Where does the graph of y = sin x from x = 0 to x = 2π start?

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Where does the graph of y = sin x from x = 0 to x = 2π start?

Mathematics
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A simple way to approach this question is to simply plug in "x = 0" into "y = sin x" so work out what y = sin 0 equals to and this will tell you where the graph begins :)
the answer is at its maximum?
hmm.. what do you mean?

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Do you know what your standard y = sinx graph looks like? :)
im not sure what that is I just got into trig today
hmm.. okay. Well here's what your standard y = sinx graph will look like : http://www.wolframalpha.com/input/?i=y+%3D+sinx+ And here's what your y= sinx graph will look like from x = 0 to x = 2pi http://www.wolframalpha.com/input/?i=y+%3D+sinx+from+x+%3D+0+to+x+%3D+2pi
Since they've told us the boundaries (so the start and end of your graph) are x = 0 and x = 2pi. I guess you could say that the graph starts at (0,0) right? Since x = 0, and if we sub that into our equation : y = sinx... we get: y = sin 0 = 0
thank you for the help much appreciated!
You're welcome ^_^!

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