1. ## Re: function point

How to look graphics functions

a) y=|a-x|
b) y=-|a-x|
c) y=a-x
d) y=x-a
e) y={|a-x|}$\displaystyle \cup${-|a-x|}

2. ## Re: function point

Originally Posted by point1967
How to look graphics functions

a) y=|a-x|
b) y=-|a-x|
c) y=a-x
d) y=x-a
e) y={|a-x|}$\displaystyle \cup${-|a-x|}
I am not sure what you mean by this but if you mean "What do the graphs look like?" then
a) y= |a- x|= |x- a| is a straight line with slope -1 for x< a, ending at (a, 0) then a straight line with slope 1 for x> a, starting at x= a. A "V" shape.
b) y= -|a- x|= -|x- a| is a straight line with slope 1 for x< a, ending at (a, 0) then a straight line with slope -1 for x> a, starting at x= a. An upside down "V" shape.
c) y= a- x. A straight line with slope -1 passing through (0, a) and (a, 0).
d) y= x- a. A straight line with slope 1 passing through (0, -a) and (-a, 0).

e) is meaningless- the union is defined for sets, not expressions. If you mean the union of the graphs (which are sets of points) of (a) and (b) then it is two straight lines, with slopes 1 and -1, both passing thorough (a, 0). An "X" shape. It is also the union of the graphs in (c) and (d).

There is nothing new here. This is all pretty standard secondary school math.

3. ## Re: function point

a) y=|2-x|
graph, the red surface
https://cfxpzq.bn1302.livefilestore..../01.png?psid=1

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

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

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

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

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

4. ## Re: function point

Each of these is wrong.

Originally Posted by point1967
The red area (extending infinitely) is the graph of the inequality y> |2- x| The graph of y= |2- x| is the boundary of the red area.

b) y=-|2-x|
graph, the red surface
https://nq6hfq.bn1302.livefilestore..../02.png?psid=1
The red area is the graph of the inequality y< -|2- x|. The graph of y=-|2- x| is the boundary of the red area.

The graph of y= 2- x is the slanting boundary. The red area shown above the x-axis is the graph of y> 2- x for x< 2. Below the x-axis, it is y< 2- x for x> 2.

The graph of y= x- 2 is the slanting boundary. The red area, for y> 0 is the set of y> x- 2 for x>2. For y< 0, it is the set of y< x- 2 for x< 2.

e) y={|2-x|}$\displaystyle \cup${-|2-x|} or y={2-x}$\displaystyle \cup${x-2}
graph, the red surface
https://qhdsnq.bn1302.livefilestore..../05.png?psid=1
The graph of that y is the union of the two lines forming boundary. your red are is y> (2-x)(x- 2) (again, NOT "union") for y> 0 and y< (2- x)(x-2) for y< 0.

which are geometric objects obtained for valuesx and y , shape a≥x≥b ( a≥y≥b ) ? , you have a graph
I wonder why you keep posting these when you are paying no attention to the responses.

5. ## Re: function point

Originally Posted by HallsofIvy
Each of these is wrong.

The red area (extending infinitely) is the graph of the inequality y> |2- x| The graph of y= |2- x| is the boundary of the red area.
it seems that you are not reading the rules I set , mapped straight line (a and x) not mapped (x) , for your test to see whether or not an innovative mathematician ,
that geometric object in the plane meets these requirements
a ) has six angles
b )the sum of the angles $\displaystyle 1080^o$

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

3≥y≥1 ( The general form c≥y≥b , b>0 , c>0 ) regular trapeze
https://dc4d8a.bn1302.livefilestore..../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..../y3.png?psid=1

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

6. ## Re: function point

Operations on sets - difference, this operation returns a new geometric objects
{ 5≥ x ≥0 }$\displaystyle \setminus{$ {1≥y≥0} , hexagon
https://nq6ifq.bn1302.livefilestore..../a1.png?psid=1
{ 3≥y≥0}$\displaystyle \setminus${1≥x≥0} heptagon
https://0niwia.bn1302.livefilestore..../a2.png?psid=1
{5≥x≥-1}$\displaystyle \setminus${2≥y≥1} trapezoid and triangle together
https://d6pfkg.bn1302.livefilestore..../a3.png?psid=1

7. ## Re: function point

The symmetry of geometric object

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

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

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

8. ## Re: function point

$\displaystyle a_{x.}\rightarrow a_{y.},y_{x.}\rightarrow y_{y.}, (y_{y.} x_{x.}a_{y.})\rightarrow (a_{xy.}x_{xy.})$

$\displaystyle y_{x.}=|2_{x.}-x_{x.}|$

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