Hybrid Functions Help

[Asx4Life]

How would you get the hybrid function for y=lx-3l+3lx+2l ?
I know I have to consider different scenarios but I don't know which 

[Greatness]

Highly unlikely i reckon, but possible. And yes split them up into the hybrid functions then consider each of the possible domains.

[Asx4Life]

I think I got it now, thanks guys

[Phy124]

Hi guys, this is my own solution to the question which I had began writing for Asx4Life before I realized you had helped him.

I was curious whether this is a sufficient method to undertake the question and also whether it was correct? 

http://i40.tinypic.com/2enyik8.jpg

edit: The points x = -2 and x = 3 should be inclusive.

[Asx4Life]

Hi guys, this is my own solution to the question which I had began writing for Asx4Life before I realized you had helped him.

I was curious whether this is a sufficient method to undertake the question and also whether it was correct? 

http://i40.tinypic.com/2enyik8.jpg

Yeah that's right I think. The graph you drew cleared up any confusions I had as to the -2<x<3 component. I didn't know absolute function to take as positive and which one to take as a negative.
Is there any way that we could know this without drawing a graph?

Also how would we go about this question
Determine the coefficient of x^6 in the expansion of (x^2+2x)^4
The answer says 4C2 x (x)^2 x (2x)^2=24x^6 hence coefficient=24. I don't understand this?

[Greatness]

Hi guys, this is my own solution to the question which I had began writing for Asx4Life before I realized you had helped him.

I was curious whether this is a sufficient method to undertake the question and also whether it was correct? 

http://i40.tinypic.com/2enyik8.jpg

Yep that's correct.

You can write it into the hybrid functions but keep the 2 sides separate. f(x)={lx-3l + {3|x+2|      (theyre suppose to be hybrid functions xD)
Then consider the domains for which they're similar then add them together, but the graph is safer i think.
This would probably be exam 2.

[Phy124]

Thanks for the clarification swarley, as you say I believe this would be more likely to appear on exam 2.

Hi guys, this is my own solution to the question which I had began writing for Asx4Life before I realized you had helped him.

I was curious whether this is a sufficient method to undertake the question and also whether it was correct? 

http://i40.tinypic.com/2enyik8.jpg

Yeah that's right I think. The graph you drew cleared up any confusions I had as to the -2<x<3 component. I didn't know absolute function to take as positive and which one to take as a negative.
Is there any way that we could know this without drawing a graph?

Also how would we go about this question
Determine the coefficient of x^6 in the expansion of (x^2+2x)^4
The answer says 4C2 x (x)^2 x (2x)^2=24x^6 hence coefficient=24. I don't understand this?

Are you familiar with the binomial theorem (pascal's triangle)?

I'm pretty sure that's how it is done...

I can elaborate if you would like.

Edit:

Here is a sheet I did earlier in the year regarding the binomial theorem:

http://i40.tinypic.com/19kylu.jpg

Here is my working out for your question (parts of it are not necessary but I did them any way just to show you)

http://i43.tinypic.com/3178u87.jpg

[Asx4Life]

Thanks Phy124, the worksheet was very helpful!!

[Phy124]

No problem, happy to help.

[dc302]

Hi guys, this is my own solution to the question which I had began writing for Asx4Life before I realized you had helped him.

I was curious whether this is a sufficient method to undertake the question and also whether it was correct? 

http://i40.tinypic.com/2enyik8.jpg

Yeah that's right I think. The graph you drew cleared up any confusions I had as to the -2<x<3 component. I didn't know absolute function to take as positive and which one to take as a negative.
Is there any way that we could know this without drawing a graph?

Also how would we go about this question
Determine the coefficient of x^6 in the expansion of (x^2+2x)^4
The answer says 4C2 x (x)^2 x (2x)^2=24x^6 hence coefficient=24. I don't understand this?

Definitely, you can do it without drawing graphs. All you have to do is note at which points do the terms inside the abs values change signs. The first term changes at x=3, and the second term changes at x = -2. From this you can deduce 3 cases:

x < -2
-2 < x < 3
X > 3 

(also make sure to include a less than or equals sign to make the whole thing continuous, unless someone can help me out with typing equations in forums )

Once you have those three cases, it is simply a matter of swapping the signs or not inside the abs. Clearly, if x < -2, then both terms are negative etc.