# How do I graph a function?

ms.srki Auf diesen Beitrag antworten »
How do I graph a function?
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 ?
Point A=a , point B=x , straight line AB=?
Hippocampus Auf diesen Beitrag antworten »

Do you have any idea?
You should draw or at least imagine a sketch.
ms.srki Auf diesen Beitrag antworten »

Removed full quote. Steffen

solution, in my opinion

$\begin{eqnarray*} a) y_{x.}=|a_{x.}-x_{x.}| \end{eqnarray*}$
$\begin{eqnarray*} b) y_{x.}=-|a_{x.}-x_{x.}| \end{eqnarray*}$
$\begin{eqnarray*} c) y_{x.}=a_{x.}-x_{x.} \end{eqnarray*}$
$\begin{eqnarray*} d) y_{x.}=x_{x.}-a_{x.} \end{eqnarray*}$
$\begin{eqnarray*} e) y_{x.}=\{|a_{x.}-x_{x.}|\}\cup\{-|a_{x.}-x_{x.}|\} \end{eqnarray*}$

current graph is as follows $\begin{eqnarray*} (y_{y.},x_{x.})\rightarrow(x,y) \end{eqnarray*}$
to look at the graph of a level different from the above ?

notation :
$\begin{eqnarray*} x. \end{eqnarray*}$ - constants or variables are on the x - coordinate
$\begin{eqnarray*} y. \end{eqnarray*}$ - constants or variables are on the y - coordinate
ms.srki Auf diesen Beitrag antworten »

The graph function from the x-coordinates of the plane (Cartesian coordinate system)
$\begin{eqnarray*} y_{x.}=x_{x.}-a_{x.} \end{eqnarray*}$ , $\begin{eqnarray*} x_{x.} \end{eqnarray*}$and $\begin{eqnarray*} a_{x.} \end{eqnarray*}$ remain on the x-coordinate, $\begin{eqnarray*} y_{x.} \end{eqnarray*}$goes to the y-coordinate $\begin{eqnarray*} y_{y.} \end{eqnarray*}$.
$\begin{eqnarray*} y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{x.})\rightarrow (a_{xy.}x_{xy.}) \end{eqnarray*}$
$\begin{eqnarray*} xy. \end{eqnarray*}$ coordinates of the points straight line
view photo
[attach]37674[/attach]
$\begin{eqnarray*} y_{x.}=x_{x.}-2_{x.} \end{eqnarray*}$
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 $\begin{eqnarray*} x_{x.}=4 , a_{x.}=2 , y_{y.}=2 \end{eqnarray*}$
Repeat for $\begin{eqnarray*} x_{x.}=3.5 , a_{x.}=2 , y_{y.}=1.5 \end{eqnarray*}$ , view photo
formed at the intersection of real points C and D
points C and D are combined and received straight line CD
[attach]37675[/attach]

connect the dots AC (BD) straight lines AB and CD
ABDC points form the surface of $\begin{eqnarray*} 4\geq x_{x.}\geq 3.5 \end{eqnarray*}$
Draw a graph of the function at the current proceedings for
$\begin{eqnarray*} a) y_{x.}=|a_{x.}-x_{x.}| \end{eqnarray*}$
$\begin{eqnarray*} b) y_{x.}=-|a_{x.}-x_{x.}| \end{eqnarray*}$
$\begin{eqnarray*} c) y_{x.}=a_{x.}-x_{x.} \end{eqnarray*}$
$\begin{eqnarray*} d) y_{x.}=x_{x.}-a_{x.} \end{eqnarray*}$
$\begin{eqnarray*} e) y_{x.}=\{|a_{x.}-x_{x.}|\}\cup\{-|a_{x.}-x_{x.}|\} \end{eqnarray*}$
Steffen Bühler Auf diesen Beitrag antworten »

So, for y=x-a, you got the four points A(2;4-2), B(4;4-2), C(2;3.5-2), and D(3.5;3.5-2).

Now, its |a-x| instead of x-a. Here, for instance, A becomes (2;|2-4|).

See? Now calculate the three other points and draw.

Next comes -|a-x| which means A becomes (2;-2) and so on.

Regards
Steffen
ms.srki Auf diesen Beitrag antworten »

Removed full quote. Steffen
You understand the process, we need to draw a graph for each $\begin{eqnarray*} x_{x.}=(..., 2.2,2.4,2.6,...) \end{eqnarray*}$ , as a whole

Steffen Bühler Auf diesen Beitrag antworten »

So you want to assign a 2D shape to every x value? This would result in a 3D diagram, I guess.

For each x, you get a respective polygon in the yz plane. You could then render this set of polygons.

Is this what you want?
ms.srki Auf diesen Beitrag antworten »

Removed full quote. Steffen
will receive the 2D surface for each x , will not be obtained a 3D diagram , I want you able to give the graph of the conditions that have given .
Steffen Bühler Auf diesen Beitrag antworten »

I must confess I didn't understand your last post.
ms.srki Auf diesen Beitrag antworten »

Removed full quote. Steffen
I've said it looks like this

$\begin{eqnarray*} a) y_{x.}=|2_{x.}-x_{x.}| \end{eqnarray*}$
graph, the red surface
[attach]37678[/attach]

$\begin{eqnarray*} b) y_{x.}=-|2_{x.}-x_{x.}| \end{eqnarray*}$
graph, the red surface
[attach]37679[/attach]

$\begin{eqnarray*} c) y_{x.}=2_{x.}-x_{x.} \end{eqnarray*}$
graph, the red surface
[attach]37680[/attach]

$\begin{eqnarray*} d) y_{x.}=x_{x.}-2_{x.} \end{eqnarray*}$
graph, the red surface
[attach]37681[/attach]
$\begin{eqnarray*} e) y_{x.}=\{|2_{x.}-x_{x.}|\}\cup\{-|2_{x.}-x_{x.}|\} \end{eqnarray*}$
graph, the red surface
[attach]37682[/attach]

which are geometric objects obtained for valuesx and y , shape $\begin{eqnarray*} a\geq x_{x.}\geq b \end{eqnarray*}$ ($\begin{eqnarray*} a\geq y_{y.}\geq b \end{eqnarray*}$ )? , you have a graph
Steffen Bühler Auf diesen Beitrag antworten »

Do you need a mathematical formula for, e.g. this red surface?

[attach]37678[/attach]

This is the set of all pairs (x;y) where y>|2-x|:

Did I now get it right?

PS: if you answer me, don't quote me, just answer. I'll remove your full quotes for the sake of clarity.
ms.srki Auf diesen Beitrag antworten »

 Zitat: Original von Steffen Bühler

what did you draw the current graph $\begin{eqnarray*} (y_{y.},x_{x.})\rightarrow(x,y) \end{eqnarray*}$ , y=|2-x|
[attach]37678[/attach]
I am drawn by the rules
$\begin{eqnarray*} y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{x.})\rightarrow (a_{xy.}x_{xy.}) \end{eqnarray*}$ , $\begin{eqnarray*} y_{x.}=|2_{x.}-x_{x.}| \end{eqnarray*}$

graph is not as in classical mathematics, obtained points in the plane $\begin{eqnarray*} (x,y) \end{eqnarray*}$, for me it gets straight lines $\begin{eqnarray*} (a_{xy.},x_{xy.}) \end{eqnarray*}$ in the plane

I hope you realize what it is
Steffen Bühler Auf diesen Beitrag antworten »

The graph of the function y=|2-x| is only to show that the inequality y>|2-x| perhaps is what you need in this case. As you said, this is not a function (there's more than one y value for one x value), but can be written as a set:

{ (x,y) | y>|2-x| }

which is exactly the red surface in your drawing.

Regards
Steffen
ms.srki Auf diesen Beitrag antworten »

a)$\begin{eqnarray*} y_{x.}=|2_{x.}-x_{x.}| \end{eqnarray*}$
b)$\begin{eqnarray*} y_{x.}=-|2_{x.}-x_{x.}| \end{eqnarray*}$, the same graph is reversed only to $\begin{eqnarray*} 180^o \end{eqnarray*}$ , and relates to a negative value y

$\begin{eqnarray*} 2\geq y_{y.}\geq0 \end{eqnarray*}$ rectangular isosceles triangle

[attach]37696[/attach]
$\begin{eqnarray*} 3\geq y_{y.}\geq1 \end{eqnarray*}$ regular trapeze

[attach]37697[/attach]

$\begin{eqnarray*} 1\geq x_{x.}\geq-1 \end{eqnarray*}$ rectangular trapeze

[attach]37698[/attach]

$\begin{eqnarray*} 6\geq x_{x.}\geq-1 \end{eqnarray*}$ pentagon

[attach]37699[/attach]
more geometric objects that can be obtained ???
ms.srki Auf diesen Beitrag antworten »

Operations on sets - difference, this operation returns a new geometric objects
$\begin{eqnarray*} \{ 5\geq x \geq 0 \}\setminus \{1\geq y\geq 0\} \end{eqnarray*}$ , hexagon

[attach]37712[/attach]
$\begin{eqnarray*} \{ 3\geq y \geq 0 \}\setminus \{1\geq x\geq 0\} \end{eqnarray*}$ heptagon

[attach]37713[/attach]
$\begin{eqnarray*} \{ 5\geq x \geq- 1 \}\setminus \{2\geq y\geq 1\} \end{eqnarray*}$ trapezoid and triangle together

[attach]37714[/attach]

that even mathematical topic can be displayed through the graphics functions (part) ?
Steffen Bühler Auf diesen Beitrag antworten »

Like I said, I can't help you. Maybe someone else will answer here, but it's surely better to start a new thread (in German, if possible) and ask a clear question. Up to now, I still don't know what you want to do.

Best regards
Steffen
ms.srki Auf diesen Beitrag antworten »

The symmetry of geometric object

trapez - $\begin{eqnarray*} y_{x.}=\{|a_{x.}-x_{x.}|\}\cup\{-|a_{x.}-x_{x.}|\} , \{2\geq y_{y.}-2\}\setminus\{1\geq y_{y.}-1\} \end{eqnarray*}$

[attach]37735[/attach]

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

$\begin{eqnarray*} a_{x.}\rightarrow a_{z.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{z.})\rightarrow ( x , y , z) \end{eqnarray*}$
ms.srki Auf diesen Beitrag antworten »

$\begin{eqnarray*} a_{x.}\rightarrow a_{y.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{y.})\rightarrow (a_{xy.}x_{xy.}) \end{eqnarray*}$
[attach]37761[/attach]
$\begin{eqnarray*} y_{x.}=|2_{x.}-x_{x.}| \end{eqnarray*}$|
[attach]37762[/attach]
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