2004 Paper 3 Q2

Year: 2004
Paper: 3
Question Number: 2

Course: LFM Stats And Pure
Section: Curve Sketching

Difficulty: 1700.0 Banger: 1516.0

Problem

The equation of a curve is \(y=\f ( x )\) where \[ \f ( x ) = x-4-\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r} \;. \]
  1. Write down the equations of the vertical and oblique asymptotes to the curve and show that the oblique asymptote is a tangent to the curve.
  2. Show that the equation \(\f ( x ) =0\) has a double root.
  3. Sketch the curve.

Solution

  1. Clearly \(x = 0\) and \(x = 4\) are vertical asymptotes. Notice that \(\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}\) tends to \(0\) as \(x \to \infty\). Therefore the oblique asymptote is \(y = x-4\).
  2. \begin{align*} && 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\ &&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\ &&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\ &&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)} \end{align*} Therefore \(f(x) = 0\) has a double root at \(x = -2\). Notice it also has roots at \(6 \pm 2\sqrt{10}\)
  3. TikZ diagram
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1516.0

Banger Comparisons: 1

Show LaTeX source
Problem source
The equation of a curve is $y=\f ( x )$ where
\[
\f ( x ) = x-4-\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r} \;.
\]
\begin{questionparts}
\item
Write down the equations of the vertical and oblique asymptotes to the curve and 
show that the oblique asymptote is a tangent to the curve. 
\item
Show that the equation $\f ( x ) =0$ has a double root.
\item
Sketch the curve.
\end{questionparts}
Solution source
\begin{questionparts}
\item Clearly $x = 0$ and $x = 4$ are vertical asymptotes.

Notice that $\frac{16 \l 2x+1 \r^2}{x^2 \l x - 4 \r}$ tends to $0$ as $x \to \infty$. Therefore the oblique asymptote is $y = x-4$.

\item \begin{align*}
&& 0 &= \frac{x^2(x-4)^2-4^2(2x+1)^2}{x^2(x-4)} \\
&&&= \frac{(x(x-4)-4(2x+1))(x(x-4)+4(2x+1))}{x^2(x-4)} \\
&&&= \frac{(x^2-12x-4)(x^2+4x+4)}{x^2(x-4)}\\
&&&= \frac{(x^2-12x-4)(x+2)^2}{x^2(x-4)}
\end{align*}
Therefore $f(x) = 0$ has a double root at $x = -2$. Notice it also has roots at $6 \pm 2\sqrt{10}$

\item 

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){(#1)-4-(16*(2*(#1)+1)^2)/((#1)^2*((#1)-4))};
    \def\xl{-30};
    \def\xu{35};
    \def\yl{-40};
    \def\yu{100};
    \def\functiong(#1){(#1)-4-(16*(2+1/(#1))^2)/(((#1)-4))};
    
    % 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=-2:-0.1, samples=200] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=\xl:-2, samples=200] 
            plot (\x, {\functiong(\x)});
        
        \draw[thick, blue, smooth, domain=0.1:3.9, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, blue, smooth, domain=4.1:\xu, samples=200] 
            plot (\x, {\functiong(\x)});
        \draw[thick, red, dashed] (0, \yl) -- (0, \yu);
        \draw[thick, red, dashed] (4, \yl) -- (4, \yu);
        \draw[thick, red, dashed] (\xl, {\xl-4}) -- ({\xu}, {\xu-4});
    \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{questionparts}