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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

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2 answers

  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.

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  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
    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

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