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## #1 2006-05-31 01:52:13

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### graphs slid around with subtraction

I want to know WHY this works.
Take most any 2D graph with y and x in the equation, and notice that you can slide the graph up, down, left, and right by subtracting from x to move right, and subtracting from y to move up.
Adding to y moves the whole graph down, and adding to x moves the whole graph left.
For example.
y = x^3 + x^1.8
This graph has a certain shape.
To move the whole graph down 5.5 units and right 33 units, then we get this:
y + 5.5 = (x - 33)^3 + (x - 33)^1.8
I don't know if I was taught this or if I just saw it through the years and now it seems to be true.
But I'd like to know WHY it works.

Here is another example:
Move the whole graph left by 13 which essentially does nothing since it is still a horizontal line.
But where is the x?  y = 0(x + 13) + 5
y = 5 still.

Here is another one:
y = sin(x), where sine is in degrees.
Move whole graph left by 90 degrees:  y = sin(x + 90), which by the way is probably y = cos(x), but that's off the subject.

Anyone know why this simple substitution for y and x appears to move the graph, or does it move the axis the other way?

igloo myrtilles fourmis

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## #2 2006-05-31 13:46:20

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: graphs slid around with subtraction

The easiest way to see why this happens is to make a table.  In general we have:

f(x) be some function
g(x) = f(x - a)

x  |  x - a
-----------
0  |  -a
1  |  1 - a
2  |  2 - a
3  |  3 - a

So if we are looking at f(x) and g(x):

f(0) and g(a) = f(a - a) = f(0)
f(1) and g(a + 1) = f(a + 1 - a) = f(1)
f(2) and g(a + 2) = f(a + 2 - a) = f(2)

and so on.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #3 2006-06-01 08:01:01

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: graphs slid around with subtraction

Hey thanks Ricky, I like to see how other people can explain things.
Does this show why it goes in the opposite direction as the sign?
Is this example horizontal or vertical movement?

igloo myrtilles fourmis

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## #4 2006-06-01 08:45:48

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: graphs slid around with subtraction

This is just horizontal movements.

And it does explain why the sign is reversed.  If we plug in x-a like I did for g:

g(x) = f(x - a)

If x = 0, then g(0) = f(-a).  But to put it another way, g(x + a) = f(x - a + a), and so if x = 0, then g(a) = f(0).  And like I showed before, the same is true for 1, 2, 3... etc.  So even though your pugging in a negative a, it shifts it over a positions.

For verticle, it's basically the same concept:

g(x) = f(x) + b

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #5 2006-06-01 11:17:07

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: graphs slid around with subtraction

Yes but notice that if the relationship is a circle, you go up by altering the y with subtraction to move it up before the y is squared.    So g(x)=f(x) + b may not work for circles.   I originally said you get the -a right in with the y to go up by a, so if the y is not solved for and is in a mess, the vertical constant is right in tight with the y.
For example  25 = (y-6)^2 + x^2  would move the circle up vertically by 6 I think.  I might be wrong, but that's how I think it is.

igloo myrtilles fourmis

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## #6 2006-06-01 12:21:51

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: graphs slid around with subtraction

But a circle isn't a graph!  It isn't well defined.  In other words, it doesn't pass the verticle line test.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #7 2006-06-07 02:59:58

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Re: graphs slid around with subtraction

that doesn't matter, the shifting rule is not limited to functions.

for post 4, Ricky, I recommand you  using x[sub]1[/sub] and x[sub]2[/sub] instead of both xs, to make it clearer.

post 2 is very clear.

I think the shifting rule also applies to x²+y²≤4 and (x-.5)²+(y+1.5)²≤4 as well.

X'(y-Xβ)=0

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## #8 2006-06-07 04:13:00

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

### Re: graphs slid around with subtraction

Ricky wrote:

But a circle isn't a graph!

x^2 + y^2 = r^2 can be rewritten y = +/- sqrt{r^2-x^2}

which if you plot with the +/- will give you a circle

(and if you plot complex numbers aswell, you get a verticaly standing circle, and then two horizontal square root patterns)

take the x axis, as x
the y axis as the real part of the circle function
and the z axis as the imaginary part of the circle function

obviously the circle has all z = 0
and the outper parts have all y = 0

The Beginning Of All Things To End.
The End Of All Things To Come.

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## #9 2006-06-07 08:32:54

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: graphs slid around with subtraction

x^2 + y^2 = r^2 can be rewritten y = +/- sqrt{r^2-x^2}

which if you plot with the +/- will give you a circle

If you have two functions:

y1 = sqrt(r^2 - x^2)
y2 = sqrt(r^2 - x^2)

And you try to apply the shifting rule to both those functions, it will work.  But it only works when you can put a function in the form of:

y = f(x)

Which is not always possible.

Edit: And I made a minor change to my post #2 which I think makes it a bit clearer.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #10 2006-06-07 14:50:32

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Re: graphs slid around with subtraction

shifting rule can also be applied to
f(x,y)=0

Because:
Break the curve into very small pieces. Locally, you will solve out {y=g[sub]i[/sub](x)}, get one to one functions, and then you just need to connect them together.
Now the altogether f(x,y)=0 is ready for shifting rule.

Accually, that's how the theorem about implicit differential works.

Last edited by George,Y (2006-06-07 14:53:21)

X'(y-Xβ)=0

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## #11 2006-06-07 14:56:13

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: graphs slid around with subtraction

basicly the reason the shifting occurs is it puts it a few steps ahead, say for instance f(x) = 2x. To get a 4 you'd have to inset an x value of 2. But if f(x) = 2x + 1 then x can be one less and produce the same value. So the whole graph is shifted horizontally one unit. It basicly gives it a headstart.

A logarithm is just a misspelled algorithm.

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## #12 2006-06-07 15:20:33

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Re: graphs slid around with subtraction

Another proof about g(x,y)=0 qualified for shifting rule

f(x[sub]0[/sub])=y[sub]0[/sub]
then f(?-a)=y[sub]0[/sub]
The question here may seem to be solving an equation out. But not so.
First, x[sub]0[/sub]+a is qualified,
for f(x[sub]0[/sub]+a-a)=y[sub]0[/sub]
Second, every x satisfying f(x)=y[sub]0[/sub] can be grouped into {x[sub]i[/sub]}
then for each i, or each x[sub]i[/sub]
f(x[sub]i[/sub]+a-a)=y[sub]0[/sub]
here we proved x[sub]i[/sub]+a are qualified solutions for f(x-a)=y[sub]0[/sub], but are they all solutions? Is there a solution out of {x[sub]i[/sub]+a}? If there is one, the shifted function may have one more point at the altitude y=y[sub]0[/sub]

IF y[sub]0[/sub]=f(x) and y[sub]0[/sub]=f(x-a) has the same number of solutions, {x[sub]i[/sub]+a} is the complete solution set for f(x-a)=y[sub]0[/sub].

The assumption is what I cannot prove. But if anyone prove it, or has proved it, the proof is complete. Or to put it inanother way, now that f(x)=y can be applied by shifting rule, the assumption should be true.

The proof for g(x,y)=0 is the same. the same projection. the difference is minor.
for any x[sub]i[/sub] satisfying g(x,y[sub]0[/sub])=0
x[sub]i[/sub]+a satisfy g(x-a,y[sub]0[/sub])=0.

IF g(x,y[sub]0[/sub])=0 and g(x-a,y[sub]0[/sub])=0 have the same amount of solutions, the shifting rule applies.

I think the shifting rule can also be applied to inequality like g(x,y)≤0 because equalities and inequalities have tight connections. However, strict proof need rigorous knowledge on inequality theorems. So I assume it.

Last edited by George,Y (2006-06-07 15:21:52)

X'(y-Xβ)=0

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## #13 2006-06-07 15:38:47

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

### Re: graphs slid around with subtraction

Proof of the assumption
--I discovered no knowledge about solving a polynomial solution is needed

the proof before assumpition IMPLIES that f(x-a)=y[sub]0[/sub] has no less solutions than f(x)=y[sub]0[/sub], for there is a solution for f(x)=y[sub]0[/sub], there exists a corresponding one for f(x-a)=y[sub]0[/sub].

This is a corollary from the x[sub]i[/sub] and x[sub]i[/sub]+a theorem.

Note this corollary applies to any a and any function.
Name f(x-a)=g(x)
t=x-a
f(t)=g(t+a)
t=x
f(x)=g(x+a)=g(x-(-a))
Using the corollary,
g(x-(-a))=y[sub]0[/sub] has no less solutions than g(x)=y[sub]0[/sub]
that's analogous to say f(x)=y[sub]0[/sub] has no less solutions than f(x-a)=y[sub]0[/sub]

Together, f(x)=y[sub]0[/sub] and f(x-a)=y[sub]0[/sub] has the same amount of solutions, assumption PROVEN!

Should I go to  a math journal. Or was I doing some already-done work?

X'(y-Xβ)=0

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