Graphing (1 Viewer)

[Premus]

hey,

if y= f(x) ,
what would the graph of /y/ = f(x) look like? ( thats mod y = f(x) )


thanks

 

[nick1048]

\....../...........wOuldnt
.\..../..................it
..\../.................look
...\/...................like
........................dis?

 

[Archman]

take the graph of f(x), discard the bits below the x axis, now reflect and copy everything about the x axis and there u have it.

 

[Premus]

so is ....

/y/ = f(x) the same as y^2 = f(x) ?

 

[wogboy]

so is ....

/y/ = f(x) the same as y^2 = f(x) ?

No it's the same as: y^2 = (f(x))^2 (where f(x) >= 0)

Edit: Whoops error fixed

 

[Rorix]

no it's not


(by inspection)

 

[Archman]

No it's the same as: y^2 = (f(x))^2

no its not that either, y^2 = (f(x))^2 is the same as |y| = |f(x)|

 

[dawso]

yeah, aint it just like an absolute value graph like nick said, everything is positive (the distance from the x axis is constant)

 

[dawso]

actually.......if its mod y, then it aint like an absolute value, mod is the distance from the origin, not the x axis, so its like.......hm........is that the whole question or where ya given a graph 4 f(x)? if not yeah, i dunno, but remember that mod z is the distcance from the origin, vs lzl is the distance from x axis
hope this helps
-dawso

 

[Slidey]

no its not that either, y^2 = (f(x))^2 is the same as |y| = |f(x)|

Um... I would think even that is wrong. Wouldn't it be:

sqrt(y)^2=sqrt(f(x))^2 is the same as |y|=|f(x)| ?

 

[Slidey]

Say f(x)=y=|x^2-4x+3|,

When y>=0, y=x^2-4x+3. 
y is >=0 when x= 1 or 3.
So when x<=1, or >=3, y=x^2-4x+3
When y<0, y=-x^2+4x-3
So when 1<x<3, y=-x^2+4x-3

Inserted into PHP tages, since the inequality signs were screw up the formatting and making some parts disappear.

 

[Premus]

As Archman said, i think its just getting rid of the graph below the x-axis and reflecting whats left about the x-axis.

and is
|y|=|f(x)|
the same as |y|=f(x) ?

Thanks!

 

[Slidey]

That depends on whether or not you have defined y to be identical to f(x).

If, for example, f(x)=x+1, but you've said "let y=f(x)", then |y| will equal |f(x)|. However, if you've said "let |y|=f(x)", then y=f(x) or y=-f(x).

 

[Slidey]

As Archman said, i think its just getting rid of the graph below the x-axis and reflecting whats left about the x-axis.

Yes, but you have to be careful of non-absolute value parts, such as:

y=|x|-1. In this case, you WOULD have parts of the graph below the x-axis. 1 unit below, to be exact.

 

[Premus]

oh i meant....

if you had y = f(x) , where y = x^3

then, would |y| = f(x) be the same as discarding the graph in the third quadrant, and then reflecting the rest abt the x-axis?

 

[Slidey]

I think what you are saying is correct. You flip the bit below the x-axis so that it would be in the 2nd quadrant.

So, uh, yes, you are correct, if by "rest" you mean everything in the 3rd and 4th quadrants only.

 

[Archman]

Um... I would think even that is wrong. Wouldn't it be:

sqrt(y)^2=sqrt(f(x))^2 is the same as |y|=|f(x)| ?

well a^2 = b^2 implies |a| = |b|
on sqrt(y)^2=sqrt(f(x))^2, u can just go straight to y = f(x) because you cant have negatives within square roots.. well for curve sketching purposes anyway.