Reflection

so i hope everyone had a great christmas and i guess we were supposed to do another blog but i dont know. Here is some review stuff from chapter 8. We learned how to find the angle of inclination, which i found was really easy compared to some stuff we learn. We also learned about amplitudes, periods, vertical shifts, etc. There are some formulas we had to learn to be able to work these problems:

1.) For a line
m=tan alpha where m=slope and alpha=angle of inclination
2.) For a conic
tan 2 alpha=B/A-C
3.) For a conic if A=C then
a=pi/4

EXAMPLE:

Find the angle of inclination.
2x+5y=15
m=-2/5 tan alpha=-2/5 Checks are in the II and IV area and 21.801 degrees in I
alpha=tan^-1(-2/5)
180-21.801 alpha ~ 158.199 degrees, 338.199 degrees
158.199+180

reflection 18

alright, i forgot to do this sunday..so this is the blog for december 20. soo...that exam, pretty hard...but im so glad that we are off for the christmas holidays! i guess i will do my blog on 7-3, this stuff i actually understand and i think it is pretty easy. all you have to remember is the little chart. which is:

sin= y/r
cos=x/r
tan=y/x

csc=r/y
sec=r/x
cot=x/y


and to find r it is the square root of x^2 + y^2.

this chart is really easy, all you have to remember is the first part and then the second part is just the opposite.
and there is also the unit circle that goes along with this, but i can't really draw that on here..

heres an example:

find all six trig funtions (1,-1)

sin=-square root of 2/2
cos=square root of 2/2
tan=-1
csc=square root of 2
sec=square root of 2
cot=-1

REFLECTION #18

Okay so since I don't have my notes with me I'll just go over some simple things that we learned early this year. We should all know how to do these really good by now. ha. I'll explain the different types of factoring.

First there's factoring by using simple algebra. Here's an example of that:

Ex. 1.) 2x^2 + 4x = 6
*the first thing you notice that you can do is factor out a 2x like this:
2x(x + 2) = 6 >>(and what's in parenthesis is what you get after you divide by 2x)
*then you take both things and set them equal to 6 and then solve:
2x = 6 >> = 3
x + 2 = 6 >> = 4
*So your final answers in point form are (3,0) (4,0)

Next there's factoring....(I don't know exactly what it's called so I'll just say it's regular factoring)
Here's an example of those:

Ex. 2.) x^2 + 7x + 12
*to factor this equation you can say, "what are the factors of 12 that add to give me 7?"
*factors of 12 that add to give you 7 are 4 and 3
*so since both signs are addition you write it as >> (x + 4)(x + 3)
*then you solve for x by setting them both equal to zero. x + 4 = 0 x + 3 = 0
*And your final answers in point form are (-4,0) (-3,0)

Ex. 3.) 2x^2 + 5x + 3 = 0
*to factor this you have to first multiply the outer numbers (2 and 3) and you get 6. Then you can ask yourself "what are the factors of 6 that add to give me 5?"
*that would be the numbers 3 and 2
*so you rewrite your problem like this: 2x^2 + 3x + 2x + 3 = 0
*now since you have 4 terms, you can group them like this: (2x^2 + 3x) + (2x + 3) = 0
*now solve. you can take an x out of the first parenthesis. so you get x(2x + 3) + 1(2x + 3) = 0
*then you get (2x + 3)(x + 1)
*then set them both equal to zero and solve for x.
2x + 3 = 0 >> 2x = -3 >> x = -3/2
x + 1 = 0 >> x = -1
*Your final answers are (-1,0)(-3/2,0)

The other two ways to factor are the quadratic formula and completing the square. You use the quadratic formula when regular factoring doesn't work and when the linear term is odd. You use completing the square when the linear term is even.

Here's a few examples:

Ex. 4.) Complete the square. x^2 + 8x + 9 = 0
*the first thing you want to do is move the 9 over to the other side. so you subtract 9.
x^2 + 8x = -9
*then you leave a space between the 8x and the equal sign like this:
x^2 + 8x = -9
*then you take the linear term, which is 8x, and divide it by 2 and square it. so you get 16.
*then you add 16 to both sides of the equation like this:
x^2 + 8x + 16 = -9 + 16 >> then simplify it and get this:
(x + 4)^2 = 7
*then take the square root of both sides and you get this:
x + 4 = +/- square root of 7
*then subtract 4 over and get x = -4 +/- square root of 7

Now for the quadratic formula!! woot woot.

Ex. 5.) 2x^2 + 6x + 5 = 0

*the formula you use is x = -b +/- square root of b^2-4ac/2a
*so when you plug the numbers into the formula you get x = -6 +/- square root of 36-4(2)(5)/4
*then simplifying that you get: x = -6 +/- square root of -4/4
*and simplifying it again you get: x = -3 +/- i over 2

And there yah go!!!! (:
MERRY CHRISTMAS EVERYBODY (:

reflection

Im gonna show you something in chapter 2. Solving anything bigger than a quadratic usuing quadratic form. To use quadratic form you have to have 3 terms only. The first term must equal the 2nd exponentx2, and the last term must be a constant. The first thing you do is make g=x^exponent/2 so that you would get g^2+g+#. The second thing is to do the quadratic formula, factor, or complete the square. The the last thing is to plug back in for g. (Whenever you do step three you are basically just plugging back into step one. g=x^2)

An example:
x^4-4x^2-12=0

1. g=x^4/2
g=x^2
g^2-4g-12

2. (g^2-6g)+(2g-12)
g(g-6)+2(g-6)
(g+2)(g-6)
g=-2 g=6

have a good christmas guys:)

Reflection #18

Okay, how about a review on the rational root therom:

ex. x^4+2x^3-7x^2-8x+12
factors of p (constant) = +/-12, +/-1, +/-6, +/-2, +/-3, +/-4
factors of q (leading coeff.) = +/-1
find p/q=+/-12, +/-1, +/-6, +/-2, +/-3, +/-4
plug all possibilities into table on calculator and find all where y is 0.
you get 1, 2, -2, and -3.
do synthetic division with any two.
1 1 2 -7 -8 12
1 3 -4 -12
1 3 -4 -12 0
(x^3+3x^2-4x-12)(x-1)
2 1 3 -4 -12
2 10 12
1 5 6 0
(x^2+5x+6)(x-1)(x-2)
solve by factoring x^2+5x+6=
(x+2)(x+3)
x=-2, -3, 1, 2
put in point form:
(-2, 0)(-3, 0)(1, 0)(2, 0)

Simply stuff from chapter 2(?).

reflection

HAPPY HOLIDAYS! Hope everyone has a Merry Christmas and Happy New Year!!



A little review of Chapter 7.


section one.

convert degrees to radians you have to multipy the degrees by pi/180.

To convert radians to degrees you take the radian and multiply by the reciprical of how you convert to radians 180/pi.

Ex: 225 degrees x pi/180 = 5/4 pi
3pi/4 x 180/pi=135 degrees

section three.


Know the unit circle and chart. STUDYYYY!



The Chart: find all 6 trig functions.
Sin=y/r Csc=r/y
Cos=x/r Sec=r/x
tan=y/x Cot=x/y

EXAMPLE:

Find all 6 trig functions of (-3,4).
Sin=4/5 Csc=5/4
Cos=-3/5 Sec=-5/3
tan=-4/3 Cot=-3/4



andd The unit circle:
90 degrees= pi/2 and (0,1)
180 degrees=pi and (-1,0)
270 degrees=3pi/2 and (0,-1)
360 degrees= 2pi and (1,0)

that can be used when finding coterminal angles.

Reflection

Ok. Now we have trig involving lines and conics. The basic formula for a line in this section is m=tan(alpha)
m=slope
(alpha)= angle of inclination.

And don't forget that when you are finding an inverse that you are finding two angles. And don't forget that you must also check the quadrants. The positive quadrants for tangent are one and three. Making two and four the negative quadrants for tangent.

I'll use an example she gave us in class: 2x+5y=15 find the angle of inclination (alpha). Find your slope. Slope=-A/B so that is -2/5 referring to the standard formula Ax+By=C. now (-2/5)=tan(alpha) since we dont divide by trig functions you take an inverse of tangent to find your angle. With your calculator the inverse tan of -2/5 is -21.801. Ignore the negative so that you have 21.801. We need this answer in the quadrants where tan is negative. Because our slope was negative. To move 21.801 to quadrant two you make it negative then add 180. To move it to the fourth quadrant you make it negative then add 360. Those are your two angles of inclination and equal alpha.

The basic formula for a conic is tan(2x(alpha))=B/(A-C)

To make things easier you should remember that if A=C then (alpha)=Pi/4

The basic formula for a conic is Ax^2 + Bxy +Cy^2 +Dx + Ey + F=0
you should know that

Reflection 18

Okay, so I'm finally glad we're all done exams, and I'm also glad we're on Christmas break! This week we basically did nothing but take exams, and on Monday we had a review day, and we turned in our study guides. Finding symmetry is pretty easy, and I'm going to explain that.

To find symmetry on the y-axis:
you have to plug in (-x) to every x there is

Example:
y = x^2 - 7
y = (-x)^2 - 7
y = x^2 - 7
Therefore, yes, there is symmetry.

To find symmetry on the x-axis:
you have to plug in (-y) to every y there is

Example:
y = x - 2
(-y) = x - 2
y = -x + 2
Therefore, no, there is no symmetry.

To find symmetry on the origin:
you have to plug in (-y) and (-x)

Example:
y = x^2 + 4
(-y) = (-x)^2 + 4
y = -x^2 + 4
Therefore, no, there is no symmetry.

To find symmetry on y=x:
find the inverse

Example:
y = x^2 + 4
x = y^2 + 4
y^2 = x - 4
y = sqrt(x-4)
Therefore, no, there is no symmetry.

Reflection 18?

YESSSSSSSSSSS! We are finally off for the holidays and it's about time cause i'm so tired of school. Two weeks of no math :)..anyways the exam was super hard and I probably failed. But back to this blog since we didn't really learn anything something I remembered how to do was logs. Logs are pretty easy you just have to remember all the rules. Remember that if theres a minus sign in the problem your gonna divide and if theres a plus sign your going to add.

To condense a log-Example:

log 4 - log pi + 3 log r - log b

*You can pull the log out of this, and remember -/divide, +/multiply.

Answer: log(4r^3/pi b)

**Remember when theres a number infront of a log you raise it to the number or variable behind the log. ex: 3 log r...log r^3.

------------------------------------------

Since were off and I don't have anything with me, I can't think of anything specific that I don't understand.

Reflection

WOW! What a week. One word explains it, EXAMS! This week just review over stuff from chapters 1-8, which is a lot of information to review. I'll explain again from when we learned about domain, range, functions, finding any symmetry, and inverses. What I found was the easiest to learn was finding the symmetry.

EXAMPLE: y=x^3=4x

a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No

b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No

c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No

d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes

I also understood how to do domain and range very well. The types of the domain and range problems i understood the most were the fractions and polynomials. (oo => stands for infiniti

EXAMPLE: y=x^3+4x^2+12
For any type of polynomial the domain would be (-oo,oo) and for the range, odd:
(-oo,oo); quadratics: [vertex,oo] or [-oo,vertex].
The answer to this problem would be Domain: (-oo,oo) Range (-oo,oo)

EXAMPLE: 5x+4/x^2-4
First, set the bottom of the fraction equal to zero.
x^2-4=0
+4+4
x^2=4
x=+ or - 2
Your answer then comes to,
Domain: (-oo,-2)u(-2,2)u(2,oo)

So thats blog number one for the holidays. I think what i had the most trouble with on the exam was from chap. 9.

Reflection #18

WOW! that exam was really, really easy..if you studied. It think i did really good. I did all of my study guides and i understood everythign that was on them, so...and one thing that i did understand on the exam was how to find the symmetry. And by the way, the exam was on chapters 1-8. AHHHHH. PS: just because i said i think i did good on my exam, i probably failed, lol.

___________________________________________________________

EXAMPLES:


y=x^3=4x


a.) x-axis
(-y)=x^3+4x
y=-x^3-4x Reflect; No

b.) y-axis
y=(-x^3)+4(-x)
y=-x^3-4x No

c.) y=x
x=y^3+4y
y^3+4y=x
y(y^2+4)=x No

d.) origin
(-y)=(-x)^3+4(-x)
-y=-x^3-4x
y=x^3+4x Yes


^^the only thing this problem was symmetric about was the origin

_________________________________________________________________


I still do not understand how to do the domain and range. I know how to do that number line thing, but when i do not understand how to do a problem, i just put (-infinity, infinity). I know that for the fraction problems, youi just use the top part, but if someone can give me an example of the domain and range problems, i would greatly appreciate it. I really need help with these problems. And we will probably need these lessons as we go further anyway, so i might as well learn it now. THANKSSS :)