function

Thread Starter

fpoint

Joined Mar 28, 2015
11
On the x-coordinate, there is straight line AB, point A is fixed on the x-coordinate, point B is located at any point x-coordinates, to describe this function ?
 

Papabravo

Joined Feb 24, 2006
21,094
The point slope form is:

(y - y0) = m(x - x0)

(x0, y0) are the coordinates of one of the points
if (x1, y1) are the coordinates of another point then
m = (y1 - y0) / (x1 - x0)
 

WBahn

Joined Mar 31, 2012
29,932
On the x-coordinate, there is straight line AB, point A is fixed on the x-coordinate, point B is located at any point x-coordinates, to describe this function ?
You only talk about the x-coordinate. Are you working with a function in an x-y plane, or just on a number line?

A picture would really help.

As I try to interpret this, it only makes sense if it is reworded as:

On the x-y plane a function is described by a straight line AB. The point A is fixed at coordinate <x0,y0>, The point B is located at an arbitrary point, <x1,y1>, What is the function, y=f(x), that describes this function.

Does that sound close to what you are being asked for?

If so, then is the function supposed to match only the portion of the line between A and B, or the entire line that passes through A and B?
 

Thread Starter

fpoint

Joined Mar 28, 2015
11
You only talk about the x-coordinate. Are you working with a function in an x-y plane, or just on a number line?

A picture would really help.

As I try to interpret this, it only makes sense if it is reworded as:

On the x-y plane a function is described by a straight line AB. The point A is fixed at coordinate <x0,y0>, The point B is located at an arbitrary point, <x1,y1>, What is the function, y=f(x), that describes this function.

Does that sound close to what you are being asked for?

If so, then is the function supposed to match only the portion of the line between A and B, or the entire line that passes through A and B?
x-coordinate represents
\(\mathbb{R}^1\)
there is no need for labeling (x1,y1) plane represents
\(\mathbb{R}^2\)
function can be solved in the x-coordinate ( indicated in red letters )
\(A=a_{x.}\) , \(B=x_{x.}\) , AB=function ?
label (x.) - takes place in the x-coordinate
 

WBahn

Joined Mar 31, 2012
29,932
There's no red letters in your post, so I'm not sure what you were trying to indicate with them.

I think I am following what you are doing. I'm not too comfortable with the math set-theoretic notation, so I may not be following you completely.

If I am, then you are basically saying that you looking for a function f(x) such that f(x_a) = A, f(x_x) = B, and the f(x) is linear. Is that correct?

Also, you need to provide YOUR best attempt to solve YOUR homework. Not only is that expected, but it goes a long way toward making sure that we are on the same page.
 

amilton542

Joined Nov 13, 2010
497
There are two cases:

1) Any two given points that lie on the line.

2) The slope and a point on the line.

Post #3 is your solution.
 

Thread Starter

fpoint

Joined Mar 28, 2015
11
\(a) y_{x.}=|a_{x.}-x_{x.}|\)
\(b) y_{x.}=-|a_{x.}-x_{x.}|\)
\(c) y_{x.}=a_{x.}-x_{x.}\)
\(d) y_{x.}=x_{x.}-a_{x.}|\)
\(e) y_{x.}=\{|a_{x.}-x_{x.}|\}\cup\{-|a_{x.}-x_{x.}|\}\)
\(x.\) - place (label for x-coordinate)

Mathematics official says geometric objects measures must be positive numbers, say my function to geometric objects measures may be negative numbers


question - how to make it look proceedings graphics of my functions in the plane (Cartesian coordinate system)?
 

studiot

Joined Nov 9, 2007
4,998
Mathematics official says geometric objects measures must be positive numbers,
Well your mathematics teacher is wrong.
Zero is a valid value.
Zero is neither negative nor positive.

He or she may want non-negative values, but that is not the same as positive definite.

So your modulus function will do this

\({y_x} =\left| {{a_x} - \left. {{X_x}} \right|} \right.\)

alternatively you might like to know that we often use this one instead since the modulus function is not differentiable at x=0.

\({y_x} = \sqrt {{{\left( {{a_x} - {X_x}} \right)}^2}} \)

but this one is.[/quote]
 
Last edited:

Thread Starter

fpoint

Joined Mar 28, 2015
11
The mapping function from the x-coordinates of the plane (Cartesian coordinate system)
y = x-a, x and a remain on the x-coordinate, y goes to the y-coordinate.
view photo
https://pkxnqg.bn1302.livefilestore...WSh5idkVdC-swrTkqYaXV8fmts9x7Ks/ii.png?psid=1
the lines of x and a parallel to the y-coordinates
line of y parallel to the x-coordinate
formed at the intersection of real points A and B
points A and B are combined and gets straight line AB
is given by x = 4, a = 2, y = 2
Repeat for x = 3.5, a = 2, y = 1.5, view photo
formed at the intersection of real points C and D
points C and D are combined and received straight line CD
https://befwwg.bn1302.livefilestore...DKraCcJKIy-UHkR4VeCHL_PmPvJTSMeM/i.png?psid=1
connect the dots AC (BD) straight lines AB and CD
ABDC points form the surface of 4≥x≥3.5
Draw a graph of the function at the current proceedings for
a) y=|a-x|
b) y=-|a-x|
c) y=a-x
d) y=x-a
e) y={|a-x|}\(\cup\){-|a-x|}
 

Thread Starter

fpoint

Joined Mar 28, 2015
11
y=|2-x|
graph, the red surface
https://cfxpzq.bn1302.livefilestore...duf35TDz7kJFlvpinPGfiGmOhMAVbDw/01.png?psid=1

b) y=-|2-x|
graph, the red surface
https://nq6hfq.bn1302.livefilestore...xhUjRvcF6i6WXHS-qNoA1O50MwSx-Kw/02.png?psid=1


c) y=2-x
graph, the red surface
https://0nivia.bn1302.livefilestore...qqqkUyEXwbMEkel843ZQ20Rq__x6V-A/03.png?psid=1

d) y=x-2
graph, the red surface
https://d6pekg.bn1302.livefilestore...0AJlHabqjfVBAgiGORjpymT7vzmMKCA/04.png?psid=1

e) y={|2-x|}\(\cup\){-|2-x|}
graph, the red surface
https://qhdsnq.bn1302.livefilestore...HeXxWALY-BwcLW4G5ObhnQSYfKmEaVQ/05.png?psid=1

which are geometric objects obtained for valuesx and y , shape a≥x≥b ( a≥y≥b ) ? , you have a graph
 

Thread Starter

fpoint

Joined Mar 28, 2015
11
a)\(y_{x.}=|a_{x.}-x_{x.}|\)
b)\(y_{x.}=-|a_{x.}-x_{x.}|\), the same graph is reversed only to \(180^o\) , and relates to a negative value y
the scene (\(x.\))x-coordinates , (\(y.\))y-coordinates ,( \(xy.\))plane
Graph functions \(y_{x.}\rightarrow y_{y.}\) , mapped straight line \((y_{y.},a_{x.},x_{x.})\rightarrow(a_{xy.}x_{xy.})\)
2≥y≥0 ( The general form b≥y≥0 , b>0 ) rectangular isosceles triangle
https://2bl1tq.bn1302.livefilestore...kPva7zBN1RUVuKqNL12VYCgSDrGVr0A/y1.png?psid=1

3≥y≥1 ( The general form c≥y≥b , b>0 , c>0 ) regular trapeze
https://dc4d8a.bn1302.livefilestore...WL9PS8DTbYrpR0fD2Vhx9lCacxIjf8g/y2.png?psid=1

1≥x≥-1 ( The general form c≥x≥b x<a , c≥x≥b x>a ) rectangular trapeze
https://o9amca.bn1302.livefilestore...FCWt4FDy7sqTIZNX57SWFbuE4v6HR6w/y3.png?psid=1

6≥x≥-1 ( The general form c≥x≥b , b>a , c<a , |b|\(\neq\)|c|) pentagon
https://pkxoqg.bn1302.livefilestore...ThLPbQyTNzHDTKaz5o26xnOhTE2PD9Q/y4.png?psid=1
more geometric objects that can be obtained ???
 

Thread Starter

fpoint

Joined Mar 28, 2015
11
Operations on sets - difference, this operation returns a new geometric objects
{ 5≥ x ≥0 }\(\setminus{\) {1≥y≥0} , hexagon
https://nq6ifq.bn1302.livefilestore...qgLXrV5A5-e_o-SITTtYzhwpKXd4QiQ/a1.png?psid=1
{ 3≥y≥0}\(\setminus\){1≥x≥0} heptagon
https://0niwia.bn1302.livefilestore...XOyUtFLLOTWA8tFqTLypALME92OR_Jw/a2.png?psid=1
{5≥x≥-1}\(\setminus\){2≥y≥1} trapezoid and triangle together
https://d6pfkg.bn1302.livefilestore...pSGGWByH1jMIRAA6Q1VUZ2sP6cr8U0Q/a3.png?psid=1
 

Thread Starter

fpoint

Joined Mar 28, 2015
11
The symmetry of geometric object

trapez - \(y_{x.}=\{|a_{x.}-x_{x.}|\}\cup\{-|a_{x.}-x_{x.}|\} , \{2\geq y_{y.}-2\}\setminus\{1\geq y_{y.}-1\}\)

https://dc4e8a.bn1302.livefilestore...INtxRZAQz1QXkFp5aRMZDT3JqsHa5w/aa1.png?psid=1


to make it look a graph \(a_{x.}\rightarrow a_{y.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{y.})\rightarrow (a_{xy.}x_{xy.})\)

\(a_{x.}\rightarrow a_{z.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{z.})\rightarrow ( x , y , z)\)
 

WBahn

Joined Mar 31, 2012
29,932
I remember socratus used to post threads in this fashion, until no one bothered to read or answer them any more.
Ah, yes. That was the socratus that associated themselves, rather appropriately, with a particularly dense metal, right?

You lasted longer than I did.
 
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