Taylor reflection #10

For the blog for week 10 I am posting late because I missed the entire week of notes so I figured I would wait and get help with the hellacious week I expected to follow. However what came this week was surprising. Everyone was having trouble with what was learned the week I was out so we spent the majority of this week reviewing. So far I think I understand the notes. By the time I have to post a blog again I will have a very definitive idea of how I am handling the trig things we have learned so far.
As of right now I definitely understand the trig chart
TRIGCHART
0° *sin 0= 0 *cos 0= 1 *tan 0= 0 *csc 0= undefined *sec 0= 1
*cot 0= undefined

30° * sin π/6= 1/2 *cos π/6= √3/2 *tan π/6= √3/3 *csc π/6= 2 *sec π /6= 2 √3/3
*cot π/6= √3
45° *sin π/4= √2/2 *cos π/4= √2/2 *tan π/4= 1 *csc π/4= √2 *sec π/4= √2
*cot π/4= 1
60° *sin π/3= √3/2 *cos π/3= 1/2 *tan π/3= √3 *csc π/3= 2 √3/3 *sec π/3= 2
*cot π/3= √3/2
90° *sin π/2= 1 * cos π/2= 0 *tan π/2= undefined *csc π/2= 1 *sec π/2= undefined *cot π/2= 0

I also discovered an easier way to memorize the chart
First you would have to set up the chart differently than it was given
It would have to look like this
Sin Cos Tan Csc Sec Cot
30
45
60
90
Just like a graph and then you fill in the parts for each
If you can memorize the sin column for the chart you can figure out the rest of the chart
Cos is just sin backwards
Tan is just sin/cos
Csc is just 1/sin
Etc. You just have to find the relations of each

And that’s all I’ve got so far
Im still trying to learn everything else

Devin's Reflection #10

This week we learned some more trig stuff. We learned some trig functions. The sin and csc both deal with the y factor and radius. The cos and sec both deal with the x factor and radius. The tan and cot both deal with the x factor along and the y factor. Radius is the square root of x(2) + y(2).

The trig functions are:

sin 0= y/r

cos 0= x/r

tan 0= y/x

csc 0= r/y

sec 0= r/x

cot 0= x/y

Unit Circle

2-(0,1)-90-pi/2 1-(1,0)-0,360-0,2pi

3-(-1,0)-180-pi 4-(0,-1)-270-3pi/2

The Trig Chart

0 sin 0=0 cos 0-1 tan 0=0

30 sin pi/6=1/2 cos pi/6= (3)/2 tan pi/6=(3)/3

45 sin pi/4=(2)/2 cos pi/4=(2)/2 tan pi/4=1

60 sin pi/3=(3)/2 cos pi/3= 1/2 tan pi/3=(3)

90 sin pi/2=1 cos pi/2=0 tan pi/2= underined

0 csc 0= undefined sec 0=1 cot 0= undefined

30 sin pi/6=2 sec pi/6=(2) cot pi/6=(3)

45 sin pi/4=(2) sec pi/4=(2) cot pi/4=1

60 sin pi/3=2(3)/3 sec pi/3=2 cot pi/3=(3)/3

90 sin pi/2=1 sec pi/2=undefined cot pi/2=0

Reference angles must be between 0 and 90, 0 and pi/2

1. Find which quadrant angle is in

2. Determine the sign in that quadrant (+ve or -ve)

3. Subtract 180 until the angle is between 0 and 90, 0 and pi/2

We also learned inverses.

To solve for an angle

1. simplify any expression

2. get the trig function by itself

3. take the trig inverse of both sides

4. use chart unit circle or calculator to find 1 angle

5. set up and find other quadrants with the same sign as the value.

Ex. sin -1((2)/2)=45 quadrants 1 and 2

To make from quadrant to quadrant:

1-3 make it -ve and add 360

1-2 make it -ve and add 180

1-4 add 360

2-4 add 180

Alicia's reflection #10

okay so this week wasn't so bad. So far, trig is not as hard as it sounds but I still need to memorize the trig chart.

7-3

sinθ=y/r
cosθ=x/r
tanθ=y/x
cscθ=r/x
secθ=x/y
cotθ=x/y

The equation to use is r=√(x^2+y^2)

***Remember... sin relates to the y axis, cos relates to the x axis, and tan relates to both the x and y axis.

7-4

Memorize this trig chart for the ch. 7 test!!!

0° sin0=0 cos0=1 tan0=0 csc0=undefined sec0=1 cot0=undefined

30° sinπ/6=1/2 cosπ/6=√3/2 tanπ/6=√3/3 cscπ/6=2 secπ/6=2 √3/3 cotπ/6=√3

45°sinπ/4=√2/2 cosπ/4=√2/2 tanπ/4=1 cscπ/4=√2 secπ/4=√2 cotπ/4=1

60°sinπ/3=√3/2 cosπ/3=1/2 tanπ/3=√3 cscπ/3=2 √3/3 secπ/3=2 cotπ/3=√3/2

90°sinπ/2=1 cosπ/2=0 tanπ/2=undefined cscπ/2=1 secπ/2=undefined cotπ/2=0

***I have a hard time with the word problems from the homeworks. They really throw me off so if anyone can help with word problems that would really help for the test. Thankss!!



Goodluck on the test :)

I really do understand trig, I just have to take the initiative to study it all.
First:
sin= y/r csc=r/y
cos=x/r sec=r/x
tan=y/x cot=x/y
r=√x^2+y^2
*csc, sec, and cot are opposites of sin, cos, and tan.
*they are used to find θ, an angle measure. (unit circle)

Second:
Trig chart.
It is pretty self explanitory. MEMORIZING it is the hard part. Using it is pretty simple.

Third:
Reference angles.
1. find which quadrant the angle is in.
2. determine the sign in that quadrant
3. subtract 180° until the angle is between the absolute value of 0° and 90°

e.g. Find the reference angle for sin210°
1. lies in quadrant 3
2. sin is negative in quadrant 3
3. 210-180=30
sin210°= -sin30
you would then look to your trig chart and find a reference to sin 30 (π/6)
sin 210°=1/2

I pretty much understand everything. I just need to keep studying. If I have any questions, I'll be sure to ask on Monday.

Stephanie's Reflection

1)s=rθ
2)k=1/2r^2θ (k=1/2rs)
r = radius, θ = angle, s = arc length, k = area of sector

EG: A sector of a circle has an arc length of 6cm and an area of 75cm^2. Find its radius and measure of its central angle.
s=6cm
k=75cm^2
r=?
θ=?
k=1/2rs
75=1/2r6
75=3r
r=25cm
s=r
6=25θ
θ=6/25

sinθ=y/r
cosθ=x/r
tanθ=y/x
cscθ=r/x
secθ=x/y
cotθ=x/y
r=√(x^2+y^2)

EG: sin180° = y/r
at 180°, y=0
0/1 = 0
thus sin180°=0

EG: cosπ/2 = x/r
π/2 is at 90°
at 90°, x=0
thus cosπ/2=0

Trig Chart:
0°
sin0=0
cos0=1
tan0=0
csc0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0

Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)

1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°

EG: sin^-1(-√2/2) = 45°
45 + 180 = 225° = θ
I to IV
-45 + 360 = 315°
θ = 225°, 315°

I believe I mostly caught up with this stuff but I still don't quite understand the last few sections.

Stephen's Reflection

Ok this week was all trig...fun!! (not really). But anyway i think the easiest stuff was writing degrees as ' and ", converting to radians, and converting to degrees. So i guess i will talk about that. The first one im going to talk about is writing degrees as ' and ".
Ex:
write 12.3 degrees as ' and ".
what you do first is separate the numbers before and after the decimal and multiply everything after the decimal by 60
-12 degrees (.3 x 60)
Then your answer would be 12 degrees 18'
*if you multiplied the decimal by 60 and it gave you another decimal number then you would do the same thing as before but you would put " after the answer

Now to convert to radians:
Ex:
36 degrees converted to radians
first you multiply the degrees by pi/180
so you would divide 36 by 180 and you would get a decimal and then you want to convert it to a fraction.
your answer would be 1/5 pi or pi/5

Now to convert to degrees:
Ex:
3pi/4
multiply 3pi/4 x 180/pi..both pi's will cancel
then multiply 3 x 180 all over 4
then your answer would be 135 degrees

Some stuff i will have trouble with is remembering the unit circle and remembering the trig chart. I can remember the degrees part of the unit circle but with the parts with pi but i guess thats kinda my responsibility to remember it but if someone can just help me with it in class or stuff i would be thankful.

This week I was nervous because everyone was telling me that we were starting Trig. and saying that it is really hard. The week started off pretty hard and I did not really understand anything too good but it started getting easier. One thing that I really know, and I'm pretty pumped about knowing, is the Trig Chart.

0° sin0=0 cos0=1 tan0=0 csc0=undefined sec0=1 cot0=undefined

30° sinπ/6=1/2 cosπ/6=√3/2 tanπ/6=√3/3 cscπ/6=2 secπ/6=2√3/3 cotπ/6=√3

45° sinπ/4=√2/2 cosπ/4=√2/2 tanπ/4=1 cscπ/4=√2/2 secπ/4=√2/2 cotπ/4=1

60° sinπ/3=√3/2 cosπ/3=1/2 tanπ/3=√3 cscπ/3=2√3/3 secπ/3=2 cotπ/3=√3/3

90° sinπ/2=1 cosπ/2=0 tanπ/2=undefined cscπ/2=1 secπ/2=undefined cotπ/2=0


(I swear I did that all by myself off the top of my head =])

One thing that I am struggling in, however, is Sector of a Circle.
I think the thing that is messing me up is how the letters stand for completely different things, like K is the are of a sector.

Amy's Reflection #10

This week we covered quite a lot in trig and this is some of the stuff i understand:

Sector of a Circle:

1. s = rθ

2. k = θ/2 r^2

(**Note: r = radius, θ = angle, s= arc length, & k = area of sector)

Example #1:

Find the length of the arc of a circle with radius 4 cm and central angle 5.1 radians.

s = rθ

= 4 × 5.1

= 20.4 cm

Example #2:

Find the area of the sector with radius 7 cm and central angle 2.5 radians.

k = θ/2 r^2

= (2.5)/2 (7)^2

= 61.25

Reference Angles:

1. The reference angle in the 1st quad is equal to the angle measure.

Example- θ = 60 = 60

2. To find the reference angle in the 2nd quad just subtract the angle from 180°.

Example- θ = 130° = 180° -130°= 50°

3. To find the reference angle in the 3rd quad subtract 180° from the angle given.

Example- θ = 240° = 240° - 180° = 60°

4. To find the reference angle in the 4th quad simply subtracting the angle from 360° will provide the reference angle in the 4th quad.

Example- θ = 315° = 360° -315° = 45°

I need help in finding the inverse. if someone could explain that to me, i'd really apprieciate that...oh people, don't forget we have a quiz on monday & our chapter 7 test is on tuesday..

Stephanie's Reflection

TRIGONOMETRY

Angles

• measured in degrees
• to find minutes, multiply what is behind the decimal by 60
• to find seconds, multiply what is behind the decimal by 0 and divide by 300 to get decimal
• angles are measured in degrees and radian

radians = degrees times pi over 180 

degrees = radians times 180 over pi

• to find coterminal angles, add or subtract 360 degrees or 2 pi
• must use degrees symbol if in degree or its wrong

if no degree symbol, its assumed that you are in radians

EG: 12.3 degrees... write as t

12 degrees .3 times 60

12 degrees 1.8 minutes

15.36 degrees

15 degrees .36 times 60

15 degrees 21 minutes .6 times 60

15 degrees 21 minutes 36 seconds

25 degrees 20 minutes 6 seconds

25 plus 20 divided by 60 plus 6 divided by 3600

25.335 degrees

15 degrees 26 minutes 15 seconds

15 plus 26 divided by 60 plus 15 divided by 3600

15.4375 degrees

36 degrees

36 divided by 180 pi

1/5 pi

pi/5

3 pi/4 times 180/pi

135 degrees

225 degrees

225 divided by 180 pi

5/4 pi

pi/9 times 80 divided by pi

20 degrees

2.4 times 180 divided by pi

137 degrees 30 minutes 35.535 seconds

245 degrees 15 minutes 300 seconds

2445 plus 15 divided by 60 plus 300 divided by 3600

245.3 degrees

Alaina's reflection

This week, we started Trig. Lovely. For the most part, I understand it. I think that I might have trouble remembering all of the formulas, but that is just me needing to study. A lot.

Section 7.1

1. angles are measured in degrees (minutes or seconds) or radians
2. to find minutes, you multiply what is behind the origional decimal by 60
3. to find seconds, you multiply the "minutes decimal" by 60 and then divide by 3600

to change from degrees to radians, multiply degrees x pi/180

to change from radians to degrees, multiply radians x 180/pi

*NEVER PLUG PI INTO CALCULATOR & ALWAYS USE EXACT ANSWERS.

Example:

15.3°rees;

15°rees; (.3 x 60)

15°rees; 18'

90°rees; convert to radians

90 x pi/180

90/180 pi

1/2 pi=pi/2

Taylor Reflection #9

This week was kind of stressful for me
ive been spiratically checking edline and with mrs robinson for the exam grades. My advanced math grade is the only one i was worried about. However when we started trig on wednesday i was a little bit scared ill admit to that.. but when we started doing examples i realized that so far things were going smoothly. Section seven one homework was done in less then ten minutes. i realize that trig builds and it has the potential to get rediculously hard but i think im going to be the optimist and live in the now and stick with being happy that so far so good.

I can easily explain the summary of section seven one

Angles are measured in DEGREES (possibly with either minutes' or minutes' andseconds")
and RADIANS


To find minutes
multiply what is behind the decimal by 60
and take whole number as minutes'

To find seconds
multiply what is behind the decimal of minutes by 60
then whats behind that decimal by 3600 to get seconds'

To get radians

degrees x pi/180degrees

((***Remember to always ues exact answers))
((*** Remember to never plug pi into calculator))



We also learned section seven two

I feel like the formulas for section seven two are going to trip me up
if any body has an easy way to remember the formulas that would be greaatly appreciated advice.

Alicia's Reflection #9

alrighty well this week we started chapter 7: Trigonometry. So far its not hard at all but that doesnt mean it wont get harder. Mrs Robinson said that trig builds each section so make sure you understand every section or you may struggle.

Okay well to summarize section 7.1......

Angles are measured in degrees. (either minutes' or seconds") or radians

To find minutes, multiply what is behind the decimal by 60

To find seconds, multiply what is behind the decimal of minutes by 60; divide seconds by 3600 to get decimal.

radians=degrees x pi/180

degrees=radians x 180/pi

Remember to always ues exact answers. Never plug pi into calculator.

example:

12.3 o
12 o(.3 x 60)
12 o 18'

225 o convert to radians

225 x pi/180 = 225/180 pi = 5/4 pi

Reflection #9

This week we started to learn trig. I'm actually pretty excited to learn trig b/c of in geometry last year we learned SOHCAHTOA and it helped me alot. So i'm expecting the trig we learn this year to come in handy.

I'm gonna explain how to convert back and forth between rads and degrees.
1.)Rads==>Degrees
To do this I am going to multiply the radian by 180/pi.
Ex. Convert 3pi/4 to degrees.

• 3pi/4 x 180/pi
• 540pi/4pi
• 135degrees

2.)Degrees==>Rads

To do this I will simply multiply by pi/180.

Ex. Convert 135 degrees to radians.

• 135 x pi/180
• 135pi/180
• 3pi/4

And thats how you do that.

The only hard thing about trig so far is that i will have to memorize those formulas, but thats nothing really that i need help with i just need to study. Hope i helped.

Stephen's Reflection 9

Ok this week we got into trig. Im not to excited about that because of what ive heard from last year but it seems easy right now. So ill talk about the stuff on 7.2. There are 3 different formulas for this stuff: s=r(theta) , K=1/2 r^2(theta), and k=1/2rs where s= the arc lenght, r=radius, theta=angle, and k=area of circle. Now these involve word problems so its all about figuring out what is s, k, theta, and r. And once that happens, you just plug in and solve. And you cant use all these formulas because sometimes they only give u k, r and theta or s r and theta so you have to figure out which formula you have to use.

I understand alot about these two sections but i know i will forget how to convert degrees and stuff to radius. But what i could use help in is how to make it into minutes or seconds so ill be happy if someone helps.

Amy's Rleflection #9

Trigonometry

ok last week went quite smoothly. we started chapter 7. i have to say that trigonometry isn't as bad as i thought it was gonna be and plus we only did what like two sections. here is an explanation for what is covered in 7.1 :

Angles:

• measured in degrees (°), minutes ('), and seconds (")
• to find minutes multiply what is behind the decimal by 60
• to find seconds multiply what is behind decimal of minutes by 60; divide by 3600 to get decimal
• angles are measuredd in degrees & radians: radians= ° x π/180, degrees= radians x 180/π (always use the exact answer and don't plug π in calc.)
• to find coterminal angles add or substract 360° or 2π
• HINTS- must use ° symbol if in degrees or your answer is WRONG. (if there is no ° symbol then it is assumed that your answer is in radians)

Examples:

1. write as minutes & and seconds ( ' & " ) :

12.3°

12°(.36 x 60)'

12° 18'

2. find the decimal:

25° 20' 6"

25 + 20/60 + 6/3600

25.335°

3. degrees to radians:

36 x π/180 = 36/180 π = 1/5 π = π/5

4. find a negative coterminal angle

1125°

1125 - 3600 = - 315

alright now for what i don't understand is #13 on page 265. does anyone know how to did because i don't got clue...oh i might be wrong but people i think we have quiz on 7.1 & 7.2 on monday so be prepared..

Reflection: Introduction to Trigonometry

This week seems to have been quite easy. In section 1 we learned about the measurement of angles, and in section 2 we learned about sectors of circles. Today, I’ll be going over section 1.

In Advanced Math, Angles will be measures as either degrees (°) or radians (rad or no symbol).

NOTE: IF YOU DO NOT HAVE A SYMBOL, IT IS ASSUMED TO BE RADIANS. 

In Geometry, you most probably measured everything in degrees.  Degrees can be written two ways in decimal degrees or in DMS form (degrees, minutes, seconds)

It is usually preferred that you state degrees in DMS form. 

Convert 19.64° to DMS form.

There are 2 methods for converting:

METHOD 1:

1. Take your integer unitsand write them down as whole degrees.

19°

2. Multiply everything behind the decimal (including the decimal) by 60.

.64*60 =38.4

3. Write your integer units down as minutes,

19°38’

4. Repeat step 3 & 4 to find seconds.

.4*60 = 24

Final Answer: 19°38’24”

Note: If  you have decimal places after all of this, it is okay to put your seconds value as a decimal.

METHOD 2: (Will only work on TI-80’s series calculators)

1. Type in the decimal degrees:

2. Press 2nd>Apps  to get to the Angle menu, and scroll down to >DMS (0ption 4). Press enter.

3.Press enter once again and Voila!

 

--------

One Radian is 180/╥ degrees and one degree is ╥/180 but there is a similarly easy way to convert Radians to degrees.

1. Type in radians:

2. Get to Angle Menu and select degrees (option 1):

3. Press enter:

 

Hope this helps!

Dinidu Perera, the Calculator Guru

This week did not seem like it was too bad. What I'm thinking is that as long as I remember the formulas I should do pretty good. The exam on the other hand was very difficult for me, I completely bombed it. Here's what I really understood this week though.

Notes:
To find radians:

radians = degrees x pi/180

To find degrees:

degrees = radians x 180/pi

Ex:

360 degrees, convert to radians

360 x π/180 = 360π/ 180 = π/2

Convert π/10 to degrees.

π/10 x 180/π = 18 degrees

The only thing I am a little hazy on is the minutes and seconds thing. Knowing me I probably just wasn't paying attention for that part of the lesson

Stephen's Reflection 8

Ok first of all sorry im a little late. I wasnt sure if we still had to do blogs because on exams. And second, ima talk about the exponential functions. There are 3 main formulas: A(t)=Ao(l+r)^t, A(t)=Aob^t/k, and P(t)=Poe^rt. For the first formula:
Ao=# you start with
r=rate as a decimal
t=time

For the second formula:
t=time
Ao=what you start with
b=double, half, etc...(key words that tell you whether or not to use this formula)
k=time required to double, half, etc...

For the third formula:
Po=what you start with
r=rate
t=time
(use this formula when compounding continuously)

The things i do not understand is how to tell the difference between conics. And i also do not understand how they are supposed to look when they are written in standard form.

alaina's reflection

Okay, so this week I have been completely bogged down with studying for exams and keeping up with notes and homework in everything else. And, I almost forgot to do my reflection. So here goes.

First, I understand conics a lot better than I did last year, or durring the summer.
Circle
(x-h)^2+(y-k)^2=r^2
I always thought circles were easiest.

• Your center is (h,k)
• radius=r; take square root of r^2
• to graph, plot your center and count your radius all the way around.

Ellipse

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

• Your center is (h,k)
• Your major axis is the non-negative denominator
• minor is negative

Hyperbola

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

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

Parabola

y-y⊂1&sub/=m(x-x&sub/1)

**don't understand**

not much. There are certain little things that I keep blanking out on, but for the most part, I think I've got it. I'm really stressed out about this exam, though. I've been studying for 6 hours and it seems the more that I study, the more I forget. So I've put away the books for the night. Hopefully when I wake up in the morning, everything will be alright. I'll continue to study throughout the day, until the exam... I'll probably have questions in the morning. Night all.

Stephanie's Reflection

I have had many problems with the study guide this week, especially with Chapter 5 but I'll start at the beginning.

Chapter 1:
#13 - y=x+2; y=x^2+6x+8 Set them equal. x+2=x^2+6x+8
Then what? Solve for x? How do I get the answers B-Rob gave us if I do so?

Chapter 2:
#4 - The equation x^4+x^3-5x^2+x-6 has the roots x=2 and x=-3. Find the remaining roots.
Um, what? I looked it up in my book and notes but apparently can't seem to find anything on this.
#8 - Solve for x: 2x^3-5x^2-4X+3
How to get the answers she gave us, I have no idea. I tried many ways and am still baffled.
#9 - Solve for x: x^4-7x^2-8
Same as #8, I have no idea.


Chapter 3:
#4 - The equation x^4-2x^3-4x^2+2x+3 has the x-intercepts x=-1 and x=3. Find the remaining x-intercepts.
#8 - which is the same as the #8 of the previous chapter.
#9 - Solve for x: 2x^4+14x+12
#11 - Solve for x: x^3+14x^2+56x+64
#16 - Solve for x: (x+3)/8+(x-1)/4 is less than or equal to 3x/6

Chapter 4:
#3/C - Domain and range of y=(4x-3)/(7x+1)
#3/D - Domain and range of y= square root of(x-4)+3


Chapter 5:
#2 - Express y in terms of x: log(2y)=3log(x)+log(x-1)
#3/B - 2^(log 2 of 6)
#3/D - log 5 of 10
#4 - 2^x=100
#13 - 9^(x+2)=1/729
#15 - Solve for x: 3^(2x)+5(3^x)+6

Taylor Reflection #8

All right
this week can be described in many ways.
stressful. overwhelming. etc
First lets talk about the takehome test.
that test was still freaking hard. This bothers me so badly.
even with the notes infront of my im thrown off by tests. Im back to the drawing board on trying to figure out why.
Then came the study guides in need of completion. i was happy to see that for the majority of the tests i could still answer the same ones i did the day i took the actual tests. This boosted my confidence for the coming exam
there were however still a few questions i cannot find notes to answer the following questions

chapter 1 question number 19: find the perpendicular bisector of the line 4x-2y=4 through the point (1,0)

I need to know the equations of slope intercept, point slope, and standard form

what is the degree of a line

Chapter 4 number 5 letter d
is y=squareroot 9-x^2 symmetric about the line y=x

chapter 5 number 3 letter f
find the exact value of -ln e

Alicia's reflection #8

Alrighty well I am having such a difficult time studying for this exam. I have basically accepted the fact that I am going to fail :(

Parabolas have no major axis and no asymptotes.

Axis of symmetry x=-b/2a

Finding the vertex
(-b/2a, f(-b/2a))

or

complete the square to get vertex form

y=(x+a)^2+b a&b are #'s

(-a,b) vertex

focus: 1/4p= coeff of x^2 then add p.

directrix is p units behind vertex. subtract p.

EX: 1/8y^2

v(0,0)

Focus: p=2
(2,0)

directrix: x=-2

Amy's Reflection #8

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

guys, im really nervous about the exam tomorrow. i have a feeling im gonna freak out and forget half of it...btw, does anyone know how to do #12 on the chapter 5 test: solve for x: log(x+7)=2..if anyone can help me out with that problem then that would be great...

Reflection #8

So this week was a kind of laid back week for me, I don't know if it was like that for everyone else though. We did a bunch of reviewing in class and while doing that I realized that I did NOT remember too much so I am expecting a pretty bad grade on this exam. It probably will be a beast, but anyway finding the center of Conics and the radius didn't seem too hard.

Steps:
1. The equation of a circle in standard form: (x-h)^2+(y-k)^2=r^2
the center=(h,k) and r=radius
2. If not in standard form, you must complete the square to put in standard form
3. Given the center and a point, you can use the distance formula to find the radius

Ex.: (x-3)^2+(y+7)^2
Center:(3,-7)
Radius:√19

Ex. with completing the square:
x^2+y^2-6x+4y-12=0
x^2-6x+__+4^2+4y+__=12
x^2-6x+9+4^2+4y+4=12+9+4
(x-3)^2+(y+2)^2=25
Center:(3,-2)
Radius:25

Now what I am struggling with is remembering all the things for the Exam tomorrow, I'm very nervous and don't know how I'll be able to do it. If anyone has any way to help me with this little problem I'd appreciate it very much.

ATTN: Taylor

im really sorry that i didn't give it back to you in english. i honestly did not know you check out after 4th and when i realized you left, i gave your studyguide to your lil brother...so im guessing he forgot to give it to you...please let me know if you get it or not...

ATTN: AMY

I need my study guides before sunday is over!

you borrowed them and i have to complete them before monday, plus i need them to study

i have no other way to contact you

write back.

Alaina's reflection

Ahhhh! CONICS. This week was not that good for me. I do not like conics. I get all the formulas mixed up and it becomes a big jumbled mess. It is mostly ellipses and hyperbolas that I have problems with so I will explain the rules for circles (which I do get) for this reflection.

1. the eqn for a circle in standard form: (x-h)^2+(y-k)^2=r^2 where the center=(h,k) and r=radius
2. if eqn is not already in standard form, you must complete the square to put in standard form
3. given the 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 of the line
2. substitute in circle eqn
3. solve for x
4. plug x value in to get y value
*NOTE: if your x-value is imaginary, then there is no point of intersection

Given the Equation (x-3)^2+(y-7)^2=19 find th ecenter and radius
CENTER: (3,7) RADIUS: ⌋19

B. x^2+y^2+12y+16x-5
x^2+16x+64+y^2+12y+36=5+64+36
(x+8)^2+(y+6^2)=105
Center: (-8,-6) Radius: ⌋105

The main problem I have had with the homework over the weekend is where the point on the minor axis comes from. I don't quite understand it. And also, numbers 19-22 on the homework. If anyone knows how to do this, I would like to know. Thanks.

Taylor Rodriguez reflection #7

Reflection #7
This week was a hard to handle week. The week started with a test..if that doesn’t say “beware of the week to follow” I don’t know what does. To speak on the test, I wanna say that I was completely confident Monday morning going into school that I knew the test material and I would totally fly through the test. I mean, we had a quiz on each separate section that would be on the test, and I studied for each quiz and did fairly well. Plus I had the weekend to study extra for the test. I can’t say I really feel like I did badly. I was disappointed that there were a few questions I got mixed up on but I am proud to say that there was only one question that I did not answer.
Now for the basics that each blog entry needs,
I most definitely am confident in my knowledge of the steps necessary for graphing a conic equation.
STEPS FOR SKETCHING CONICS
Determine shape
Find center
Mark Major axis
Mark Minor axis
Mark Other intercepts
Mark Focus (or Foci)
Decipher Asymptotes
Mark Vertex
Complete Sketch
What I am not confident in is my knowledge of how to decipher between a circle equation and an elliptical equation. I kind of figure that I could find a way to guess between the two or I could find a way to solve the problem and when I go to sketch the conic it will appear as the shape. However I doubt this is an effective way to go about completing the Chapter 6 test. So notes on how to distinguish between the two would be great!

Stephen's Reflection 7

Ok so this week we did conics, ellipses, and hyperbolas. So i guess ill talk about the hyperbolas. For hyperbolas, the formula is (x-h)^2/(length/20^2)-(y-k)^2/(length/2)^2 =1 or -(x-h)^2/(length/2)^2+(y-k)^2/(length/2)^2 =1....i know thats a little confusing. So after u put it in that form, then you will follow steps in order to make a graph:
1. Center (h,k)
2. Major axis is non-negative
3. vertex +/- the square root of the non-negative denominator
4. asymptotes: y+/- sqr. root of y denom/sqr. root of x denom
5. focus^2=x denom + y denom
focus^2=vertex^2+other denom
6. then to sketch the graph make sure u have:
-shape
-center
-major axis
-minor axis
-other interger...if any
-focus
-asymptotes
-vertex
To sketch:
1. Draw a box using the vertex and sqr. root of other denom.
2. Draw diagonals through the box
3. Sketch parabola on each vertex
4. Label foci and asymptotes

Thats basically it. It may seem hard but thats just because there are alot of steps that you have to do. And i dont really have anything i dont understand right now but i do need a little help on changing bases from chapter 5 that i want to make sure i understand so if someone can help...thank you!!

Dustin's Reflection #(idk what number i'm on)

This week we learned about 3 conics. We learned about circles, ellipses, and hyperbola's. Last year I nearly failed the final exam b/c of conics, but now I think they are pretty simple. I am going to explain how to put a circle in standard form. To do this I will use a form of completing the square.

Ex. x^2+y^2-6x+4y-12=0
1.)Put your like terms together and leave space:
x^2-6x+__+y^2+4y+__=0
2.)Take your linear coefficients and divide by two then square. Add results to each side in respective blanks:
x^2-6x+9+y^2+4y+4=12+9+4
3.)Factor and Simplify:
(x-3)^2+(y+2)^2=25
And that is standard form of a circle.
Standard Form is: (x-h)^2+(y-k)^2=r^2; where (h,k) is the center and r is the radius.

***If you want to find the center or radius, put equation in standard form and take respective elements.****

I understood all of the conic stuff it's just remembering formulas is what will be hard. I looked through my chapter tests and one thing I'm still not sure on how to do is find inverses of functions. Any help is appreciated. Thanks.

Alicia's Reflection #6

Alrighty so this week we learned a good bit of information: circles, ellipses, and hyperbolas.

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

Devin's Reflection #7

This week we have dealt with conics, ellipses, and hyperbolas. The standard form for a circle equation is (x-h)^2+(y-k)^@=r^2. The center is (h,k).

Examples: Find the center

a. (x-3)^2+(y+7)^2=19

center (3,-7)

b. x^2+y^2-6x+4y-12=0

x^2-6x+ (9)+y^2+4y+ (4)=12+9+4

(x-3)^2+(y+2)^2=25

center (3,-2)

The r stands for the radius of the circle. When the equation is not in standard equation you have to complete the square to put the equation in standard equation. You can determine the radius of a circle, by using the distance formula and if you are given the center and a point.

Example: Find radius

a.(x-3)^@+(y+7)^2=19

radius square root of (19)




To find the intersection of a line and a cirlce

1. solve the linear equation for y
2. substitute in circle equation
3. solve for x
4. plug x-value into get y-value

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

With ellipses, the main thing to know is the steps. They are:

1. Find center
2. Find major axis - big denom.
3. Find vertex - square root of big denom.
4. Find other intercepts - square root of small denom
5. Find Focus
6. Find length of major axis
7. Find length of minor axis
8. Graph

Just like the ellipses, to sketch the hyperbolas you must follow the steps.
First find:

1. shape
2. center
3. major axis
4. minor axis
5. 0ther int.
6. vertex
7. focus
8. asymptotes
9. then sketch

To sketch a hyperbola,do the following

1. Draw a box using the vertex and square root of the other denom
2. Draw diagonals through box
3. Sketch a parabola on each vertex
4. Label focus and asymptotes

Stephanie's Reflection

CIRCLES
The equation of a circle in standard form is (x-h)^2-(y-h)^2=r^2 with the center being (h,k) and r being the radius.
Finding the intersection of a line and a circle:
1) solve linear equation for y
2) substitute in circle equation
3) solve for x
4) plug x in to get y value
(if x happens to be imaginary, there is no point of intersection)

EG:
(x-4)^2+(y+2)^2=16
center:(4,-2) radius:4

x^2+y^2+12y+16x-5=0
x^2+16x+64+y^2+12y+36=5+64+36(x+8)^2+(y+6)^2=105
center:(-8,-6) radius:square root of 105

ELLIPSES
1) (x-h)^2/(length of x/2)^2 + (y-k)^2/(length of y/2)^2 =1
2)center is (h,k)
3) major axis has larger denominator
4) vertex is on major axis
5) focus is smaller denom squared = larger denom squared - focus squared
focus is on major axis
Graphing:
1) find center
2) major axis = plus or minus the square root of the bigger denom
3) vertex
4) other intercepts
5) focus
6) length of major axis = 2 square root of
7) length of minor axis = 2 square root of
8) graph

EG:
x^2/4+y^2/1=1
1) (0,0)
2) x
3) +/-2 (2,0) (-2,0)
4) +/-1 (0,1) (0,-1)
5) 1=4-c^2 c=+/-square root of 3 (sr3,0) (-sr3,0)
6) 2 square root of 4 = 4
7) 2 square root of 1
8) graph

HYPERBOLAS
1) (x+h)^2/(length/2)^2 - (y-k)^2/(length/2)^2 =1
OR
-(x-h)^2/(length/2)^2 + (y-k)^2/(length/2)^2 =1
2) center (h,k)
3) major axis is non-negative
4) vertex is the square root of non-negative denom
5) asymptotes y=+/-(square root of y)/(square root of x)x
6) focus^2 = x denom + y denom
focus^2 = vertex^2 + other denom

to sketch:
1) shape
2) center
3) major
4) minor
5) other intercept - none for hyperbolas
6) focus
7) asymptotes y=+/-square root of y/square root of x
8) vertex
9) sketch
A) draw a box using the vertex and +/-sr of other denom
B) draw diagonal through box corners
C) sketch a parabola on each vertex
D) label focus and asymptotes

EG:
x/36-y/9=1
2) (0,0)
3) x
4) y
5) none
6) c^2=36+9 c^2=45 c=sr45 (sr45,0) (-sr45,0)
7) y=+/-square root of 5/square root of 6
8) +/-sr36 = +/-6 (6,0) (-6,0)
9) sketch

Yes, I get that if I don't understand something to look in my notes but I wasn't there to hear the explanation of circles so my notes don't really help me. I'd really appreciate it if someone would explain.

Amy's Reflection #7

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..

ok im also having some problems with hyperbolas..i know how to do #s 1-6 from the homework. i mean that's easy; all you got to do is follow your notes, but i dont know how to #s 7 & 8. if anyone can help me with those two problems, i'll be really greatful..

Reflection #7

One thing this week that i really picked up pretty well was Conics. So I'll go straight into that. 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.

So for example, say you have:
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 would 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 and you should get this:
(x+8)^2+(y+6)^2=105

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

This week I did have one little problem with Hyperbolas. I don't understand how we've been getting the foci when we sketch the problem. So if anyone can help me with that little problem I would be very grateful.