- mathmath333

How to graph this.

- jamiebookeater

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- mathmath333

\(\large \color{black}{\begin{align} \mid y\mid \geq 1\hspace{.33em}\\~\\
\end{align}}\)

- Vocaloid

\[\left| y \right| \ge 1 \] means that
\[y \ge 1\] or
\[y \le -1\]
|dw:1434816235241:dw|

- mathmath333

u mean this one |dw:1434816613473:dw|

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## More answers

- Vocaloid

yeah, that graph is lot neater, thanks!

- mathmath333

@Vocaloid but wolfram gave this as the graph
|dw:1434816760211:dw|

- hartnn

1. |y| >= 1
is a one variable inequality.
so you'd use a number line.
2. If you want a xy plot, your plot should show regions for all values of y...
like if y = -2, would your inequality satisfy ?? is it shown by that plot?

- hartnn

point 1,2 satisfies your inequality, but the wolf graph says it doesn't :)

- hartnn

wolf graph is wrong.
it plots x =|y| and y =1

- mathmath333

how can i plot 1 in (1,2) there is no space for x co-ordinate

- hartnn

it should have plotted x=1 instead of y=1

- hartnn

|dw:1434817459914:dw|

- mathmath333

|dw:1434817543444:dw|

- ganeshie8

how do you sketch y = 1 in 1D, 2D and 3D ?

- mathmath333

|dw:1434817611958:dw|

- hartnn

the wolf graph took vertical axis as 'x' ! lol

- hartnn

if you want a 2D graph for |y|>=1, then the first graph is correct

- ganeshie8

Haha wolfram is trolling us
|dw:1434817710812:dw|

- hartnn

|y|â‰¥1
means that
yâ‰¥1
or
yâ‰¤âˆ’1
is absolutely correct

- mathmath333

but why is that desmos graphing calculator also cannot graph
\(|y|\geq 1\) when i input the syntax.

##### 1 Attachment

- ganeshie8

As the question stands, nobody can stop me from thinking of \(y\) as distance from origin (polar) ;p
|dw:1434817996160:dw|

- hartnn

"We don't solve complicated single variable equations yet" lol
true ganeshie!
it depends on how we want to graph...multiple correct answers.

- hartnn

although the best thing to do here is to plot on the number line!

- mathmath333

so this below graph correct for \(|y| \geq 1\)
|dw:1434818180900:dw|

- ganeshie8

yeah some are ridiculous and some are more or less obvious from the context

- hartnn

yes, it is math.
verify by taking different values of y

- mathmath333

ok thnx

- Vocaloid

hm, I suppose graphing calculators/graphing software is usually designed to handle x and y together
think of it this way:
we can graph y = 1, right? it's just a horizontal line through y = 1
y > 1 is just everything shaded above that line.

- hartnn

y=1 is a point on a number line.
y= 1 is a line in xy plot: 2D
y=1 is a plane in xyz plot: 3D

- Vocaloid

^ that, too :)

- mathmath333

i have checked in geogebra too, it also can't plot \(|y|\geq 1\)

- ParthKohli

But you can.

- ganeshie8

try below in geogebra
abs(y) >= 1

- hartnn

so you > desmos/geogebra
:P :)

- ParthKohli

We created Desmos and Geogebra.
Humans 1 - 0 Computers

- mathmath333

tried this "abs(y) >= 1" , nothing happened from geogebra

- mathmath333

after all man made computer ,
computer did'nt made man

- ganeshie8

id love to be in a matrix for a change

- hartnn

|y| >= 1
the first thought that came to mind,
all values of y, that are greater than 1,
irrespective of the sign of the number....

- ParthKohli

There's really not much to think about this problem. You just find the set of points with the absolute value of their y-coordinate either 1 or greater than that.
So the "base"-set of points would be points with y-coordinate = 1.
I just realised that it's exactly what hartnn explained but I don't want to remove this reply so yeah

- hartnn

|dw:1434819148818:dw|

- ganeshie8

|dw:1434819259110:dw|

- ParthKohli

|dw:1434819280155:dw|

- ganeshie8

|dw:1434819455607:dw|

- ParthKohli

close enough

- ganeshie8

your painting looks way prettier than the original smith !

- ParthKohli

Of course it does - where do you think Smith was inspired from?

- ganeshie8

|dw:1434819600999:dw|

- ganeshie8

i remember something like matrix was inspired from Gita

- ParthKohli

Yes, and that is obviously not true...

- mathmath333

if i have to plot the graph for this
\(y\geq |1|\)
then will it also be same as \(|y|\geq 1\)

- hartnn

|1| is just 1, right ?

- mathmath333

but |-1| is also 1

- hartnn

but your question does not have |-1|

- hartnn

y >=|1|
is same as y>=1
all y values greater than 1.
so y will take only positive values, not negative

- mathmath333

solving \(y>=|1|\)this by algebra will give me this this two inequalities
\( y>=1\) and \(y>=-1\) right ?

- ganeshie8

It might help to think of \(|y|\) as \(|y-0|\), the distance between a point \(y\) on number line.and origin \(0\)

- ganeshie8

\(|1|\) is same as \(|1-0|\)
the distance between points \(1\) and \(0\) on number line

- hartnn

if there is a variable inside |...|
only then it would lead to 2 posibilities
if there is a constant, then its a unique value...thats why its called a constant

- mathmath333

ohk

- ganeshie8

@ParthKohli
https://www.youtube.com/watch?v=re7nmrDbAG4

- mathmath333

i got a question based on this one
If \(|y|\geq 1\) and \(x=-|a|y\) , then which one of the following is true?
\(\large \color{black}{\begin{align} a.)\ a-xy<0\hspace{.33em}\\~\\
b.)\ a-xy\geq 0\hspace{.33em}\\~\\
c.)\ a-xy>0\hspace{.33em}\\~\\
d.)\ a-xy\leq 0\hspace{.33em}\\~\\
\end{align}}\)

- ParthKohli

Since all of the questions involve the expression \(a - xy\), let's try to check its nature.\[a - xy\]\[=a - |a|y^2 \]Now \(y^2 \ge 1 \) so this expression is \(0 \) for \(y = \pm 1\) and positive \(a\). As soon as you change \(y\), it would become negative.

- hartnn

**
a + |a|y^2

- ParthKohli

Oh wow, skipped that negative sign. =_=

- ParthKohli

OK thanks, then it should be positive. Looks good.

- hartnn

|a|y^2 is always positive

- hartnn

yes,
just because |y| >=1
we can say a will be less than or = |a|y^2 numerically
and that would make
a+ |a|y^2 always positive :)

- ParthKohli

OK, a better word would be "nonnegative".

- hartnn

can be 0 ?
y can't be 0

- hartnn

a =0 :P

- mathmath333

so only options 'b' and 'd' are valid

- ParthKohli

Ew, what am I even talking about

- hartnn

b says
a-xy is 0 or positive (non-negative)
d says
a-xy is 0 or negative (non-positive)

- mathmath333

can \(a\) be 0

- hartnn

sure

- mathmath333

so option be would be correct

- hartnn

\(\Huge \checkmark \)

- mathmath333

thnx!

- hartnn

wlcmx!

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