How would I rewrite these functions in "standard form"?

Dec 2018
29
2
USA
The functions are: |x+4| and 2|x+4|.
The "standard form" I am talking about involves constants A,B,C, and D: A|B(x-C)|+D.

How can I mold these two functions into a different majestic outcome?
 

topsquark

Forum Staff
Jan 2006
11,569
3,453
Wellsville, NY
The functions are: |x+4| and 2|x+4|.
The "standard form" I am talking about involves constants A,B,C, and D: A|B(x-C)|+D.

How can I mold these two functions into a different majestic outcome?
Are you trying to think of something like, for |x + 4|: \(\displaystyle (1) |(1)(x - (-4))| + (0)\)?

-Dan
 
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Plato

MHF Helper
Aug 2006
22,475
8,643
The functions are: |x+4| and 2|x+4|.
The "standard form" I am talking about involves constants A,B,C, and D: A|B(x-C)|+D.
How can I mold these two functions into a different majestic outcome?
Dear bossbasslol, you have shown with this post that you are not playing with all the marbles available.
Neither $|x+4|$ nor $2|x+4|$ is a function. The first is a simple expression: the distance $x$ is from $-4$.
The second is twice that. There is nothing more to either. Please learn the vocabulary so as not to embarrass yourself.
 
Dec 2018
29
2
USA
Ah, I see that my superfluous way of thinking directed my question to a bundle of confusion. To simply fix the |x+4| and 2|x+4| not being functions, add f(x)= to the left of them. Lack of clarity tends to throw off even the brightest of mathematicians. Anyways, I was confused because the "standard form" resembled the vertex formula for a quadratic. I thought that there was some sort of unknown form that applied to functions like these two. Now that I think of it, this seems pretty simple, like what Admin Topsquark suggested. If this is true, I think I can apply this method to functions like f(x)= -2|x+4|+1. Expect more brain-scrambling questions to come!