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I know how to graph a cosine equation. Basically the amplitude (minus/plus the vertical shift) is the upper bound, and I know how to get the lower bound. For sine equations I can find the upper lower bounds and the sinuoidal axis. BUT, HOW DO I FIND WERE TO EXACTLY START THE EQUATION ACCORDING TO THE PERIOD?
upper and lower bounds of sin x are -1 and 1 and sin 0 = 0 so the curve goes through the origin
I know that, i'm just confused on the graphing of it. For example: Y=20sin1/2(x-120 degree) -10
If it was a cosine equation, I could just graph the point on (120 deg, 10) and that would be the upper bound point.
But for sine, were exactly should I start at the x axis? Basically I now the Y axis and how high/low it would be, but were do I start is my question? Thanks.
y = sin x passes through the origin. Use the phase shift value to shift the function horizontally. If that's what you're asking
So basically were on x would the point be located at? Since the phase shift is 120, should I do 120-90 or 120+90
I can attach the book answer, the graph approximately starts at 180 degrees horizontally. Even thought the phase shift is 120
Not sure where the 90 is coming from. Shift it 120° to the right This is y = sin x|dw:1442167291189:dw|
so this would be sin (x - 120°) |dw:1442167349251:dw|
yes I figured that too, but please look the the image, I will attatch it now
Problem: 3. Again the equation is: 20sin1/2(x-120deg) -10
ok. I see. The midline of the sine function passes through the origin. that's what's at 120°. so your first point should be (120°, -10)
The point that the graph passes through is 180 not 120. Or is it 120?
nevermind, I see it
Oh that makes sense now
i recall this as \[A\sin(\psi(t-\phi))+B\]
yes, so the point that is at the origin in y = sin x is translated to \((\phi,B)\)