Question

graph the function y=(3/2)cos (x+2pi/3) ?

Answer 1

well there are two types of method

solving method 1

3cos(2pi/3x-pi/6), factor out the lcf in order to get x alone. 3cos[2pi/3(x-1/4)], where b is 2pi/3. The formula to find the period of a sine or cosine equation is 2pi/B, so 2pi/2pi/3=3. The period is 3. The Amplitude for the cosine is [-3,3] because the coefficiant tells us to multiply the Range of the Cosine function [-1,1] by 3. In order to find the Y-int let x=0 to get 3cos(-pi\6), then you get 2.59807. When y=0, x=1



solving method 2
Begin with a simpler function and subject it to transformations. Leave out the y-intercept until the end.

y = 3cos(2πx/3 - π/6) = y = 3cos[(2π/3)(x - 1/4)]

y = cos(x)
x-intercepts: π/2, 3π/2, 5π/2, 7π/2, 9π/2,...
maxima: 0, 2π, 4π,...
minima: π, 3π, 5π,...

y = cos[(2π/3)x]
The graph is scaled horizontally by ratio 3/(2π).
x-intercepts: 3/4, 9/4, 15/4, 21/4, 27/4,...
maxima: 0, 3, 6,...
minima: 3/2, 9/2, 15/2,...

cos[(2π/3)(x - 1/4)]
The graph is translated to the right by 1/4 unit.
x-intercepts: 1, 5/2, 4, 11/2, 7,...
maxima: 1/4, 13/4, 25/4,...
minima: 7/4, 19/4, 31/4,...

3cos[(2π/3)(x - 1/4)]
This final transformation gives you an amplitude of 3. The x-intercepts and extrema have the same horizontal position. The maxima have y-coordinate 3. The minima have y-coordinate -3.
Now for the y-intercept, substitute x = 0.
y = 3cos[(2π/3)(0 - 1/4)] = y = 3cos(-π/6) = 3√(3)/2

Answer 2

can u show me the graph?