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how do i graph the following problem: y=-2sin2(x-3pi/4)
\n" ); document.write( "i also need to find the period, phase shift, amplitude, left, quarter, mid, three-quarter and right end points
\n" ); document.write( "**
\n" ); document.write( "Standard form of equation for sin function: y=Asin(Bx-C), A=amplitude, Period =2π/B. Phase shift=C/B.
\n" ); document.write( "For given sin function: y=-2sin2(x-3pi/4)
\n" ); document.write( "Can be rewritten: y=-2sin(2x-3π/2
\n" ); document.write( "B=2
\n" ); document.write( "Period=2π/B=2π/2=π
\n" ); document.write( "C=-3π/2
\n" ); document.write( "Phase shift=C/B=(3π/2)/2=3π/4 (to the right)
\n" ); document.write( "Amplitude=2
\n" ); document.write( "..
\n" ); document.write( "Graphing for one period:
\n" ); document.write( "On the x-axis make tick marks at: 0, π/4, 2π/4, 3π/4, 4π/4, 5π/4, 6π/4, 7π/4, 8π/4 (For clarity, I left the fraction in 4th's, but in the final graph they should be reduced to lowest terms)
\n" ); document.write( "At these x-values, y=-2, 0, 2, 0,-2, 0, 2, and 0 respectively.
\n" ); document.write( "You now have these points to graph the function: (0,-2), (π/4,0), (π/2,2), (3π/4,0), (π,-2), (5π/4,0), (3/π/2,2), (7π/4,0), (2π,-2)
\n" ); document.write( "note: Because of the phase shift, the curve started at 3π/4 and ended at 2π which covered one period. The graph was extended back to zero to show the y-intercept at -2. \n" ); document.write( "