This week we have gone over logarithems and functions. Logs have many dfferent principles and methods to solving them. For converting I have come up with my own littl system.
For logs to express them or simplify them, alll you do is rotate them. To express them all you do is rotate the variables and intergers to the right. The last number will be the exponent.
Example:
x^2=5
2log(2) 5
We have also gone threw the process of expanding and condensing logarithmic equations. We were also introduced to other symbols in which these equations can contain.
Example Expanding:
log2(x^2 y c^3/x y^4)
2log(2)x + log(2)y + 3log(2)c - log(2x - 4log(2)y
Example Condensing:
4logx - log2 - 2logc + logy + 3log4 - 2log6
log(x^4 y 4^3/2 6^2)
I do not really understand the correct way to do the functions so I am open to anyone that can help. Thanx
Sunday, September 27, 2009
Taylor Reflection #6
This week was an okay week for me. When we got tests back I was excited to see that my grades have started t o improve, not that I was making bad grades but A’s are better than B’s. Let’s see, Chapter 5 I feel like I get most of it I definitely know how to work logs. Of course solving logs are my favorite because they are a matter of switching around what is already given to you. Changing bases really isn’t hard either. Condensing logs isn’t hard for me to grasp, even when there are awkward symbols and to be honest; I really appreciate Mrs. Robinson explaining that the symbols mean nothing concerning the solving of the problems. I know that’s mostly a common sense thing but I know it would have tripped me up on the ACT.
The steps for condensing logs are easy
First you have to remember the relations
Mn = m+n
m/n = m-n
m^k = k log M
sub b B^k = k
b^log sub b^k = K
so any problem will fit into one of these relations
Ex: expand log sub b MN^2
Log sub b M + 2 Log sub b N
What I do not understand is how to work is still the solving for exponent part that includes
Sandwiching and flipping the fraction
For example
X^5 + X^-2 / X ^-3
So now I don’t understand how in my notes the next step is
X^2/X^2 times X^5 + 1/X \^3 times 1/1 all over 1/X^3
I really think I need the rules for flipping fractions and when to sandwich explained simply to me
Any help though would be great!
The steps for condensing logs are easy
First you have to remember the relations
Mn = m+n
m/n = m-n
m^k = k log M
sub b B^k = k
b^log sub b^k = K
so any problem will fit into one of these relations
Ex: expand log sub b MN^2
Log sub b M + 2 Log sub b N
What I do not understand is how to work is still the solving for exponent part that includes
Sandwiching and flipping the fraction
For example
X^5 + X^-2 / X ^-3
So now I don’t understand how in my notes the next step is
X^2/X^2 times X^5 + 1/X \^3 times 1/1 all over 1/X^3
I really think I need the rules for flipping fractions and when to sandwich explained simply to me
Any help though would be great!
Reflection 6 by: Stephen
This week was pretty easy i guess. I understand alot of it really so....yeah. The thing that is probably the easiest is exponential functions. All you have to do is basically plug things in and then ur set. There are 3 different formulas:
A(t)=Ao(l+r)^t
A(t)=Aob^t/k
Ao=what you start with
b=double, half, etc (use this if u see double, half, etc in the problem)
k=time reg. to double, half, etc
t=time
P(t)=Poe^rt
Po=what you start with
r=rate
t=time
(only use this when compounding continuously)
All you basically have to do is plug the numbers in and solve if it says so.
I still dont understand the exponent thing so if someone can please help me then id be really happy.
A(t)=Ao(l+r)^t
A(t)=Aob^t/k
Ao=what you start with
b=double, half, etc (use this if u see double, half, etc in the problem)
k=time reg. to double, half, etc
t=time
P(t)=Poe^rt
Po=what you start with
r=rate
t=time
(only use this when compounding continuously)
All you basically have to do is plug the numbers in and solve if it says so.
I still dont understand the exponent thing so if someone can please help me then id be really happy.
Alaina's reflection
I'm happy about this past week. It was quite easy because I already understand logs.
Expanding logs:
1) logAB=logA+logB
2)logA/B=logA-logB
3)logA^B=BlogA
Combining logs is pretty much the opposite.
Also, if a base for the log is not indicated, it is understood to be 10
You cannot combine two logs if the base is not the same.
Changing Base:
1) write in exponential form
2) take the log of both sides
3) solve
Although I understand logs, I do not understand the formulas using Ao and Po. I know you use them when trying to solve word problems, but I don't understand when I am supposed to use which formula. I know that Po is used when the problem states that something is being compounded continually, but as for the two problems that both concern Ao, I do not know. I also do not know how to discern what t, r, and k are supposed to be. If anyone understands this completely and wouldn't have a problem explaining it to me, I would be very grateful.
Expanding logs:
1) logAB=logA+logB
2)logA/B=logA-logB
3)logA^B=BlogA
Combining logs is pretty much the opposite.
Also, if a base for the log is not indicated, it is understood to be 10
You cannot combine two logs if the base is not the same.
Changing Base:
1) write in exponential form
2) take the log of both sides
3) solve
Although I understand logs, I do not understand the formulas using Ao and Po. I know you use them when trying to solve word problems, but I don't understand when I am supposed to use which formula. I know that Po is used when the problem states that something is being compounded continually, but as for the two problems that both concern Ao, I do not know. I also do not know how to discern what t, r, and k are supposed to be. If anyone understands this completely and wouldn't have a problem explaining it to me, I would be very grateful.
Stephanie's Reflection
This week was probably the easiest week yet. Lets see... We learned how to change bases:
1)rewrite problem in exponential form
2)take the log of both sides
3)move the variable to the front
4)solve
Eg. log5of10=x
5^x=10
log5x=log10
xlog5=1
x=1/log5
We also learned to graph exponential functions (ab^x):
if b is greater than 1, the graph goes up whereas if it is less than one, the graph goes down
Eg. f(x)=5(3)^x the graph would go up because b is greater than one
The formulas are pretty ease but remembering them will be pretty hard and using them in the correct problem will also be kinda hard.
A(t)=Ao(1+r)^t
Ao = what you start with
r = rate
t = time
A(t)=Aob^t/k
Ao = what you start with
t = time
b = double, half, etc.
k = regular time to double, half, etc.
Eg. half-life of 5 days
b = ½
k = 5
A(t)=Ao(1/2)^t/5
P(t)=Poe^rt
Po = wha tyou start with
r = rate
t = time
you only use the problem with p when it says compounding continuously in the problem
then theres the limit thing and rule of 72 (72 / r% = how long it takes to double)
As for the test on Friday, I didnt completely get it because of the fact that it was a quiz... and, like many others, I completely forgot everything once I saw it.
For the test tomorrow, I'm probably not ready but I'm trying to study as much as I can before then and if anyone could help with remembering the formulas if anyone has any tricks or anything, I'd appreciate help.
1)rewrite problem in exponential form
2)take the log of both sides
3)move the variable to the front
4)solve
Eg. log5of10=x
5^x=10
log5x=log10
xlog5=1
x=1/log5
We also learned to graph exponential functions (ab^x):
if b is greater than 1, the graph goes up whereas if it is less than one, the graph goes down
Eg. f(x)=5(3)^x the graph would go up because b is greater than one
The formulas are pretty ease but remembering them will be pretty hard and using them in the correct problem will also be kinda hard.
A(t)=Ao(1+r)^t
Ao = what you start with
r = rate
t = time
A(t)=Aob^t/k
Ao = what you start with
t = time
b = double, half, etc.
k = regular time to double, half, etc.
Eg. half-life of 5 days
b = ½
k = 5
A(t)=Ao(1/2)^t/5
P(t)=Poe^rt
Po = wha tyou start with
r = rate
t = time
you only use the problem with p when it says compounding continuously in the problem
then theres the limit thing and rule of 72 (72 / r% = how long it takes to double)
As for the test on Friday, I didnt completely get it because of the fact that it was a quiz... and, like many others, I completely forgot everything once I saw it.
For the test tomorrow, I'm probably not ready but I'm trying to study as much as I can before then and if anyone could help with remembering the formulas if anyone has any tricks or anything, I'd appreciate help.
Alicia's Reflection #6
alrighty well the beginning of this week i was pretty excited because i actually understand logs. If anyone needs help with logs just ask!!!
the log properties are pretty easy to understand and remember because they are basically the same as properties of exponents.
expanding logs:
logbMN^2 = logbM+2logN..... when expanding, the exponent always goes in front of the log.
condensing logs:
log 45-2 log 3= log (45/9)= log 5... 3 to the 2nd power is 9...thats where the 9 came from.
write y in terms of x if lny= 1/3lnx+ln4:
lny=ln4x^1/3....always put the number in front of the log or ln as the exponent and combine the other terms.
condensing and expanding logs with symbols is the exact same so dont freak out when you see symbols!!!
remember positive terms are the numerator and negative terms are the denominator of the fraction.
I could use some help with the word problems where you use the formulas we took notes on in class on thursday!
the log properties are pretty easy to understand and remember because they are basically the same as properties of exponents.
expanding logs:
logbMN^2 = logbM+2logN..... when expanding, the exponent always goes in front of the log.
condensing logs:
log 45-2 log 3= log (45/9)= log 5... 3 to the 2nd power is 9...thats where the 9 came from.
write y in terms of x if lny= 1/3lnx+ln4:
lny=ln4x^1/3....always put the number in front of the log or ln as the exponent and combine the other terms.
condensing and expanding logs with symbols is the exact same so dont freak out when you see symbols!!!
remember positive terms are the numerator and negative terms are the denominator of the fraction.
I could use some help with the word problems where you use the formulas we took notes on in class on thursday!
Reflection #6
This week was really easy. I'm glad we're learning logs because I'm very good at it. I did have one little problem with our last quiz though. I completely blanked when I got the quiz. I guess it was because we had just finished the best pep rally ever right before this class! One thing I couldn't think of how to do was "Changing of Bases". But now I know how to do it. The steps for this are:Take the log (base what you want of both sides), write as an exponential, move exponent to the front, solve for variable, write as a fraction or whole number if possible. If not possible leave in log form.
Ex: 1. 2^x = 10
2. log 2^x = log 10
3. xlog 2 = 1
4. Answer: x = 1/log 2
The only thing that I'm really having trouble with is "Logarithm Properties". I get confused on what to do when I have to do the problems on my own.
Ex: 1. 2^x = 10
2. log 2^x = log 10
3. xlog 2 = 1
4. Answer: x = 1/log 2
The only thing that I'm really having trouble with is "Logarithm Properties". I get confused on what to do when I have to do the problems on my own.
Reflection on Exponential Functions
This week wasn’t so bad for me, and some of the only problems that I really had were with exponential growth and decay. Logs and exponent properties seem quite easy, and the word problems are easy to pick out the information you need quickly. Something that I feel that I’ve just gotten is Exponential Functions, so I’m going to try to expain it to you:
Formula Needed: f(x) = ab^x
Goal: Use the functions to find the value of the exponential function so that you have a formula to show the relation of change in the solution to the change in the x value.
A is the value of your first function, typically f(0) = a
B is the value you need to find. It is raised to the x value.
Example:
Find an exponential function if: f(0) = 3, f(1) = 15
1. Remember the formula: f(x) = ab^x
2. Let’ s start with f(0) = 3
3. Raise b to the x power: f(0) = ab^0
4. Since x is equal to zero, b is canceled leaving you with: f(0) = a
5. Therefore, a=3
6. Continue with the second part: f(1) = 15
7. Plug in the a value from the first part and raise b to the x value: f(1) = (3)b^1
8. 3b^1 = 15, So, b = 5.
9. Replace the x values with the variable, x: f(x) = (3)(5)^x
Saturday, September 26, 2009
Amy's Reflection #6
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
(remeber b-rob might use random symbol so don't panic)
ok i know i just explained logarithms but im nervous about monday's test...i really don't like the word problems on the chapter test on page 209...if anyone one can explain an easy way to solve them i'll be really greatfull...
Sunday, September 20, 2009
Devin's Reflection #4
Week by week it is getting easier for me. We covered a good bit of informatio this week. I really thought the exponents was easy. The hardest part is just finding the same base. When it is b^x*b^y= b^x+y. When it s b^x/b^y= b^x-y. When it is (ab)^x= a^xb^y. When it is (a/b)^x= a^x/b^x. When it is (b^x)^y= b^xy. When it is b^x/y= y{b^x}.
To solve for an exponent
a. write as the same base
b. set exponents equal
c. solve for x
Simplfy
(b^2/a) ^-2
-b^-4/a^-2
-1/b^41/a^2
= a^2/b^4
With double fraction, you have to multiply the outsides by each other, and the insides byeach other.
Ex. (a^-2+b^-2)^-1
-(1/a^2+1/b^2)^-1
-(b^2/b^2*1/a^2+1/b^2*a^2/a^2)^-1
-(b^2/a^2b^2+a^2/a^2b^2)^-1
-(b^2+a^2/a^2b^2)^-1
=a^2b^2/b^2+a^2
This week we also covered logirythms (I doubt I spelled that correctly).
logb x=a
- b^a=x
Ex. log2 8=x
-2^x=8
x= 3
Domain of logs and ln = (0,infinity)
Range of logs and ln = (-infinity, infinity)
I do not fully understand the natural log thing (e) so if anyone can assist me, it will be well appreciated.
To solve for an exponent
a. write as the same base
b. set exponents equal
c. solve for x
Simplfy
(b^2/a) ^-2
-b^-4/a^-2
-1/b^41/a^2
= a^2/b^4
With double fraction, you have to multiply the outsides by each other, and the insides byeach other.
Ex. (a^-2+b^-2)^-1
-(1/a^2+1/b^2)^-1
-(b^2/b^2*1/a^2+1/b^2*a^2/a^2)^-1
-(b^2/a^2b^2+a^2/a^2b^2)^-1
-(b^2+a^2/a^2b^2)^-1
=a^2b^2/b^2+a^2
This week we also covered logirythms (I doubt I spelled that correctly).
logb x=a
- b^a=x
Ex. log2 8=x
-2^x=8
x= 3
Domain of logs and ln = (0,infinity)
Range of logs and ln = (-infinity, infinity)
I do not fully understand the natural log thing (e) so if anyone can assist me, it will be well appreciated.
Taylor's reflection #5
This week was a confident week for me. Each section of chapter four was pretty straight forward. As I said in my last reflection I definitely understand how to find domain and ranges even how to solve for domain for absolute values I simplified the steps so they were easier to remember.
Steps for solving domain and range of absolute values
The domain is always (-infinity, infinity)
The range is always (shift, infinity)
Ex: /x+4/-1 D=(-infinity, infinity) R= (-1, infinity)
I was so happy that for the test I knew how to do everything. I knew how to work the things from previous tests like finding parallel and perpendicular slopes. I knew how to find the sum of roots as well as the product of roots. I also knew how to work the problems using skills from chapter 2, skills necessary for finding the vertex, skills necessary for finding the axis of symmetry, and skills necessary for finding the X-intercepts. I was proud of myself for not just memorizing the necessary skills for just the chapter test but I actually learned and retained the necessary skills and carried them on to this last test. I want to say that I found all problems with F(x) and G(x) to also be easy. I also want to say the I have finally been able to complete a test. Actually, I finished a test with time to spare.
Now for what I don’ t understand.. I don’t even know what the first sections of chapter five are called.
I get really mixed up on all the switching around of numbers.
I think I need a simple way to remember all the changes necessary to be made
I have examples worked out in my notes but it’s really hard for me to decide when to flip a fraction or when to sandwich or when to multiply a fraction by its denominator
Any notes to remember would be great
Steps for solving domain and range of absolute values
The domain is always (-infinity, infinity)
The range is always (shift, infinity)
Ex: /x+4/-1 D=(-infinity, infinity) R= (-1, infinity)
I was so happy that for the test I knew how to do everything. I knew how to work the things from previous tests like finding parallel and perpendicular slopes. I knew how to find the sum of roots as well as the product of roots. I also knew how to work the problems using skills from chapter 2, skills necessary for finding the vertex, skills necessary for finding the axis of symmetry, and skills necessary for finding the X-intercepts. I was proud of myself for not just memorizing the necessary skills for just the chapter test but I actually learned and retained the necessary skills and carried them on to this last test. I want to say that I found all problems with F(x) and G(x) to also be easy. I also want to say the I have finally been able to complete a test. Actually, I finished a test with time to spare.
Now for what I don’ t understand.. I don’t even know what the first sections of chapter five are called.
I get really mixed up on all the switching around of numbers.
I think I need a simple way to remember all the changes necessary to be made
I have examples worked out in my notes but it’s really hard for me to decide when to flip a fraction or when to sandwich or when to multiply a fraction by its denominator
Any notes to remember would be great
Alaina's reflection
I am super excited with LOGS!!!!! They are my favorite. I completely don't understand how people have trouble with them.
log24=2
it implies that 2^2=4
the subscript 2 is your base
4 is your answer
2 is the exponent
I still don't really understand proving something is an inverse. I get that you have to plug in a -x for the x-axis, a (-x) for the y-axis, but i don't understand what I am supposed to do afterwards. I know that I have to plug in the what I have as the inverse into f(-f(x)) and -f(f(x)). I guess it is just a matter of me understanding the notes so I'll probably read over them and try to figure it out on my own, but if I have any questions, I will be asking in class.
log24=2
it implies that 2^2=4
the subscript 2 is your base
4 is your answer
2 is the exponent
I still don't really understand proving something is an inverse. I get that you have to plug in a -x for the x-axis, a (-x) for the y-axis, but i don't understand what I am supposed to do afterwards. I know that I have to plug in the what I have as the inverse into f(-f(x)) and -f(f(x)). I guess it is just a matter of me understanding the notes so I'll probably read over them and try to figure it out on my own, but if I have any questions, I will be asking in class.
Amy's Reflection #5
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:
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.
i hope this wasn't too confusing and i was actually able to help someone...
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.
i hope this wasn't too confusing and i was actually able to help someone...
Stephanie's Reflection...Again
First off, I am so happy we are on logs! I LOVE logs! The number one thing you should know when solving a log is the basic formula (exponential form) which happens to be b^a=x while log for would be log b^x=a. When a problem does not have anything on the opposite side of the equal sign (opposite of log), put an x. If a log problem doesn't have a base, it is understood to be 10 because 10 is the default base.
Examples:
log3^9=2
In order to solve, you need to put this into exponential form...
3^2=9
...and that is your answer.
log2^16=4
2^4=16
log5^125=x
5^x=125 (if it asks for exponential form)
x=3 (if it says to solve)
log x=2
10^2=x
x=100
logx^8=3
x^3=8
x=2
Examples:
log3^9=2
In order to solve, you need to put this into exponential form...
3^2=9
...and that is your answer.
log2^16=4
2^4=16
log5^125=x
5^x=125 (if it asks for exponential form)
x=3 (if it says to solve)
log x=2
10^2=x
x=100
logx^8=3
x^3=8
x=2
Stephen's Reflection 5
This week has been pretty easy for me. We got to logs and we did some exponent thing. Im really happy that we got to logs because im really good at it. When solving logs, you have to put the log in the form b^a=x. This is exponential form
Ex: log(2)4=x
Put the log in the form b^a=x. So it would look like 2^x=4
Then you see how many times the 2 can go into the 4 which is 2 times so your answer will be x=2.
We also learned how to put it into log form. All you have to do is do the exponential form backwards starting with b^a=x.
Ex: 3^x=6
The 3 will be ur subscript, the x will be what ur log equals and the 6 will be the number infront of the log. It will look like log(3)6=x.
What i dont understand is the exponent things we did. I just cant understand it. But when i really get confused is when you have to do the "sandwich" thing and you have a fraction over a fraction. I dont understand how to get it in that form so if anyone can help me with that, I'd be grateful
Ex: log(2)4=x
Put the log in the form b^a=x. So it would look like 2^x=4
Then you see how many times the 2 can go into the 4 which is 2 times so your answer will be x=2.
We also learned how to put it into log form. All you have to do is do the exponential form backwards starting with b^a=x.
Ex: 3^x=6
The 3 will be ur subscript, the x will be what ur log equals and the 6 will be the number infront of the log. It will look like log(3)6=x.
What i dont understand is the exponent things we did. I just cant understand it. But when i really get confused is when you have to do the "sandwich" thing and you have a fraction over a fraction. I dont understand how to get it in that form so if anyone can help me with that, I'd be grateful
Alicia's Reflection #5
Okay so this week was pretty good for me. I really understand logarithms so I will try to explain them as best as I can.....
the basic formula is log b^x=a
this is how you solve b^a=x
All you do is switch your a and x term. the base b does not change, only the a and x term.
example: log 5^25=2
5^2=25
example: if your problem does not have a term on the other side of the = sign, add x
log 2^8 =x
2^x=8 ****when the directions ask for exponential form, this is your answer
x=3 ****this is your answer if the directions say solve. always solve for x.
REMINDER : logs are implied to have a base of 10. if there is no base, it is understood to be 10.
Okay I hope that helped. I am having difficulty with the exponent problems. if anyone could help me with problems like # 17-28 on page 178 it would help me greatly!!! thanksss :)
the basic formula is log b^x=a
this is how you solve b^a=x
All you do is switch your a and x term. the base b does not change, only the a and x term.
example: log 5^25=2
5^2=25
example: if your problem does not have a term on the other side of the = sign, add x
log 2^8 =x
2^x=8 ****when the directions ask for exponential form, this is your answer
x=3 ****this is your answer if the directions say solve. always solve for x.
REMINDER : logs are implied to have a base of 10. if there is no base, it is understood to be 10.
Okay I hope that helped. I am having difficulty with the exponent problems. if anyone could help me with problems like # 17-28 on page 178 it would help me greatly!!! thanksss :)
Yeah so this week was pretty easy to me. The only thing I really understood last year in Algebra II was Logs so I understand that pretty well. It's really a very easy thing to do.
If you have Log(5)25=2 you just switch everything around to make sense. The base (5) goes first and this is what the answer should be: 5^2=25. If a log has no base the base is understood to be 10. If there is an e in the log then it is a natural log. Now on to what I don't understand.
Last week I missed Thursday and Friday and I believe ya'll went over axis of symmetry and inverses. Inverses seem the easiest but I don't exactly understand any of it. Dale was suppose to show me how to do this stuff but he says he doesn't feel comfortable showing me because he thinks he might teach me wrong.
If you have Log(5)25=2 you just switch everything around to make sense. The base (5) goes first and this is what the answer should be: 5^2=25. If a log has no base the base is understood to be 10. If there is an e in the log then it is a natural log. Now on to what I don't understand.
Last week I missed Thursday and Friday and I believe ya'll went over axis of symmetry and inverses. Inverses seem the easiest but I don't exactly understand any of it. Dale was suppose to show me how to do this stuff but he says he doesn't feel comfortable showing me because he thinks he might teach me wrong.
Sunday, September 13, 2009
Stephanie's Reflection
So this week we went over domain & range and inverse functions. I'm still kinda iffy on the domain & range portion of everything but I'm going to see if my sister can help me with that quickly before tomorrow's test. Anyhow, inverse functions are pretty simple. Probably the simplest thing I will learn in this class.
- First you switch x and y then you solve for y but before you do this, you have to try out the horizontal line test. When graphing the equation, if the x axis is touches more than once, it does not have an inverse.
- After you find an inverse, you have to prove it actually is an inverse. Do this by plugging the inverse into the original equation and vise versa.
f(x) f^-1(x) - If both problems come out equalling x, you did it correctly and that is indeed an inverse.
- Graph it and see that it only touches the x axis once.
- Switch the x and y:
x=5y-2
Solve for y:
x-2=5y
x+2/5=y - Prove:
y=5(x+2/5)-2
=x+2-2
=x
y=(5x-2)/5+2
=x-2+2
=x
Alaina's reflection
This week was alright I guess. I figured out the domain and range thing, yay. Learning, more like reviewing, functions was really easy. They are very easy to do and I really understand them.
functions:
f(x)=x^2+1 g(x)=-x+3
a)find f(1)
f(1)=1^2+1
f(1)=2
b)find g(2)
g(2)=-(2)+3
g(2)=1
composite functions:
c)using f(x)=x^2+1 and g(x)=-x+3, find (f o g)(x)
f(-x+3)
(-x+3)^2+1
x-6x+10
However, I do not understand reflections. I know that you plug into the formulas in the notes, but I don't understand what I am supposed to be plugging in to find, if that makes sense. I know Alicia said she was good with these so I will probably ask her for help with this. Hope everyone has a good upcoming week.
functions:
f(x)=x^2+1 g(x)=-x+3
a)find f(1)
f(1)=1^2+1
f(1)=2
b)find g(2)
g(2)=-(2)+3
g(2)=1
composite functions:
c)using f(x)=x^2+1 and g(x)=-x+3, find (f o g)(x)
f(-x+3)
(-x+3)^2+1
x-6x+10
However, I do not understand reflections. I know that you plug into the formulas in the notes, but I don't understand what I am supposed to be plugging in to find, if that makes sense. I know Alicia said she was good with these so I will probably ask her for help with this. Hope everyone has a good upcoming week.
Dustin's Reflection #4
This week was pretty crazy, considering that we had a four day week and two of them were with subs. Thurs. and Fri. were the only days we learned this week (Not considering the fact I figured out domain and range while B-Rob was gone). It was pretty easy for the most part, but one thing that kinda confused me was reflections. It wasn't all that hard to do i just have a problem with memorizing all of the steps. One thing that i'm really good with is functions. For example:
Find f(u(c(1)))+1
f(x)=2x+1
u(x)=3x+2
c(k)=4x+3
1.) c(1)=7
2.) u(7)=23
3.) f(23)=47
4.) 47+1=48
To work problems like these you work from the inside out. If you have a number in the closed parenthesis, like 1 in the example, your answer will be a number.
Find f(u(c(1)))+1
f(x)=2x+1
u(x)=3x+2
c(k)=4x+3
1.) c(1)=7
2.) u(7)=23
3.) f(23)=47
4.) 47+1=48
To work problems like these you work from the inside out. If you have a number in the closed parenthesis, like 1 in the example, your answer will be a number.
Kane's Reflection
This week went by pretty fast, probably because I was only there three days. We did a lot of domain and range when Mrs. Robinson was out. Thursday we went over functions and reflections. I don't really know what we did Friday. I guess it might be good to figure that out. I'm not really too confident with reflections. Really, I just need to memorize the formulas for that. I do know how to work the functions, though. For example:
(f-g)(x) is f(x)-g(x)
If f(x)= 2x-1 and g(x)=3x then we get 2x-1-3x= -x-1
(f+g)(x)= 2x-1+3x= 5x+1
(f o g)(x)- this just means f(g(x))
So if f(x)=x+2 and g(x)= 4x+1 then we get (f o g)(x)= (4x+1)+2= 4x+3
In this case the answer would be 4x+3.
Once again, I'm not really sure about reflections. Also, I don't know what we went over Friday. So any help would be appreciated.
(f-g)(x) is f(x)-g(x)
If f(x)= 2x-1 and g(x)=3x then we get 2x-1-3x= -x-1
(f+g)(x)= 2x-1+3x= 5x+1
(f o g)(x)- this just means f(g(x))
So if f(x)=x+2 and g(x)= 4x+1 then we get (f o g)(x)= (4x+1)+2= 4x+3
In this case the answer would be 4x+3.
Once again, I'm not really sure about reflections. Also, I don't know what we went over Friday. So any help would be appreciated.
Alicia's Reflection #4
Okay well this week was one of the easier weeks for me because I had more time to review domain and range. I understand the functions and reflections because all you have to do is know the equations from the notes and plug in for what the problem is asking. If you are having trouble with this, let me know.
example: (f+g)(x)= add f(x) & g(x)
this equation is also used when subtracting, multiplying, and dividing.
If you see (FoG)(x)=f(g(x))=composite.
**Reflecting
To reflect on the x axis, put a - in front of the equation
example: y=x^2
1. x axis--- y= -x^2 No
To reflect on the y axis, plug in (- )
2. y axis---y= (-x)^2
= x^2 Yes
To reflect on y=x find the inverse
a. switch x & y
b. solve for y
3. y= x
x=y^2
y= square root of x No
To reflect on the origin, find x & y axis.
4. origin
y=-(-x^2)
y=-x^2 No
I am having trouble with 4.5. -finding inverses and proving them. If anyone can help with inverses it would help alot!!!
example: (f+g)(x)= add f(x) & g(x)
this equation is also used when subtracting, multiplying, and dividing.
If you see (FoG)(x)=f(g(x))=composite.
**Reflecting
To reflect on the x axis, put a - in front of the equation
example: y=x^2
1. x axis--- y= -x^2 No
To reflect on the y axis, plug in (- )
2. y axis---y= (-x)^2
= x^2 Yes
To reflect on y=x find the inverse
a. switch x & y
b. solve for y
3. y= x
x=y^2
y= square root of x No
To reflect on the origin, find x & y axis.
4. origin
y=-(-x^2)
y=-x^2 No
I am having trouble with 4.5. -finding inverses and proving them. If anyone can help with inverses it would help alot!!!
Reflections, Reflections, Reflections
This week has been okay for me, I kind of get domain, so i'll try to explain that, in the meanwhile, can someone help me out with reflections? I know it seems so easy, but i get confused when i'm trying to work a problem on paper!
Here's the explanation, hope it helps:
There are 2 rules you should know when trying to find the domain of a function :
1: Never try to divide by zero, it just doesn't work. Only Chuck Norris can divide by zero.
2: Don't leave negative's under even powered roots.
Example:1/(x-2)
First, set the denominator equal to 0
Second, solve for x=2
Since fractions form L-shaped graphs, use the root to figure out that the L's curve at 2, Therefore, your answer is (infinity, 2),(2,infinity)
Here's the explanation, hope it helps:There are 2 rules you should know when trying to find the domain of a function :
1: Never try to divide by zero, it just doesn't work. Only Chuck Norris can divide by zero.
2: Don't leave negative's under even powered roots.
Example:1/(x-2)
First, set the denominator equal to 0
Second, solve for x=2
Since fractions form L-shaped graphs, use the root to figure out that the L's curve at 2, Therefore, your answer is (infinity, 2),(2,infinity)
Stephen's Reflection #4
Ok so this week, I was kinda relieved. We didn't have a test this week so im kinda happy about that mainly cause I need some help finding domain and range. What I understand is finding reflections. Its really simple cause all you have to do is get an answer that matches the original equation. When finding reflections, you have to follow some steps:
To reflect the x-axis, you put a negative infront of the equation
To reflect on y-axis, you plug in (-x).
To reflect on y=x, find the inverse:
a. switch x & y
b. solve for y
and to reflect on the origin, reflect the y and x axis again.
Some of the things i dont understand is working f(x) stuff. I understand it when we learn it that day and then after that it just ends. I dont know how to set up the problems. So if someone could help me with (f o g)(x) and stuff like that, i would be grateful
To reflect the x-axis, you put a negative infront of the equation
To reflect on y-axis, you plug in (-x).
To reflect on y=x, find the inverse:
a. switch x & y
b. solve for y
and to reflect on the origin, reflect the y and x axis again.
Some of the things i dont understand is working f(x) stuff. I understand it when we learn it that day and then after that it just ends. I dont know how to set up the problems. So if someone could help me with (f o g)(x) and stuff like that, i would be grateful
Amy's Reflection #4
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
- write the function as an equation: y = 2x + 3
- solve for x: x = (y - 3)/2
- now write f-1(y) as follows .
f -1(y) = (y - 3)/2 or f -1(x) = (x - 3)/2 - 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
Eample 2 - f(x) = √x + 4
- (x)^2 = (√y + 4)^2
- x^2 = y + 4
- y = x^2 - 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
ok im kinda nervous about tomorrow's test; i understand most of it but im afraid that im gonna freak out and forget...hopefully not. ok..umm..one question: we only have to the range for polynomails & absolute values, right?
OK so this week was a lot different than a normal week. For the first two days of school Mrs. Robinson was not at school and we had to do class work on Domain and Range. Then the last two days of school I'm not sure what went on yet because I missed those days and we probably learned something new, which means I'm going to be lost on Monday when I get back so I'm not really looking forward to that.
Well, this week since we were doing classwork with Domain and Range I guess I'll talk about that. For Domain and Range I understand how to find the domain of a polynomial and how to find the domain of an equation with absolute value. For both of these the Domain is always (- ∞ , + ∞). But for the rest of the work that you do for this I kind of struggle. Last year we never really talked about this stuff in Algebra II and I don't really understand it. I mean I can do it if I look at my notes for about an hour and use them to do the work, but for some reason I just cant get when you use the Us to dived the the answers you get and when you get three answers how you get them and what not. I don't know, I'm just confused a little.
Well, this week since we were doing classwork with Domain and Range I guess I'll talk about that. For Domain and Range I understand how to find the domain of a polynomial and how to find the domain of an equation with absolute value. For both of these the Domain is always (- ∞ , + ∞). But for the rest of the work that you do for this I kind of struggle. Last year we never really talked about this stuff in Algebra II and I don't really understand it. I mean I can do it if I look at my notes for about an hour and use them to do the work, but for some reason I just cant get when you use the Us to dived the the answers you get and when you get three answers how you get them and what not. I don't know, I'm just confused a little.
Taylor Rodriguez Reflection #4
This week flew by since we only had four days. Plus Mrs. Bandy was out two of the four days. I was so relieved to have our quiz on domain and range moved to Monday because I really didn’t get a chance to understand it since I missed the original day of the domain and range lesson. My plan for this coming week is to dominate. Haha. I plan to study hard and make sure I grasp the domain and range enough to fly though the quiz. As for the test on Wednesday, I will finish it. I refuse to go through my fourth test of this year still unable to finish it. I plan to work the problems by listening to my gut instead of my doubt. I’m going to quickly complete the problem and if it looks wrong but if I can’t exactly tell why I’m just going to move on and come back to it if I have time. I absolutely hate that I haven’t finished a test yet.
So I definitely understood the chapter four lessons two and three because they were pretty straight forward. I’m excited to say that I understand the process of finding the domain of a square root.
The steps were easy for me to memorize
1. Set equation under the root sign to zero.
2. Form a number line
3. Then using the values on either side of the number on the number line try with f(x)
4. Look for a positive answer.
5. Set up the intervals.
EX: square root of x+4
1. X+4= 0 therefore X= -4
2. ß--------(-4)---------à
3./4. F(-5)= square root -5+4 = Square root of negative one
5. (-4, Infinity)
What I don’t understand is how to find the domain an absolute value
I understand the steps for the most part but I get lost when I go to work problems
Ex: # 11 letter c
H(x)= / x/ - 2
Any help would be awesome
I think I just need a simple view on how to solve.
So I definitely understood the chapter four lessons two and three because they were pretty straight forward. I’m excited to say that I understand the process of finding the domain of a square root.
The steps were easy for me to memorize
1. Set equation under the root sign to zero.
2. Form a number line
3. Then using the values on either side of the number on the number line try with f(x)
4. Look for a positive answer.
5. Set up the intervals.
EX: square root of x+4
1. X+4= 0 therefore X= -4
2. ß--------(-4)---------à
3./4. F(-5)= square root -5+4 = Square root of negative one
5. (-4, Infinity)
What I don’t understand is how to find the domain an absolute value
I understand the steps for the most part but I get lost when I go to work problems
Ex: # 11 letter c
H(x)= / x/ - 2
Any help would be awesome
I think I just need a simple view on how to solve.
Tuesday, September 8, 2009
So I don't think many people quite understood what we were supposed to do in the worksheet today. I would try to explain, but we had to turn in the worksheet. I remember there being everything we took notes on, but I still did not quite understand how to do it. As for the worksheet we were supposed to turn in today: I thing I got the hang of it, but when I saw w8hat other people 1put as the answers, I got really confused. I know we were supposed to figure out which ones were supposed to use the rational root theorem, quadratic form, and factor by grouping but I still didn't quite know how to do them. Examples:
5) y=2x^3-7x+2
13) y=x^4-7x^2-8
18) y=x^3-6x^2
22) y=x^5-3x^4+3^3-5x^2+12
These are just a select few that I had no clue what to do or even where to start.
5) y=2x^3-7x+2
13) y=x^4-7x^2-8
18) y=x^3-6x^2
22) y=x^5-3x^4+3^3-5x^2+12
These are just a select few that I had no clue what to do or even where to start.
Monday, September 7, 2009
Alicia's Reflection #3
Heyy, so this week wasn't so bad for me. I seem to understand domain and range when it comes to the domain of all polynomials. All you do for the range is use x= -b/2a which gives you your point.
The fractions are pretty easy to deal with. Just set the denominator = 0. Solve for X. Then set up intervals.
The number behind the absolute value tells you where to shift. For example: /x+4/-1. The negative 1 means that you are going to shift down 1. Your range is going to be the number outside the absolute value,00. such as [-1,00)
*** You use a bracket when you have a square root or absolute value.
I could use some help with the square roots. On the homework page 122 #12 a,b,&c are giving me trouble. if anyone finished the homework and knows how to do this problem please show me how you did it! Thaanks so much.
The fractions are pretty easy to deal with. Just set the denominator = 0. Solve for X. Then set up intervals.
The number behind the absolute value tells you where to shift. For example: /x+4/-1. The negative 1 means that you are going to shift down 1. Your range is going to be the number outside the absolute value,00. such as [-1,00)
*** You use a bracket when you have a square root or absolute value.
I could use some help with the square roots. On the homework page 122 #12 a,b,&c are giving me trouble. if anyone finished the homework and knows how to do this problem please show me how you did it! Thaanks so much.
Reminder
Blogs should be 250 words, EXPLAIN something and ASK a question. You will not receive full credit if you are not abiding by this. If you are having trouble be sure to work a problem for each blog.
Reflection
Well, this week has been interesting. Not only did I fail another test, but I failed it twice. As for the new thing we learned, I understand completely everything we learned on I don't know what day. I know that you put brackets around the absolute points and parentheses go around the infinity (because it is not a set point). The rest of the domain and range stuff seems pretty simple.
For polynomials, the domain is always (-∞,∞). For fractions, the bottom is set to 0 then you solve for x. For square roots, you set the inside equal to 0, you create a number line, you use the values on each side of the numbers on that number line, you rid the problem of negatives, then set up intervals, and graph.
I know you factor by grouping if there are even terms and you use quadratic form when there are three terms. Also, you don't use group factoring if there is no number in front. What I don't know is when you plug in the problem to you calculator and have to check the x's, how do you know if it is correct or not (I tried asking nunu but that didn't go over so well)?
(sorry I posted this kinda late >.<)
For polynomials, the domain is always (-∞,∞). For fractions, the bottom is set to 0 then you solve for x. For square roots, you set the inside equal to 0, you create a number line, you use the values on each side of the numbers on that number line, you rid the problem of negatives, then set up intervals, and graph.
I know you factor by grouping if there are even terms and you use quadratic form when there are three terms. Also, you don't use group factoring if there is no number in front. What I don't know is when you plug in the problem to you calculator and have to check the x's, how do you know if it is correct or not (I tried asking nunu but that didn't go over so well)?
(sorry I posted this kinda late >.<)
Sunday, September 6, 2009
Amy's Reflection #3
ok since almost everyone did inequalities...i thought i'd do something on domain & range...
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:
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:
√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]
*im pretty sure that's the range but...i might be reading the graph wrong
i hope i was able to help someone with this...
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
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
- graph
√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]
*im pretty sure that's the range but...i might be reading the graph wrong
i hope i was able to help someone with this...
Reflection
Inequalities are the easiest thing ever at the moment. I mean, I completely understand them. When you absolute value, you set up two inequalities the first the same as the original, the second with the opposite sign and the opposite of the constant (in terms of +/-).
Now what I don't get is what we did on Friday. The whole domain and range thing threw me off. I never learned it last year, I must have missed that day or something because i'm completely lost. I know that your domain=x-values and your range=y-values. From the homework, I don't get how you decide if it is a function or not from a graph, and if it is a function, how do you find the domain and range??????
That is pretty much the just of my problem. I didn't learn it last year and it finally caught up to me. Anybody that can help with the domain and range thing, that would be amazing.
Now what I don't get is what we did on Friday. The whole domain and range thing threw me off. I never learned it last year, I must have missed that day or something because i'm completely lost. I know that your domain=x-values and your range=y-values. From the homework, I don't get how you decide if it is a function or not from a graph, and if it is a function, how do you find the domain and range??????
That is pretty much the just of my problem. I didn't learn it last year and it finally caught up to me. Anybody that can help with the domain and range thing, that would be amazing.
Devin's Reflection #3
This week has been rather fustrating. I try and try but I steadily misinterpret information. Like with the quadratic form.
Ex. 2x^4-x^2-3=0
1. 2g^2-g-3=0
2. g(2g-3) 1(2g-3)
3. (g+1)(2g-3)
4. g=-1, 2g-3=0
5. g=-1, g=3/2
Normally I would stop right here I wouldn't plug the g's back into the equation.
6. g=x^2
7. x^2= -1, x^2= 3/2
8. x= +-i, x= (square root of 6)/2
And it took me the longest to understand the rational root theorem.
I'm not gone do an equation yall know it already.
And o yeah, I don't know how to do the backwards synthetic division. And I don't know when I need to use it. So if anyone can help me with that I will appreciate it.
I think I understand how to do inequalities. Its pretty simpe.
1. When its greater than
= -#<_<#
2. When its lesser than
= _<-# or _>#
The Domin and Range thing wasn't to difficult. You just have to remember to put the brackets where they should be.
1. brackets around shift
2. brackets arouund square roots with the limited domain and range
So I will keep on trying. Next week. Enjoy the 3 day weekend.
And if anyone can help me with the synthetic division thing, get at me.
Ex. 2x^4-x^2-3=0
1. 2g^2-g-3=0
2. g(2g-3) 1(2g-3)
3. (g+1)(2g-3)
4. g=-1, 2g-3=0
5. g=-1, g=3/2
Normally I would stop right here I wouldn't plug the g's back into the equation.
6. g=x^2
7. x^2= -1, x^2= 3/2
8. x= +-i, x= (square root of 6)/2
And it took me the longest to understand the rational root theorem.
I'm not gone do an equation yall know it already.
And o yeah, I don't know how to do the backwards synthetic division. And I don't know when I need to use it. So if anyone can help me with that I will appreciate it.
I think I understand how to do inequalities. Its pretty simpe.
1. When its greater than
= -#<_<#
2. When its lesser than
= _<-# or _>#
The Domin and Range thing wasn't to difficult. You just have to remember to put the brackets where they should be.
1. brackets around shift
2. brackets arouund square roots with the limited domain and range
So I will keep on trying. Next week. Enjoy the 3 day weekend.
And if anyone can help me with the synthetic division thing, get at me.
Stephen's Refelction 3
This week was pretty easy for me. Chapter 3 kinda got me scared a little bit. What i found out to be super easy was inequalities. Of course i had that one question about how the equal sign replaced the <> sign but it was a question just to make sure i knew how to do the problem. But one thing i might have trouble with is figuring out which sign is "and" and which sign is "or". Of course i know right now that <> means "or". And i know that you have to switch the first sign for >.
What i dont understand is what we learned in Chapter 4 with the infinity stuff. I dont get the overall point of it and what to do. I mainly dont know how to solve the problems and how find domain and range so i will have problems with that. But other than that, it was pretty easy. Now im just praying that i got at lease a C on the test.
What i dont understand is what we learned in Chapter 4 with the infinity stuff. I dont get the overall point of it and what to do. I mainly dont know how to solve the problems and how find domain and range so i will have problems with that. But other than that, it was pretty easy. Now im just praying that i got at lease a C on the test.
Dustin Reflection #3
This week was pretty easy for me. Since it was mostly review i understood just about everything. One thing we learned that I found to be pretty easy was inequalities. Since we've been working inequalities since algebra 1 I can do most of them with no problem. Say you have -8x>4. All you do is solve for x as if you had an equals sign. The one difference is, whenever you divide by a negative number the sign flips. So for this equation:
- -8x>4 (divide by -8 on both sides)
- x>1/2 (flip your sign)
- x<1/2
Now for the stuff i still don't understand. Going into this test, I new how to work every type of equation, but when i got the test and looked at equations like the word problem, i just forget what to do. I'm pretty sure the main two problems I had trouble with on the test are the first two in the free response, which was the max and min one and the word problem. For the max and min one everything was easy, except for whenever i sketched my graph it looked nothing like the one on the calculator so i didn't know what to do. For the word problem I got down to the part where you have it plugged into the area formula and i just forget everything. I guess thats about it for this week.
Reflection #3
Yeah, so this week seemed a lot better than last week. Because of the fact that it was mostly all review from last week made it pretty easy. I understood just about everything, but the thing that I understood the most by far would still have to be factoring by grouping. Say the problem is 12x^3-3x^2-8x+2. All you have to do is first, group the problem into two halves (12x^3-3x^2)(8x-2). Then, you would reduce the problem 3x^2(4x-1) -2(4x-1). Next, you would group the problem and get (3x^2-2)(4x-1). Last, you would just solve for x and get x=1/4 and x=square root of 2 over three and -square root of 2 over 3. The only thing I had trouble with this week was completing the square. I just completely forgot how to do it.
Taylor Reflection #3
Since this week was mainly a review week for us and I also missed school on Friday .. I don’t really have too much to reflect on.. but I’ll give it a shot. For the test this week I feel like I was less thrown off by the problems. However after three tests I still have yet to complete a test. This is killing me because I have no idea what the problem is. I though the main problem was deciphering which method to use to solve the problems on the test as I had stated in my last reflection. However, prior to the test this week I was taught how to decipher which method to use with each set of directions. This is partially what I understood from this week. I also understand the method used for solving inequalities. Secretly I don’t really think I had a problem with solving inequalities but I know that this week I definitely grasped how to solve inequalities with ease.
As always, the negative side of the week is up next for discussion.
On this test I realized somewhere along the lines im still having trouble with sketching a polynomial.
I know all the steps necessary. I even gave detailed directions as a comment. When it comes to the test I work the problem to the best of my ability and I get stuck. Something just doesn’t look right or work out right but I can’t pinpoint the problem so I wind up erasing the entire work I had just done only to re work it and wind up stopping at the same point.
I have a few theories on what the issue could be.
Its possible that im making the same stupid simple mistake that causes the problem to work out looking weird.
Its also possible that im working the problem correctly and im just being neurotic about getting the answer that paranoia is scaring me out of seeing the problem as done correctly
Or
by the time im getting to the test im having a mental block that is just causing me to skip a step or something.
When we get the tests back ill be sure to show my exact process and that will help yall to give advice on where im messing up
That’s all for now!
As always, the negative side of the week is up next for discussion.
On this test I realized somewhere along the lines im still having trouble with sketching a polynomial.
I know all the steps necessary. I even gave detailed directions as a comment. When it comes to the test I work the problem to the best of my ability and I get stuck. Something just doesn’t look right or work out right but I can’t pinpoint the problem so I wind up erasing the entire work I had just done only to re work it and wind up stopping at the same point.
I have a few theories on what the issue could be.
Its possible that im making the same stupid simple mistake that causes the problem to work out looking weird.
Its also possible that im working the problem correctly and im just being neurotic about getting the answer that paranoia is scaring me out of seeing the problem as done correctly
Or
by the time im getting to the test im having a mental block that is just causing me to skip a step or something.
When we get the tests back ill be sure to show my exact process and that will help yall to give advice on where im messing up
That’s all for now!
Tuesday, September 1, 2009
Explanation of Window Settings
For those of you who are still having trouble with Window Settings or have no idea what changing the numbers in 'Window' really does. (I think Fizgig and Alaina were having trouble with this.) Idk if this counts as two comments or one, so i'll probably come up with another post soon.
P.S. I'm posting my comments as blog entries because it's easier to attach pictures this way.
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