Problem source

The function $\f(x)$ is defined by $$ \f(x) = \frac{x( x - 2 )(x-a)}{  x^2 - 1}. $$ 
 
 
Prove algebraically that the line $y = x + c$ intersects the curve  $y = \f ( x )$  if  $\vert a \vert \ge1$, but there are values of $c$ for which there are no points of intersection if $\vert a \vert <1$. 
 
Find the equation of the oblique asymptote of the curve $y=\f(x)$. 
Sketch the graph in the two cases 
\begin{enumerate}
\item $a<-1$
\item $-1 < a < -\frac12$
\end{enumerate}
(You need not calculate the turning points.)

Solution source

\begin{align*}
&& x+ c &= f(x) \\
\Rightarrow && (x+c)(x^2-1) &= x(x-2)(x-a) \\
\Rightarrow && x^3 + cx^2-x-c &= x^3-(2+a)x^2+2ax \\
\Rightarrow && 0 &= (c+2+a)x^2-(1+2a)x-c \\
&& 0 &\leq \Delta = (1+2a)^2 + 4(2+c+a)c \\
&&&= 4c^2+(4a+8)c + (1+2a)^2 \\
&& \Delta_c &= 16(a+2)^2-16(1+2a)^2 \\
&&&= 16(1-a)(3a+3) \\
&&&= 48(1-a^2)
\end{align*}

Therefore if $|a| \geq 1$ we must have $\Delta_c \leq 0$ which means $\Delta \geq 0$ and so there are always solutions. 

If $|a| < 1$ there are values for $c$ where $\Delta < 0$ and there would be no solutions.

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

therefore the oblique asymptote has equation $y = x - (2+a)$

\begin{enumerate}
\item 
\begin{center}
    \begin{tikzpicture}
    \def\a{-1.5};
    \def\functionf(#1){(#1)*((#1)-2)*((#1)-\a)/((#1)^2-1)};
    \def\xl{-10};
    \def\xu{10};
    \def\yl{-10};
    \def\yu{10};
    
    % 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[blue, smooth, thick, domain=\xl:-1.05, samples=101]
            plot({\x}, {\functionf(\x)});
        \draw[blue, smooth, thick, domain=-0.99:0.99, samples=101]
            plot({\x}, {\functionf(\x)});
        \draw[blue, smooth, thick, domain=1.05:\xu, samples=101]
            plot({\x}, {\functionf(\x)});

        \draw[red, dashed, thick] (\xl, {\xl - (2 + \a)}) -- (\xu, {\xu - (2+\a)}) node[pos=0.8, above, sloped] {\tiny $y = x - (2+a)$};
        % \filldraw (0.5, 0) circle (1.5pt) node[below right]  {$a$};
        % \draw[blue, thick] ({\b*\t}, {-\t}) -- ({-\b*\t}, {\t});

        \filldraw (\a, 0) circle (1.5pt) node[below] {$a$};
        \filldraw (2, 0) circle (1.5pt) node[below] {$2$};
        
    \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}

\item 
\begin{center}
    \begin{tikzpicture}
    \def\a{-0.8};
    \def\functionf(#1){(#1)*((#1)-2)*((#1)-\a)/((#1)^2-1)};
    \def\xl{-5};
    \def\xu{5};
    \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[blue, smooth, thick, domain=\xl:-1.05, samples=101]
            plot({\x}, {\functionf(\x)});
        \draw[blue, smooth, thick, domain=-0.99:0.99, samples=101]
            plot({\x}, {\functionf(\x)});
        \draw[blue, smooth, thick, domain=1.05:\xu, samples=101]
            plot({\x}, {\functionf(\x)});

        \draw[red, dashed, thick] (\xl, {\xl - (2 + \a)}) -- (\xu, {\xu - (2+\a)}) node[pos=0.8, above, sloped] {\tiny $y = x - (2+a)$};
        % \filldraw (0.5, 0) circle (1.5pt) node[below right]  {$a$};
        % \draw[blue, thick] ({\b*\t}, {-\t}) -- ({-\b*\t}, {\t});

        \filldraw (\a, 0) circle (1.5pt) node[below] {$a$};
        \filldraw (2, 0) circle (1.5pt) node[below] {$2$};

        \draw[red, dashed, thick] (1, \yl) -- (1, \yu) node[pos=0.8, below, sloped] {\tiny $x=1$};
        \draw[red, dashed, thick] (-1, \yl) -- (-1, \yu) node[pos=0.8, above, sloped] {\tiny $x=-1$};
    \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}
\end{enumerate}