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Thread: Graphing absolute value with trigonometric functions

  1. #1
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    Graphing absolute value with trigonometric functions

    Hey, I just wanted to verify that my answer was correct with someone. If not, can someone show me how I am supposed to actually do it? Any clarification would be awesome, thanks.

    The question states to sketch a graph of the absolute value transformation from g(x) = sin(x) to g(x) = |-2+sin x|
    Attached in the image is my answerGraphing absolute value with trigonometric functions-img_0260.jpg
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  2. #2
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    Re: Graphing absolute value with trigonometric functions

    your graph is almost correct
    Have graphed y = -2 + sin x in red and then |-2 + sin x| in blue to help display the transformation... the absolute value essentially just 'flips' it in a reflection about the x-axis, all the previous negative values for g(x) will become positive

    Graphing absolute value with trigonometric functions-sin.jpg

    hope that helps a little
    Last edited by jacs; Jan 11th 2017 at 02:44 PM.
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  3. #3
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    Re: Graphing absolute value with trigonometric functions

    You are nearly right. Note that

    -1 \le \sin x \le 1 \implies -3 \le -2+\sin x \le -1

    which is always negative so

    |-2+\sin x|=-(-2+\sin x)=2-\sin x

    You have sketched

    2+\sin x
    Last edited by Archie; Jan 11th 2017 at 02:48 PM.
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    Re: Graphing absolute value with trigonometric functions

    Quote Originally Posted by thebestrose828 View Post
    Hey, I just wanted to verify that my answer was correct with someone. If not, can someone show me how I am supposed to actually do it? Any clarification would be awesome, thanks.

    The question states to sketch a graph of the absolute value transformation from g(x) = sin(x) to g(x) = |-2+sin x|
    Attached in the image is my answerClick image for larger version. 

Name:	IMG_0260.jpg 
Views:	8 
Size:	1,002.7 KB 
ID:	36861
    Not quite. You have graphed y=2+sin x

    Firstly graph y = -2 + sin x , ie translate the graph of y=sin x down 2 units.
    This will be totally below the x-axis.

    Now apply the absolute value, by reflecting all points below the x-axis in the x-axis.
    So, the graph will pass through (0,2) (as yours does), but will fall first rather than rise(as you move to the right from (0,2)).

    You can check this by letting x=pi/2. (pi/2, 1) is a point not (pi/2, 3) as in your graph.
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