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\lhead{\footnotesize\color{ibblue}\bfseries Modelling with Quadratic Functions}
\rhead{\footnotesize\color{accent}IBDP Prep $\cdot$ Sheet 2}
\cfoot{\footnotesize\thepage}
\rfoot{\footnotesize J. Murphy}

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\begin{document}

\begin{center}
  {\color{ibblue}\Large\bfseries Modelling with Quadratic Functions}\\[0pt]
  {\color{accent}\footnotesize Completing the square \,$\cdot$\, vertex form \,$\cdot$\, roots \,$\cdot$\, model interpretation \& limitations}
\end{center}
\vspace{-2pt}
\small

% ===================== Q1 =====================
\Q{Q1}{7}
\begin{minipage}[t]{0.60\linewidth}
\vspace{-2pt}
An events company organises outdoor music concerts. The weekly revenue is modelled by
\[ R = 180 - 7.2(p-10)^2 \]
where $R$ is the revenue in thousands of Swiss francs and $p$ is the ticket price in Swiss francs. A sketch of $R$ against $p$ is shown.\\[3pt]
Using the model,
\begin{enumerate}[label=(\alph*)]
  \item explain why \textup{CHF}\,20 is not a sensible ticket price.\m{2}
\end{enumerate}
Given that the company needs a weekly revenue of more than \textup{CHF}\,150\,000,
\begin{enumerate}[label=(\alph*),start=2]
  \item find, according to the model, the least possible ticket price.\m{3}
\end{enumerate}
The company wishes to maximise its weekly revenue. State, according to the model,
\begin{enumerate}[label=(\alph*),start=3]
  \item[(c)] (i) the maximum possible weekly revenue;\\
             (ii) the ticket price that maximises the weekly revenue.\m{2}
\end{enumerate}
\end{minipage}\hfill
\begin{minipage}[t]{0.37\linewidth}
\vspace{0pt}
\centering
\begin{tikzpicture}[scale=0.82]
  \draw[-{Stealth[length=2mm]}] (-0.2,0) -- (5.6,0) node[right] {$p$};
  \draw[-{Stealth[length=2mm]}] (0,-0.8) -- (0,3.5) node[above] {$R$};
  \node[below left] at (0,0) {$O$};
  \draw[ibblue,thick,smooth,domain=1.06:4.94,samples=80]
        plot (\x,{3 - 0.8*(\x-3)^2});
  \node[ibblue,anchor=west,font=\scriptsize] at (1.9,3.25) {$R=180-7.2(p-10)^2$};
\end{tikzpicture}\\[2pt]
{\footnotesize\itshape Figure 1}
\end{minipage}

% ===================== Q2 =====================
\Q{Q2}{6}
A solar farm near Lausanne began generating electricity on 1st January 2015. A model for the total energy generated is
\[ E = 850 - 4(y-15)^2 \]
where $E$ MWh is the total energy generated in the $y$ years since the farm opened. Using this model,
\begin{enumerate}[label=(\alph*)]
  \item calculate the total energy generated up to 1st January 2016;\m{1}
  \item deduce the maximum total energy that could be generated;\m{1}
  \item calculate the energy generated during 2017;\m{2}
  \item state, giving reasons, the limitation on the values of $y$.\m{2}
\end{enumerate}

% ===================== Q3 =====================
\Q{Q3}{9}
A drone is launched from the roof of a building. The height $h$, in metres, of the drone above ground level after $t$ seconds is modelled by
\[ h(t) = 80 + 18t - 4.5t^2. \]
\begin{enumerate}[label=(\alph*)]
  \item Give a physical interpretation of the meaning of the constant term $80$ in the model.\m{1}
  \item Write $h(t)$ in the form $A - B(t-C)^2$, where $A$, $B$ and $C$ are constants to be found.\m{3}
  \item Using your answer to part (b), or otherwise, find, with justification,
  \begin{enumerate}[label=(\roman*)]
    \item the time taken after launch for the drone to reach ground level;\m{3}
    \item the maximum height of the drone above the ground and the time at which this maximum height is reached.\m{2}
  \end{enumerate}
\end{enumerate}

% ===================== Q4 =====================
\Q{Q4}{6}
\begin{minipage}[t]{0.58\linewidth}
\vspace{-2pt}
Mia and Lars are analysing water-level data from a tidal gauge. The table shows their measurements of water depth $y$ (metres) at time $x$ (hours).\\[3pt]
\renewcommand{\arraystretch}{1.15}
\begin{tabular}{|c|c|c|c|c|}
\hline
\rowcolor{ibblue!12}$x$ & $0$ & $2$ & $3$ & $6$\\\hline
$y$ & $12$ & $0$ & $1.5$ & $0$\\\hline
\end{tabular}\\[4pt]
Lars proposes the model $y = k(x-2)(x-6)$, shown in Fig.~10.
\begin{enumerate}[label=(\roman*)]
  \item Find the value of $k$ for which the points $(0,12)$, $(2,0)$ and $(6,0)$ satisfy this equation.\m{2}
\end{enumerate}
Mia proposes a different model, using $y = c(x-2)^2(x-6)$.
\begin{enumerate}[label=(\roman*),start=2]
  \item Find the value of $c$ for which the points $(0,12)$, $(2,0)$ and $(6,0)$ satisfy this equation.\m{2}
  \item Which is the better model for Mia and Lars's results? Give a reason for your answer.\m{2}
\end{enumerate}
\end{minipage}\hfill
\begin{minipage}[t]{0.39\linewidth}
\vspace{0pt}
\centering
\begin{tikzpicture}
\begin{axis}[
  width=5.6cm, height=4.7cm,
  axis lines=middle,
  xmin=-0.4, xmax=8.0, ymin=-4.0, ymax=14.0,
  xtick={1,2,3,4,5,6,7}, ytick={-3,-2,-1,1,2,3,4,5,6,7,8,9,10,11,12,13},
  tick label style={font=\scriptsize},
  xlabel={$x$}, ylabel={$y$},
  xlabel style={right}, ylabel style={above},
  clip=false]
  \addplot[ibblue,thick,smooth,domain=0:7.4,samples=80]
        {(x-2)*(x-6)};
\end{axis}
\end{tikzpicture}\\[1pt]
{\footnotesize\itshape Fig.~10}
\end{minipage}

% ===================== Q5 =====================
\Q{Q5}{7}
\begin{minipage}[t]{0.60\linewidth}
\vspace{-2pt}
Figure 2 shows the trajectory of a ski jumper. The height of the jumper above the ground, $H$ metres, is plotted against the horizontal distance, $x$ metres, measured from the end of the ramp. The jumper travels in a vertical plane.\\[2pt]
The jumper reaches a maximum height of $18$ metres above the ground and lands at a point $50$ metres horizontally from the end of the ramp.
\begin{enumerate}[label=(\alph*)]
  \item Find a quadratic equation linking $H$ with $x$ that models this situation.\m{3}
\end{enumerate}
The trajectory passes over a timing gate, which is a horizontal bar perpendicular to the direction of travel. The bar is $6$ metres above the ground.
\begin{enumerate}[label=(\alph*),start=2]
  \item Use your equation to find the greatest horizontal distance of the bar from the end of the ramp.\m{3}
  \item Give one limitation of the model.\m{1}
\end{enumerate}
\end{minipage}\hfill
\begin{minipage}[t]{0.37\linewidth}
\vspace{0pt}
\centering
\begin{tikzpicture}[scale=0.8]
  \draw[-{Stealth[length=2mm]}] (-0.2,0) -- (5.6,0) node[right] {$x$};
  \draw[-{Stealth[length=2mm]}] (0,-0.2) -- (0,3.3) node[above] {$H$};
  \node[below left] at (0,0) {$O$};
  \draw[ibblue,thick,smooth,domain=0:4.72,samples=80]
        plot (\x,{-0.5817*(\x-2.36)^2+3.24});
\end{tikzpicture}\\[2pt]
{\footnotesize\itshape Figure 2}
\end{minipage}

\end{document}