Find the period P, amplitude A, and phase shift S, and graph at least two periods of the equation. $y = 5 \sin \left( \frac { 1 } { 2 } x + 120 ^ { \circ } \right)$
A) P = 720°, A = 5, S = 240°
B) P = 720°, A = 5, S = -240°
C) P = 72°, A = 0.5, S = 24°
D) P = 72°, A = 0.5, S = 24°
E) P = 72°, A = 0.5, S = 24°

Question

Find the period P, amplitude A, and phase shift S, and graph at least two periods of the equation.   $y = 5 \sin \left( \frac { 1 } { 2 } x + 120 ^ { \circ } \right)$
A) P = 720°, A = 5, S = 240° 
B) P = 720°, A = 5, S = -240° 
C) P = 72°, A = 0.5, S = 24° 
D) P = 72°, A = 0.5, S = 24° 
E) P = 72°, A = 0.5, S = 24°

Quizplus · Accepted Answer

The general form of a sinusoidal function is y = A sin(Bx - C) + D, where A is the amplitude, B is the period (the horizontal length of one cycle), C is the phase shift (horizontal shift), and D is the vertical shift (if any).

Comparing this to the given equation, we see that the amplitude A = 5. To find the period, we use the formula P = 360°/|B|. Here, B = 1/2, so P = 360°/|1/2| = 720°. Therefore, the period is 720°.

Finally, to find the phase shift, we set the argument of the sine function equal to zero and solve for x:

Bx - C = 0 
1/2 x - 120° = 0 
x = 240°

So the phase shift is S = 240°.

To graph the function, we start at the phase shift (x = 240°) and go one period to the right (adding 720° to x), then repeat the same pattern for the next period. The graph should look like a sine curve with a maximum value of 5 and a minimum value of -5, shifted 240° to the right.

Find the Period P, Amplitude A, and Phase Shift S y=5sin⁡(12x+120∘)y = 5 \sin \left( \frac { 1 } { 2 } x + 120 ^ { \circ } \right)y=5sin(21​x+120∘)

Find the Period P, Amplitude A, and Phase Shift S $y = 5 \sin \left( \frac { 1 } { 2 } x + 120 ^ { \circ } \right)$