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Explanation:

Step1: Recall sine - function formula

For a sine function (y = A\sin(Bx - C)), the amplitude (A) is given by (|A|), the period (T=\frac{2\pi}{|B|}), and the phase - shift (P.S.=\frac{C}{B}). For (y = 3\sin(6x-\frac{\pi}{4})), (A = 3), (B = 6), (C=\frac{\pi}{4}).

Step2: Calculate amplitude

(A=|3| = 3)

Step3: Calculate period

(T=\frac{2\pi}{|6|}=\frac{\pi}{3})

Step4: Calculate phase - shift

(P.S.=\frac{\frac{\pi}{4}}{6}=\frac{\pi}{24})

Step5: Find maximum and minimum points

The maximum value of (y = 3\sin(6x-\frac{\pi}{4})) occurs when (\sin(6x-\frac{\pi}{4}) = 1). (6x-\frac{\pi}{4}=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}), (6x=\frac{3\pi}{4}+2k\pi), (x=\frac{\pi}{8}+\frac{k\pi}{3}). In one period ((k = 0)), the maximum point is ((\frac{\pi}{8},3)). The minimum value occurs when (\sin(6x - \frac{\pi}{4})=-1). (6x-\frac{\pi}{4}=-\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}), (6x=-\frac{\pi}{4}+2k\pi), (x =-\frac{\pi}{24}+\frac{k\pi}{3}). In one period ((k = 0)), the minimum point is ((-\frac{\pi}{24}, - 3)).

Step6: Find x - intercepts

Set (y = 0), then (\sin(6x-\frac{\pi}{4})=0), (6x-\frac{\pi}{4}=k\pi,k\in\mathbb{Z}), (6x=k\pi+\frac{\pi}{4}), (x=\frac{k\pi}{6}+\frac{\pi}{24}). In one period ((k = 0)), (x=\frac{\pi}{24}); (k = 1), (x=\frac{5\pi}{24}).

For the cosine function (y=-2\cos(2x - \frac{\pi}{2})), where (A=-2), (B = 2), (C=\frac{\pi}{2}).

Step7: Calculate amplitude

(A = |-2|=2)

Step8: Calculate period

(T=\frac{2\pi}{|2|}=\pi)

Step9: Calculate phase - shift

(P.S.=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4})

Step10: Find maximum and minimum points

The maximum value of (y=-2\cos(2x - \frac{\pi}{2})) occurs when (\cos(2x - \frac{\pi}{2})=-1). (2x-\frac{\pi}{2}=\pi + 2k\pi,k\in\mathbb{Z}), (2x=\frac{3\pi}{2}+2k\pi), (x=\frac{3\pi}{4}+k\pi). In one period ((k = 0)), the maximum point is ((\frac{3\pi}{4},2)). The minimum value occurs when (\cos(2x - \frac{\pi}{2}) = 1). (2x-\frac{\pi}{2}=2k\pi,k\in\mathbb{Z}), (2x=\frac{\pi}{2}+2k\pi), (x=\frac{\pi}{4}+k\pi). In one period ((k = 0)), the minimum point is ((\frac{\pi}{4},-2)).

Step11: Find x - intercepts

Set (y = 0), then (\cos(2x-\frac{\pi}{2}) = 0), (2x-\frac{\pi}{2}=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}), (2x=\pi + k\pi), (x=\frac{\pi}{2}+\frac{k\pi}{2}). In one period ((k = 0)), (x=\frac{\pi}{2}); (k = 1), (x=\pi).

Answer:

1. For (y = 3\sin(6x-\frac{\pi}{4})):
   • Amplitude (A = 3)
   • Period (T=\frac{\pi}{3})
   • Phase - shift (P.S.=\frac{\pi}{24})
   • Maximum point: ((\frac{\pi}{8},3))
   • Minimum point: ((-\frac{\pi}{24},-3))
   • x - intercepts in one period: (x=\frac{\pi}{24},\frac{5\pi}{24})
2. For (y=-2\cos(2x - \frac{\pi}{2})):
   • Amplitude (A = 2)
   • Period (T=\pi)
   • Phase - shift (P.S.=\frac{\pi}{4})
   • Maximum point: ((\frac{3\pi}{4},2))
   • Minimum point: ((\frac{\pi}{4},-2))
   • x - intercepts in one period: (x=\frac{\pi}{2},\pi)