High School Correlation between shifting graph of a function and shifting the axes

[JC2000]

1.To shift the graph of a function :

Vertical Shifts : $y=f(x) +h$ where the graph shifts $k$ units up if $k$ is positive and downwards when $k$ is negative.
Horizontal Shifts : $y=f(x+h)$ where the graph shifts to the left by $h$ units when positive and to the right when $h$ is negative.

2.Upon shift the coordinate axes :

$x=X+h$ encapsulates the x-coordinate of any point w.r.t original axes ($x$), the coordinate of the point w.r.t 'new' axes ($X$) and the x-coordinate of the new origin w.r.t original axes ($h$). Likewise for the y-coordinate we have $y=Y+k$.

3.I understand that shifting the axes to the left is the equivalent of shifting the graph of a function to the right.

My Question : Given the above points, I believe there should be some sort of correlation between the two sets of formulas(?).

My Guess : Since $y=f(x+h)$ where shifts the function to the left, this ought to be the same as $y=f(X)$ where $X=x+h$, yet this does not conform to the derived formula. (Maybe I am making an error by not clearly labelling each variable?)

 

[fresh_42]

My Guess : Since $y=f(x+h)$ where shifts the function to the left, this ought to be the same as $y=f(X)$ where $X=x+h$, yet this does not conform to the derived formula. (Maybe I am making an error by not clearly labelling each variable?)

This is trivially true. If $y=f(x+h)$ and I substitute $x+h$ by a $\text{shrubbery}$, then $y$ becomes $y=f(x+h)=f(shrubbery)$ only that your shrubbery is called $X$.In general it is easiest to write the entire graph as a set: $y=f(x)$ corresponds to the graph
$$
\Gamma_f := \{\,(x,y)\in \mathbb{R}^2\,|\,y=f(x)\,\} = \{\,(x,f(x)\,|\,x\in \mathbb{R}\,\}
$$
Your shifts are now the shift of a graph by a vector $\vec{v}=(h,b)$ which is pointwise:
$$
\Gamma'_f = \Gamma_f + \vec{v} = \{\,(x,y)+(h,b)\in \mathbb{R}^2\,|\,y=f(x)\,\} = \{\,(x+h,f(x)+b\,|\,x\in \mathbb{R}\,\}
$$
and the shifted function is $F_s(x)=f(x+h)=f(x)+b=y_s$ which corresponds to a change in coordinates by $x'=x+h\, , \,y'=y+b$ so that we have $$y'=y+b=f(x)+b=y_s=F_s(x)=f(x+h)=f(x')$$
We can now either consider the shifted function $y_s=F_s(x)$ in the same coordinate system $(x,y)$ or the same function $y'=f(x')$ in a different coordinate system $(x',y')\,.$

 

[JC2000]

Your explanation did make things a lot clearer, could you please clarify a few things though...

$F_s(x)=f(x+h)=f(x)+b=y_s$

Is this the same as $F_s(x)=y_s=f(x+h)+b$?

or the same function $y′=f(x′)$ in a different coordinate system $(x′,y′)$.

Wouldn't substituting $y'=y+b$ give $y=f(x')-b$ instead of what should be $y=f(x')+b$?

 

[fresh_42]

Your explanation did make things a lot clearer, could you please clarify a few things though...

There you ask the right one. I'm notoriously bad with those left - right things and coordinate transformations. Shifting a function and shifting the coordinate axis is basically the same so the main difficulty is to distinguish the two. I had tried only to use equations derived from $\Gamma' = \Gamma + \vec{v}$ because I know of my weakness with those things. Let's see if I confused something.

Let's see.

Case 1:
We keep the coordinates $(x,y)$ and only shift the function.
The original function is $y=f(x)$ and $\Gamma_f = \{\,(x,f(x))\,\}$ its graph.
The shifted function is a new function in the same coordinate system. We call it $F_s(x)=y\,.$
($y_s$ was confusing, as we still have the same $y$.)

The graph of $y=F_s(x)$ be $\Gamma'_{F_s}=\{\,(x,F_s(x)\,\}$. Now this is the same graph and function if we write $\Gamma'_{F_s}=\{\,(x+h,F_s(x+h))\,\}$ because we run through all $x\in \mathbb{R}$ and to run through all $x+h \in \mathbb{R}$ doesn't make a difference: the second coordinate is still the evaluation of $F_s$ at the first coordinate. But then
$$
\Gamma' = \Gamma + \vec{v}=\{\,(x+h,f(x)+b)\,\} = \{\,(x+h,F_s(x+h))\,\}
$$
which means $F_s(x+h)=f(x)+b$ or equivalently $F_s(x)\stackrel{(*)}{=}f(x-h)+b$.

(Seems I made a mistake as $(*)$ is the notation of the shifted function within the same coordinate system.)

Case 2:
Here we keep the function and change the coordinates.
This means there is no second function $F_s$, only the function $y=f(x)$. Instead we have a pair of new coordinates: $x'=x+h$ and $y'=y+b$. So we shifted the old coordinates by the same vector $\vec{v}=(h,b)$ to get the new coordinates: $(x',y')=(x,y)+\vec{v}=(x+h,y+b)\,.$
Thus we still have $\Gamma'_f = \{\,(x',y')\,\}= \{\,(x+h,y+b\,\} = \Gamma_f + \vec{v}$ only, that the new graph is still from the same function. Hence in the new coordinate system we get
$$
y' = f(x') \Longleftrightarrow y+b=f(x+h) \Longleftrightarrow f(x)+b=f(x+h) \Longleftrightarrow f(x)=f(x+h)-b
$$

(I hope this is now without any confusion of left - right or plus- minus.)

 

[JC2000]

Thank you for your painstaking response.

1)Why is $y_s$ ignored in Case 1? Shouldn't the entire function be different?

$\Gamma'_{F_s}=\{\,(x+h,F_s(x+h))\,\}$ because we run through all $x\in \mathbb{R}$ and to run through all $x+h\in \mathbb{R}$ doesn't make a difference: the second coordinate is still the evaluation of $F_s$ at the first coordinate.

2) Wouldn't $x+h$ give a different output than $x$ (hence making a difference to the second coordinate)? Or have I misunderstood the statement?

3)The way I look at it is as follows :

Moving the coordinate axes to the right gives : $x=X+h$ or, $X=x-h$
Which is the same as shifting the function to the left : $y=f(x+h')$ or, $y=f(X)$ where $h' = -h$. (The opposite signs indicate the fact that movement of axes/function in opposite directions results in the same movement?).

4)Likewise for moving vertically?

Though my explanation(if correct) seems to be very crude compared to how rigorously you draw the correlation.

 

[fresh_42]

Why is $y_s$ ignored in Case 1? Shouldn't the entire function be different?

Yes, and that is why I called it $F_s(x)$. You can of course set $y_s:=F_s(x)$. The problem is, that we still are in the $(x,y)$ world, where the variable comes from the horizontal $x-$axis and the function value is scaled along the vertical $y-$axis. I thought it would lead to more confusion if we drew the $y_s-$ values on the $y-$axis. That's why I thought it is better to avoid the $y_s$ notation and only write $F_s(x)$ for it.

Wouldn't $x+h$ give a different output than $x$ (hence making a difference to the second coordinate)? Or have I misunderstood the statement?

Sure. But as we consider the entire set of points, the $x$ output will be achieved at $x-h$ by $(x-h)+h$.
It is $(x,f(x)) \neq (x+h,f(x+h))$ but $\{\,(x,f(x))\,|\,x \in \mathbb{R}\,\} = \{\,(x+h,f(x+h))\,|\,x\in \mathbb{R}\,\}\,.$

Though my explanation (if correct) seems to be very crude compared to how rigorously you draw the correlation.

Yes, it's ok. You can see it in the final equations of my explanations, too. Look at the inverse signs!

Function shifted by $+ (h,b)\, : \, F_s(x){=}f(x-h)+b$

Coordinate system shifted by $+ (h,b)\, : \, f(x)=f(x+h)-b$