Question

Graphing question: cos, sine.

Answer 1

Just to graph them both?

Answer 2

I'm checking your answers right?

Answer 3

amp is great!
this looks like one cycle of the sin graph
the length of that cycle (period) by doing b-a where b>a
you know where b is the right endpoint
and a is the left endpoint

phaseshift number is great

Answer 4

\[y=8\sin(k[x+\frac{1}{2}])\]
everything you wrote here is good except your k

Answer 5

well and the period as mentioned above

Answer 6

2pi/k=period
once you find the period you can replace that word with period with the number you found for it and solve for k

Answer 7

Sorry, went to go eat dinner :P But i'm back!

Answer 8

ok well let me know what you think if you have any thoughts when you get done reading

Answer 9

So.. period = 3/2
so k =3pi?

Answer 10

period is 1/2+1 right
so yes period is 3/2

now k checking k....
\[\frac{2\pi}{k}=\frac{3}{2} \\ 2 \pi(2)=3(k)\]
oops looks you right on k to begin with

Answer 11

so your period is 3/2
and your k is 4pi/3 as you got before
I image you just said period was 4pi/3 since k was 4pi/3
but that isn't the relationship between k and period

Answer 12

ah, so
Amp=8
prd: 1.5
Phase s: -1/2

y=8sin(4pi/3(x+1/2))

Answer 13

yes I agree with that
and we do not have to change the sign on the 8 this time
since our max's and min's occur in the order of the max's and min's of the parent function
y=8sin(x)
you know we have the max 8 then the min -8

unlike the question before
we had the max and min occurring in a different order then the parent function :)

Answer 14

anyways I have to do some things that involve very little math

Answer 15

unless you have any questions about this question

Answer 16

right, right.
Nope it's all good :)
Good luck with your maths!

Answer 17

Thanks ^-^

Answer 18

in the words of jennifer garner
"good luck with fractions"

Answer 19

or at least her character on 13 going on 30

Answer 20

hahah.
ey, I take that back actually. The homework thing marked the answer as wrong.
y=8sin4pi/3(x+1/2) was marked wrong :|

Answer 21

the amp is right
hmm
the phase shift is at -1/2
let me check the k again
\[\frac{2\pi}{k}=\frac{3}{2} \\ 2 \pi(2)=3k \\ \frac{4 \pi}{3}=k\]
hmmm

Answer 22

http://www.wolframalpha.com/input/?i=y%3D8sin%284pi%2F3%28x%2B1%2F2%29%29
your graph matches this

Answer 23

there is nothing wrong with
\[y=8 \sin(\frac{4\pi}{3}(x+\frac{1}{2}))\]

Answer 24

but hmm
let me try to write it differently

Answer 25

one sec

Answer 26

actually no using cos won't work because it looks like they want sin because they were talking about sin
but we could write it using cos

Answer 27

Right, I think it will mark it wrong if I were to use cos. but i'm not sure why it's not accepting it -.-

Answer 28

how many guesses can we have?
I'm asking because maybe we can try to guess what the teacher meant

but with the graph given this is definitely the answer

Answer 29

Er ,5. But I think I only have 3 left.

Answer 30

the only thing I can think I guess is maybe she meant to put pi's next to the numbers on the x-axis

Answer 31

but wait you said the period was not marked wrong right?

Answer 32

just the equation we got?

Answer 33

yeah
amp=8
period=1.5
ps.=-1/2
are all correct.
only the equation is wrong

Answer 34

ok just so you know y=8sin(4pi/3(x+1/2))
was the correct answer

but what happens if you have write y=8sin(4/3(x+1/2))
don't try that if you already have though ?

Answer 35

I could try putting it in like: y=8sin(4 times pi/3(x+1/2)) ?

Answer 36

that would be weird if that made a difference

Answer 37

is it giving you an error message ?

Answer 38

like it is saying anything about how you are entering it

Answer 39

yeah it said it's wrong.
the cos one is coming out wrong too :|

Answer 40

No it's not, it just marks it wrong.

Answer 41

the cos one we did before?

Answer 42

Yeah

Answer 43

Ah, I'm sorry this is taking so long :/

Answer 44

http://openstudy.com/users/babynini#/updates/5530703de4b0f0b1ce3ba56e
but you said it was right here?

Answer 45

so the computer software changed its mind?

Answer 46

haha no, I didn't notice it had marked it wrong until now. Sorry about that.

Answer 47

oh! I found it out. It wanted a parenthesis around the k as well. So y=8sin(4pi/3(x+1/2))

Answer 48

computers are so dumb

Answer 49

like you know the interesting thing is that is now how they have their thingy above
like they write y=a sin k(x-b)
and not y=a sin(k(x-b))

Answer 50

not*

Answer 51

Dang right they are.

Answer 52

that is how I write mine as you see above
but I think it is totally weird they wouldn't accept y=a sin k (x-b)
when that is what they have

Answer 53

you know instead of y=a sin(k(x-b))

Answer 54

Yeah ._. I'm so sorry that took waay longer than it should have.

Answer 55

well it wasn't your fault
it was the software your school or whoever chose
or who ever programmed it to be weird like that

Answer 56

anyways peace

Answer 57

"whoever programmed it to be weird" xD
Cya. You're the bomb!

Answer 58

lol later

Answer 59

:)