Question 665614
y=cot(3x-pi/2)+5
equation of cot function: y=cot(Bx-C), period=π/B, phase shift=C/B
For given trig function: y=cot(3x-pi/2)+5
B=3
Period=π/B=π/3
1/4 period=π/12
Phase shift=C/B=(π/2)/3=π/6 (shift to the right)
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Graphing:(3 steps) (x-axis in radians, y-axis in units)
coordinates of basic cot curve cot3x for one period:[0,π/3]
(0,asymptote), (π/12,1), (π/6, 0), (3π/12,-1), (π/3, asymptote)
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phase shift π/6 to the right: (add π/6≈.524 to x-coordinate)
(0.524,asymptote), (.786,1), (1.048, 0), (1.309,-1), (1.571, asymptote)
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bump up curve 5 units (final configuration)(add 5 to y-coordinates)
(0.524,asymptote), (.786,6), (1.048, 5), (1.309,-4), (1.571, asymptote)