horizontal transformation of functions

Observe two functions, translated by some factor but otherwise identical:

\documentclass[tikz]{standalone}
\usepackage{pgfplots}
\usetikzlibrary{arrows.meta}
%
\definecolor{firstfunctioncolor}{HTML}{8A5CF5}
\definecolor{secondfunctioncolor}{HTML}{E0DE71} % lighter
% \definecolor{secondfunctioncolor}{HTML}{E0AC00} % darker
%
\begin{document}
\begin{tikzpicture}
    \begin{axis}[
        width=10cm,
        height=10cm,
        axis lines = left,
        enlargelimits=0.05,
        xticklabels = {\empty},
        yticklabels = {\empty},
        legend pos = north west,
        legend image post style = -,
        legend cell align = left,
        xtick = {-4,-3,-2,-1,...,6,7,8,9},
        ytick = {-52,-27,0,27,52},
        ]
	%
    \addplot [
        Latex-Latex,
        domain=-5:5, 
        samples=101, 
        color=firstfunctioncolor,
        very thick,
        ]
        {x^3};
    \addlegendentry{\(f(x)\)}
	%
    \addplot [
        Latex-Latex,
        domain=0:10, 
        samples=101, 
        color=secondfunctioncolor,
        very thick,
        ]
        {(x-5)^3)};
    \addlegendentry{\(g(x)=f(x-h)\)}
	%
    \draw [dashed] (axis cs:10,27) -- (axis cs:0.5,27) node[left] {\(f(d-h)=g(d)\)};
    \draw [dashed] (axis cs:-5,-27) -- (axis cs:3.5,-27) node[right] {\(f(c)=g(c+h)\)};
	%
    \begin{scope}[every node/.style={circle, fill, inner sep=1.5pt}]
        \node[color=firstfunctioncolor, pin={[pin edge={firstfunctioncolor!50}]105:\small $d-h$}] at (axis cs:3,27) {};
        \node[color=firstfunctioncolor, pin={[pin edge={firstfunctioncolor!50}]285:\small $c$}] at (axis cs:-3,-27) {};
        \node[color=secondfunctioncolor, pin={[pin edge={secondfunctioncolor!50}]105:\small $d$}] at (axis cs:8,27) {};
        \node[color=secondfunctioncolor, pin={[pin edge={secondfunctioncolor!50}]285:\small $c+h$}] at (axis cs:2,-27) {};
    \end{scope}
	%
    % \node at (axis cs:5.5,14) {\(h\)};
    % \node at (axis cs:-0.5,-14) {\(h\)};
	%
    \end{axis}
\end{tikzpicture}
\end{document}

But why, when shifting $f (x)$ $h$ units to the right, is it $f (x - h)$ ? Why is it minus if the shift is positive?

First, let $f (x)$ be our main function, and let $g (x)$ be the function $f$ shifted to the right by $h$ . Then, for each point $(x, y)$ in $f$ , where $y = f (x)$ , our shifted function $g$ has the same output at an input shift of $h$ ; in other words, $(x + h, y)$ is in $g$ , i.e. $g (x + h) = y = f (x)$ .
This can be interpreted as:


$f (c) = g (c + h)$	Whenever I input some value $c$ into $f$ , I have to input a value $h$ more than $c$ into $g$ to get the same output.
Whenever I input some value $d$ into $g$ , I have to input a value $h$ less than $d$ into $f$ to get the same output.	$g (d) = f (d - h)$
That 'less' is why we see a minus sign when we move a function to the right. It's because we are relating the 'new' function back to the original function.

In fact, when we have a function $y = f (x)$ , we typically see a translational shift expressed as $y = f (x - h) + k$ , where a positive $k$ means a shift in the positive $y$ direction. However, using some algebraic manipulation, it's clear to see that an equivalent interpretation of a general translation is

(y - k) = f (x - h)

where $k$ is the shift in the positive $y$ direction and $h$ is the shift in the positive $x$ direction. The reasoning for why $k$ has a minus sign is exactly the same as the reasoning above for why $h$ has a minus sign: we are relating the 'new' function back to the original function.