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Trigonometry 15 Online
OpenStudy (anonymous):

finding equations for trig graphs?

OpenStudy (anonymous):

it's these four and all I know how to do is find the amplitude which is the easiest part. Could someone help me out?

OpenStudy (misty1212):

what else do you need ?

OpenStudy (anonymous):

Finding the function associated with the graph

OpenStudy (misty1212):

ok wanna do the first one?

OpenStudy (anonymous):

If you're up for it yeah

OpenStudy (misty1212):

what did you get for the amplitude?

OpenStudy (anonymous):

+2

OpenStudy (anonymous):

if that's wrong i s2g.

OpenStudy (misty1212):

it is 2, right

OpenStudy (misty1212):

it goes from \(-5\) to \(-1\) which has length 4 so the amplitude is 2 to make it go from \(-2\) to \(2\) you are going to have to lift it up 3 units

OpenStudy (misty1212):

did you get the period?

OpenStudy (anonymous):

I think it's either 6 or 12

OpenStudy (misty1212):

oh no

OpenStudy (misty1212):

look at the top most point of the graph, where the curve is at \(-1\) one is where \(x=\frac{\pi}{4}\) where is the next one?

OpenStudy (anonymous):

5x/4

OpenStudy (misty1212):

right well \(\frac{5\pi}{4}\)

OpenStudy (anonymous):

new to openstudy my bad

OpenStudy (misty1212):

no problem in any case the period it the length of that interval, the length of \[[\frac{\pi}{4},\frac{5\pi}{4}]\] aka \(\frac{5\pi}{4}-\frac{\pi}{4}\)

OpenStudy (misty1212):

on other words the period is \(\pi\)

OpenStudy (anonymous):

okay good, I actually understand that

OpenStudy (misty1212):

you know \(\sin(bx)\) has period \(\frac{2\pi}{b}\) set \[\frac{2\pi}{b}=\pi\] and solve for \(b\)

OpenStudy (misty1212):

let me know when you get \(b=2\)

OpenStudy (anonymous):

so, \[b = 2\]

OpenStudy (anonymous):

oh, yeah I got that

OpenStudy (misty1212):

right now we have the amplitude is 2, \(b=2\) and it looks like sine shifted down 3 units

OpenStudy (misty1212):

final answer \[y=2\sin(2x)-3\] btw i guess we should say why it is sine and not cosine

OpenStudy (misty1212):

if you shift this one up 3 units it goes right through \((0,0)\) and \(\sin(0)=0\) so it is sine shifted down

OpenStudy (misty1212):

as compared to the next one, the one below it, that looks like cosine shifted down

OpenStudy (anonymous):

so, if I understand this, sine cannot go through \[(0,0)\] but cosine can

OpenStudy (anonymous):

if I'm wrong I apologize I'm a little slow with this

OpenStudy (misty1212):

sine goes through \((0,0)\)

OpenStudy (misty1212):

that is how i know this is sine it is sine shifted down

OpenStudy (anonymous):

but it's shifted down 3 okay I get it

OpenStudy (misty1212):

here is a graph of both \(y=2\sin(x)-3\) and \(y=\sin(x)\) http://www.wolframalpha.com/input/?i=2sin%282x%29-3%2Csin%28x%29

OpenStudy (misty1212):

the next one is cosine, because cosine has its maximum at 0, as does this graph the amplitude is easy to find, the period is easy too if you understand how to see it

OpenStudy (misty1212):

gotta run to class, good luck!

OpenStudy (anonymous):

okay thank you so much I really appreciate the help!

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