## anonymous one year ago Can someone help me with determining the equation for a trig function? I'm aware of how to find the amplitude and period, i'm just confused with this graph!

1. anonymous

2. MTALHAHASSAN2

wait how can we find the amplitute

3. anonymous

it's usually the "a", I know how to determine it by looking at an equation

4. jim_thompson5910

$\large \text{amplitude} = \frac{|m-n|}{2}$ m = largest y coordinate of a point on the curve n = smallest y coordinate of a point on the curve

5. jim_thompson5910

|dw:1444532222391:dw|

6. jim_thompson5910

|dw:1444532254763:dw|

7. anonymous

in this case it would be 2+4/2

8. anonymous

which would be 3..

9. jim_thompson5910

yes

10. jim_thompson5910

the midline is found by averaging the two extreme values $\Large D = \frac{m+n}{2}$ y = D is the equation of the midline |dw:1444532397834:dw|

11. anonymous

the midline is -1

12. jim_thompson5910

|dw:1444532428898:dw|

13. jim_thompson5910

the midline is -1 yes

14. anonymous

so for the period since the points aren't actually on the radians how would I calculate where it starts and where it ends?

15. jim_thompson5910

look at the peak that lies on the y axis this is when x = 0 the next peak over to the right is at x = pi so the period is |pi - 0| = |pi| = pi units

16. anonymous

I'm a little confused..

17. anonymous

when x=0 y=2, is that the peak that you're talking about?

18. jim_thompson5910

see the attached image

19. anonymous

ok thanks! I got that. I'm just confused why we are subtracting the pi-0. I thought the formula for period was 2pi/k

20. jim_thompson5910

you might be thinking of T = 2pi/B and that is useful when you have an equation of this form y = A*cos(Bx - C) + D

21. jim_thompson5910

I subtracted the x values to find the distance between them

22. jim_thompson5910

the absolute value is to ensure the distance is positive

23. anonymous

so for this graph we don't need to use that form. We just find distance simply by finding out the x values from the y and subtract them from one another?

24. jim_thompson5910

you can pick on any 2 neighboring peaks and do the same thing

25. jim_thompson5910

we can't use that form because we don't have the equation

26. anonymous

oh ok! That's where I was confused at :)

27. jim_thompson5910

you might have learned how to pick apart y = A*cos(Bx - C) + D (eg find the amplitude to be |A|, etc) but now we're going in reverse: finding the key elements and using them to construct the equation

28. anonymous

so our equation would be y=3sin(x)-1?

29. anonymous

y=sin(pi)-1

30. jim_thompson5910

what is y when x = 0 ?

31. anonymous

2

32. jim_thompson5910

but does plugging in x = 0 give y = 2 when you use y = 3*sin(x) - 1 ?

33. anonymous

no

34. jim_thompson5910

so that equation doesn't fit

35. jim_thompson5910

let's try cos instead of sin

36. jim_thompson5910

if x = 0, does y = 2 for y = 3*cos(x) - 1 ?

37. anonymous

it does fit

38. jim_thompson5910

yep, ok how about if x = pi/2, does y = -4 like it should?

39. anonymous

we know that cos(x)=1

40. anonymous

I don't think so

41. anonymous

cos of pi/2 is 1

42. jim_thompson5910

cos(pi/2) = 0 actually

43. anonymous

and 3-1=2, and 2-1=1

44. jim_thompson5910

recall that the period is pi units so T = pi Solve the equation T = 2pi/B for B to get... T = 2pi/B pi = 2pi/B B*pi = 2pi B = 2pi/pi B = 2 so the equation should be y = 3*cos(2x) - 1

45. jim_thompson5910

now if we try x = pi/2, then y = 3*cos(2x) - 1 y = 3*cos(2*pi/2) - 1 y = 3*cos(pi) - 1 y = 3*(-1) - 1 y = -3 - 1 y = -4 and it fits

46. anonymous

I'm a little lost. Why are we using t=2pi/3?

47. anonymous

B*

48. jim_thompson5910

The period is T = pi going from peak to peak is pi units across

49. jim_thompson5910

plug T = pi into T = 2pi/B and solve for B to get B = 2

50. jim_thompson5910

T = 2pi/B is the period formula If you know B, you can find the period T (or vice versa)

51. anonymous

so we just knew the T, but in order to find b we had to plug in pi for t

52. jim_thompson5910

exactly

53. anonymous

and how did you know that y=-4?

54. anonymous

just the point on the graph?

55. jim_thompson5910

56. jim_thompson5910

Yes the point (pi/2, -4) is on the graph Notice how in my work above I replaced x with pi/2 in the equation y = 3*cos(2x) - 1 Then I evaluated/simplified to get y = -4. So that confirms the equation works for x = pi/2

57. anonymous

oh so you just replaced it to check to see if the equation that we wrote was correct.. makes sense. :) Thank you lots!

58. jim_thompson5910

yes I recommend you check x = pi as well. Make sure that the output is y = 2

59. anonymous

Thanks again :) You're a life saver! I honestly had a lot of trouble with this problem

60. jim_thompson5910

I'm glad it's making sense now