Question

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

Answer 1

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.

Answer 2

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
<b>so (7π/6 , 2) is a maximum and</b>
(13π/6 , -2 ) is a minimum , but beyond 2π
another minimum : at 13π/6 - 2π = π/6
<b>at (π/6, -2) we have a minimum</b>

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