2004 Paper 2 Q3

Year: 2004
Paper: 2
Question Number: 3

Course: LFM Stats And Pure
Section: Curve Sketching

Difficulty: 1600.0 Banger: 1600.7

Problem

The curve \(C\) has equation $$ y = x(x+1)(x-2)^4. $$ Determine the coordinates of all the stationary points of \(C\) and the nature of each. Sketch \(C\). In separate diagrams draw sketches of the curves whose equations are:
  1. \( y^2 = x(x+1)(x-2)^4\;\);
  2. \(y = x^2(x^2+1)(x^2-2)^4\,\).

Solution

\begin{align*} && y &= x(x+1)(x-2)^4 \\ \Rightarrow && y' &= (x+1)(x-2)^4+x(x-2)^4+4x(x+1)(x-2)^3 \\ &&&= (x-2)^3 \left ( (2x+1)(x-2)+4x(x+1) \right) \\ &&&= (x-2)^3 \left (2x^2-3x-2+4x^2+4x \right) \\ &&&=(x-2)^3(6x^2+x-2) \\ &&&=(x-2)^3(2x-1)(3x+2) \end{align*} Therefore there are stationary points at \((2,0), (\frac12, -\frac{625}{64}), (-\frac23, -\frac{4078}{81})\) \((0,2)\) is a minimum by considering the sign of \(y'\) either side. \( (-\frac23, \frac{2560}{729})\) is a minimum, since it's the first stationary point. \( (\frac12, \frac{243}{64})\) is a maximum since you can't have consecutive minima and the second derivative is clearly non-zero.
TikZ diagram
  1. \(\,\)
    TikZ diagram
  2. \(\,\)
    TikZ diagram
Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1600.7

Banger Comparisons: 10

Show LaTeX source
Problem source
The curve $C$ has equation
$$
y = x(x+1)(x-2)^4.
$$
Determine the coordinates of all the stationary points of $C$ and the nature of each. Sketch  $C$.
In  separate diagrams draw  sketches of the curves whose equations are: 
\begin{questionparts}
\item $ y^2 = x(x+1)(x-2)^4\;$;
\item $y = x^2(x^2+1)(x^2-2)^4\,$. 
\end{questionparts}
Solution source
\begin{align*}
&& y &= x(x+1)(x-2)^4 \\
\Rightarrow && y' &= (x+1)(x-2)^4+x(x-2)^4+4x(x+1)(x-2)^3 \\
&&&= (x-2)^3 \left ( (2x+1)(x-2)+4x(x+1) \right) \\
&&&= (x-2)^3 \left (2x^2-3x-2+4x^2+4x \right) \\
&&&=(x-2)^3(6x^2+x-2) \\
&&&=(x-2)^3(2x-1)(3x+2)
\end{align*}

Therefore there are stationary points at $(2,0), (\frac12, -\frac{625}{64}), (-\frac23, -\frac{4078}{81})$

$(0,2)$ is a minimum by considering the sign of $y'$ either side.
$ (-\frac23, \frac{2560}{729})$ is a minimum, since it's the first stationary point. 
$ (\frac12, \frac{243}{64})$ is a maximum since you can't have consecutive minima and the second derivative is clearly non-zero.


\begin{center}
    \begin{tikzpicture}
    \def\a{-0.8};
    \def\functionf(#1){(#1)*((#1)+1)*((#1)-2)^4};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-15};
    \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[blue, smooth, thick, domain=-2:3, samples=101]
            plot({\x}, {\functionf(\x)}); 
        % \draw[red, smooth, thick, domain=1:2, samples=101]
            % plot({\x}, {1+3-1}); 
        % \draw[red, smooth, thick, domain=2:3, samples=101]
            % plot({\x}, {4+6-1}); 
        
        \node[blue, above, rotate=-87.5] at (-1.1, {\functionf(-1.1)}) {\tiny $y=x(x+1)(x-2)^4$};

        \filldraw (2, 0) circle (1.5pt) node[below] {$2$};
        \filldraw (-1, 0) circle (1.5pt) node[below] {$-1$};
        \filldraw ({-2/3}, {\functionf(-2/3)}) circle (1.5pt) node[below] {$(-\frac23, -\frac{4078}{81})$};
        \filldraw ({1/2}, {243/64}) circle (1.5pt) node[above] {$(\frac12,\frac{243}{64})$};
    \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 $\,$ 
\begin{center}
    \begin{tikzpicture}
    \def\a{-0.8};
    \def\functionf(#1){(#1)*((#1)+1)*((#1)-2)^4};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-15};
    \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[blue, smooth, thick, domain=-2:-1, samples=101]
            plot({\x}, {sqrt(\functionf(\x))}); 
        \draw[blue, smooth, thick, domain=-2:-1, samples=101]
            plot({\x}, {-sqrt(\functionf(\x))}); 
        \draw[blue, smooth, thick, domain=0:\xu, samples=101]
            plot({\x}, {sqrt(\functionf(\x))}); 
        \draw[blue, smooth, thick, domain=0:\xu, samples=101]
            plot({\x}, {-sqrt(\functionf(\x))}); 
        % \draw[red, smooth, thick, domain=1:2, samples=101]
            % plot({\x}, {1+3-1}); 
        % \draw[red, smooth, thick, domain=2:3, samples=101]
            % plot({\x}, {4+6-1}); 
        
        \node[blue, above, rotate=-80.5] at (-1.5, {sqrt(\functionf(-1.5))}) {\tiny $y^2=x(x+1)(x-2)^4$};

        \filldraw (2, 0) circle (1.5pt) node[below] {$2$};
        \filldraw (-1, 0) circle (1.5pt) node[below] {$-1$};
        % \filldraw ({-2/3}, {\functionf(-2/3)}) circle (1.5pt) node[below] {$(-\frac23, -\frac{4078}{81})$};
        \filldraw ({1/2}, {sqrt(243/64)}) circle (1.5pt) node[above] {$(\frac12,\frac{9\sqrt{3}}{8})$};
        \filldraw ({1/2}, {-sqrt(243/64)}) circle (1.5pt) node[below] {$(\frac12,-\frac{9\sqrt{3}}{8})$};
    \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)+1)*((#1)-2)^4};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-15};
    \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[blue, smooth, thick, domain=-2:2, samples=101]
            plot({\x}, {\functionf(\x*\x)}); 
        % \draw[red, smooth, thick, domain=1:2, samples=101]
            % plot({\x}, {1+3-1}); 
        % \draw[red, smooth, thick, domain=2:3, samples=101]
            % plot({\x}, {4+6-1}); 
        
        \node[blue, above, rotate=-85] at (-1.65, {\functionf(1.73*1.73))}) {\tiny $y=x^2(x^2+1)(x^2-2)^4$};

        \filldraw ({sqrt(2)}, 0) circle (1.5pt) node[below] {\tiny $\sqrt{2}$};
        \filldraw ({-sqrt(2)}, 0) circle (1.5pt) node[below] {\tiny $-\sqrt{2}$};
        % \filldraw (-1, 0) circle (1.5pt) node[below] {$-1$};
        % \filldraw ({-2/3}, {\functionf(-2/3)}) circle (1.5pt) node[below] {$(-\frac23, -\frac{4078}{81})$};
        \filldraw ({1/sqrt(2)}, {243/64}) circle (1.5pt) node[above] {\tiny $(\frac1{\sqrt{2}},\frac{243}{64})$};
        \filldraw ({-1/sqrt(2)}, {243/64}) circle (1.5pt) node[above] {\tiny $(-\frac1{\sqrt{2}},\frac{243}{64})$};
    \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}