This is a blog for my second semester pre-calculus 30s students. Have a blast!

Tuesday, February 28, 2006

Fill-In Scribe Post

Since this is just a fill-in scribe post, I'll just make it as straightforward as possible and since most of us understand how to do this.


How many possible triangles?

* 1 triangle
* the angle is opposite the longer side

sin Y / 5 = sin 42 / 3

sin y = 1.1

* no
* /sin x/ > 1.


Find all possible values for side c.

5 / sin 28 = 6 / sin B

5 sin B / 5 = 6 sin 28 / 5

sin B = 6 sin 28 / 5

angle B = 34.3 degrees

To find the related angle:

Br = 180 - 34.3
B r = 145.7 degrees

There are two possible answers for angle C:

angle C = 180 - 28 - 34.3

angle C = 117.7 degrees

for the related angle:

angle Cr = 180 - 28 - 145.7
angle Cr = 6.3 degrees

Since there are two possible answers for ang
le C, we organize our work into a table.

when angle C = 117.7 degrees:
c / sin 117.7 = 5 / sin 28
c sin 28 = 5 sin 117.7
c = 9.4

when angle C = 6.3 degrees:
c / sin 6.3 = 5 / sin 28
c sin 28 = 5 sin 6.3
c = 1.2

last example: Find all possible angles for angle C.

19 / sin 19 = 23 / sin B
19 sin B =
23 sin 16
sin B = 23 sin 16 / 19
angle B = 19.5 degrees

Solving for the related angle:

angle Br = 180 - 19.5
angle Br = 160.5 degrees

Solving for angle C:

when angle B = 19.5 degrees:
angle C = 180 - 16 - 19.5
angle C = 144.5 degrees

when angle B = 160.5 degrees
angle C = 180 - 160.5 - 16
angle C = 3.5 degrees

And that's the end of today's lesson. I hope that helps.

Er.. do I have to pick a scribe for tomorrow? If I have to.. well, I pick:

rus-L. You can do it!

Scribe Responsibilities

Hey guys,

Remember the "scribe responsibilities" that we talked about before. It's been two days now with no-one filling in as the scribe and noone taking that responsibility.

Sunday, February 26, 2006


Here's #3. Is this what you're talking about? So, this is how I drew my diagram, hoping it's correct.
You find the sides 'c' and 'a' by using sine law.
You have to add all of the sides together to get the total distance that the person traveled.
I hope this helps. If it's wrong, I'm sorry. HAHA!

The homework

Can anyone help me on # 3?? Do I use Tan or Sine to figure out the side?

oh my goodness ....


I am having the hardest time with the assignment that is due tomorrow. I have literally been staring at it, and I have no idea what to do, or where to start. Can someone help me? If not, I know I'm going to fail......yikes..

I'm super frustrated....

Friday, February 24, 2006

First Weekend Scribe Post

My first time doing the scribe post and thankfully I got the weekend job. It's not going to be a very long one since we did have the quiz in the middle of class.

After the usual mental math we started to correct the homework sheets that we had for homework the previous night, we also got some hints for the hand-in assignment that's due on Monday at 9:00:

(My computer's been acting nutty with any kind of pictures lately so you'll just have to bare with my descriptions for now, sorry guys.)

#5) For the diagram you drew one large square and a smaller one in the center of the first. Inside the lengths are of course 30 and 40 meaning the area is 1200. on the outside the width and length is representing by X with the area being 984. The equation to use is y=(40+2X)(30+2X).

#6) Replace TAN and SIN with X and Y:


Since the question is worth 5 marks that meant that there are 5 answers, so all you have to do now is solve the equation.

And that's all I can remember from today, oh yeah as for the quiz the only thing I can say is that it's not hard but then again it's not easy, can't give you much more then that since I don't remember much of it, it's how my mind works once a test or semester is over my mind gets wiped blank except for the things that were really pounded into my head, it only starts to come back when I'm back in class.

The next scribe will be, Jonathon!

LOST: episode 2

I am SCREWED! I'm okay with the lessons I missed on Tuesday since they were fairly easy; also with wednesday's topic - about related angle -- sin, cos, tan and stuff, then the quadrants. Those are some easy stuff. But wut are these 'bquadrantrent' and so forth?! I'm totally lost! And I'm missing that quiz! I am screwed! Then that assignment is pretty hard. I don't get the last two questions. I amy be slow at understanding Dino's post but it's hard to understand especially if I missed the class. It would be easy for those who actually witnessed how things happen. Someone help me! UH! I hate getting sick!

Question... again....

Mr. M... is there any way to solve for sine, cosine or tangent manually?...

Also, i wanted to ask if:

tan=cos/sin or tan= sin/cos

Which one is right??

I think this actually has something to do wwith the assignment... since if tan= cos/sin... i could cancel the sin and simplify the problem...

this post was made late coz i went to work...

Thursday, February 23, 2006

Solving Trig. Equations

To Solve Trig. Equations

There are two simple rules:
  1. Find values which satisfy the equation.
  2. The domain will always be from [0°,360°], unless stated.

Here are some examples to help you:

ex. sinø = 0.78615 Solve!

To solve this question, you first take your (TI 83) calculator and press the (2nd) button, then the (sin-1) button, and type in the number 0.78615. then the calculator will give you a number of 51.83° (round of to two decimal places). Then you need to find out which quadrants have a sin value that is positive (Wednesday's class), which is bquadrantrent 1 quadrantrent 2. So you already hquadrantrent 1 done, it is 51.83° , so to figure out the degree valuequadrantdrent 2 you take 180° ( 180° is the maximum degree value you can have in Quad. 2) and subtract 51.83° from it . The value for Quad. 2 is 1that's°, thats it your done.

sinø = 0.78615

ø= 51.83°

ør = 180 - 51.83 = 128.17°

ex. cosø = .43214 ( for (cos) use the same method as above but press the (cos-1) button instead of the (sin-1) button)

ø = 64.40°

ø= 360 - 64.40 = 295.6° ( you have to subtract it from 360° , because cosine is positive in quad.1 and quad. 4).

Now try this one on your own. (answer at bottom of page)

2 sinø-1=0 ( hint move the 1 to the other side)

For homework do exercise #9 and ask Mr. Malandrakis for a sheet for review, there is also a quiz tommorow on quad. Equations and trig. equations. So study and Good luck.

Answer to question uptop is ø = 30° , 150°

Tommorow's scribe is Jason.

By: Berardino Petrelli (Dino)

Wednesday, February 22, 2006

Angles in standard position, reference(related) angles, and the like.

Helloo fellow scribes ...

I am your scribe for today, and hopefully this will start the scribe program up again, and no one will get confused and mxed up again.
Today in class we learned about Angles in the standard position. We started off with this recall about quadrants: the top right section of a coordinate plain is the first quadrant. Moving counter clockwise from the first quadrant makes the top left section of the coordinate plain the second quadrant, and so on.

When the postive x axis is the first side of an angel, that is called the inital side or arm. When a ray is drawn through the origin between 0 and degrees, this ray is called the terminal side or arm. The angle that the initial and terminal sides make is called an angle in standard position.

When an angle is in standard postion:

1) it's initial side is the x axis, and the terminal side is formed by a ray extending out from the origin.
2)the angle is measured in counter clockwise direction


Remember : r= the square root of x^2 +y^2
if (x,y) is any point on the terminal side of an angle in standard position:
The SIN of the angle is y/r
The COS of the angle is x/r
The Tan of the angle is y/x

This diagram lays out just what you have to do to find the angle you are looking for.

Remember : unknown angles are usually expressed with the greek letter theta.

Some angles have special relationships with each other. This relationship could be referred to as a Reference/related angle.

A related/reference angle is the acute angle formed between the terminal side of the angle and the nearest x axis.

We were also given a summary of reference angles, and basic facts about SINE and COSINE graphs. This I do not need to type here because I am sure you all have it. Also, there is a project due on monday, about quadratic functions, trigononmetry, and a little bit of mental math. You can work with anyone, just make sure you hand it in on monday!

Homework for tonight is Exercise #8....

If anyone has any questions, or comments, feel free to contact me. I may have made a mistake on something, so feedback is always appreciated. I am kind of in a hurry, but I hope this post does justice to all the ones that have come before it!

And the scribe for tomorrow is....Dino!

Good luck man, I know you will do well! Keep up the good work guys!


Tuesday, February 21, 2006

sine and cosine graphs...

This is something I got from my trigonometry class last year in the Philippines...

I hope this can help most of you guys in figuring out how the graphs work...

Image Hosting by

sin graphs always start at zero (only if there is no vertical shift)

+ cos graphs start at the MAX point

-cos graphs start at the MIN point

These graphs have zero vertical shift and zero horizontal shift... so it's just the basic form of the graphs...


sin -360= 0
sin -270= 1
sin-180= 0
sin -90= -1
sin 0= 0
sin 90= 1
sin 180= 0
sin 270= -1
sin 360= 0

cos -360= 1
cos -270= 0
cos -180 = -1
cos -90= 0
cos 0= 1
cos 90= 0
cos 180= -1
cos 270= 0
cos 360= 1

I know this isn't that much of a help since these are obvious data in periodic graphs...

y= asinx + k y= acosx + k

k= vertical shift
a= amptitude... (did i spell it right?) or also known as the stretch factor of a periodic graph

Well... that's how i understood the stuff...

Hope this can actually help...

Sorry if the post was made this late coz i was finishing some other stuff...

What happened..

I'm getting really anxious. Where's the scribe post? I have no idea what's going on. I and Vince are worried. If this goes on...

Mr. M, I'm going to miss tomorrow's class too because I have to host games for the junior high kids.

Scribe responsibilities

This is a quick comment based on what I've seen the last couple of days. Yesterday Jojo, who was the assigned scribe, didn't get a chance to make a post because his internet wasn't working. That's no problem but he assigned Andrew to be today's scribe and it appears that he perhaps forgot to make a post and Candace made an open post as you can see wondering what hapenned in class today.

Keep in mind folks, the main reason this blog exists is FOR REASONS LIKE THIS, when people miss class or are not understanding what went on in class. I have a challenge for, from what I can tell so far to be a very mature, intelligent, and motivated class (that's you guys!). If one of you dosen't see a fresh post by let's say around 8:30 or so at the latest, please take the mature and responsible initiative, for your fellow class members, and voluntarily be the "fill-in" scribe for the night. If it even makes a difference to one or two people, like Candace tonight, then it's all worth while.

Remember you're only "required" to post on class lessons once every 25 classes or so, that's once every month and a half or so. So please adhere to this very miniscule responsibility as some of your classmates truly depend on the scribe to do their job as it just might make the difference for them.


Hello everyone...
ok, well I was just wondering what happened in class today?? Also who is the scribe for today??


Are TI-89 calculators the same as the calculators we use?? or even somewhat similar...

Monday, February 20, 2006

Sunday Funday

Well, well, well, it's the end of another weekend, and it's the start of a new week. Why don't we start a new week with games, hey? (I got the idea from Mr.K ",)

This is a game from last semester's Pre-Cal 20S blog. This is a hard one, I tell you. It took me weeks to figure it out. First person to get out of the room gets a free pack of gum from me. =)

Saturday, February 18, 2006

Scribe # 4: Weekend Scribe

Scribe #4 : Weekend Scribe
Hey guys! So.. a test on Monday, right? I'm actually kind of panicking right now. I might not do so well on the test. Well, before I state what we did on Friday, let's review some important things that we have learned in the past several days. These terms will be essential for Monday's test.
1) Quadratic Function - any function or equation with degree 2
f(x) = x2 + 2
2) Parabola - the graph of a quadratic function

3) General Form - equation of a quadratic function written in: f(x) = ax2 + bx + c
y = 4x2 + 5x + 15

4) Standard Form - equation of a quadratic function written in: g(x) = a(x - h)2 + k
g(x) = 4(x - 3)2 + 7

5) Zero Product Property - one of the many methods of determining the roots of a parabola
y = x2 - 3x - 4
0 = x2 - 3x - 4 [ make y = 0 ]
0 = ( x + 1 ) ( x - 4 ) [ factor ]

[ solve for x ] 0 = x + 1 x - 4 = 0
-1 = x x = 4

[ state the roots of the equation ] roots = -1, 4

6) Completing the Square - a method used to transform an equation of a parabola from general form to standard form
- shown by Alge-tiles

y = x2 + 4x - 5
y + 5 = x2 + 4x [ move 'c' over to the other side (do not forget to change its sign) ]
y + 5 + 4 = x2 + 4x + 4 [ add 'c' to both sides by finding the half of 'b' and the square of whatever the value might be (ex. half of 4 is 2... 22 is 4... therefore, 'c' is 4) ]
y + 9 = ( x + 2 )2 [ factor ]
y = ( x + 2 )2 + 9 [ move 9 over to the right side, this will be your standard form of the equation: y = x2 + 4x - 5 ]

7) Characteristics of a Quadratic Function:

a. VERTEX - a point (x, y) directly through the center of the parabola
b. AXIS OF SYMMETRY (a.o.s. - Mr. Malandrakis' version) - equation of a line passing through the vertex
c. OPENING - if a > 0, parabola opens up (has a MIN)
- if a < color="#3333ff">MAX)
e. ROOTS - where the parabola crosses OR touches the x - axis [ x-intercept, zero ]
f. DOMAIN and RANGE - limits of x (domain) and y (range) axis
+ ways to express DOMAIN and RANGE:
Set Notation
Number Line
Interval Notation
g. WIDTH - how wide the parabola is
- depends on the coefficients of 'x'
- if coefficient > 0, parabola is THIN
- if 0 > coefficient <>
8) Vertical Shift - identified by the letter 'c' of the general form
- if 'c' > 0, parabola shifts up from the vertex
- if 'c' <>
9) Horizontal Shift - graph is shifted h (absolute value of 'h') units horizontally

10) 'a' controls the width and the direction of the
11) 'h' controls the horizontal shift

12) 'k' controls the vertical shift

NOW... this is what we did on Friday (that took a while)..
A pre-test for Monday about Quadratic Functions1) An archer shoots an arrow into the air that its height at any time( t ) is given by the function h(t) = -16t2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?
h = -k / 2a
4 = -k / 2 9 16 0
4 (32) = -k / -32 (-32)
-128 = -k
-128 / -1 = -k / -1
128 = k

2) [pretty straight forward]

3) What is the y-coordinate of the vertex for the function f(x) = x2 - 12x + 7?

SOLUTION: use complete the square or the 'h' formula...
f (x) = f(x) = x2 - 12x + 7
y - 7 = x2 - 12x
y - 7 + 36 = x2 - 12x +36
y +29 = (x - 6)2
y = (x - 6)2- 29


4) Find the exact values of the root of the quadratic equation: x2 + 8x + 2 = 0

SOLUTION: complete the square
x2 + 8x + 2
x2 + 8x = -2
x2 + 8x + 16 = -2 +16
x2 + 8x + 16 = 14
root of ( x + 4 )2 = root of 14
x + 4 = + or - root of 14
x = -4 + or - root of 14

AND THE REST ARE JUST ON YOUR SHEET. I WAS GOING TO WRITE MORE BUT I THINK IT'S POINTLESS BECAUSE EVERYONE HAS THE ANSWERS AND SOLUTION. YES, I DO GET LAZY (sometimes). Well, good luck to you guys! And good luck to me too! HAHA! I'm done for now... I hope this post helped some of you. I'm so tired. It took me 2 and half hours to do this thingy. Perhaps, it's also because my computer is slow.

The NEXT SCRIBE is JOJO! Sorry, man... Just do it...

PS.. DONT FORGET OUR HOMEWORK! > exercise 7.. and the sheet Mr. Malandrakis gave us..

Friday, February 17, 2006


Hey guys,

I just found some errors on the answer key on the Quadratic Functions review Mr. Malandrakis gave us. The answer for 1a is the answer for 1b, and the answer for 1b is the answer for 1a. I hope this reduces the confusion.

Have a jolly weekend!

Thursday, February 16, 2006

I'm the scribe for Friday.

So guys, I'm the scribe for Friday. But is it okay to do it sometime on Saturday since it's the weekend? It's because I signed up for leadership tomorrow and I won't be home until 9:30. So I'm wondering if it's okay to do it the next day. I don't consider Leadership more important than Pre-cal. But, I didn't know I'm going to be the scribe for tomorrow. So yeah.. can I post it on Saturday?

Scribe # 3 "Applications of Quad. Functions"

Okay lets get down to what we learned today in class. Most of what we did today in class was based on what we did this whole week with completing the square and changing expanded form into standard form using the skill of completing the square. All we did today was applied it to real life situations.
We were given a sheet and Mr. M walked us through it.

Before I get to the questions I'd like to point out some of the instructions that were given to us to help us understand. Such as read the question carefully. We were also instructed to look for any words which could help make the question clearer such as MAXIMUM or MINIMUM. And drawing a diagram can also prove useful too.

Question #1

What is the maximum rectangular area that can be enclosed by 120 m of fencing, if one of the sides of the rectangle is an existing wall?
*NOTE: the word "maximum" was used

Work and Explanation

First we must come up with a base equation.
since we know the perimeter is 120 and that the sum of the fenced sides must equal 120 we come up with equation: 2w+l= 120

Now we make a statement for l.
we do this by moving the 2w over.
l= -2w+120

Now to find w we let y = area.
y= (-2w + 120) (w)

Convert into Standard form.
To do this we will use the skill of completing the squares to convert this into standard form.
-1800=-2(w^2-6Ow + 9oo)
=-2(w-30)^2 + 1800

Understanding the vertex.
The "k" value in the vertex is the area of the rectangle.
Therefore, max area = 18OO m squared

The homework for tonight is Exercise 6. The scribe for tommorow is vincent.

Wednesday, February 15, 2006

The Scribe List

Now Mr. Malandrakis has implemented the "scribe" posts, we should have a Scribe List. I'm making this, and I will update it everyday so that we will not lose track of who will be the scribe. If you see someone's name crossed off on this list you CANNOT pick them as scribe again, unless we start a new cycle. We start a new cycle when everyone has been picked to be a scribe.

You can access this by the Links section, and you may want to take a look at the scribe list before you pick a scribe.

Cycle 2

John D. - #12
Jamilyn G.

Scribe Post #2

Well hello everyone.. I'm Tim-Math-y your scribe for today. I'm going to try and do this quick so everyone can read it and so that it won't take up too much of my time.

Today, we started off class with Mental Math which included 11 questions. Next we did a 10-15 minute review of last night's homework: Ex. 3 (questions 5a, 5b, 7, 8, 4c, and 11).

Now onto today's lesson!


*This proccess is referred to: completing the square.

When thinking about this, we think back upon "Alge-tiles".

---Sigh.. unfortunately, it says error when I try to add images.. so please bear with me on this Scribe post holding no visual pieces... (-.- )---

y = ax2 + bx + c ---> General / Expanded Form

- a square with length of (x+1) and a width of (x+1)
- the area would be expressed as (x+1)(x+1) --> x^2 + 2x + 1

- x^2 + 2x + _?_
- to complete the square, the unknown is equal to 1.

- x^2 + 4x + _?_
- to complete the square, the unknown is equal to 4.

Do you see how to find the unknown value of C by looking at what's givin?

*In a quadratic func in general form:
C = (b/2)^2 ... is used to complete the square

>>>a perfect square trinomial is where a trinomial has 2 equal factors.

Example: x^2 + 6x + 9 = (x + 3)^2
Example: x^2 + 8x + 16 = (x + 4)^2

>>>convert from general form ----> standard form by completing the square.

y = x^2 + 4x - 5
Step (1.) y + 5 = x^2 + 4x
Step (2.) y + 5 + 4 = x^2 + 4x + 4
Step (3.) y + 9 = (x + 2)^2
Step (4.) y = (x + 2)^2 -9
V = (-2, -9)

y = 3x^2 - 12x + 16
y - 16 = 3x^2 - 12x
y - 4 = 3(x^2 - 4x)
y - 4 = 3(x^2 - 4x + 4)
y = 3(x - 2)^2 + 4
V = (2, 4)

Now that those two are done, Mr. Malandrakis decides to blow our minds with the so called "friendly" fractions..

y = -2x^2 - 5x + 3
y - 3 = -2x^2 - 5x
y - 3 = -2(x^2 + 5/2 x)
y - 3 - 50/16 = -2(x^2 + 5/2 x + 25/16)
y - 98/16 = -2(x + 5/4 )^2
y = -2(x + 5/4)^2 + 49/8
V = (-5/4, 49/8)

__________ / / __________

*In any quadratic function in standard form:
h = -b/2a
K = 4ac - b^2/ 4a

y = x^2 + 4x - 5

h = -4/2
h = -2

K = (-2)2 + 4(-2) - 5
K = 4 + 8 - 5
K = -9

V = (-2, -9)

And that is all folks! Wow.. I can smell dinner from downstairs.. I'm sure HUNGARY :). But before I am done, have to mention some things ^^.

Mr. Malandrakis: "It is very important to take some time to read what's on the blog."
Tomorrow's Scribe: (sorry no drum roll :P) It is.. MR. "StevenF" because he said it's fine by him ^.-
Homework!: (which is essential for this course) EXERCISE #5

And for sure now, I think that is all.. see you guy tomorrow! I'm going to eat dinner :P

Tuesday, February 14, 2006

Valentine's Scribe


It's true that we learn new stuff everyday. Today in Pre-Cal, we learned about determining characteristics of a quadratic function algebraically, or in other words, without you using a graphing calculator.


First of all, we now know what is the standard form of a quadratic function. Now on this lesson, we used the expanded form mostly.

The expanded form of a quadratic function is:

y = ax2 + bx + c

1) a represents the width and direction of the opening of the parabola.
b creates an oblique translation (a translation of a line that is neither parallel nor perpendicular)
3) c is basically the y-intercept of the parabola.


Recall: Find the roots and the vertex of f(x) = 6x2 + x -12 using zero product property.

Factoring the expression:
f(x) =
6x2 + x -12
f(x) = (2x + 3)(3x - 4)

Ah.. this involves fractions. So this is how we set it up to find the roots:
2x + 3 = 0
| 3x - 4 = 0

(note: ALWAYS organize your solutions.)

Solving for the roots:
2x + 3 = 0 | 3x - 4 = 0
2x = -3 | 3x = 4
x = -3/2 | x = 4/3

So the roots of the equation f(x) = 6x2 + x -12 are:
x = -3/2, 4/3.

* the x coordinate of the vertex a.k.a horizontal shift is found by finding the midpoint of the roots:

h = (x1 + x2) / 2

To find the horizontal shift, we just need to substitute x values in the equation with the roots.

h = (-3/2 + 4/3) / 2

Ah... dissimilar fractions. Now we need to find the LCD of the equation so we can solve them. LCD = 6.

h = (-9/6 + 8/6)/ 2
h = (-1/6) / 2
h = -1/12

Now we have found the x coordinate a.k.a horizontal shift of the vertex. We just need to find the y coordinate a.k.a vertical shift so that we can find the vertex. To find the horizontal shift, we substitute
"h" back into f(x).
[*Note: This involves ugly fractions for this particular solution, so please just bear with me.]

And oh, don't make it hard for yourself, just use the factored form of the expanded form of the parabola since it's easier. This is one handy tip coming from me. ;)

f(-1/12) = [2(-1/12) + 3][3(-1/12) + 4]
f(-1/12) = [-1/6 + 3][-1/4 + 4]
f(-1/12) = [-1/6 + 18/6][-1/4 + 16/4]
f(-1/12) = [17/6][17/4]
f(-1/12) = -289/29

Now we have the y coordinate of the vertex a.k.a horizontal shift. Therefore, we can state that:
vertex: (
-1/12, -289/29)

ex. Find the roots, vertex, axis of symmetry of f(x) =
x2 + 4x -5

Solving for the roots:
x2 + 4x -5
0 = (x + 5)(x - 1)

x + 5 = 0 | x - 1 = 0
x = -5 | x = 1

Therefore, we can say that the roots:
-5, 1.

Finding the x coordinate of the vertex a.k.a horizontal shift:

h = (
x1 + x2) / 2
h = (-5 + 1) / 2
h = (-4) / 2
h = -2

Now that we know the x coordinate of the vertex a.k.a horizontal shift, we can also state the axis of symmetry which is:
x = -2

Finding the y coordinate of the vertex a.k.a vertical shift:

f(-2) = (-2 + 5)(-2 -1)
f(-2) = (3)(-3)
f(-2) = -9

And there you have it! We can now state the coordinates of the vertex:
-2, -9)

We also came up with the rules of finding the roots, axis of symmetry, and vertex of a quadratic function in expanded form.

1) To find
- factor
y = ax2 + bx + c using zero product property

2) To find axis of symmetry, find the midpoint of the two roots. MAKE SURE to always write it like this:
x = h

3) To find vertex, substitute h for f(x) to get the k value. MAKE SURE to enclose the points of the vertex in round brackets as it is a point in the line.
h, k)

And there you have it! That's today's lesson. Sorry it took a long time to post. Homework for today is Exercise 4, and tomorrow's scribe is..... (drumroll please)......
tim-Math-y. Haha.

Have a great evening people! Get out there and commit heinous acts of kindness.. Cheers!

Sunday, February 12, 2006

the quizz! :( the quiz..yeah i didn't do so well on that..hahh. i knew how to like do all the other stuff. that was easy, but the last page screwed me over. i like didn't know how to do the last three or four questions, which is yenno...bad.hahah...well i hope when we go over it i'll be able to understand...this is my first real blog, if this is a blog. i don't know what i'm doing or if it is right, so can someone tell me how i'm supposed to used this....hahh i'm a BIG noob. hah yupp..well see you guys at school tomorrow! bye:D

Saturday, February 11, 2006

Quiz correction

Hey there folks,

A minor correction from Friday's quiz, which we'll go over in class on Tuesday (Monday is the assembly) anyways but I'll give you a heads was the last question on the quiz where you were asked to write the equation of the graph of the parabola. You were asked to leave your answer in expanded fom, and the question was supposed to ask you to leave it in expanded form. If you left it in standard form you received full marks. Just a little "FYI" for everyone who checks out the blog on the weekend. Also another "FYI" the marks (overall) on the quiz were great. A 72% class average, which is rare and outstanding. Great job overall and keep up the fabulous posting. case you guys haven't realized, if you look in the links section on the side, you'll find some great sites that really explain quadratic functions in full colour and worked out examples. You should check them out if you're still having trouble.

Friday, February 10, 2006

Blog, Quiz


* I think this blog is cooler than i thought. I just regret that I didn't view it the day before the quiz (yesterday). Man, seriously. How can you guys spend so much time on this?! I used to update my AA page like every time. But I got lazy. I know this is not that much different form those pages but I think I lost my touch on web designs. You guys did a good job on your posts.

* I have a question. Do we have to write like this in here every time? proper spelling, capitalization, and punctuations? coz i wanna write like this... its so much easier..

* Anyways, I decided I'm going to go here every night to see the updates that might help me dig deeped into the topic.


* That quiz SHOOK me! I thought it was pretty easy even though I kind of screwed up on the last page. I didn't understand the EXPANDED FORM. Is it this one? :

y = ax2 + bx + c

* I didn't know what to do then. I just left my answer in standard form:

y = a(x - h)2 + k

* I think I will get that wrong.

* Also, about Zero Product Property... Are you supposed to enclose the points in BRACKETS? I didn't do that actually. I thought it would be okay just to leave it the way you factored it. AHHHH! I hope I did well on that quiz.

* I think QUADRATIC FUNCTIONS is easy overall. I just have to study more. Pay attention to little details since they might play an important role on the answer/equation.

Look I'm Back...

Hi fellow bloggers. If you don’t know my name… My name is Richard. If you don’t know why I’m here… Then… I don’t either. But anyways. The main reason I’m like on your blog is that I’m here to help you guys. Yes as impossible as it may seem. I’m here on the blog. So live with it. I actually passed this course last semester if that’s so hard to believe. I know it is for me. But in other cases I’m just around to help the fellow bloggers blog and to give at least a tiny bit of incite into every thing you do on the blog. Yes I’ll be commenting allot and allot and that means allot. I’ll also be posting up allot of blogging tips and crazy help from myself the remnants of the first semester bloggers. Who knows… A study group may get back together again. Hahaha. Well anyways. I just wanted to say hi and bye. I’ll be back…

Oh yeah. Here’s a tip for Mr. Malandrakis… Make a chat box. Because it’s actually a good study tool and it creates a classroom outside a classroom. ;) And... If you want to read the course... Then.. Read PC30S first Sem. It's all there. There's some intresting posts there too. You know... Pop goes the weasel... Hahahahaha

Richy Out.

Some Blogger Help

Hello there!

Now, making a post with a lot of detail takes time, like the one I just did. Thankfully, I got some help from the post that I found in the last semester's Pre-Cal 30S class of Mr. K. It was created by Graeme and Richard and it sure was a lot of help. This post contains a lot of stuff you can do in Blogger e.g. creating tables, making a marquee, stuff like that. I hope this helps.

Have a good weekend!

Edit: Mr. Malandrakis, when can we have a chatbox? It'll be really helpful and marvelous to have one. It'll encourage student-to-teacher and student-to-student interaction a lot. And a tag can help us too. In case you don't know, when we are bored, we go to Google to find some sites that may be able to help us out in the course. Cheers!

Remember: You can find everything in the Internet. If you can't find it, you are not looking hard enough. ;)

And I found another post that is really important. NO, NO, NO, NO, NO! We don't ever do this. Never. And I'm sure that some of us from Mr. K's class can remember this. =D happy times


The quiz was actually longer than i expected. I'm used to doing quizzes with only 5 questions that's usually out of 14 marks. But i guess i did pretty good. I just had trouble with question no. 3. But i got it after i handed in my paper. I was surprised that i answered all those questions. I guess i understand it now.

my first post

This is my first post, just seeing if it works. Great job guys!

Thursday, February 09, 2006

Scribe Post?!

Hi everyone!

I just decided to make a scribe post about what we learned today and I have nothing to do. Still concerns quadratic functions, it's turning out to be alright.

Anyway, we started the class with Mental Math. It wasn't that bad.. Then we moved on to homework. I asked how to solve number 11 on Exercise 2. Everyone didn't get that question, so we went into small groups trying to figure out how to solve it. The question is:

Jim and Kim each have money to buy ice-cream cones. Unfortunately, Jim is 24 cents short of the price of a cone and Kim is 2 cents short. They decide to pool their money and buy a single cone, but they discover they still don't have enough money. What is the cost of an ice cream.
We set it up like this:

let x = cost of ice cream
(x-2) = Jim's money
(x-24) = Kim's money

Then, we combine similar terms.
(x-2)+(x-24) = 2x-26

This is not a linear equality, this is a linear inequality as it is said in the problem that when they pool their money, they are still short. So we set it up like this:

2x-26 < (6)

We solve the equation like a linear equation.

Putting the xs on one side and the constants on t
he other side:

2x-x < (6)

Combining like terms:

x < (6)

Now, if Kim has (x-2) cents (
24 cents) and it's not enough, then one ice cream cone must be 25 cents. Let's see if that's true:

If x=25;
(25-2) = 23 cents (Kim's money)
(25-24) = 1 cent (Jim's money)
Total: 24 cents

Therefore we can say that one ice cream cone must be 25 cents. This may become useful in future encounters, so take note.

Today's lesson: More on Quadratic Functions

The standard form of a quadratic function is:
y = a(x-h)2 + K

1) The vertex of this equation is: (h,k)
2) Axis of symmetry: x = h
3) If a > 0, width is thin.
a < style="color: rgb(204, 51, 204);">a controls width and opening.
4) "h" controls horizontal shift.
k" controls vertical shift.
ex (Sorry for the poor graph, I'm a noob at MS Paint)

Here, we plot a parabola with the coordinates (3,0), (1,4), (0,3) and (-1, 0).
We need to

state all eight characteristics of this parabola and create an equation(y = a(x-h)2 + K ).

Recall: The eight characteristics of a parabola are:

1) Vertex
2) Axis of symmetry
3) Domain
4) Range
5) Roots
6) Opening
7) Max/Min
8) Width

Now we state the characteristics of the parabola above.
1) Vertex: (1, 4)
2) Axis of symmetry: x = 1
3) Domain:
(-∞, ∞)
4) Range: (-∞, 4)
5) Roots: (3, 0), (-1, 0)
6) Opening: down
7) Max/Min: max
8) Width: none, since it's the base width (note: we can either write a dash (-) or none if it's the base width.
If a = 1, it's the base width.

Since now we have enough information, we can create the equation.
(y = a(x-h)2 + K )

Substituting the vertex coordinates:
y = a(x-1)2 + 4

To complete the equation, we need to find a. Any point from the parabola won't matter. I'm picking (0, 3) for this one.

Substituting x and y coordinates:

= a(0-1)2 + 4

3 = a(1)2 + 4
3 = a + 4
-4+3 = a
a = -1

Final equation:
y = -(x-1)2 + 4

ex If (1, 3/2) lies on this parabola
y = (x-2)2 + k, find the minimum value.
Substituting the x and y coordinates:
3/2 = (1-2)2 + k

Solving the equation (k):
3/2 = (-1)2 + k
3/2 = 1 + k
-2/2 + 3/2 = k
1/2 = k
Now we have the minimum value, we can complete the equation.
y = (x-2)2 + 1/2

Phew. Now, on to the last example.
Find the root(s) of
y = x2 - x - 2.

What we would do normally when we see x2-x-2 is factor them out. So we do that to find the roots.
0 = (x - 2) (x + 1)
x = 2, -1

* Note: This is called zero product property.
ex. Use
zero product property to find the roots of: y = x2 + 5x + 6
0 = (x + 2) (x + 3)
x = -3, -2
End of lesson. Phew, that took a long time. And oh, we have a quiz tomorrow, and homework is exercise 3.

Wednesday, February 08, 2006

Great Posts!!!!!

Katrina, that was a fabulous post. I'm not sure if you wrote this blog tonight or back in semester one but it was great. The diagrams were out of this world, excellent job on that. Carla, I hope that helped a bit with your trouble with understanding interval notation.

Just remember that interval notation is just a simpler way of expressing domain and range. You use round brackets when that value is NOT included and square brackets when it is included.

If it still dosen't make sense by tomorrow, remember to bring it up after mental math! Great job again!!!