finding equations for trig graphs?
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?
what else do you need ?
Finding the function associated with the graph
ok wanna do the first one?
If you're up for it yeah
what did you get for the amplitude?
+2
if that's wrong i s2g.
it is 2, right
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
did you get the period?
I think it's either 6 or 12
oh no
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?
5x/4
right well \(\frac{5\pi}{4}\)
new to openstudy my bad
no problem in any case the period it the length of that interval, the length of \[[\frac{\pi}{4},\frac{5\pi}{4}]\] aka \(\frac{5\pi}{4}-\frac{\pi}{4}\)
on other words the period is \(\pi\)
okay good, I actually understand that
you know \(\sin(bx)\) has period \(\frac{2\pi}{b}\) set \[\frac{2\pi}{b}=\pi\] and solve for \(b\)
let me know when you get \(b=2\)
so, \[b = 2\]
oh, yeah I got that
right now we have the amplitude is 2, \(b=2\) and it looks like sine shifted down 3 units
final answer \[y=2\sin(2x)-3\] btw i guess we should say why it is sine and not cosine
if you shift this one up 3 units it goes right through \((0,0)\) and \(\sin(0)=0\) so it is sine shifted down
as compared to the next one, the one below it, that looks like cosine shifted down
so, if I understand this, sine cannot go through \[(0,0)\] but cosine can
if I'm wrong I apologize I'm a little slow with this
sine goes through \((0,0)\)
that is how i know this is sine it is sine shifted down
but it's shifted down 3 okay I get it
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
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
gotta run to class, good luck!
okay thank you so much I really appreciate the help!
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