hey guys, a question on graphs (1 Viewer)

[bleakarcher]

http://community.boredofstudies.org/attachment.php?attachmentid=24811&d=1333966172

mirakon posted earlier some notes and the above is one link is one set of those. for the graphs of abs(y)=f(x) and abs(y)=abs[f(x)] can someone explain the logic behind the procedure?

 

[Aysce]

For abs(y) = f(x) all negative values of y are disregarded because absolute value is all about taking the positive y-values in this case, then you continue with reflecting and etc. Im unsure atm with the second one based on his notes, sorry.

EDIT:

For the second graph, break it up into separate steps.

So firstly, you'd do:

y= l f(x) l - Absolute the whole graph above the x-axis as no value that is positive can be under the x-axis.

then move on to do,

y= l f(l x l) l - This is when you take your current graph, reflect the graph which lies within the positive range of x-values and reflect it under the x-axis.

then finally,

lyl = lf(lxl)l which is taking the positive y-values and reflecting this as well.

I am pretty sure this is how it works but yes, the logic of it should be correct.

 

[bleakarcher]

okay i understand now that abs(y)=f(x) so f(x)>=0 and so you disregard what is below the x-axis but why do you flip what remains about the x-axis?

 

[deswa1]

Whatever is negative becomes positive due to the absolute value. For example, if you look at the line y=x, the line y=|x| is exactly the same for x>0, but for x is less than zero, for every negative value it becomes positive. Consider the point (-2,-2). Under y=|x|, it becomes (-2,2) which is the same as reflecting in the x axis.

 

[Carrotsticks]

okay i understand now that abs(y)=f(x) so f(x)>=0 and so you disregard what is below the x-axis but why do you flip what remains about the x-axis?

Much like why if |x| = 2, then x = plus/minus 2 instead of just 2.

 

[bleakarcher]

alright thanks guys, my mind is a bit down today

 

[bleakarcher]

basically since abs(y)=f(x) then f(x)=+/-y?

 

[Carrotsticks]

basically since abs(y)=f(x) then f(x)=+/-y?

Essentially, yes.

 

[bleakarcher]

guys one last question on this. by flipping what remains above the x-axis arent we no longer satisfying f(x)>=0?

 

[Aysce]

No because you are flipping what remains above the x-axis for all POSITIVE VALUES, so you're still satisfying f(x)>=0. The graph over negative values has been erased and all is left is the positive side.

 

[bleakarcher]

^ah right, thanks

 

[Aysce]

Anytime.

PS: First time I've helped someone on 4u forum, YES! Not that anyone cares haha.

 

[Drongoski]

Anytime.

PS: First time I've helped someone on 4u forum, YES! Not that anyone cares haha.

Feels good, eh.

Now remember to thank whoever helps you on this forum. In my view, "thanks in advance" is not good enough.

 

[goobi]

http://community.boredofstudies.org/attachment.php?attachmentid=24811&d=1333966172

mirakon posted earlier some notes and the above is one link is one set of those. for the graphs of abs(y)=f(x) and abs(y)=abs[f(x)] can someone explain the logic behind the procedure?

Hey bleakarcher,
Can you please link me to the thread where mirakon posted those notes?
Thanks in advance!

 

[AAEldar]

Hey bleakarcher,
Can you please link me to the thread where mirakon posted those notes?
Thanks in advance!

http://community.boredofstudies.org/showthread.php?t=280515&p=5797394&viewfull=1#post5797394

They're in there.

 

[goobi]

http://community.boredofstudies.org/showthread.php?t=280515&p=5797394&viewfull=1#post5797394

They're in there.

Much appreciated

 

[Shadowdude]

Graphs is by nature, a visual topic. I think it'd be better to ask a teacher or ask someone in RL so you can see the diagrams they draw... etc. etc.

 

[Aysce]

Feels good, eh.

Now remember to thank whoever helps you on this forum. In my view, "thanks in advance" is not good enough.

I will keep that in mind, thanks.

 

[bleakarcher]

Graphs is by nature, a visual topic. I think it'd be better to ask a teacher or ask someone in RL so you can see the diagrams they draw... etc. etc.

This is true. I reckon, because I had to learn it on my own, it was one the hardest topics. I've covered it for now but it was really all with the help of Carrotsticks who basically got me started with the videos he made for me.