Problem source

The curve $C$ has equation
$$ 
y = \frac x {\sqrt{x^2-2x+a}}\; ,
$$
where the square root is positive.
Show that, if $a>1$,  then $C$ has exactly one stationary point. 
Sketch $C$ when \textbf{(i)} $a=2$ and \textbf{(ii)} $a=1$.

Solution source

\begin{align*}
&& y &= \frac x {\sqrt{x^2-2x+a}} \\
&& y' &= \frac{\sqrt{x^2-2x+a} - \frac{x(x-1)}{\sqrt{x^2-2x+a}}}{x^2-2x+a} \\
&&&= \frac{-x+a}{(x^2-2x+a)^{3/2}}
\end{align*}

Since the denominator is always positive, the only stationary point is when $x = a$

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)/sqrt((#1)^2-2*(#1)+2)};
    \def\xl{-20};
    \def\xu{20};
    \def\yl{-2};
    \def\yu{2};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=\xl:\xu, samples=100] 
            plot ({\x}, {\functionf(\x)});

        \draw[red, dashed, thick] (\xl,1) -- (\xu, 1);
        \draw[red, dashed, thick] (\xl,-1) -- (\xu, -1);
    \end{scope}

    \filldraw (0,0) circle (1.5pt) node[above left] {$(0,0)$};
    \filldraw (2,{\functionf(2)}) circle (1.5pt) node[above] {$(2,\sqrt{2})$};

    \node[left] at (0,1) {$1$};
    \node[left] at (0,-1) {$-1$};
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};

    
    \end{tikzpicture}
\end{center}



\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)/sqrt((#1)^2-2*(#1)+1)};
    \def\xl{-20};
    \def\xu{20};
    \def\yl{-5};
    \def\yu{5};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=\xl:0.9, samples=100] 
            plot ({\x}, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=1.1:\xu, samples=100] 
            plot ({\x}, {\functionf(\x)});

        \draw[red, dashed, thick] (\xl,1) -- (\xu, 1);
        \draw[red, dashed, thick] (\xl,-1) -- (\xu, -1);
        \draw[red, dashed, thick] (1,\yl) -- (1, \yu);
    \end{scope}

    \filldraw (0,0) circle (1.5pt) node[above left] {$(0,0)$};
    % \filldraw (2,{\functionf(2)}) circle (1.5pt) node[above] {$(2,\sqrt{2})$};

    \node[left] at (0,1) {$1$};
    \node[left] at (0,-1) {$-1$};
    \node[below] at (0,1) {$1$};
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};

    
    \end{tikzpicture}
\end{center}