## anonymous 4 years ago Find the amplitude of the sine curve shown below.

1. anonymous

2. anonymous

amplitude is just the height of the sine curve

3. anonymous

How do i do that? :/

4. anonymous

Maximum value - minimum value divided by 2

5. anonymous

How do i work this out? I have no clue

6. anonymous

I.e., A regular sine curve has a minimum value at -pi/2, equal to -1. At +pi/2, it's value is 1. Thus, you have $\frac{1-(-1)}{2}=1$For your curve, what're the max and min values?

7. anonymous

4, -4?

8. anonymous

Yes. Now what do you do?

9. anonymous

the amplitude is 4, because if you look at the graph.. the amplitude is from zero to the highest point.

10. jim_thompson5910

or amplitude = distance from middle to either extreme

11. anonymous

@bronzegoddess "from zero" is not a solution for all possible sine curves. For example, sin(x)+1, "from zero" has an amplitude of 2, whereas in reality it's 1. @jim_thompson5910 yeah that's right but not rigidly defined.

12. jim_thompson5910

true, but it's a good way to think about it

13. jim_thompson5910

it's a basic way at least, if you get too rigorous, then you may confuse things (at least in my opinion)

14. anonymous

Well that's why I didn't start getting into domains and formal declarations and such...

15. jim_thompson5910

lol yeah, don't need to get into those ideas (just yet)

16. anonymous

@vf321 its a simple way to think about it sorry if I am wrong.. if I was given an example like sin(x)+1 I would know that the amplitude is 1 because Asinx(Bx-C)+D, the amplitude is the number before the sin.

17. anonymous

@bronzegoddess yes of course no attack intended just didn't want @EmilyJernigan to use that method since she doesn't have the same intuition about it that you do.

18. anonymous

its okay, no harm intended on my part too :)

19. anonymous

@vf321 do you think you could explain relations to me?

20. anonymous

realtions? What do you mean?

21. anonymous

Make a question about it and address all points. It's a bit informal doing it on someone else's question.

22. anonymous

okay'll switch, I just wanted to know if you could help.