Precal Blog

Today the class learned more about rational functions.
A rational function is a polynomial over a polynomial.
rational functions create vertical asymptotes and removable discontinuity we all know what a asymptote is. A removable discontinuity is a point in a rational function curve that is removed from that curve with out interrupting the natural curve of the graph. this is an undefined y in a function,
http://en.wikipedia.org/wiki/Image:Discontinuity_removable.eps.png
X sub 0 is the removable discontinuity. this point appears to change the curve of the graph but the graph just naturally curves at around that point.
http://upload.wikimedia.org/wikipedia/commons/thumb/e/e6/Discontinuity_jump.eps.png/180px-Discontinuity_jump.eps.png
to be clear removable discontinuity might move the graph but it doesn’t interrupt the curve like an asymptote.
vertical asymptotes when the function reaches 1/0 and removable discontinuity occurs when th function hits o/o.
To take this further if you find the point of were it is removed from the graph that is the limit of the function.
iIdon’t know if this is something we learned about in class but this is what i learned.

Next scribe is….?

We started out with Doug working over problem number 4 with a couple of right hooks. After he was done getting an A, Mr. B threw a sucker punch at Ethan. Blindsided by the cheap shot, Ethan stumbled and couldn’t produce results on # 19. Red quickly proceeded to work over #37 like it was his job. First he did it just like the book told us, got some stupid pi fraction, and then forged his own way. I have no clue how he did it, but he got the right answer, harder, better, faster, and stronger. Mr. B decided no one else was worthy of claiming, so he started up his spiel.

We got a sheet on the new options for the semester portfolio, which every started questioning immediately. Nothing anybody said is really worth mentioning, you should just know there’s another way to do the semester portfolio.

We started out talking about cubics and stuff. Since they have three x’s, you have to take their 3rd common difference to find their pattern. Just like the stuff we did in the last unit.

Tuesdays scribe is jon.

So we didn’t really do a lot today in class.

No one claimed #37, so there was no presentation on that.

Then we were handed out a paper with the two options we have for our semester portfolio on it.
The first option, the exercise portfolio, was a new one with only three parts. First is the table of contents. Next is completing three problems per section with clear step by step solutions and brief justifications. The problems should represent the entire section. The final part for this option is chapter summaries which is preparing a one to two page typed summary of the mathematics in that chapter.
The second choice was the term portfolio, aka the one we had to do last semester.

Next in class we went over some new material, but apparently we arent being tested on it.
It involved using the tables we did previously with functions to find the differences between the numbers with a 3rd difference. If you remember from the last test, quadratic functions have a second difference, and today we saw cubic functions have a third difference.
We then solved the cubic polynomial (ax^3 + bx^2 + cx + d) for a, b, c, &d. To do this you can use matrices or in list. If you wanted to solve using list, you have to enter the numbers from the given chart into list 1 &2 then go to cubic regression under STAT, CALCULATE, cubic regression. It wasn’t too difficult, and like i said we arent being tested on it.

After that we still had a good 20 minutes left, but we spent this time critiquing mr b..i guess this is confusing because i cant use names but there are two mr. b’s. Well the student teacher one. I guess we came to the conclusion that he should grow his beard and go to second grade to learn how to write…oh yeah and he changed his last name. haha ok thats it.

Scribe for tomorrow will be Blair.

Hello, Hannah was not able to blog for today. So i volunteered to help her out and do it.

Friday’s class we started out with claims.
However, since more than half the class was gone for the Ap History test, we decided it best to actually present the claims on Monday.
This does NOT mean that those of you who did NOT claim any problems, can claim some on monday.
Because that really wouldnt be fair to the rest of the class.
And Like Mr. Byrne told me, it’s nobody’s fault but my OWN that i didnt do any claims, which is true.
HOWEVER, #37 will still be claimable on monday.

Lizzie present #4 to us, because she will not be here on monday.
Number 4 dealt with looking at a polynomial’s graph and telling the degree (number of critical points), the number of real zeroes, and the number of nonreal zeroes. This particular problem had a vertex as a zero, which counts as two real zeroes (a double zero).

Then, Mr. Byrne went over this concept a little more, dealing with imaginery numbers.

We were reminded again that Imaginery numbers are not really imaginery. THEY DO EXIST.

This led us in to discussing the history of irrational numbers.
In the time of Pythagorus (of The Pythagorean Theorem), there were no such thing as irrational numbers. And so Pythagorus’ theorem, was a very radical idea.
Apparently, Pythagorus and his “followers” were thought of as a “math cult.”

Last but not least, we had a class discussion about how our class has been working so far, and talked about ideas to help us all improve for the rest of the year.

I hope everyones been having a great weekend.

Monday’s scribe will be TYLER.

PS. is imaginery spelt with an E or an A?

Because i cannot remember the online textbook username, you are going to have to use your imagination because i do not have any visuals.

On Friday we started off class with claims for problems 2, 4, 5, 8, 9, and 19.

Rose started with the presentation for number 2. This problem asked you to graph the function `f(x)=x^3-3x^2+9x+13`. After that, Rose found a real zero of the equation by using her graphing calculator. That zero was at (-1,0). She then plugged it into our new method of division, and found that it did divide in evenly without a remainder. After dividing the original equation by (x+1) we were left with the quadratic function `f(x)=x^2-4x+13`. Because this equation is not easily factored, we were forced to use the quadratic formula. The quadratic forumula was then simplified to `(4+sqrt(-36))/2` and `(4-sqrt(-36))/2`. After some help from the class, Rose was able to simplify `sqrt(-36)` to equal `sqrt(36)*sqrt(-1)`. This can then be simplified to 6`i`. Simplifying the rest of the equation leaves us with the zeroes of the equation:

`-1, 2+3i, 2-3i`

Nicholas was then called on to present #4. This problem showed a graph of a polynomial function and asked for the degree of the polynomial, the number of real zeroes and the number of nonreal zeroes. Nicholas pointed out that the degree of the polynomial can be found ONLY by counting the extreme points and adding 1. Therefore the degree of the polynomial in #6 is 7. Since the function crossed the x axis once and was tangent to the x axis once, Nicholas pointed out that there were 3 real zeroes (1 from the crossing of the axis, and two zeroes when the graph is tangent). Nicholas also told us that the number of nonreal zeroes is equal to the degree of the polynomial minus the number of real zeroes. Therefore, there are (7-3) nonreal zeroes, or 4. Overall, Nicholas’s presentation was solid and concise.

Julie then presented #5. This problem was similiar to Nicholas’s, and asked for the same criteria. The only difference in this problem was the graph of the function. This function, Julie pointed out, was a degree 8 polynomial. Unlike Nicholas’s polynomial, Julie’s did not cross the x axis. However, Julie pointed out that since it was thrice tangent to the x axis, it had 6 real zeroes. Since it was a degree 8 polynomial, Julie decided that it had 2 nonreal zeroes. Another solid presentation.

Stephanie then presented #9. To solve this problem, Stephanie had to draw a graph of a polynomial function that fit the given criteria. Stephanie’s presentation was thorough and extremely short, making for an overall great day of presentations.

The homework for Monday is to study our notes about finding nonreal zeroes because we are going to continue using them in the future.

The scribe for Monday will be Alyssa.

So… Here’s the 4th hour’s attempt to be back on the 1st page of the blog again! And an attempt to make our quarter journal grades not die. Ok, here goes…

Today we did a sketchpad project. The basis of it is that if you have a polynomial with clearly defined zeros because it’s in factored form, you can figure out the multiplicities of those zeros, and therefore the degree of the polynomial, by looking at the shape of the graph as it crosses those zero points. If it is simply a line, it is a multiplicity of 1. If it is the vertex of a parabola, it’s 2. If it’s the “center” of a `x^3` curve, it’s 3. I’m not sure what it is if it’s greater than that, but I’m sure even numbers have similar shapes to the 2nd degree shape, and odd numbers have similar shapes to the 3rd deree one. The other, non “m or n” values were just either translations or dilations, and were nothing new.

On the back side of the sketchpad sheet, there were some practice problems. The top section was pretty self-explanatory, but the later parts of the sheet were somewhat difficult, involving finding zeros with imaginary numbers. All you have to do to solve these is set the zero equal to x, and solve for zero by moving the imaginary number over to the x side. After that, you need to make sure your equation includes “x plus or minus (imaginary number)”. After this, you need to multiply these two factors together, and you will get a quadratic equation to solve for more zeros. Pretty easy, no?

Well, that’s all we did today. Pretty good class, huh? Oh, cookie if anyone noticed the somewhat out of place message at the back of the computer lab. Well, tomorrow’s scribe is Hannah. Have a good night, and see you all tomorrow!

~~~~~~Andy~~~~~~

LAB DAY

Today for math we went to the computer lab where we were given a worksheet to work on. The work sheet basically just consisted of example polynomials, where we had to factor out the equation, find the “zeros,” as well as negative and imaginary roots. If you missed out today you can pick up an extra sheet in class, it makes good practice..

Tomorrows scribe will be Ian

Today we talked about positive, negative, real, and imaginary roots. I was confused at first but Kendall got me through it. Here is the problem that we did:

First we were given a polynomial,

Q(x)= x^3 – 4x^2 + x + 26

Then we have to find the number of sign changes (positive to negative or negative to a positive). In this problem, the sign changes twice. It changes from positive x^3 to negative 4x^2 and again from negative 4x^2 to positive x. Since there is two changes and imaginary roots come in multiples of two because the plus and minus inthe quadratic formula we know there is two positive and zero imaginary or zero positive and two imaginary roots. Then to find the number of negative roots you have to count the number of sign changes after negating the coefficients of odd power terms.

The equation will then look like this:

-x^3 – 4x^2 - x + 26

The sign changes once from negative x to positive 26. So, We know there is one negative root. To find it we use that chart thing that we used earlier in the week using multiples of 26 because of the rational root theorem. It is a negative root so we only had to use the negative ones -1,-2,-13, and -26. After doing -2 we got the ramainder of zero. We know that (x+2) is the one negative root now.

After we did the long division dividing by (x+2) we got (x^2 – 6x + 13). Then, to find whether the two are imaginary or positive you put what you got into the quadratic formula with x^2 = a , -6x = b, and 13 = c. After putting the numbers into the equation we got -16 inside the square root. Since you cant have negative square roots, you can change the negative to an i. After working through that you find out there are 2 imaginary roots and zero positive. You get (3+2i) and (3-2i).

Tonights homework is to try to work our answers of (3+2i)(3-2i) back to (x^2 – 6x+13).

I took Tony’s day so he is goig tomorrow.

Today we spent the time in class reviewing what we have been working on with polynomials since yesterday we had a test on explicit formulas, recursion formulas, and partial sums. I will list some of the notes I took in class today. However, I really don’t know if they are helpful or just more confusing.

Rational Root Theorems:

If the rational number x=b/c is a zero of the nth degree polynomial, P(x)=sx^n + … + t where all the coefficients are integers, then b will be a factor of t and c will be a factor of s.

For Example:
If you have a function that is K(x)=4x^2 + 29x + -24 and one of your zeros is 3/4, you know this is true because 3 is a factor of -24 and 4 is a factor of 4. Looking at the statement above and relating it to this example you can tell that x=3/4, s=4, and t=-24. If you’re wondering where I found 3/4 it is found when I factored and got K(x)=(4x-3)(x+8).

The next thing we did in class was fill out a worksheet for 15.2.

For #1 the number of zeros for the function is the same as the highest exponent of the function. The extreme point is n-1 where n is the highest exponent in the function.

#2 refers to multiplicities of the zeros. when factored if there is an exponent outside the parenthesis you know that this zero occurs that many times. If there is just say x^4, like in letter d, this means the zero x=0 occurs 4 times.

For #3 and #4 the first thing you need to do is find the factors of the first and last integers of the function. Then you try each using the tricky little chart we learned to see which one gives you a final answer of zero. Once you find this number you plug it into x in an expression that will equal zero when using the number you found. Next, you divide this expression from the initial function to simplify it. You should then be able to factor to find the other zeros for #3. #4 is a little harder because you must use the quadratic equation. You will notice that this formula cannot be solved using real numbers because you end up with the square root of a negative. In this case you should substitute “i” in for the square root making, what was the square root of -23, 23i instead.

I hope this helped clear up confusion using polynomials.
Tomorrow’s scribe will be Tony.

Today was a test day, so there really is nothing substancial to blog about. I did collect a number of responses about the test though, continuing the test day tradition started by Liza. These quotes are a combination of 4th and 6th hour, and the quotes will remain anonymous.

“I dominated that thing!”
“I am soo confused right now.”
“What is a partial sum?”
“It was hard.”
“Cake; owned it 100%, well 94%, I didn’t know one.”
“I love Math Man.”
“I think I got a lot of P’s. Not sure how many A’s.”
“Next to one of the questions I wrote, ‘Is this a joke?’ and drew a smiley face.”
“It was okay, but the two learning targets I wasn’t here for were rough.”
AND last, but not least, “silence.”

Since no one is really crossed off on the scribe list, I am picking Kendall to scribe tomorrow. Hopefully that isn’t a repeat.