document.write( "Question 517613: How do u graph y=3cos(2(x+pi/4))+2?? \n" ); document.write( "
Algebra.Com's Answer #344858 by lwsshak3(11628)\"\" \"About 
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How do u graph y=3cos(2(x+pi/4))+2??
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\n" ); document.write( "y=3cos(2(x+pi/4))+2
\n" ); document.write( "Equation for graphing sin function: y=Acos(Bx-C), with A=amplitude, Period=2π/B,
\n" ); document.write( "Phase shift=C/B.
\n" ); document.write( "For given sin function: y=3cos(2(x+pi/4))+2=3cos(2x+pi/2))+2
\n" ); document.write( "Amplitude=3
\n" ); document.write( "B=2
\n" ); document.write( "Period=2π/B=2π/2=π
\n" ); document.write( "1/4 period=π/4
\n" ); document.write( "Phase shift=C/B=(π/2)/2=π/4 (shift to the left)
\n" ); document.write( "Curve is shifted vertically 2 units
\n" ); document.write( "y-intercept
\n" ); document.write( "set x=0
\n" ); document.write( "y=3cos(2x+pi/2))+2=3cos(π/2)+2=0+2=2
\n" ); document.write( "..
\n" ); document.write( "Graphing for one period:
\n" ); document.write( "Without any phase shift or vertical shift, you can plot the given cos curve (with amplitude=3) on an (x,y) coordinate system with the following coordinates:
\n" ); document.write( "(0,3), (π/4,0), (π/2,-3), (3π/4,0), (π,3)
\n" ); document.write( "With a phase shift of π/4 to the left, the x-coordinates change to read as follows:
\n" ); document.write( "(-π/4,3), (0,0), (π/4,-3), (π/2,0), (3π/4,3)
\n" ); document.write( "Adding a vertical shift of 2 units cause the y-coordinates to change, resulting in the final configuration of the graph as follows:
\n" ); document.write( "(-π/4,5), (0,2), (π/4,-1), (π/2,2), (3π/4,5)
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