Find the period P and amplitude A and graph at least two periods of the equation. $y = 4 \cos \frac { 4 } { 3 } x$
A) P = 480°, A = 4
B) P = 90.0°, A = 1.33
C) P = 270°, A = 4
D) P = 270°, A = 4
E) P = 480°, A = 4

Question

Find the period P and amplitude A and graph at least two periods of the equation. $y = 4 \cos \frac { 4 } { 3 } x$
A) P = 480°, A = 4 
B) P = 90.0°, A = 1.33 
C) P = 270°, A = 4 
D) P = 270°, A = 4 
E) P = 480°, A = 4

Quizplus · Accepted Answer

The general form of the equation for a cosine function is given by:

y = A cos(Bx + C)

where A is the amplitude, B determines the period, and C is the phase shift.

In this case, we have:

y = 4 cos (4/3 x)

Comparing with the general form, we can see that:

- A = 4
- B = 4/3

The period P of a cosine function is given by:

P = (2π)/B

Substituting B = 4/3, we get:

P = (2π)/(4/3) = (3/4) x 2π = (3/2)π radians

Converting to degrees, we get:

P = (3/2)π x (180/π) = 270°

So the correct answer is C, P = 270° and A = 4.

To graph the function, we can first find the x-values for one period. From the equation y = 4 cos (4/3 x), we know that one period is:

(2π)/(4/3) = (3/4) x 2π = (3/2)π radians

Dividing this by 4, we get the x-values for one quarter of a period:

(3/8)π, (5/8)π, (7/8)π, (9/8)π

Using these x-values, we can calculate the corresponding y-values:

y = 4 cos (4/3 x)

y = 4 cos (4/3 x) = 4 cos [4/3 x (180/π)] = 4 cos (240/π x)

y = 4 cos (240/π x)

Plugging in the x-values, we get:

y = 4 cos (240/π x) ≈ 2.928, 0, -2.928, 0

So the graph of two periods of the function looks like this:

graph{4cos(240x/pi) [-3.707, 3.707, -3.707, 3.707]}

Find the Period P and Amplitude a and Graph at Least