Problem source

Sketch the curve with cartesian equation 
\[
y = \frac{2x(x^2-5)}{x^2-4}
\]
and give the equations of the asymptotes and of the tangent to the curve at the origin.
Hence determine the number of real roots of the following equations:
\begin{questionparts}
\item  $3x(x^2-5)= (x^2-4)(x+3)\,$;
\item $4x(x^2-5)= (x^2-4)(5x-2)\,$;
\item  $4x^2(x^2-5)^2= (x^2-4)^2(x^2+1)\,$.
\end{questionparts}

Solution source

\begin{align*}
&& y &= \frac{2x(x^2-5)}{x^2-4} \\
&&&= 2x(x^2-5)(-\tfrac14)(1-\tfrac14x^2)^{-1} \\
&&&= \tfrac52x + \cdots \\
&&&= \frac{2x(x^2-4)-2x}{x^2-4} \\
&&&= 2x - \frac{2x}{x^2-4}
\end{align*}


\begin{center}
    \begin{tikzpicture}
    \def\a{2};
    \def\functionf(#1){2*(#1)*((#1)^2-5)/((#1)^2-4)};
    \def\xl{-10};
    \def\xu{10};
    \def\yl{-20};
    \def\yu{20};
    
    % 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);
        


        % --- Outer Branches (Exist for all x) ---
        % \addplot[thick, blue, domain=0:3.5] {sqrt((5 + sqrt(\disc(\x)/2)))};
        % \addplot[thick, blue, domain=-3.5:3.5] {-sqrt((5 + sqrt(\disc))/2)};

        % --- Inner Loop (Exists only for -2 <= x <= 2) ---
        % \addplot[thick, red, domain=-2:2] {sqrt((5 - sqrt(\disc))/2)};
        % \addplot[thick, red, domain=-2:2] {-sqrt((5 - sqrt(\disc))/2)};
        
        \draw[thick, blue, smooth, domain=\xl:-2.05, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1.95:1.95, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=2.05:\xu, samples =100]
            plot (\x, {\functionf(\x)});
        

        \draw[dashed, red] (-2, \yl) -- (-2, \yu) node[pos=0.25, sloped, above] {\tiny $y = -2$};
        \draw[dashed, red] (2, \yl) -- (2, \yu)node[pos=0.75, sloped, below] {\tiny $y = 2$};;
        \draw[dashed, red] (\xl, {5/2*\xl}) -- (\xu, {5/2*\xu}) node[pos=0.25, sloped, below] {\tiny $y = \frac52x$};

        \draw[dashed, red] (\xl, {2*\xl}) -- (\xu, {2*\xu}) node[pos=0.95, sloped, above] {\tiny $y =2x$};
    \end{scope}
    
    % 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{questionparts}
\item We are looking for the intersections of $y = \frac23(x+3)$ and $y = f(x)$

\begin{center}
    \begin{tikzpicture}
    \def\a{2};
    \def\functionf(#1){2*(#1)*((#1)^2-5)/((#1)^2-4)};
    \def\xl{-10};
    \def\xu{10};
    \def\yl{-20};
    \def\yu{20};
    
    % 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);
        


        % --- Outer Branches (Exist for all x) ---
        % \addplot[thick, blue, domain=0:3.5] {sqrt((5 + sqrt(\disc(\x)/2)))};
        % \addplot[thick, blue, domain=-3.5:3.5] {-sqrt((5 + sqrt(\disc))/2)};

        % --- Inner Loop (Exists only for -2 <= x <= 2) ---
        % \addplot[thick, red, domain=-2:2] {sqrt((5 - sqrt(\disc))/2)};
        % \addplot[thick, red, domain=-2:2] {-sqrt((5 - sqrt(\disc))/2)};
        
        \draw[thick, blue, smooth, domain=\xl:-2.05, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1.95:1.95, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=2.05:\xu, samples =100]
            plot (\x, {\functionf(\x)});
        

        \draw[dashed, red] (-2, \yl) -- (-2, \yu) node[pos=0.25, sloped, above] {\tiny $y = -2$};
        \draw[dashed, red] (2, \yl) -- (2, \yu)node[pos=0.75, sloped, below] {\tiny $y = 2$};;
        \draw[dashed, red] (\xl, {5/2*\xl}) -- (\xu, {5/2*\xu}) node[pos=0.25, sloped, below] {\tiny $y = \frac52x$};

        \draw[dashed, red] (\xl, {2*\xl}) -- (\xu, {2*\xu}) node[pos=0.95, sloped, above] {\tiny $y =2x$};

        \draw[thick, green] (\xl, {2/3*(\xl+3)}) -- (\xu, {2/3*(\xu+3)}) node[pos=0.8, sloped, above] {\tiny $y =\frac23(x+3)$};
    \end{scope}
    
    % 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}

Therefore 3 real roots.

\item We are looking for intersections of $y = \frac12(5x-2)$ and $y = f(x)$

\begin{center}
    \begin{tikzpicture}
    \def\a{2};
    \def\functionf(#1){2*(#1)*((#1)^2-5)/((#1)^2-4)};
    \def\xl{-10};
    \def\xu{10};
    \def\yl{-20};
    \def\yu{20};
    
    % 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);
        


        % --- Outer Branches (Exist for all x) ---
        % \addplot[thick, blue, domain=0:3.5] {sqrt((5 + sqrt(\disc(\x)/2)))};
        % \addplot[thick, blue, domain=-3.5:3.5] {-sqrt((5 + sqrt(\disc))/2)};

        % --- Inner Loop (Exists only for -2 <= x <= 2) ---
        % \addplot[thick, red, domain=-2:2] {sqrt((5 - sqrt(\disc))/2)};
        % \addplot[thick, red, domain=-2:2] {-sqrt((5 - sqrt(\disc))/2)};
        
        \draw[thick, blue, smooth, domain=\xl:-2.05, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1.95:1.95, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=2.05:\xu, samples =100]
            plot (\x, {\functionf(\x)});
        

        \draw[dashed, red] (-2, \yl) -- (-2, \yu) node[pos=0.25, sloped, above] {\tiny $y = -2$};
        \draw[dashed, red] (2, \yl) -- (2, \yu)node[pos=0.75, sloped, below] {\tiny $y = 2$};;
        \draw[dashed, red] (\xl, {5/2*\xl}) -- (\xu, {5/2*\xu}) node[pos=0.25, sloped, below] {\tiny $y = \frac52x$};

        \draw[dashed, red] (\xl, {2*\xl}) -- (\xu, {2*\xu}) node[pos=0.95, sloped, above] {\tiny $y =2x$};

        \draw[thick, green] (\xl, {1/2*(5*\xl-2)}) -- (\xu, {1/2*(5*\xu-2)}) node[pos=0.8, sloped, above] {\tiny $y =\frac12(5x-2)$};
    \end{scope}
    
    % 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}

so one solution.

\item We are looking for intersections of $y = f(x)^2$ and $y = x^2+1$, or $y = \sqrt{x^2+1}$ and $y = f(x)$ where $f(x) \geq 0$


\begin{center}
    \begin{tikzpicture}
    \def\a{2};
    \def\functionf(#1){2*(#1)*((#1)^2-5)/((#1)^2-4)};
    \def\xl{-10};
    \def\xu{10};
    \def\yl{-20};
    \def\yu{20};
    
    % 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);
        


        % --- Outer Branches (Exist for all x) ---
        % \addplot[thick, blue, domain=0:3.5] {sqrt((5 + sqrt(\disc(\x)/2)))};
        % \addplot[thick, blue, domain=-3.5:3.5] {-sqrt((5 + sqrt(\disc))/2)};

        % --- Inner Loop (Exists only for -2 <= x <= 2) ---
        % \addplot[thick, red, domain=-2:2] {sqrt((5 - sqrt(\disc))/2)};
        % \addplot[thick, red, domain=-2:2] {-sqrt((5 - sqrt(\disc))/2)};
        
        \draw[thick, blue, smooth, domain=\xl:-2.05, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=-1.95:1.95, samples =100]
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=2.05:\xu, samples =100]
            plot (\x, {\functionf(\x)});
        

        \draw[dashed, red] (-2, \yl) -- (-2, \yu) node[pos=0.25, sloped, above] {\tiny $y = -2$};
        \draw[dashed, red] (2, \yl) -- (2, \yu)node[pos=0.75, sloped, below] {\tiny $y = 2$};;
        \draw[dashed, red] (\xl, {5/2*\xl}) -- (\xu, {5/2*\xu}) node[pos=0.25, sloped, below] {\tiny $y = \frac52x$};

        \draw[dashed, red] (\xl, {2*\xl}) -- (\xu, {2*\xu}) node[pos=0.95, sloped, above] {\tiny $y =2x$};


        \draw[thick, green, smooth, domain=\xl:\xu, samples =100]
            plot (\x, {sqrt((\x)^2+1)});
        % \draw[thick, green] (\xl, {1/2*(5*\xl-2)}) -- (\xu, {1/2*(5*\xu-2)}) node[pos=0.8, sloped, above] {\tiny $y =\frac12(5x-2)$};
    \end{scope}
    
    % 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}

So $3$ solutions.

\end{questionparts}