## lilsis76 Group Title Determine Amplitude/period/phase shift /graph function. over 1 period. Indicate xintercepts and coordinates of Hi/low points on graph. y = cos (2x - pi/3) +1 y = A ( Bx - C ) so Amplitude is 1 Period: 2pi/B--> 2pi/2 = Pi Okay is my phase shift right? Phaseshift = C/B pi ---- 3 --------------- = pi/6? 2 one year ago one year ago

1. satellite73 Group Title

looks good to me i never remember the formula, so i set $$2x-\frac{\pi}{3}=0$$ and solve for $$x$$

2. satellite73 Group Title

this also help me remember whether the graph has moved left or right. in this case you get $$x=\frac{\pi}{6}$$ so it is shifted to the right

3. lilsis76 Group Title

oh!, okay so since it was positive it goes right, right? I thought if if was negative it goes right?? i am still confused on how they shift.

4. lilsis76 Group Title

|dw:1349892894075:dw|

5. satellite73 Group Title

think of it this way for regular old cosine, you know $$\cos(0)=1$$ now you solve $$2x-\frac{\pi}{3}=0$$ and you see $$x=\frac{\pi}{6}$$ which means instead of $$(0,1)$$ on the graph, you have $$(\frac{\pi}{6},1)$$ on the graph, so it is shifted right

6. satellite73 Group Title

hold on, i did not say what you wrote set it equal to zero $2x-\frac{\pi}{3}=0$ $2x=\frac{\pi}{3}$ $x=\frac{\pi}{6}$

7. satellite73 Group Title

i meant set $2x-\frac{\pi}{3}=0$ not $2x-\frac{x}{3}=0$

8. lilsis76 Group Title

haha okay, i thought = to zero

9. satellite73 Group Title

whew

10. lilsis76 Group Title

11. lilsis76 Group Title

okay so to figure the points, i start at pi/6 and move right by pi/6?

12. satellite73 Group Title

lets go slow because i do not want to confuse you

13. satellite73 Group Title

you know $$\cos(0)=1$$ right?

14. lilsis76 Group Title

okay then. well then what i know know is Amp = 1 period 2pi/b--> 2pi/3 phase shift---> 2x-pi/3=0--> 2x=pi/3 divide both by 2--> = pi/6 Goes Right And yes Cos(0)=1 means at the point zero it goes up to 1 which is the highest point

15. satellite73 Group Title

right, so in this case you have to make $$x=\frac{\pi}{6}$$ in order to get $$2x-\frac{\pi}{3}=0$$ so for your graph, instead of $$(0,1)$$ you have $$(\frac{\pi}{6},1)$$ on the graph

16. satellite73 Group Title

all you wrote is correct, here is a nice picture http://www.wolframalpha.com/input/?i=+y+%3D+cos+%282x+-+pi%2F3%29+%2B1

17. lilsis76 Group Title

OH! i forgot i have to move it up 1

18. lilsis76 Group Title

okay so then i have it like this: |dw:1349893756732:dw|

19. lilsis76 Group Title

then i move up 1

20. lilsis76 Group Title

|dw:1349893880189:dw|

21. lilsis76 Group Title

@satellite73 Like this? so amplitude is 1 period is Pi Phase Shift is pi/6 I graphed the function the x intercepts: there are not any. the lowest point is (pi/2,0), and the highest point is (pi/6,2)

22. lilsis76 Group Title

|dw:1349894136637:dw|

23. lilsis76 Group Title

@swissgirl @marcoduuuh checked the answers in the back of my book. How did they come up with the x intercept of 2pi/3? the hight point i got is (pi/6,2) but how do i get (7pi/6, 2)? the low point of (2pi/3, 0 ) ?

24. phi Group Title

this curve just touches the x-axis when y=0 cos (2x - pi/3) +1 = 0 cos(2x- pi/3)= -1 arc cos both sides 2x - pi/3 = pi 2x= pi+pi/3 = 4/3 pi x= 2/3 pi

25. lilsis76 Group Title

how did you do that?

26. phi Group Title

you know the period is pi, so every time you increase x by pi you will get the same value as x

27. lilsis76 Group Title

sorry, im new to precalculus. all this is super new to me

28. phi Group Title

the hight point i got is (pi/6,2) but how do i get (7pi/6, 2)? pi/6 + pi = pi/6 + 6pi/6 = 7pi/6

29. phi Group Title

how did you do that? are you asking how to get the lowest point at x= 2/3 pi ?

30. lilsis76 Group Title

oh....okay, now is that going to be the same step for getting the HIghs on all problems like that? you get the Period and add it too the phase shift? Yes and I cant get the lowest point either, i end up getting pi/2, 0

31. phi Group Title

Let's look at How did they come up with the x intercept of 2pi/3? First, you know x-intercept means the x value where the curve touches or crosses the x-axis?

32. lilsis76 Group Title

yes how did they come up with 2pi/3, because i got... pi/6 pi/3 pi/2 and wait i guess i got the 2pi/3. but okay i know and see my question. how did they get 2pi/3 an point 0, because when i graphed it by hand it shows that 2pi/3 is at 1

33. lilsis76 Group Title

|dw:1349897328597:dw|

34. lilsis76 Group Title

this is the graph i got, but i got pi/2 as my lowest. did i do something wrong?

35. phi Group Title

type y = cos (2x - pi/3) +1 in the google search window, it should graph it

36. lilsis76 Group Title

okay im lookin at the graph

37. lilsis76 Group Title

so how do i figure the lowest with that graph?

38. phi Group Title

You see the curve just touches the x-axis. This happens when y=0 we start with y = cos (2x - pi/3) +1 cos (2x - pi/3) +1 = 0 (look for where the curve crosses or touches the x-axis) use algebra to solve for x: write -1 on both sides of the = cos(2x -pi/3) +1 -1 = 0-1 simplify to get cos(2x-pi/3) = -1

39. lilsis76 Group Title

cos(2x-pi/3) = -1 okay i get this like u did

40. phi Group Title

cos(2x-pi/3) = -1 take the inverse cosine of both sides 2x -pi/3 = acos(-1) type in google acos(-1)= you get pi radians (3.14159....) or 180 degrees. you now have 2x -pi/3 = pi can you finish?

41. lilsis76 Group Title

i end up getting 2pi/3. and that graph on google is confusing me with different numbers

42. phi Group Title

The graph is show the curve with x in radians. the region from x=0 to 2pi (6.283...) is the region your question is asking about. you see 2 peaks in this interval. Decimals make it hard to match these spots with x= pi/6 and x= 7pi/6 but using a calculator or google we see x= 0.523... and x=3.665...

43. phi Group Title

*The graph shows the curve with x in radians.

44. lilsis76 Group Title

oh....okay, i understand that now, thank you. for your help. Imma close this question and fix my homework. Ill be back later. so thank you

45. phi Group Title

OK. For graphing, I would pick a few x values and use a calculator to find the corresponding y value. Then plot that point. After a few points you can sketch in the curve.

46. lilsis76 Group Title

okay, i would have to try that, I probably wont beable to do that on a quiz tomorrow. she is not allowing calculators. but if i know the steps i know i can do the question or problem