SOLUTION: I need help with solving and graphing one cycle of y= 2sin(3x-pi) and y = -3 cos(1/2x)-4. So far for the first question I have gotten really mixed up but I got 3x-pi =o x=pi/

Algebra ->  Trigonometry-basics -> SOLUTION: I need help with solving and graphing one cycle of y= 2sin(3x-pi) and y = -3 cos(1/2x)-4. So far for the first question I have gotten really mixed up but I got 3x-pi =o x=pi/      Log On


   



Question 171418: I need help with solving and graphing one cycle of y= 2sin(3x-pi) and y = -3 cos(1/2x)-4.
So far for the first question I have gotten really mixed up but I got
3x-pi =o
x=pi/3
x=pi/2

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I need help with solving and graphing one cycle of y= 2sin(3x-pi) and y = -3 cos(1/2x)-4.
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y= 2sin(3x-pi)
Period = (2/3)pi
That tells you the distance from peak to peak.
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Amplitude = 2
That tells you the height of each peak
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x-intercept
2sin(3x-pi) = 0
sin(3x-pi) = 0
3x-pi = 0
x = pi/3
Mark that point on the x-axis.
It is the point where the curve crosses the x-axis
and it is the middle of one period of the curve.
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Maybe with that info you can sketch the period centered at pi/3
and then duplicate that period to the left and to the right.
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Cheers,
Stan H.
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