# 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.

So, if your question is how to express the graph mathematically, this is the answer. Otherwise, please tell us your question, maybe in another thread, and maybe in German. I hope someone else can help you here.

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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