Monday, May 10, 2010

Alicia's Final Reflection :)

Okay so the senior's final exam is tuesday... chapters 1-6 and 13!!! I am going to review some stuff from a few of these chapters:

Exponents:

* b^x * b^y = b^x + y

*  b^x/b^y = b^x - y

* (ab)^x = a^xb^x

* (a/b)^x = a^x/b^x

*  (b^x)^y = b^xy

*  b^x/y = y^√b^x

Changing Bases:

* Rewrite it as an exponential
* Take the log of both sides
* Move the variable to the front
* solve


Ch. 13

* Arithmetic- tn*t1+(n-1)d

n=term # t1=first term d=what you add

* Geometric- tn=t1*r^(n-1)

r= what you multiply by  t1= first term


Examples:

1. Find the formula for the nth term of the arithmetic sequence: 3,5,7,...

tn = 3 + (n-1) (2)

tn = 3 +2n - 2

tn = 1 + 2

2. Find the formula for the nth term of the sequence: 3,4.5,6.75,..

* Don't forget to divide the 2nd term by the 1st term to find r

4.5/3 = 3/2 = r

tn = 3 (3/2)^(n-1)


taylor rodriguez reflection 10 may 2010

soving and sketching parabolas

write this in your notes as you see it posted. it should help your graphing problem.

**#1
you need to see if the parabola will open up or down. think of it this was: if the first thing you see in the equation is a negative sign relate that to which way negative numbers go on a graph or think "if some thing is negative you get a thumbs down" like wise "if something is positive it gets a thumbs up"

so first thing you see at the front of the equation is a negative sign? thumbs down therefore the parabola opens down. If the first thing you see is a positive number? thumbs up therefore the parabola opens up.

(using analogies like this is good for memory. If you start thinking in terms of analogies you get faster at retaining information)

**#2
deciding the number of X intercepts is also an easy remembering problem to fix.
first you need to answer the problem

bsquared - 4(a)(c)

as you said you are very good at plugging in this formula because you have remembered it well.
look at your answer to that and

remember: positive answer is two x intercepts
negative answer is none
zero for an answer is one X intercept

its better to have two than none
so POSITIVE thing to have TWO
NEGATIVE thing to have NONE

(i dont have a trick to remember zero.. i think its just a process of elimination thing.. if i didnt get a positive answer or a negative anser that means its not two x intercepts nor is it no X intercepts,, well that means its one X intercept)

**#3
to find an x intercept you solve for X

it says that in your question

"find X intercept"
remember "find X"

(dont forget to put answer into point form. when solving for x you will always wind up having to square root. you know this meas the answer will be +/-. be sure to show this when convertine to point form. {I.E. (#,0) & (-#,0)} in many of the problems we had there was also a matter of carring a number to the other side. this is no big deal you just tack it on also. for example... if you ended with

X-2= +/- square root 6/2

you would add 2 to both sides and put in point form. therefore you'd have
(squareroot 6/2+2,0) & (- squareroot 6/2+2,0))


**#4
y- intercept is just taking the 0 in the y spot for the last answer and plugging it into the x spot in the equation. which then leaves you only the Y variable to solve for.

Remember: "Find Y intercept"
"find Y"

common sense will tell you the only way to do that is to plug something into the X spot.. and i told you what to plug in

**#5
Axis of semmatry is a simple conversion formula you'll have to memorize the same way you did for the quadratic formula. by writing it down everytime you solve for axis of semmatry until you see the formula in your sleep.

the Formula (in case it isnt written down) is

X= -b/2(a)
(the a and b plug ins of course come from the original equation)

your answer will be the point to put your DOTTED LINE on. because this formula solves for X you know it will pass through that point on the X line. You also know its a vertical line. so no worries.

**#6
the vertex is also just a matter of plugging in
remember this step follows the step ahead of it so it retains the answer for X

that means half of your vertex is solved
you already have your X point for the vertex answer

that answer is also plugged into the original equation which again leaves you to solve for y.

this means you now have your vertex point
because you solved for X in step 5
and your answer after plugging that in gave you the Y

FINALLY! now that you have turned everything into points its just a matter of locating them and marking them all on your graph.

After each point is marked connect the dots.

just as a quick check look back and see if your parabola is supposed to open up or down if your graph matches then

congratulations! everything seems to have gone
well.i know everything i've given you is alot to remember. just write down the hints and keep the sheet as a reference. it doesnt have to be word for word. just putting "step one- opens up or down? work {b-4ac} *positive thumbs up *negative thumbs down"

reading and rereading tricks to help you remember will pay off i promise

i need help on finding inverses like those is chapter four

Sunday, May 9, 2010

Reflection on old stuff

Conics:
The steps to find the intersection of a line and a circle are: solve the linear equation for y, next substitute in circle equation, after this you solve for x, and last you plug x value in to get y value **If your x value is imaginary, then there is no point of intersection.

Example:
x^2+y^2+12y+16x-5=0

First you rewrite the problem in order with x's in front and y's in back, or vice versa, and you get this:
x^2+16x__+y^2+12y__=5

Next you would fill in the blanks with the number that belongs, for this you divide the x and y by 2 and then square it. For this problem you would use 16x and 12y, and you would get 64 and 36. So the answer would be:
x^2+16x+64+y^2+12y+36=5

After this you add the new numbers to the other side of the problem and you would get this:
x^2+16x+64+y^2+12y+36=5+64+36
or
x^2+16x+64+y^2+12y+36=105

Then you factor out the x's and y's:
(x+8)^2+(y+6)^2=105

In the end you would get:
Center=(-8,-6) Radius=square root of 105

alaina's blog, 9 May 2010

There are 2 main types of sequences:

1.) Arithmetic- tn*t1+(n-1)d

n=term # t1=first term d=what you add

2.) Geometric- tn=t1*r^(n-1)

r= what you multiply by

Example: find the formula for the nth term of the arithmetic sequence

3,5,7

tn= 3+(n-1)(2)
tn=3+2n-2
tn=1+2n

Example: find the formula for the nth term of the sequence

3, 4.5, 6.75

divide the second term by the first to get your r.

4.5/3= 3/2 r= 3/2

tn=3(3/2)^n-1

I could use some help on the problems that ask you specifically to find the 200th term for example. I could also use some help with the recursive definitions. THANKSS!!!

ALSO, if anyone can exlpain how to find the equation of a graph when given a graph, i'd like your help.

Dustin's Blog

Ok, haven't done this in a while. I figured my grade is terrible and this is how I can raise it up. My blog won't be ridiculously long like Amy's, but I'll try to teach how to do something. One of my favorite things is...................................(can't think of anything).......................Sigma Notation.

Let's say the equation is 2x+5. Underneath the sigma is x=2 and above the sigma is 5. What this means is that you have to solve 2x+5 for numbers from 2-5.

So, first we plug in 2 and get 9.
Then, we plug in 3 and get 11.
Next, we plug in 4 and get 13.
Finally, we plug in 5 and get 15.

Our expanded answer is 9, 11, 13, 15.

Thats about it to expanding sigma notation.

Some things I don't understand are how to use some of the formulas on sequences and series. I also forgot how to graph conics. Thats about it for this blog i guess.

Amy's Reflection #38

our last reflection!! okay anyway, y'all here are some stuff from chapters 1 - 6 & 13..im gonna print it out and use it to study..i hope y'all will do the same :) good luck

Completing the Square:

You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².

For example:

x² + 6x - 2 = 0

* anytime you are solving a quadratic you’re finding x-intercepts

* Move the constant term to the right side:

x² + 6x = 2

* Take half of the coefficient on the x-term (divide it by two, and keeping the sign), and then square it. Add the squared value to both sides of the equation:

x² + 6x + 9 = -2 + 9

* Convert the left-hand side to squared form. Simplify the right-hand side:

(x + 3)² = 7

* the # half of the coefficient goes in the parentheses.

* Square-root both sides:

x + 3 = √7

* Solve for "x =". Remember to put the "±" on the right side and that it gives you two solutions.

x = -3 ± √7

* The two points for this solution are:

(-3 + √7,0) , (-3 -√7,0)

Rational Root therom

Example: f(x)= 2x^3 + 3x^2 - 8 + 3

Step 1: find all possible roots..

p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2

*p is the leading constant term & q is the leading coefficient

possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2

Step 2: now you can plug all of the possible roots in your calculator to find the roots that work

* the zero will be: 1, 1/2, -3

Step 3: use synthetic division to factor all of the roots that work

you should get: (x - 1) (2x^2 + 5x + 3)

Domain & Range of functions:

Polynomials:

the domain of all polynomials is (−∞, ∞).

For example:

f(x) = x^2 - 3x^2 + 2x - 1

D: (−∞,∞ )

f(x) = x^2 + 3

D:(−∞,∞ )

Fraction:

* you set the bottom to zero
* solve for x
* then set up intervals

For example:

f(x) = 1/x-2

x-2=0

x=2

D: (-∞ , 2) (2, ∞ )

Absolute Value:

D:(- ∞ , + ∞)

R: [0 , + ∞)

For example:

f(x) = |x + 8| - 9

D: (- ∞ , + ∞)
R: (-9, ∞)

f(x) = |x -7| + 5

D: (- ∞ , + ∞)
R: (5, ∞)

Square Roots:

to find the domain:

* set the inside = to zero
* then set a # line
* try values on either side of each #
* get ride of the negatives
* set up intervals

to find the range:

* graph

For example:

√9 - x^2

(solve for x...)

9 - x^2 = 0

-x ^2 = -9

√x^2 = √9

x = ±3

(# line)

(#s on either side..)

f(-4) = √9 - (-4)^2 = √-7
f(0) = √9 - (-4)^2 = √9
f(x) = √4 - (-4)^2 = √-7

√9 = ±3 so...

D: [-3, 3]

(graph....)

R: [0, 3]

How to Find the Inverse of a Function:

* Replace f(x) with y
* Reverse the roles of x and y
* Solve for y in terms of x
* Replace y with f-1(x)

Example 1 - f(x) = 2x + 3

1. write the function as an equation: y = 2x + 3
2. solve for x: x = (y - 3)/2
3. now write f-1(y) as follows .
f -1(y) = (y - 3)/2 or f -1(x) = (x - 3)/2
4. Check:

* f(f -1(x))=2(f -1(x)) + 3
=2((x-3)/2)+3 =(x-3)+3 =x
* f -1(f(x))=f -1(2x+3)
=((2x+3)-3)/2 =2x/2 =x

Example 2 - f(x) = √x + 4

1. (x)^2 = (√y + 4)^2
2. x^2 = y + 4
3. y = x^2 - 4
4. f-1(x) = (x^2 - 4)

* f(f-1(x)) = f(x^2 - 4) = √(x^2 - 4) + 4 = x
* f-1(f(x)) = f-1(√x + 4) = (√x + 4)^2 - 4 = x + 4 - 4 = x

Exponents:

1. b^x * b^y = b^x + y....example: 2^3 * 2^5 = 2^8

2. b^x/b^y = b^x - y....example: 5^7/5^4 = 5^3

3. (ab)^x = a^xb^x....example: (3 * 7)^3 = 3^3 * 7^3

4. (a/b)^x = a^x/b^x....example: (3/5)^3 = 3^3/5^3

5. (b^x)^y = b^xy....example: (2^2)^3 = 2^6

6. b^x/y = y^√b^x....examples: 5^3/4 = 3^√5^3

7. to solve for exponents:

* write as the same base
* set exponents equal
* then solve for x

here are some examples:

(a). 5^3x = 5^7x - 2

In this first part we have the same base on both exponentials so there really isn’t much to do other than to set the two exponents equal to each other and solve for x.

3x = 7x - 2

2 = 4x

x = 1/2

So, if we were to plug x = 1/2 into the equation then we would get the same number on both sides of the equal sign.

(b). 4^t^2 = 4^6 - t

t^2 = 6 - t

t^2 - t - 6 = 0

(t - 2) (t + 3) = 0

t = -3, t = 2

In this case we get two solutions to the equation. That is perfectly acceptable so don’t worry about it when it happens.

(c). 3^z = 9^z + 5

Now, in this case we don’t have the same base so we can’t just set exponents equal. However, with a little manipulation of the right side we can get the same base on both exponents. To do this all we need to notice is that 9 = 3^2. Here’s what we get when we use this fact:

3^z = (3^2)^z + 5

Now, we still can’t just set exponents equal since the right side now has two exponents.

3^z = 3^2(z + 5)

We now have the same base and a single exponent on each base so we now set the exponents equal. Doing this gives us....

z = 2(z + 5)

z = 2z + 10

-10 = z

...a solution of z = -10.

Step 4: solve further

(this can be factored...)

= (x - 1) (2x^2 + 5x + 3)

= (x - 1) (2x - 1) (x + 3)

(set x = 0 )

x = 1, 1/2, -3

Logarithm Properties:

* logb MN = logb M + logb N
* logb M/N = logb M - logb N
* logb M^K = K logb M
* logb b^k = k (this one i don't get..maybe i copied it wrong)
* b^logb^k = k

Here are some examples:

1. log 2 + log 3 + log 4 = log 24 (mulitply: 2 x 3 x 4)

2. log 8 + log 5 - log 4 = log 10 (mulitply: 8 x 5 then divide: 40/4)

3. 2 ln 6 - ln 3 = ln 12 (raise 6 to the 2nd power = 36 the divided by 3 = 12)

4. log M - 3 log N = log M/ N^3

5. ln 2 + ln 6 - 1/2 ln 9 = ln 12/3 = ln 4

6. Expand logb MN^2....logb M + 2 logb N

7. Condense log 45 - 2 log 3....log (45/9) = log 5

8. Rewrite in exponetial form: log36 6 = 1/2....36^1/2 = 6

9. Rewrite in logarithmic form: 2^2 = 4....log2 4 = 2

Changing Bases: (Done when you can't solve a log)

* Rewrite it as an exponential
* Take the log of both sides
* Move the variable to the front
* then solve

(use the same steps when solving for x as an exponent when you can't write them as the same base)
examples:

1. log5 10 = x

5^x = 10

log 5^x = log 10

x log 5 = 1

x = 1/log 5

2. 2^x = 7

log 2^x = log 7

x log 2 = log 7

x = log 7/log 2

(remember b-rob might use random symbol so don't panic)

Conics

Ellipses

Steps:

1. find the center
2. determine the major axis
3. find the vertex (± √big denom)
4. find the other intercept ( ± √small denom)
5. find the focus (c^2 = a^2 + b^2)
6. determine the length of the major axis (2√big denom)
7. find the length of the minor axis (2√small denom)
8. finally graph

Example 1: Graph the following ellipse. Find its major intercepts, length of the major axis, minor intercepts, length of the minor axis, and foci.

x^2/4 + y^2/9 = 1

This ellipse is centered at (0, 0). Since the larger denominator is with the y variable, the major axis lies along the y-axis.

Since a^2 = 9 then a = 3 & Since b^2 = 4 then b = 2

Major intercepts: (0, 3), (0, –3)

Length of major axis: 2 √9 = 6

Minor intercepts: (2, 0), (–2, 0)

Length of minor axis: 2√4 = 4

c^2 = a^2 + b^2

= 9 - 4

= 5

= √5

Foci: (0, √5) , (0, -√5)

then you graph your points..

Parabolas:

how to find the axis of symmetry, vertex, focus, & directrx??

1.) to find the axis of symmetry: x = -b/2a

2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:

y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex

3.) to find the focus: 1/4p= the coefficient of x^2 and then add p

Note:

*If opens up, add to y value from vertex, if opens down, subtract

*If opens right, add to x value to vertex, if opens left, subtract)

4.) directrix: is p units behind the vertex

Note:

*If opens up, subtract; if opens down, add from y-value of vertex.
*If opens right, subtract x-value
*If opens left, add x-value

Example: x^2 + 1

~vertex:

x = -b/2a

x = 0/2(1) = 0

0^2 + 1 = 1

(0,1)

~Focus:

1/4p = 1

4p = 1

p = 1/4

(0, 1 + 1/4)

(0, 5/4)

~directrix:

y = 1 - 1/4

y = 3/4

CIRCLES
The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)

If the equation is not in standard form, you must complete the square to put it in standard form.

If you are given a center and a point, you can use the distance formula to find the radius.

To find the intersection of a line and a circle:

1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.

***Reminder. If your x value is imaginary, then there is no point of intersection.

EX: find the center and radius.

(x-3)^2+(y+7)^2=19 c:(h,k)

center: (3,-7) radius: square root of 19

EX: find the eqn of the circle with the center (1,4) through (3,7)

in the problem you are given a center and a point so you would plug into the distance formula.

square root of (3-1)^2+(7-4)^2= square root of 4+9=square root of 13. **13 has no root.

Your answer should be (x-1)^2+(y-4)^2=13

Hyperbola

(x^2-h/a^2)+(y^2-k/b^2)=1

* Your center is (h,k)
* your major axis has the larger denominator

13-1

1. sequence-list of numbers

2. two main types: 1). arithmetic-add or subtract 2).geometric-multiply

Formulas:

1. arithmetic-used to find a term: tn . t1 + (n-1)d

**n=term #, t1=first term, d=what you add, tn=term #

2. geometric: tn=t1 . r^(n-1)

**r=what you multiply by..

Examples:

1. Find the formula for the nth term of the arithmetic sequence: 3,5,7,...

tn = 3 + (n-1) (2)

tn = 3 +2n - 2

tn = 1 + 2

2. Find the formula for the nth term of the sequence: 3,4.5,6.75,..

**divide the 2nd term by the 1st term to find r

4.5/3 = 3/2 = r

tn = 3 . (3/2)^(n-1)

13-2

Formula for a sequence that involves the previous term: (an - 1)

Examples:

1. Find the recursive definition of: 81, 27, 9,3,...

an = an - 1/3

2. 1, 2, 6, 24, 120, 720, ....

n = 1: 1

n= 2: 2

n = 3: 6

an = n . an - 1

13 -3

Series-List of added or subtracted numbers

**Leave it as a list: do NOT add

Formulas:

1. Arithmetic: Sn = n(t1 + t2)/2

**Finds the sum of the first n terms

2. Geometric: Sn = t1 (1 -r^n)/1-r

Examples:

1. Find the sum of the first 25 terms of the series: 11 + 14 + 17 + 20 + ....

Sn = n (t1 + tn)/2

t25 = 11 + (24)(3)

Sn = 25 (11 + 83)/2

= 1175

2. Find the sum of the first 10 terms of the series: 2-6 + 18 - 54 +...

**This is a geometric sequence and that is because you have to add or subtract the same number for it to be an arithmetic sequence, got it??

r = -6/2 = -3

Sn = t1(1 - r^n)/1-r

= 2(1 -(-3)^10)/1 - (-3)

= 2(-59048)/2

= -29524

Tuesday, May 4, 2010

Make up Blogs