Problem source

Sketch the graph of $8y=x^{3}-12x$ for $-4\leqslant x\leqslant4$, marking the coordinates of the turning points. Similarly marking the turning points, sketch the corresponding graphs in the $(X,Y)$-plane,
if
\begin{alignat*}{3}
\rm{(a)} & \quad &  & X=\tfrac{1}{2}x, & \qquad & Y=y,\\
\rm{(b)} &  &  & X=x, &  & Y=\tfrac{1}{2}y,\\
\rm{(c)} &  &  & X=\tfrac{1}{2}x+1, &  & Y=y,\\
\rm{(d)} &  &  & X=x, &  & Y=\tfrac{1}{2}y+1.
\end{alignat*}
 Find values for $a,b,c,d$ such that, if $X=ax+b,$ $Y=cy+d$, then the graph in the $(X,Y)$-plane corresponding to $8y=x^{3}-12x$ has turning points at $(X,Y)=(0,0)$ and $(X,Y)=(1,1)$.

Solution source

$8\frac{\d y}{\d x} = 3(x^2-4)$ so the turning points are at $(\pm 2, \mp 2)$

\begin{center}
    \begin{tikzpicture}

        \draw[->] (-4, 0) -- (4,0);
        \draw[->] (0, -2) -- (0, 2);

        \node at (4,0) [right] {$x$};
        \node at (0,2) [above] {$y$};
    
        \draw[domain = -4:4, samples=180, variable = \x]  plot ({\x},{1/8 * (\x*\x*\x - 12*\x)});


        \node at (2, -2) [below] {$(2, -2)$};
        \node at (-2, 2) [above] {$(-2, 2)$};

        \node at (0,0) [above, right] {$(0,0)$};
        \node at ({-sqrt(12)},0) [below right] {$(-\sqrt{12},0)$};
        \node at ({sqrt(12)}, 0) [below right] {$(\sqrt{12},0)$};

        \node at (-2.5,-2) {$8y = x^3 - 12x$};
    \end{tikzpicture}
\end{center}

\begin{center}
    \begin{tikzpicture}

        \draw[->] (-2, 0) -- (2,0);
        \draw[->] (0, -2) -- (0, 2);

        \node at (2,0) [right] {$X = \frac12 x$};
        \node at (0,2) [above] {$Y = y$};
    
        \draw[domain = -2:2, samples=180, variable = \x]  plot ({\x},{1/8 * (8*\x*\x*\x - 2*12*\x)});


        \node at (1, -2) [below] {$(1, -2)$};
        \node at (-1, 2) [above] {$(-1, 2)$};
    \end{tikzpicture}
\end{center}

\begin{center}
    \begin{tikzpicture}

        \draw[->] (-4, 0) -- (4,0);
        \draw[->] (0, -1) -- (0, 1);

        \node at (4,0) [right] {$X = x$};
        \node at (0,1) [above] {$Y = \frac12y$};
    
        \draw[domain = -4:4, samples=180, variable = \x]  plot ({\x},{1/16 * (\x*\x*\x - 12*\x)});


        \node at (2, -1) [below] {$(2, -1)$};
        \node at (-2, 1) [above] {$(-2, 1)$};

    \end{tikzpicture}
\end{center}

\begin{center}
    \begin{tikzpicture}

        \draw[->] (-2, 0) -- (2,0);
        \draw[->] (-1, -2) -- (-1, 3);

        \node at (2,0) [right] {$X = \frac12 x+1$};
        \node at (0,3) [above] {$Y = y$};
    
        \draw[domain = -2:2, samples=180, variable = \x]  plot ({\x},{1/8 * (8*\x*\x*\x - 2*12*\x)});


        \node at (1, -2) [below] {$(2, -2)$};
        \node at (-1, 2) [above] {$(0, 2)$};


    \end{tikzpicture}
\end{center}



\begin{center}
    \begin{tikzpicture}

        \draw[->] (-4, -1) -- (4,-1);
        \draw[->] (0, -1) -- (0, 1);

        \node at (4,-1) [right] {$X = x$};
        \node at (0,1) [above] {$Y = \frac12y+1$};
    
        \draw[domain = -4:4, samples=180, variable = \x]  plot ({\x},{1/16 * (\x*\x*\x - 12*\x)});


        \node at (2, -1) [below] {$(2, 0)$};
        \node at (-2, 1) [above] {$(-2, 2)$};

    \end{tikzpicture}
\end{center}


We need either

\begin{align*}
&& \begin{cases}
-2a+b &= 0 \\
2c+d &= 0 \\
2a+b &= 1 \\
-2c+d &= 1
\end{cases} && \text{ or } && \begin{cases}
-2a+b &= 1 \\
2c+d &= 1 \\
2a+b &= 0 \\
-2c+d &= 0
\end{cases} \\
\Rightarrow &&  \begin{cases}
-2a+b &= 0 \\
2a+b &= 1 \\
2c+d &= 0 \\
-2c+d &= 1
\end{cases} && \text{ or } && \begin{cases}
-2a+b &= 1 \\
2a+b &= 0 \\
2c+d &= 1 \\
-2c+d &= 0
\end{cases}\\
\Rightarrow && \begin{cases} (a,b) = (\frac14,\frac12)  \\ 
(c,d) = (-\frac14, \frac12)\end{cases} && \text{ or } &&   \begin{cases} (a,b) = (-\frac14,\frac12)  \\ 
(c,d) = (\frac14, \frac12)\end{cases}
\end{align*}

So either $X = \frac14 x + \frac12, Y = -\frac14 y + \frac12$ or $X = -\frac14x + \frac12, Y = \frac14y + \frac12$