Graph the function y=2sin(x-2pi/3). To draw the graph, plot all x-intercepts, minima and maxima within on period.

I need to graph at least 5 points. I have -pi, 0, pi, 2pi, 3pi, and 4pi along the x-axis

so, plot those points and draw the graph. Should look something like this:

http://www.wolframalpha.com/input/?i=plot+y%3D2sin%28x-2pi%2F3%29+and+y%3D2sin%28x%29

Since there's that pesky 2pi/3 in there, I'd pick points where you expect a min/max/zero. Since sin(x) has those at
x = 0,π/2,π,3π/2,2π, Id offset those by 2π/3, and plot y at
x = 2π/3,7π/6,5π/3,13π/6,8π/3
plus and minus.

Or, you could just plot y=2sin(x) and then draw another y-axis at x = 2π/3 and note the shifted coordinates.

http://www.wolframalpha.com/input/?i=y%3D2sin%28x-2pi%2F3%29

for x-intercepts, y = 0
2sin(x-2pi/3) = 0
sin(x-2pi/3)=0
but sin 0 = 0 and sin π =0 and sin 2π = 0
so x - 2π/3 = 0 and x -2π/3 = π and x - 2π/3 = 2π
x = 2π/3 or x = π + 2π/3 = 5π/3 or x = 2π+2π/3 = 8π/3

x - intercepts: 2π/3 and 5π/3 and 8π/3 , for 0 ≤ x ≤ 2π

for max/min, I will assume you know Calculus
dy/dx = 2cos(x-2π/3)
=0 for a max/min
we know cos π/2=0 and cos 3π/2=0
so x-2π/3 = π/2 and x-2π/3 = 3π/2
x = π/2+2π/3 = 7π/6 or x = 3π/2+2π/3 = 13π/6
so (7π/6 , 2) is a maximum and
(13π/6 , -2 ) is a minimum , but beyond 2π
another minimum : at 13π/6 - 2π = π/6
at (π/6, -2) we have a minimum

for your graph
x ... y
-π 1.72 or √3
0 -1.72 or -√3
π √3
2π -√3
etc

To graph the function y = 2sin(x - 2π/3), we'll plot several points to get a sense of the graph.

Let's start with the x-values you provided: -π, 0, π, 2π, 3π, and 4π.

1. When x = -π:
y = 2sin(-π - 2π/3)
y = 2sin(-π/3)
From the unit circle, sin(-π/3) = -√3/2
Therefore, y = 2(-√3/2) = -√3.

2. When x = 0:
y = 2sin(0 - 2π/3)
y = 2sin(-2π/3)
From the unit circle, sin(-2π/3) = -√3/2
Therefore, y = 2(-√3/2) = -√3.

3. When x = π:
y = 2sin(π - 2π/3)
y = 2sin(π/3)
From the unit circle, sin(π/3) = √3/2
Therefore, y = 2(√3/2) = √3.

4. When x = 2π:
y = 2sin(2π - 2π/3)
y = 2sin(4π/3)
From the unit circle, sin(4π/3) = -√3/2
Therefore, y = 2(-√3/2) = -√3.

5. When x = 3π:
y = 2sin(3π - 2π/3)
y = 2sin(7π/3)
From the unit circle, sin(7π/3) = √3/2
Therefore, y = 2(√3/2) = √3.

6. When x = 4π:
y = 2sin(4π - 2π/3)
y = 2sin(10π/3)
From the unit circle, sin(10π/3) = -√3/2
Therefore, y = 2(-√3/2) = -√3.

Now, let's plot these points on the graph.

(-π, -√3) (0, -√3) (π, √3) (2π, -√3) (3π, √3) (4π, -√3)

The graph will have a pattern that repeats every 2π, so you can extend the graph accordingly. Make sure to connect the points with smooth curves.

To graph the function y = 2sin(x - 2π/3), let's first find the x-intercepts, minima, and maxima within one period.

The period of the sine function is 2π (so one full cycle completes in 2π radians). Therefore, we can find the x-intercepts by solving for x when sin(x - 2π/3) = 0. Since we are plotting points up to 4π, it means we are considering one full period (2π) and an additional half period (π).

1. x-intercept: Set sin(x - 2π/3) = 0 and solve for x.
sin(x - 2π/3) = 0
x - 2π/3 = 0 (since sinθ = 0 when θ = 0)
x = 2π/3

So, an x-intercept lies at x = 2π/3.

Next, let's find the minima and maxima points. Since sin(x) is always between -1 and 1, multiplying it by 2 gives us a range between -2 and 2. So, the minima occur when sin(x - 2π/3) = -2, and the maxima occur when sin(x - 2π/3) = 2.

2. Minima: Set sin(x - 2π/3) = -2 and solve for x.
sin(x - 2π/3) = -2 (no solution exists, since sin() values lie between -1 and 1)

3. Maxima: Set sin(x - 2π/3) = 2 and solve for x.
sin(x - 2π/3) = 2 (no solution exists, since sin() values lie between -1 and 1)

Since there are no minima or maxima within this particular period, we can plot some additional points.

4. Points from your list:
For x = -π:
y = 2sin(-π - 2π/3) = 2sin(-π/3) = 2(√3/2) = √3
So, we have the point (-π, √3).

For x = 0:
y = 2sin(0 - 2π/3) = 2sin(-2π/3)
To find sin(-2π/3), you can recognize that this angle lies in Quadrant III, where sine is negative. So, sin(-2π/3) = -sin(2π/3) = -√3/2
Therefore, the point is (0, -√3).

For x = π:
y = 2sin(π - 2π/3) = 2sin(π/3) = 2(√3/2) = √3
So, the point is (π, √3).

For x = 2π:
y = 2sin(2π - 2π/3) = 2sin(4π/3)
To find sin(4π/3), you can recognize that this angle lies in Quadrant III, where sine is negative. So, sin(4π/3) = -√3/2
Therefore, the point is (2π, -√3).

For x = 3π:
y = 2sin(3π - 2π/3) = 2sin(5π/3) = -√3

For x = 4π:
y = 2sin(4π - 2π/3) = 2sin(6π/3) = 2sin(2π) = 0

So, the points we have are:
(-π, √3), (0, -√3), (π, √3), (2π, -√3), (3π, -√3), and (4π, 0).

Now, you can plot these points on the graph and connect them to observe the shape of the graph. Keep in mind that the graph will repeat itself for every full period.