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Pre Calculus 12

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Graphing trigonometric functions

A sinusoidal function has a maximum at (2,10) and its next minimum (5,-2) find an equation that represents this situation

A sinuosoidal function has a zero at (5,0) and its next minimum is (7,-4). find an equation that represents this situation.


I really don't get how to come up with an equation, this chapter was really the hardest for me. I don't get how to find amplitude, vertical displacement, phase shift

Can someone please explain step by step

  • Pre Calculus 12 -

    Notice the max is 10 and the min is -2, for a range of 12
    which makes the amplitude = 6
    the axis would be 4 units high

    so far we have something like

    y = 6 sin (.......) + 4

    the distance from the max to the min along the horizontal is 3 (from 2 to 5 is 3 units)
    but the period would be twice that
    so the period is 6

    remember 2π/k = period
    2π/k = 6
    6k = 2π
    k = 2π/6 = π/3

    Ok, now also have the period
    and our equation must look something like this:

    y = 6 sin (π/3)(x + d) + 4

    almost done....
    we just have to sub in one of our points to find d , the phase shift
    let's use (2,10)

    10 = 6sin π/3(2 - d) + 4
    1 = sin π/3(2-d)
    I know that sin π/2 = 1
    so π/3(2-d) = π/2
    divide both sides by π
    (1/3)(2-d) = 1/2
    times 6
    2(2-d) = 3
    2-d = 3/2
    -d = 3/2 - 2 = -1/2
    d = 1/2

    y = 6 sin (π/3)(x + 1/2) + 4

    check for the other point:
    if x = 5, we should get y = -2

    y = 6 sin (π/3)(5+1/2) + 4
    = 6 sin ((π/3)(11/2)) + 4
    = 6 sin (11π/2) + 4
    = 6(-1) + 4 = -2 , YEAHHHHH

    I chose to use a sine curve for no specific reason
    We could have used a cosine as well.
    The period and amplitude and vertical shift would be the same
    the change would be in the value of d

    Try the second one, it is slightly different.
    Make a sketch to see what you are doing.

  • Pre Calculus 12 -

    so you added 10 and 2 to get 12, and amplitude is the middle of max and min?

    how did you get the axis 4 units?

    a period is always twice of something?

    the phase shift is sort of confusing me, because isnt that just between the x values? 2, and 5? is there some other way to figure it out?

  • Pre Calculus 12 -

    "how did you get the axis 4 units? "
    where is the middle value between -2 and 10
    sketch on a number line to see.

    the amplitude, or the a value, is 1/2 the difference between the max and the min
    difference = 10 - (-2) = 12
    half of that is 6

    I illustrated with detailed calculations, how I got the phase shift d

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