Question

Are the properties of the following periodic function correct? (I need help with 2 properties)

Function: -2*sin(2*(t+1)/3)

Answer 1

horizontal shift is 2/3
vertical shift is zero.

Answer 2

Are all the other properties correct?
How do I find the horizontal shift fot this?
And the vertical shift? how do i determine it.

Answer 3

2*(t+1)/3 = 2t/3 + 2/3 (horizontal shift)

I think 2/3 might be 2/3 * pi instead of 2/3
vertical shift: -2*sin(2*(t+1)/3 + 0 <--- vertical shift is zero.

Answer 4

\[y=-2\sin\frac{2}{3}(t-(-1))\]

Answer 5

Amplitude = 2
Period = 3 pi
Phase Shift = -1 ie 1 to the left.

Answer 6

Im confused, is the phase shift 2/3*pi? or is it -1?

Answer 7

Vertical displacement is 0

Answer 8

I am not sure ..

Answer 9

The phase shift is -1

Answer 10

Ok. I didnt see what you did up there:). that makes sense. Are the other properties correct?

Answer 11

I have listed all the properties. Please read the responses to your questions.

Answer 12

Btw, isnt the horizontal phase shift usually in degrees? So is it -1degrees?

Answer 13

http://www.wolframalpha.com/input/?i=-2*sin%282*%28t%2B1%29%2F3%29

Answer 14

@Mertsj, when I look at the graph of the function, The horizontal distance between cycles is 3. if i do 3*pi, then that gives me 9.42

Answer 15

is it? im not sure:(

Answer 16

is 1 in degree or radian ?

Answer 17

Thats why I asked if it was in degrees. Im a little bit confused

Answer 18

@experimentx, Btw, when I look at the graph of the function, The horizontal distance between cycles is 3. if i do 3*pi, then that gives me 9.42. So is it 3 or 3*pi?

Answer 19

IM just confused because my book tells me something, but I know that Merstj is never wrong.

Answer 20

You better look at the graph again.

Answer 21

What does your book tell you?

Answer 22

Im not allowed to use wolfram, So i cannot even know about the pis in the graph. i use the traditional graphing calculator. And for this, what we do to measure the horizontal distance between cycles is to merely get the exact value of the peak of one cycle and subtract it by the previous one.

Answer 23

In this graph, the peak of one cycle is -6, and the previous one is -3. therefore, we get 3, and thats the horizontal distance between cycles

Answer 24

Can you give me 3*pi by looking at the graph of a T1-83+?

Answer 25

I graphed this at Wolfram and on a graphing calculator. The period is 3 pi.
Does your graphing calculator also solve equations?

Answer 26

Yes

Answer 27

Then set this equation to 2 which is the y value of the maxima and see what the solutions are. They should be 3 pi units apart.

Answer 28

Oh ok, I see it now. So sorry! I was just really confused:(
Thank you so much @Mertsj!

Answer 29

yw