factors of the leading coefficient:

We then list all the possible rational roots, eliminating any duplicates:

We can then test out these roots by using synthetic division and the factor theorem to turn the function into a quadratic (Remember: If the remainder is 0, then it is a root):

This then gives you:

Now you just factor the equation and find the roots:

And there you have it! That's about all we did for today, tomorrow's scribe will be...Niwatori-san

Today's Slides: January 14

noreply@blogger.com (dkuropatwa) — Wed, 14 Jan 2009 13:43:34 PST

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Pre-Cal 30S January 14, 2009

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noreply@blogger.com (Zawadi) — Wed, 14 Jan 2009 23:17:44 PST

Yesterday we did more examples on synthetic division but we learn't something new.

Here is an example:

From the above we can observe that the remainder is the same with the rational root .

Another example was about finding out if a fraction is a factor or not.

If a fraction is a factor, it's remainder and rational root must be eqaul to o

Here's an example:

Today's Slides: January 13

noreply@blogger.com (dkuropatwa) — Tue, 13 Jan 2009 15:55:44 PST

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Pre-Cal 30S Jauary 13, 2009

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January 12

noreply@blogger.com (jad) — Tue, 13 Jan 2009 11:07:12 PST

Sorry for the delay...

Anyways, we did the rational roots theorem and the synthetic divison.

The following questions for the rational roots theorem were:

Find each of the following and their domains: fOg, gOf, fOf, gOg.

f(x) = 2x+3, g(x) = 4x-1

fOg(x) = f(g(x))

They're the same equation. What fOg(x) means is that g is fed up inside of f which is the same as f(g(x)). Same goes for the other functions gOf = g(f(x)), fOf = f(f(x)), gOg = g(g(x)).

Now lets try fOg(x):
f(g(x)) = 2(4x-1)+3
= 8x-2+3
= 8x+1

Another function: f(x) = √x-1, g(x) = x²

fOg(x) = f(g(x))

= √x2-1

We cant simplify this any longer, so we leave it at that. Lets move on to...

Synthetic Division.

I hate dividing because it was too long. It eats up half of your paper when you do division but Mr. K showed us the easier way to divide.

For example:

The 7 on the left side is the root of x - 7. The 2 was brought down. Now 7 * 2 = 14, we put the 14 under 6. Then 14 + 6 = 20. We are now adding instead of subtracting compared to the long division. Subtraction can lead to different answers. We then multiply 7 * 20 = 140. We put the 140 on the next, under -40. Now we can add (-40) + 140 = 100. So we multiply 700 by 100 and we get 700, we add 700 to 1 and we get 701.

Now we have 2 20 100 701. We start on the very right side. We have 701 which is a remainder. So we put that as 701/x-7. Next up is 100. After the remainded we have the constant. So 100 is our constant. Next up is 20 and it becomes 20x and 2 becomes 2x² . So we have 2x² + 20x + 100 + 701/x-7.

Today's Slides: January 12

noreply@blogger.com (dkuropatwa) — Mon, 12 Jan 2009 09:20:19 PST

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Pre-Cal 30S January 12, 2009

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function inverse and composition

noreply@blogger.com (ale) — Sun, 11 Jan 2009 14:11:32 PST

On Friday continued our new topic of Function. We first did a quick review on how to compose a function and then we learn how decompose it.
If we can compose a function. We must also be able to decompose it.for function decomposition we have to understand the "inner function" and the "Outer function"
for example if we have f(x)= 1-x^2 in order to decompose the function F(x)= x^2 Inner function
g(x)= x+1 Outer function
f(x)=1/x^3
F(x)= x^3 Inner function
g(x)=1/x Outer function
for more information about the function composition visit
http://www.purplemath.com/modules/fcncomp.htm
http://oregonstate.edu/instruct/mth251/cq/Fie ldGuide/composition/lesson.html
http://www.youtube.com/watch?v=S4AEZElTPDo
After the function decomposition we talked about the function of type
many to one function.
Different from one to one function where every one element of the input or the domain correspond to one element of the output or range. A many to one function is a function where every one element of the input domain correspond to more than one element of the output or range.

many to one function
Slide 10
We have to remember that
Input = Domain which is a set of all the first elements of the ordered pairs of a function
Output = Range which is a set of all the second elements of the ordered pairs of a function.

one to one function

One-to-one function are special because they are invertible

The we learned the horizontal line test which is a test use to determine if a function is a one-to-one. we the horizontal line test when we graph a function
If a horizontal line intersects a function's graph. more than once, then the function is not one-to-one.

Passes the test

Fail the test

Finally we learned the inverse of a function. Considering f(x) the function, the inverse of function f(x) will undo what f(x) did.
if ƒ is a function from A to B then an inverse function for ƒ is a function in the opposite direction, if an input x into the function ƒ produces an output y, then inputting y into the inverse function ƒ–(read f inverse) the inverse of a one-to-one function is obtained by switching the role of x and y.ƒ– the inverse of a one-to-one function is obtained by switching the role of x and y.

the inverse of a function is defined by

Not all function has an inverse. those that do are called invertible the one-to-one are special because they are invertible.

General procedure for finding the inverse of a function:

Interchange the variables - First, we will exchange the variables. We do this because we want to find the function that goes the other way, by mapping the old range onto the old domain. So our new equation is x=2y-5.
Solve for y -The rest is simply solving for the new y, which gives us:
2y-5 = x
2y = x+5
y = (x+5)/2
Hence, y^-1(x) = (x+5)/2

Today's Slides: January 9

noreply@blogger.com (dkuropatwa) — Fri, 09 Jan 2009 09:53:39 PST

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Pre-Cal 30S January 9, 2009

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Functions

noreply@blogger.com (Nohom) — Tue, 20 Jan 2009 19:51:15 PST

Hi every one!

I hope everyone did well on the test. Today we started a new unit called Functions. Function is a mathematical relation such that each element of a given set (the domain of the function) is associated with an element of another.

We can express functions in ways such as:-

1. sets:- For example A:{2,3,4,5} and set B:{6,7,8,9}.

2. Graphs:- For example this is a graph of the equation f(x)=x^2+2.

3. Formula:- For example f(x)=x^2+2.

We have also talked about relations. Relation is a set of order pairs.
Types of relation functions

1. One-to-one Relation:- this relation is a function because only one element of set A is corresponding another element in set B.

2. One-to-many Relation:- this is not an a function because the element 1 from set is corresponding to 2,4 in set B.

3. Many-to-one Relation:- 3 elements from set A are corresponding to the element 4 in set B which is many to one.

We have also talked about the vertical line test. It is a vertical line that touches the graph only once.
For example this graph is a Function because it touches the vertical line test once.

This one is not a function because it touches the vertical line test twice.

Sources :- http://www.tutorvista.com/content/math/algebra/function/function.php
http://www.purplemath.com/modules/fcns.htm

Next Scribe :- Ale ...

Today's Slides: January 8

noreply@blogger.com (dkuropatwa) — Thu, 08 Jan 2009 09:31:27 PST

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Pre-Cal 30S January 8, 2009

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reflection

noreply@blogger.com (Jamie) — Tue, 06 Jan 2009 23:19:19 PST

I enjoy this topic so much but then again there are some areas where in i am having difficulties. However, i will do my best and aim high for this test. goodluck everyone ! goodnight

REFLECTION

noreply@blogger.com (larlyn014) — Tue, 06 Jan 2009 21:27:09 PST

Hi everyone,

tomorrow will be our test on Logic.. In this unit I'm pretty sure that I can answer them on the test....although I don't get a chance to review this morning with all of the class, because I got really sick and won't be able to go to class.. I think I'm going to do fine...anyways.. I think this unit is okay and I think I can handle them...LOL!!...;) anyways...good luck to all of you...^_^

Reflection .

noreply@blogger.com (Jeamille) — Wed, 21 Jan 2009 15:46:56 PST

Hi everyone! The Logic Unit was tolerable :) I was pretty much okay with the Converse, Inverse, Contrapositive, Inductive Reasoning and Deductive Reasoning. I should say that the only thing I'm worried for tomorrow's test is the Proof by Contradiction (Direct and Indirect Reasoning). That's about it.

Good Luck!

Reflection!

noreply@blogger.com (Nohom) — Tue, 20 Jan 2009 19:50:07 PST

Hi everyone

Tomorrow is the test, and I am quite ready but the only thing that I don't understand is the difference between inductive and deductive reasoning.