Sine graph with absolute value

[MC3]

Hi, I know how to draw a regular sine graph e.g. f(x)=3sin(5/6x-3)-5, but I'm not sure how to start with an absolute value within the sine: f(x)=sin|x|
How do I take into account that x>0 and draw that as a sine graph? Where do I even start?
Thank you for any help.
(It's my first time posting here and I'm not a native speaker so apologize for any mistakes.)

 

[Steven G]

Just draw f(x)=3sin(5/6x-3)-5. Now wherever f(x) is negative, just draw it positive. That is reflect all parts of the graph below the x-axis across the x-axis.

 

[LCKurtz]

Hi, I know how to draw a regular sine graph e.g. f(x)=3sin(5/6x-3)-5, but I'm not sure how to start with an absolute value within the sine: f(x)=sin|x|
How do I take into account that x>0 and draw that as a sine graph? Where do I even start?
Thank you for any help.
(It's my first time posting here and I'm not a native speaker so apologize for any mistakes.)

If [imath]x\ge 0[/imath] then [math]\sin(x) = \sin|x|[/math] so the graphs are identical there. Then reflect your graph in the y axis because it is an even function.

 

[Steven G]

Ooops, you want sin|x|, not |sinx|

 

[LCKurtz]

Ooops, you want sin|x|, not |sinx|

Please quote the post to which you are responding. If it is mine, you might want to think about it a bit more.

 

[MC3]

If [imath]x\ge 0[/imath] then [math]\sin(x) = \sin|x|[/math] so the graphs are identical there. Then reflect your graph in the y axis because it is an even function.

Thank you ?

 

[JeffM]

How do I take into account that x>0

I am not sure why you are having trouble, but it may be in the false inference quoted above.[imath]|x|[/imath] does not entail that [imath]x > 0[/imath]. Nor does it entail that [imath]x \ge 0.[/imath]

[math]y = \sin ( |x|) \implies \\ y = \sin (x) \text { if } x \ge 0, \text { and }\\ y = \sin(-x) \text { if } x < 0.[/math]
I have just restated LCKurtz’s answer in different words.

 

[MC3]

I am not sure why you are having trouble, but it may be in the false inference quoted above.[imath]|x|[/imath] does not entail that [imath]x > 0[/imath]. Nor does it entail that [imath]x \ge 0.[/imath]

[math]y = \sin ( |x|) \implies \\ y = \sin (x) \text { if } x \ge 0, \text { and }\\ y = \sin(-x) \text { if } x < 0.[/math]
I have just restated LCKurtz’s answer Than

I am not sure why you are having trouble, but it may be in the false inference quoted above.[imath]|x|[/imath] does not entail that [imath]x > 0[/imath]. Nor does it entail that [imath]x \ge 0.[/imath]

[math]y = \sin ( |x|) \implies \\ y = \sin (x) \text { if } x \ge 0, \text { and }\\ y = \sin(-x) \text { if } x < 0.[/math]
I have just restated LCKurtz’s answer in different words.

Thank you ?

 

[The Highlander]

Hi, I know how to draw a regular sine graph e.g. f(x)=3sin(5/6x-3)-5, but I'm not sure how to start with an absolute value within the sine: f(x)=sin|x|
How do I take into account that x>0 and draw that as a sine graph? Where do I even start?
Thank you for any help.
(It's my first time posting here and I'm not a native speaker so apologize for any mistakes.)

The attached illustrates nicely what LCKurtz (& JeffM) have been telling you.

(A picture's worth a thousand words. )