Solve and show work for: y > or equal to |x - 1| - 2 be????
Wow smart thinking
;)
I need to know also :)
;) thanks man! :D
@Chlorophyll
A?
@Firejay5, did you compute A yet?
Could you help answer this?
For me also. not just Fire :S
y ≥ |x-1| -2 Add 2 to both sides to get y + 2 ≥ |x - 1| Take the inverse absolute value of both sides to get ± (y + 2) ≥ x - 1 Add 1 to both sides to get ± (y + 2) + 1 ≥ x
@jiji501
How would you graph it though?
If you have to graph it, just leave it in the form given to you and use a graphing calculator to graph it
Ughh, ok.
1. Graph y = x 2. Shift the graph to the right 1 unit 3. Take only the positive segment, then take the negative segment make symmetry over x axis 4. Shift the whole graph down 2 units
@Firejay5 does it answer your post, too?
yea I guess
Actually, if he followed your steps precisely @chlorophyll, he would not have the right graph. There are a couple of mistakes with them.
For example, if he graphed y = x, he would have |dw:1347503678508:dw| to begin with. But let's assume he already knows to start with |dw:1347503738442:dw| Well, then you tell him to move it over one, then flip it over the x axis. If he did that, he would end up with |dw:1347503826640:dw|
Your graph is wrong, because when you shift the down, the graph is still in V shape :(
I know it is wrong. I was following YOUR steps. Don't you understand?
NO, I don't think my graph wrong! Could you show me why???
I already attempted to show you.
Then show the CORRECT one :P
I don't want to post the answer on here
I know what it should look like however.
I was just saying that he would not get the right answer using your provided steps.
It was a good attempt though
It's up to you! I just try to help the poster. I'm still thinking about the how to explain the region, as the matter of fact I'm not articulate type as you are :)
I think it would be easier if you started by attempting to explain what y = |x| means and what it looks like, and then going from there.
I don't think he has problem graphing y = x!
When he's online, he'll surely ask more though!
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