# Thread: Graphing absolute value with trigonometric functions

1. ## 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 answer

2. ## Re: Graphing absolute value with trigonometric functions

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

hope that helps a little

3. ## 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

$2+\sin x$

4. ## Re: Graphing absolute value with trigonometric functions

Originally Posted by thebestrose828
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 answer
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.