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Math Help - Graphing trig function

  1. #1
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    Graphing trig function

    Hey guys!
    So for an assignment, we're given some trig functions and asked to graph it, from it's base graph to it's final state: but i'm having problems with these two:

    b) y = 3cos2x + 1 for 0 < x < 360
    d) y = -sin(x/3) - 2 for -pi/2 < x < pi/2


    Solution for b):
    So I know the base graph is y=cos(x), so the first graph I draw is y=cos(x). The second graph would be y=cos(x) but all the x-values are multiplied by 3. I don't know what to do with the 2x part. The last graph would be the whole thing, shifted up one unit.

    Solution for d):

    The base graph is y=sin(x), so the first graph to draw is sin(x). The second graph to draw is -sin(x), which is basically sin(x) flipped on the x-axis. I don't know what to do for the (x/3) part. The last graph would be the whole thing shifted downwards.

    the parts i'm not sure what to do is the ones italicized above. Any help would be appreciated !
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  2. #2
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    Quote Originally Posted by snypeshow View Post
    Hey guys!
    So for an assignment, we're given some trig functions and asked to graph it, from it's base graph to it's final state: but i'm having problems with these two:

    b) y = 3cos2x + 1 for 0 < x < 360
    d) y = -sin(x/3) - 2 for -pi/2 < x < pi/2

    Solution for b):
    So I know the base graph is y=cos(x), so the first graph I draw is y=cos(x). The second graph would be y=cos(x) but all the x-values are multiplied by 3. I don't know what to do with the 2x part. The last graph would be the whole thing, shifted up one unit.

    Solution for d):
    The base graph is y=sin(x), so the first graph to draw is sin(x). The second graph to draw is -sin(x), which is basically sin(x) flipped on the x-axis. I don't know what to do for the (x/3) part. The last graph would be the whole thing shifted downwards.

    the parts i'm not sure what to do is the ones italicized above. Any help would be appreciated !
    HI

    the 2x here implies that the sin graph will go 2 rounds in one cycle .

    1/3 x implies that the sin graph will go 1/3 round (stop at x=120 degree) in one cycle .

    take a look at the graphs i attached .
    Attached Thumbnails Attached Thumbnails Graphing trig function-sin-graph-100.jpg   Graphing trig function-sin-graph-200.jpg  
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  3. #3
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    thanks alot for those graphs!!!

    so i'm guessing all the y-values will be multiplied by the coefficient of x?

    for b) the y-values would be multiplied by 2
    for d) the y-values would be multiplied by 1/3

    right?
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  4. #4
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    Quote Originally Posted by snypeshow View Post
    Hey guys!
    So for an assignment, we're given some trig functions and asked to graph it, from it's base graph to it's final state: but i'm having problems with these two:

    b) y = 3cos2x + 1 for 0 < x < 360
    d) y = -sin(x/3) - 2 for -pi/2 < x < pi/2

    Solution for b):
    So I know the base graph is y=cos(x), so the first graph I draw is y=cos(x). The second graph would be y=cos(x) but all the x-values are multiplied by 3. I don't know what to do with the 2x part. The last graph would be the whole thing, shifted up one unit.

    Solution for d):
    The base graph is y=sin(x), so the first graph to draw is sin(x). The second graph to draw is -sin(x), which is basically sin(x) flipped on the x-axis. I don't know what to do for the (x/3) part. The last graph would be the whole thing shifted downwards.

    the parts i'm not sure what to do is the ones italicized above. Any help would be appreciated !
    hey , for (d) i made a mistake , i din notice the range given . It should look sth like this .
    Attached Thumbnails Attached Thumbnails Graphing trig function-sin-sin-graph.jpg  
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