Tuesday, February 20, 2007

QUADRATIC FUNCTIONS

"Parabola" what a funny word. At least, that's what I thought of at the beginning of this unit. Now I know what it is, how it can be affected, and how to represent it with equations. This unit was, on a scale of 1 - 10, where 10 is HARD and 1 is EASY, i'd say it was about a 6.5, 7. I struggled at first, but I guess I kind of got used to it, so it's not that bad. The only thing I can think of to improve on this unit, is to do more problems with it, and I'm hoping it'll become easier. Okay, thanks for reading my reflection (:

- Jho.

Quadratic functions and equations was an easy unit. I didn't have any difficulty on the concept.

Heyy..it's Chris. This unit I found rather easy, despite all the nessesary steps you needed to take to solve the problem. I found that after you had some practice with it, it became incredibly easy to understand. Now I have to go before I miss my bus, but as a final thought: good luck on the test today even though I'm sure most of us will at least pass it. ;)

Reflection on Quadratics

Hey guys, it's Jackie. Aaaand .. I guess that it's time to do a personal reflection on our current unit, which is Quadratic Functions. Quadratics, quadratics .. how i loathe thee. :\ But honestly, all in all, this unit was fairly simple to understand until we started getting into the complicated fractions and all that jazz. I'm still kinda fuzzy on finding the zeros and whatnot, actually. However, it was easy enough to figure out how parabolas worked in regards to whether they opened up or down, if they were wider or narrowed, as well as finding the vertex of a parabola on a graph. Learning how to complete a square was tricky at first, but later on became a very simple and easy process. Finally, the word problems were definitely something that many of us need to continue to work on - in regards to setting up and figuring out which information goes where and how to organize our givens to solve for Y. But hey, all in all, it was an uncomplicated unit to understand. Alright, enough ranting, good luck to everyone on their test tomorrow and to all, a good night. :)

Monday, February 19, 2007

quadratics

Hi everyone. I'm Marianne. Quadratics was pretty easy. I think the only trouble I had was with word problems. One example of a word problem that i had trouble with was the third one on the pre-test(i dont want to type it cuz its long). Mostly everything on the pre-test was basically completing the square. I'm understanding the concept way better now, after we went over the pre-test today. Maybe i just needed to study thats all. Good luck on the test everyone!!!

Quadratiiiics !

Everyone else practically said what I was gonna say. But I have no choice but to repeat it to get marks. The unit was quite alright. Like the others, I get confused the most when it comes to fractions. Why won't you accept decimals Mrs. Ingram? That's how I figured out one of the questions from the pretest today - by using decimals. I guess I can live with it. Good luck tomorrow guys! Especially me.

My Reflection

Hey everyone! This is Lisa. For me, quadratic functions was slightly easy at some points but overall, a pain in the neck. My particular problem was figuring which was negative and which was positive. Word problems were also agony. Well, not really agonizing but I am really bad at them because I can never figure which is negative or positive. Now that the unit is done, I hope everyone does really well on tomorrow's test. Wish me luck!

Reflection

Hello, my name is Kimberley and I think that this unit was overall okay. If I were to say, the most challenging part of it are the word problems. Other than that everything else is pretty basic and straight forward. Oh and I don't like completing the squares when they have fractions either, but i suppose its not that bad. So anyways good luck on our test tomorrow guys (:

Reflection

Hello, I am Luis. Pre-Calculus is an interesting subject though it's pretty hard. I didn't take grade 10 Pre-Calculus a reason for me of having some troubles especially with the word problem. Anyways, study hard! and goodluck to our test tomorrow.

REFLECTION ON QUADRATIC EQUATIONS
So overall I thought this unit was pretty easy, the concept of everything was pretty straight forward. I think the only thing I would need improvement on is probably completing the square method, when it is a fraction . I still do not understand it clearly, and maybe one of the word problems. But aside from that the unit was easy, and easy to understand at some points.
-Sheena

Hey, hey.. it's Jennie..
I found the unit, quadratic equations okay. I just hate finding the vertex when the equation has a fraction. That's the part I found confusing. The rest was .. okay. Good luck on the test tomorrow!

I am Haiyan. This is my first time blogging. I found the unit challenging and there were some things that I did not understand. I have trouble especially with the word problems. Other than that everything was okay. I hope everyone does well on the test.

-qUad. Functions-

Hi... I'm Djanine. Pre-cal is hard and everyone knows that. I took grade 10 pre-cal last year. It was hard and I’m having trouble on some topics. Quadratic Functions is confusing. I’m having a hard time doing those “word problems”. I don’t get it actually. I need to look at my notes first before doing the given problems. But sometimes, it is easy if the given problem is almost the same of the examples that I have on my notes. Well, I hope that this unit will be fun for all of us. I hope that everyone will pass the unit test. Good luck to all.

Review on Quad. Fuctions

Hello, I'm Shelly, and this is my first blog of the semester. In this unit I think it was fairly easy at some points, but I do need to work on my understanding on "Completing the Squares " a bit more. Also I'm having a little difficulty on the "Word Problems"... At times I can't seem to do it without looking at the examples on the notes. Now I'll be busy trying to finish the Quad. Function dictionary and studying for the test, which I'll probably hate. Note: I Also hate graphing

Sunday, February 18, 2007

REVIEW ON QUADRATIC FUNCTIONS

Hello class, I'm Ivanna and this is my first crummy blog. I thought this first unit "quadratic functions" was easy because I took gr.11 pre-cal last year so I remembered most of it. I hate word problems, I hate completing the square of a fraction, and I hate doing exercises. Good luck on the test everybody, because I'll probably need it too.

blogging on blogging

hello,i'm john.this is my 1st blog. In this unit, quadratic function, it was easy for me. i'm having trouble with the word problem but im trying to understand it before the unit ends. actually, i hate graphing but its all about precal 11.. there's nothing i can do except to follow and answer it. i hope everyone was having a good time in quadratic functions. good luck on the unit test for the quadratic functions.

Labels:

Wednesday, February 14, 2007

Today in class we focused on Quadratic Word Problems. There are three types of the word problems.

We did the first kind of word problem:

1.) One thing will increase invalue while another thing will decrease.

Ex. Computer software sells for $20.00 each, 300 people will buy. For every $5 increase in price 30 less people will buy. What is maximum revenue?


Revenue = Price x number sold

Price -> 20 + 5x
number sold -> 300 - 30x

Revenue = Price x number sold
Revenue = (20 + 5x)(300 - 30x) <---fill in the variables
Revenue = 6000 - 600x + 1500x - 150x² <--- factor it out
Revenue = 6000 + 900x - 150x²
Revenue = -150x² + 900x + 6000 <---put in correct order
Revenue = -150(x² - 6x + ___) + 6000 <--- factor out -150
^to find the value of the 3rd term you would use the complete square method
Revenue = -150(x² - 6x + 9 ) + 6000 + 1350 <--- again use complete square method to balance out the equation
Revenue = -150(x - 3)² + 7350 <--- the result in completing the square

Vertex = ( 3, 7350)

Price = 20 + 5x
Price = 20 + 5(3)
Price = 20 + 15
Price = 35

Number sold = 300 - 30x
Number sold = 300 - 30(3)
Number sold = 300 - 90
Number sold = 210

Revenue = Price x Number sold
Revenue = 35 x 210
Revenue = 7350

The second type of word problem:

2.) Finding an enclosed area.

ex. What is the maximum rectangular area that can be enclosed by 120m of fencing if one side of the rectangle is an existing wall?

Area = length x width
Area = (120 - 2w)w
Area = 120w - 2w²
Area = -2w² + 120 <--rearrange in correct order
Area = -2(w² - 60w + __ ) <--- complete square
Area = -2(w² -60w + 900) + 1800
Area = -2(w - 30)² + 1800

Vertex = (30, 1800)

Width = 30
Length = 120 - 2x
Length = 120 -2(30)
Length = 120 - 60
Length = 60

Maximum Area = 1800 <-- max area is always the y- value
check by using the formula A=LxW

A=LxW
A= 60 x 30
A= 1800


The third type of word equation :

3.) Given an equation.

ex. A projectile is shot straight up from a height of 6m with initial velocity of 80 m/s. It's height is given by the equation :
h = 6 + 80t - 5t²
After how many seconds will it reach it's maximum height? What is that height?
Maximum height = y value of the vertex.
h = 6 + 80t - 5t²
h = -5t² + 80t + 6 <-- rearrange in correct order
h = -5 (t² - 16t + __) + 6 <--- complete square
h = -5(t² - 16t + 64) + 6 +320
h = -5 (t - 8)² + 326
Vertex = (8,326)
Maximum height = 326
Seconds = 8
Well that's what we focused on for today. The next scribe is Kirstie.

Monday, February 12, 2007

So today in class, for those of you who were not there, or were there and are still lost about the new concept we went over which was Completing a Square, here is a little summarization of what had happened.

We were first given equations that needed to be factored out, these were the examples:

x² + 2x + 1 --> (x-1)²
x² - 2x + 1 --> (x-1)²
x² + 4x + 4 --> (x-2)²
x² + 10x + 25 --> (x+5)²
x² -8x +16 --> (x-4)²

the solutions that are on the right hand side of the equation is the result you get from using Completing a Square method. By using the method which is take the middle term divide it by two and then square it.

Notice that square root of the third term in the equation is result you get when you factor using completing the square. For example, for the third equation the square root of 4 is 2.

Now, converting Y = ax² + bx + c into Y= a(x-h)² + k

For example:
Y = x² + 4x -5
Y = x² + 4x + ___ - 5
^ Notice when the equation is rewritten there is a blank space, that is for the unknown value to complete the square.
Y = (x² + 4x + _4_) -5-4
^ since you added 4 in the brackets you would need to subtract 4 from the outside of the bracket so it is balanced, and it won't change the outcome of the equation. Having said that, the answer that you would get would be:
Y = (x+2)² - 9

Now since your equation is converted into the Y = a(x-h)² + k, it will be easier for you to find these:

Vertex : (-2, -9)
A.O.S. : x = -2
Domain : (-oo, oo)
Range: [-9, oo)

Oh btw, the "oo" are supposed to represent infinity, sorry I'm new at this . okay anyways, back to the topic.

Okay, now let's try this example:

Y = 3x² - 12x + 16

Since the leading coefficient is more than 1, you need to factor this out. Which will give you an equation of:

Y = 3(x² - 4x + ___ ) + 16
Now use the completing the square method, so you can fill in the blank space.
Y = 3(x² - 4x + _4_) 16 - 12
The same method applies, and since you've added 12 in the brackets you would subtract 12 from the outside to balance it. The end result will give you,
Y = 3(x-2)² + 4

Now everything will be easier, when asked to find the vertex, axis of symmetry, domain and range of this equation.

The last example which was given to us was a tricky one, and this is how it looked like

Y = -2x² -5x + 3 ; if you already noticed as 5 being the middle term, it is not easy to divide it by 2. Since the leading coefficient of this equation is negative you would need to factor it out. Giving you this:

Y = -2(x² + 5/2x + ___) + 3 .

Now you would divide 5/2 by 2. By doing this is it the same as multiplying it which looks like this:

5/2 * ½ = 5/4
(5/4 * 5/4) = 25 / 16
Now you would multiply the 25 / 16 by 2/1 . Or if it is easier for you, you would right away notice that if you were to cross the two 16 divided by 2 would give you 8 . and now it would look like this:

25 / 8 * 1 / 1 which is 25 / 8. Now you would plug everything that you had done into the equation.

Y = -2(x² + 5/2+ _25/16_) + 3 + 25/8

24/8 + 25/8 = 49/8

you would end up with the equation looking like this:

Y = -2(x² + 5/4)² +48/8.

So far, I hope all of you reading this are getting me so far, If not I apologize because I'm not sure if I summarized it good enough. But, Basically it is taking the middle term , dividing it by 2 and squaring it. Then whatever you get you would put that as the third term in the equation inside of the bracket. Whatever you had added in the bracket you would do the opposite of it outside, just to balance the equation. Well, I think this about wraps about what we had learnt today in class. Yes, I know its quite long. Oh and due for tomorrow in class is Excercise 5, Questions 1 and 2 .

And the Scribe for tomorrow is ROSLYN =)=)

Thursday, February 08, 2007

Using the graphing calculator (TI-83)

Since I know how much of a pain these calculators are - and yes they are confusing to - I feel that posting a short little tutorial on some of the basics will help.

For starters, there are your basic functions you should all know from grade 10 pre cal using a scientific calculator, so I'm not going to mess with those. Those use the second function key which is the big yellow button on the top left hand side. There's also the addition of the Alpha key which gives you more options. And yes there are letters involved so you can type messages but I haven't figured out the space key - not that this is relavent to using it in class I was just having fun at the time and it is slight amusement if you're REALLY bored.

On a side note, if you go into an application you dont want and you have NO clue how to get out, just go [2nd] and [mode] to quit.

Few other keys to keep in mind: Clear key does as it says, it clears the screen. If you have multiple lines of problems you were solving, pressing it once clears the current line, pressing it twice or again clears the entire thing. If you make a mistake and dont want to clear the entire thing you can use the arrow buttons to move the flashing cursor over until its over the item you don't want, and pressing delete will get rid of that specific spot. If you missed something, put the cursor over where you want it to go, press [2nd] then the [del] key (if you look at it, "ins" is in yellow on top of the del key, hence pressing the 2nd key first) and then the key you want in there and it will be placed where the cursor is and push everything over. You CAN do this without using the insert key, but it just means you have to input every single thing after that that you want again. And if you want to replace something with something else, don't bother with the delete/insert thing, just press the key and its replaced.

If you need to change from Radiants to Degree's or back and forth, press Mode, there's an area with Radiants and Degree's, just highlight the one you want and press enter.

If you're looking for the "to the power of", you will find an x² that will only do a to the power of 2, if you want to do say to the power of 13, you need to put in the number, press the ^ key just beneath the clear key, and then press the number you want it to be to the power of, so on your calculator it will look something like this: 3x^5 for example, which you know what it is, I would write it out with the little 5 above the x but I can't find that particluar icon in my character map, only a 2 and 3.

Sin Cos and Tan are also in there, but if your used to typing the number in THEN tan or something, you're going to have to switch it, because the way it works is when you press one of those buttons it shows up as SIN( and inside the ( you put in your numbers, and if you're multiplying it by something else you need to close it with a ) otherwise it gets really messed up. Same thing with the inverses.

And before I forget, if you want to multiply something like (2 + 3)(5-4) or something like that you can put it in exactly like that in the calculator and it will multiply it out, thats what the brackets are for.

If you want to put an x into the thing, you dont need to go [ALPHA] + [STO ->], you can just press the [X,T,(theta),n] button - used (theta) cause thats another symbol I can't find - and an X will pop up.

Now on to using it for doing graphs.

As Mrs. Ingram showed us today in class, pressing the y= button takes you to a funky thing where you can input a y value. By pressing graph, that y value is displayed on the graph on the calculator. If you can't see the line or graph, pressing ZOOM then the number for zoom out/zoom in then enter TWICE (mine is one of the picky ones, some you only need to press once others you need to press twice) and the view will change. Going back to its original would be ZOOM -> standard.

If you want to make a table, you can go [2nd] + [GRAPH] and it will take you to a table. If you want to plot the points on the graph, you need to hit [2nd] [y=] for the stat plot, hit ENTER on the first one, then enter again to turn that one on. You can change the x list and y list, they correspond to the lists you just put in on the table. To change it from L1 to L2 for example, highlight that part, hit [2nd] then on numbers 1 to 6 you have L1 to L6, depending on which one you want. Once thats done, hit Graph and it should be displayed on your graph, if you cant see it press ZOOM and STAT (you have to scroll down a bit to see it) and it shold show up - I'm also going off memory for this and I can't remember for sure if thats it.

Pie key is [2nd] and [^]

Don't ask me what the [math], [apps], or [prgm] buttons do, I don't know.

In the y= area, if you want to change how your line looks on the graph, just arrow over to the / thing before the y= part and press enter, it changes to different modes.

If your too lazy to type out your answer after making a calculation, just pressing + or - or something like that automatically inserts ANS-> before hand, which means whatever is after the + is added on to the answer from your last equation.

Thats about all I can remember. This will get added on to as we go along with new stuff from classes but these are some of the basics and some that I have learned. Keep in mind these calculators take 4 AAA batteries so have a stash handy. To turn it off before it does automatically, just hit [2nd] and [on] and it will shut off. Hope this helps you guys out in getting the hang of things and give you a bit of a boost in class.

Wednesday, February 07, 2007

Pre-cal 30S 1st Blog

Here's the first blog of the semester:

In class today, we reviewed the stuff we were supposed to learn in Grade 10.. If you forgot.. here it is..

If you are given:
y=2x+15 (which is a straight line, when you draw it on a graph)
the slope is -> m=2/1 (m=rise/run)
just remember to use y=mx+b to help you find the slope which is represented by the letter "m" & y-intercept which is represented by the letter "b"
y-intercept: 15 <- to figure this out just set "x" to 0 ...
y=2(0)+15
y=0+15
y=15
Do the opposite for the x-intercept:
0=2x+15
-15=2x
-15/2=2x/2
x= -15/2
& just so you remember, x-intercepts, zeros & roots are the same thing.. so don't get confused.

So the Domain would be {x׀x € R}which means "x such that, is a member of real numbers"
The Range would be {y׀y € R} which means "y such that, is a member of real numbers"
Writing things such as {y׀y € R} or {x׀x € R} is called using "Set Notation," it pretty much just needs the 2 brackets {} around it to be considered "Set Notation".

Okay, well I can't get the pictures of the parabolas up but I think you guys pretty much know what it is.. it's like a line that curves and when you do the vertical line test it only touches ONE point of the line. The lines go on forever so you should be putting arrows on the ends of these lines.. other wise Mrs. Ingram said we could lose 1/2 marks on tests if we don't. + u should also label the graph for example, show where the "x", "y" is on the graph or you could lose marks for that, too. The general parabola is y= x²

The vertex is where the curve changes its direction, it'll have an "x" and "y" coordinate
The axis of symmetry is the "line" going through the curve the same amount of units or spaces or whatever you call it, from the middle of the curve. It's pretty much the line that separates the parabola from the midpoint or cuts the parabola in half so that both sides would be even. You can find if by setting x = 0

The hw that's due 4 2morrow is Exercise 1: Quadratic Functions, all the questions except # 12

Thursday, February 01, 2007

Welcome!!

Welcome to your class blog.
This is an online tool that is set up to help you help each other through this course.
Good luck in the course, and enjoy.

Mrs. Ingram