%\documentclass[mathserif]{beamer}
\documentclass[mathserif, handout]{beamer}

\input{/home/janine/Dropbox/Travail_S/Divers/Preambule/Preambule_Cours}


\def\mySample{100}

\begin{document}

\titreCours{8}{Fonctions Quadratiques}{}{}

\begin{frame}{1. Fonctions quadratiques et paraboles}

Une fonction quadratique est une fonction de la forme $$\boxed{f(x)=\bl{a}x^2 + \ve{b}x + \ro{c}}$$ avec $\bl{a}\neq0$. \uncover<2->{La courbe représentative d'une fonction quadratique est une \textbf{parabole}.} \\

\begin{minipage}{0.49\textwidth}
\uncover<3->{
\begin{center}
\begin{tikzpicture}[scale=0.25]

\coordinate (O) at (0,0);
\def\max{8};
\def\min{-8};
\def\qmax{7};
\def\qmin{-7};

\draw[gray, thin] (-7.9, -7.9) grid (7.9,7.9);

\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 


\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};


\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample, violet] plot(\x,{(\x)^2-2*(\x)-3});
\end{tikzpicture}

\hunc{4}{Si $\bl{a}>0$ la parabole\\ est \textbf{convexe}}}
\end{center}
\end{minipage} \begin{minipage}{0.49\textwidth}

\begin{center}
\uncover<5->{
\begin{tikzpicture}[scale=0.25]

\coordinate (O) at (0,0);
\def\max{8};
\def\min{-8};
\def\qmax{7};
\def\qmin{-7};

\draw[gray, thin] (-7.9, -7.9) grid (7.9,7.9);

\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};


\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample, violet] plot(\x,{-(\x)^2+2*(\x)+3});

\end{tikzpicture}

\hunc{6}{Si $\bl{a}<0$ la parabole\\ est \textbf{concave}}}
\end{center}

\end{minipage}
\end{frame}

\begin{frame}

\underline{Exercice 1.1} Les paraboles suivantes sont-elle convexes ou concaves?


\begin{minipage}{0.49\textwidth}

\begin{center}
\begin{tikzpicture}[scale=0.2]

\coordinate (O) at (0,0);
\def\max{8};
\def\min{-8};
\def\qmax{7};
\def\qmin{-7};
\draw[gray, thin] (-7.9, -7.9) grid (7.9,7.9);
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};

\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample, violet] plot(\x,{-0.3*(\x)^2+ (\x)+3});

\end{tikzpicture}

\hunc{2}{\bl{Concave}}
\end{center}

\end{minipage} \begin{minipage}{0.49\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.2]

\coordinate (O) at (0,0);
\def\max{8};
\def\min{-8};
\def\qmax{7};
\def\qmin{-7};

\draw[gray, thin] (-7.9, -7.9) grid (7.9,7.9);

\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};

\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample, violet] plot(\x,{(\x)^2-6*(\x)-2});
\end{tikzpicture}

\hunc{3}{\bl{Convexe}}
\end{center}
\end{minipage}

\vspace{0.2cm}
\uncover<4->{
\underline{Exercice 1.2} Les fonctions suivantes sont-elles quadratiques? Si oui, la parabole correspondante est-elle convexe ou concave?
\vspace{-0.2cm}
\begin{multicols}{2}
\begin{enumerate}[a)]
\item $-2x^2 + 5x - 21$\\
\hunc{5}{\bl{Oui / Concave ($a = -2 < 0$)}}
\item $4 -x^2$\\
\hunc{6}{\bl{Oui / Concave ($a = -1 < 0$)}}
\item $\dfrac{1}{5x^2 + 2x +1}$ 
\hunc{7}{\ro{Non}}
\item $3x + 1$ 
\hunc{8}{\ro{Non}}
\item $3x + x^2$\\
\hunc{9}{\bl{Oui / Convexe ($a = 1 < 0$)}}
\item $\sqrt{x^2 + 3x -2}$ 
\hunc{10}{\ro{Non}}
\end{enumerate}
\end{multicols}}


\end{frame}


\begin{frame}{2. Méthode du discriminant}

L'ensemble des solutions de l'équation $\bl{a}x^2 + \ve{b}x + \ro{c} = 0$ dépendent de la valeur de $\ora{\Delta}= \ve{b}^2 - 4 \bl{a} \ro{c}$ (le discriminant):

\uncover<2->{
\begin{minipage}{0.7\textwidth}
(1) Si $\boxed{\ora{\Delta>0}}$, il y a deux solutions: 
$$\vi{x_1 = \frac{-\ve{b} + \sqrt{\ora{\Delta}}}{2\bl{a}}} \mbox{ et } 
\vi{x_2 = \frac{-\ve{b} - \sqrt{\ora{\Delta}}}{2\bl{a}}}$$
\end{minipage}\begin{minipage}{0.28\textwidth}

\uncover<3->{
\begin{tikzpicture}[scale=0.17]
\coordinate (O) at (0,0);
\def\max{6};
\def\min{-4};
\def\qmax{5};
\def\qmin{-3};

\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\uncover<4->{
\fill[violet] (3,0) circle (0.5cm) node[below right]{};
\fill[violet] (-1,0) circle (0.5cm) node[below left]{};}

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};


\clip (\min,\min-0.5) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x)^2-2*(\x)-3});

\end{tikzpicture}}
\end{minipage}}

\uncover<5->{
\begin{minipage}{0.7\textwidth}
(2) Si $\boxed{\ora{\Delta=0}}$, il y a une seule solution: 
$$\vi{x_1 = \frac{-\ve{b}}{2\bl{a}}}$$
\end{minipage}\begin{minipage}{0.28\textwidth}
\uncover<6->{
\begin{tikzpicture}[scale=0.17]
\coordinate (O) at (0,0);
\def\max{6};
\def\min{-4};
\def\qmax{5};
\def\qmin{-3};

\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};

\uncover<7->{\fill[violet] (1,0) circle (0.5cm) node[below]{};}

\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x)^2-2*(\x)+1});

\end{tikzpicture}}
\end{minipage}}

\uncover<8->{
\begin{minipage}{0.7\textwidth}
(3) Si $\boxed{\ora{\Delta<0}}$, il n'y a pas solutions.\\[0.5cm]
\end{minipage}\begin{minipage}{0.28\textwidth}
\uncover<9->{
\begin{tikzpicture}[scale=0.17]
\coordinate (O) at (0,0);
\def\max{6};
\def\min{-4};
\def\qmax{5};
\def\qmin{-3};

\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};


\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x)^2-2*(\x)+3});

\end{tikzpicture}}
\end{minipage}}

\end{frame}

\begin{frame}

\underline{Exemple 2.1} Résoudre l'équation suivante 
$$ \bl{3}x^2 \ve{+ 6}x \ro{- 24} = 0$$

\hunc{2}{
\uncover<2->{On a donc $\bl{a=\uncover<3->{3}}$, $\ve{b=\uncover<4->{6}}$ et $\ro{c=\uncover<5->{-24}}$.}

\uncover<6->{ On calcule 
$\ora{\Delta}= \ve{b}^2 - 4 \bl{a} \ro{c}
\uncover<7->{=(\uncover<8->{\ve{6}})^2 - 4\cdot(\uncover<9->{\bl{3}})\cdot(\uncover<10->{\ro{-24}})} 
\uncover<11->{= 36 - (-288) }\uncover<12->{= \ora{324}}$}

\uncover<13->{$\ora{\Delta} >0 \Rightarrow$\uncover<14->{ Il y a deux solutions:}}

\uncover<15->{$$\vi{x_1} = \frac{-\ve{b} + \sqrt{\ora{\Delta}}}{2\bl{a}}
\uncover<16->{= \frac{-(\uncover<17->{\ve{6}}) + \sqrt{\uncover<18->{\ora{324}}}}{2\cdot\uncover<19->{\bl{3}}}}
\uncover<20->{= \frac{-\ve{6} + 18}{6} }
\uncover<21->{= \frac{12}{6}}
\uncover<22->{= 2}
$$}

\uncover<23->{$$\vi{x_2} = \frac{-\ve{b} - \sqrt{\ora{\Delta}}}{2\bl{a}}
\uncover<24->{= \frac{-(\uncover<25->{\ve{6}}) - \sqrt{\uncover<26->{\ora{324}}}}{2\cdot\uncover<27->{\bl{3}}}}
\uncover<28->{= \frac{-\ve{6} - 18}{6} }
\uncover<29->{= \frac{-24}{6}}
\uncover<30->{= -4}
$$}

\uncover<31->{On a donc $S=\{\vi{-4};\vi{2}\}$}}

\end{frame}

\begin{frame}
\underline{Exercice 2.1} Résoudre l'équation suivante $$2x^2 + 8x = -8$$

\hunc{2}{
On met l'équation sous la forme $ax^2 + bx + c = 0$:
\begin{equa}
2x^2 + 8x &=& -8 & \hunc{3}{+8}\\
\hunc{4}{\LRq 2x^2 + 8x + 8 &=& 0} \uzero
\end{equa}
\hunc{5}{ On a donc $\bl{a=\hunc{6}{2}}$, $\ve{b=\hunc{7}{8}}$ et $\ro{c=\hunc{8}{8}}$.}\uncover<9->{ On calcule 
$\ora{\Delta}= \ve{b}^2 - 4 \bl{a} \ro{c}
\uncover<10->{=(\uncover<11->{\ve{8}})^2 - 4\cdot(\uncover<12->{\bl{2}})\cdot(\uncover<13->{\ro{8}})} 
\uncover<14->{= 64 - 64 }\uncover<15->{= \ora{0}}$}

\uncover<16->{$\ora{\Delta} = 0 \Rightarrow$\uncover<17->{ Il y a une solution:}}
\uncover<18->{$$\vi{x_1} = \frac{-\ve{b}}{2\bl{a}}
\uncover<19->{= \frac{-(\uncover<20->{\ve{8}})}{2\cdot\uncover<21->{\bl{4}}}}
\uncover<22->{= \frac{-\ve{8}}{8} }
\uncover<23->{= 1}
$$}
\uncover<24->{ On a donc $S=\{1\}$.} 
 }
 
\vfill 
\end{frame}

\begin{frame}

\underline{Exercice 2.2} Résoudre l'équation $$5x^2+ 7 =0$$

\hunc{2}{
On a $\bl{a=\hunc{3}{5}}$, $\ve{b=\hunc{4}{0}}$ et $\ro{c=\hunc{5}{7}}$. \uncover<6->{ On calcule 
$\ora{\Delta}= \ve{b}^2 - 4 \bl{a} \ro{c}
\uncover<7->{=(\uncover<8->{\ve{0}})^2 - 4\cdot(\uncover<9->{\bl{5}})\cdot(\uncover<10->{\ro{7}})} 
\uncover<11->{= 0 - 140 }\uncover<12->{= \ora{-140}}$}

\uncover<13->{$\ora{\Delta} = -140 \Rightarrow$\uncover<14->{ Il n'y a pas de solution: $S=\emptyset$.}}  

}

\vfill

\vspace{5cm}

\end{frame}

\begin{frame}{3. Points caractéristiques} 
Soit la parabole d'équation $y=x^2 + 4x - 5$. Ses points caractéristiques sont les suivants.
\begin{center}
\begin{tikzpicture}[scale=0.25]
\footnotesize
\coordinate (O) at (0,0);
\def\max{10};
\def\min{-10};
\def\qmax{9};
\def\qmin{-9};
\foreach \x in {\qmin,..., \qmax}{
\draw[black!60] (\x, \min) -- (\x, \max);}
\foreach \y in{\qmin,..., \qmax}{
\draw[black!60]  (\min , \y) -- (\max, \y );}

\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};

\begin{scope}
\clip (\min,\min) rectangle (\max,\max);
\uncover<1-9>{\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x+5)*(\x-1)});}
\uncover<11->{\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x+5)*(\x-1)});}
\end{scope}
\hunc{2}{
\uncover<9->{\draw[violet, thick] (-2, \min) -- (-2,\max); \node[left, violet, fill=white] at (-2.1,8) {Axe de symétrie $x=-2$};}
\uncover<2->{\fill[dblue] (-2,-9) circle (0.2cm) node[below left, fill=white]{Sommet $\mathcal{S}\only<3->{(-2,-9)}$};}
\uncover<3->{
\draw[thick, dashed, dblue] (0,-9) node[right]{-9} -- (-2,-9);
\draw[thick, dashed, dblue] (-2,0) node[above]{-2} -- (-2,-9);}
\uncover<4->{\fill[dgreen] (0,-5) circle (0.2cm) node[below right, fill=white]{Ordonnée à l'origine $\mathcal{H}\only<5->{(0,-5)}$};}
\uncover<6->{\fill[dred] (1,0) circle (0.2cm) node[below right, fill=white]{Zéro $\mathcal{Z}_1\only<7->{(1,0)}$};}
\uncover<6->{\fill[dred] (-5,0) circle (0.2cm) node[below left, fill=white]{Zéro $\mathcal{Z}_2\only<8->{(-5,0)}$};}}
\end{tikzpicture}
\end{center}

\hunc{12}{Une fonction quadratique a toujours un sommet et une ordonnée à l'origine; elle peut avoir 0, 1 ou 2 zéros.}
\end{frame}

\begin{frame}
Soit $y=\bl{a}x^2 + \ve{b}x + \ro{c}$ l'équation d'une parabole. 

\vfill
\uncover<2->{
\underline{Coordonnées du sommet} $\boxed{\mathcal{S}=\left(-\frac{\ve{b}}{2\bl{a}}; -\frac{\ora{\Delta}}{4\bl{a}}\right)}$
avec $\ora{\Delta} = \ve{b}^2 - 4\bl{a}\ro{c}$.}

\vfill
\uncover<3->{
\underline{Equation de l'axe de symétrie} $\boxed{x= -\frac{\ve{b}}{2\bl{a}}}$ (droite verticale passant par le sommet)} 

\vfill
\uncover<4->{
\underline{Exemple 3.1} Soit la parabole d'équation $y=-\dfrac{1}{2}x^2 - x + 4$. Calculer les coordonnées du sommet et l'équation de l'axe de symétrie. 

\vfill

\hunc{5}{On a $\bl{a}= \hunc{6}{\bl{-\frac{1}{2}}}$, $\ve{b}=\hunc{7}{\ve{-1}}$ et $\ro{c}=\hunc{8}{\ro{4}}$.}\\[0.2cm]

\hunc{9}{ On calcule: 
$\ora{\Delta} = \hunc{10}{(\ve{-1})^2 - 4 \cdot \left(\bl{-\frac{1}{2}}\right) \cdot \ro{4} =} \hunc{11}{1 + 8}\hunc{12}{=\ora{9}}$}\\[0.2cm]

\hunc{13}{On remplace:
$
\mathcal{S}=
\hunc{14}{\left(\hunc{15}{-\frac{\ve{-1}}{2\cdot\left(\bl{-\frac{1}{2}}\right)}}; \hunc{16}{-\frac{\ora{9}}{4\cdot\left(\bl{-\frac{1}{2}}\right)}} \right)}
\hunc{17}{=\left(-1, \frac{9}{2}\right)}
$\\[0.2cm]
} 

\hunc{18}{L'équation de l'axe de symétrie est donc $x=-1$. }

}
\end{frame}


\begin{frame}

\underline{Exemple 3.2} Soit $f(x)=\bl{a}x^2 + \ve{b}x + \ro{c}$ une fonction quadratique. Calculer $f(0)$.

\hunc{2}{$$\hunc{2}{f(0) =  \uncover<3->{\bl{a}\cdot 0^2 + \ve{b}\cdot 0 + \ro{c}}\uncover<4->{=\ro{c}}}$$

\uncover<5->{Le point $\mathcal{H}(0;\ro{c})$ fait donc partie du graphe de la fonction (i.e, de la parabole).}
\uncover<6->{On appelle ce point l'ordonnée à l'origine car il correspond à la valeur de l'ordonnée ($y$) lorsque $x=0$.}} \\[0.2cm]

\vfill

\uncover<7->{\underline{Ordonnée à l'origine} $\boxed{\mathcal{H}= (0;\ro{c})}$}\\

\vfill

\uncover<8->{
\underline{Exemple 3.1 (suite)} Calculer l'ordonnée à l'origine de la parabole d'équation 
$y=-\frac{1}{2}x^2 - x + 4$. Placer ce point ainsi que le sommet et l'axe de symétrie sur le graphique. }\\[0.2cm]

\hunc{9}{On a $\mathcal{H}=(0;\ro{c}) \hunc{10}{= (0;4)}$.}

\end{frame}


\begin{frame}

\begin{center}
\begin{tikzpicture}[scale=0.8]
\footnotesize
\coordinate (O) at (0,0);
\def\max{5};
\def\min{-5};
\def\qmax{4.9};
\def\qmin{-4.9};

\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[very thick] (-0.1, 1) -- (0.1, 1); 
\draw[very thick] (1,-0.1) -- (1, 0.1); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, thick] (90:\min) -- (90:\max) node[above]{y};

\begin{scope}
\clip (\min,\min) rectangle (\max,\max);
\hunc{5}{\draw[samples=\mySample, domain=\min:\max, dorange, very thick] plot(\x, {-0.5*((\x)^2) -(\x) + 4});} 
\draw[gray] (\qmin,\qmin) grid[step=1] (\qmax,\qmax);
\end{scope}

\small
\hunc{2}{\fill[dblue] (-1,4.5) circle (3pt) node[above left]{ Sommet $\mathcal{S}\left(-1;\frac{9}{2}\right)$};}
\hunc{3}{\draw[thick, dblue] (-1, \min) -- (-1, \max); \node[right, dblue] at (-1, -3.75) {Axe de symétrie $x=-1$};}
\hunc{4}{\fill[dgreen] (0,4) circle (3pt) node[above right]{Ordonnée à l'origine  $\mathcal{H}\left(0;4\right)$};}

\end{tikzpicture}
\end{center}

\end{frame}

\begin{frame}

Soit $f(x)=\bl{a}x^2 + \ve{b}x + \ro{c}$. Les zéros de la fonction $f(x)$ correspondent aux solutions de l'équation $\bl{a}x^2 + \ve{b}x + \ro{c} = 0$.\\[0.2cm]

\uncover<2->{
\underline{Zéros} 
\begin{minipage}{0.7\textwidth}
(1) Si $\boxed{\ora{\Delta>0}}$, il y a deux intersections: 
$$\vi{Z_1\left(\uncover<3->{\frac{-\ve{b} + \sqrt{\ora{\Delta}}}{2\bl{a}} ; 0}\right)} \mbox{ et } 
\vi{Z_2\left(\uncover<4->{\frac{-\ve{b} - \sqrt{\ora{\Delta}}}{2\bl{a}}; 0}\right)}$$
\end{minipage}\begin{minipage}{0.28\textwidth}

\begin{tikzpicture}[scale=0.17]
\footnotesize
\coordinate (O) at (0,0);
\def\max{6};
\def\min{-4};
\def\qmax{5};
\def\qmin{-3};

\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 


\fill[violet] (3,0) circle (0.5cm) node[below right]{$Z_1\only<4->{(x_1;0)}$};
\fill[violet] (-1,0) circle (0.5cm) node[below left]{$Z_2\only<4->{(x_2;0)}$};

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};


\clip (\min,\min-0.5) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x)^2-2*(\x)-3});

\end{tikzpicture}
\end{minipage}}

\uncover<5->{
\begin{minipage}{0.7\textwidth}
(2) Si $\boxed{\ora{\Delta=0}}$, il y a une seule intersection: 
$$\uncover<6->{\vi{Z_1\left(\frac{-\ve{b}}{2\bl{a}};0\right)}}$$
\end{minipage}\begin{minipage}{0.28\textwidth}
\begin{tikzpicture}[scale=0.17]
\footnotesize
\coordinate (O) at (0,0);
\def\max{6};
\def\min{-4};
\def\qmax{5};
\def\qmin{-3};

\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};

\uncover<7->{\fill[violet] (1,0) circle (0.5cm) node[below]{$Z_1(x_1;0)$};}

\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x)^2-2*(\x)+1});

\end{tikzpicture}}
\end{minipage}

\uncover<8->{
\begin{minipage}{0.7\textwidth}
(3) Si $\boxed{\ora{\Delta<0}}$, il n'y a pas intersections.\\[0.5cm]
\end{minipage}\begin{minipage}{0.28\textwidth}
\uncover<9->{
\begin{tikzpicture}[scale=0.17]
\coordinate (O) at (0,0);
\def\max{6};
\def\min{-4};
\def\qmax{5};
\def\qmin{-3};

\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, very thick] (90:\min) -- (90:\max) node[above]{y};


\clip (\min,\min) rectangle (\max,\max);
\draw[very thick, domain=\min:\max, samples=\mySample] plot(\x,{(\x)^2-2*(\x)+3});

\end{tikzpicture}}
\end{minipage}}

\end{frame}

\begin{frame}

\underline{Exemple 3.1 (suite)} Calculer les zéros de la fonction $f(x)=-\frac{1}{2}x^2 - x + 4$. Compléter le graphique précédent. 
\vfill

\hunc{2}{On résoud l'équation $-\frac{1}{2}x^2 - x + 4 = 0$.

\uncover<3->{
\begin{equa}
-\frac{1}{2}x^2 - x + 4 &=& 0 & \hunc{4}{\mbox{MEE}}\\
\hunc{5}{\LRq -\frac{1}{2}(x^2 +2x - 8) & = & 0 }  \hunc{6}{& SP}\\
\hunc{7}{\LRq -\frac{1}{2}(x + 4)(x -2) & = & 0 }  \hunc{8}{\uzero \qqR S=\{-4; 2\}}\\
\end{equa} 

\hunc{9}{Il y a deux solutions, il y aura donc deux zéros: 
$$\hunc{10}{Z_1(-4;0)} \hunc{11}{ \mbox{  et  } Z_2(2;0)}$$  }}}

\end{frame}

\only<beamer>{
\begin{frame}

\begin{center}
\begin{tikzpicture}[scale=0.8]
\footnotesize
\coordinate (O) at (0,0);
\def\max{5};
\def\min{-5};
\def\qmax{4.9};
\def\qmin{-4.9};

\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[very thick] (-0.1, 1) -- (0.1, 1); 
\draw[very thick] (1,-0.1) -- (1, 0.1); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, thick] (0:\min) -- (0:\max) node[right]{x};
\draw[-latex, thick] (90:\min) -- (90:\max) node[above]{y};

\begin{scope}
\clip (\min,\min) rectangle (\max,\max);
\hunc{7}{\draw[samples=\mySample, domain=\min:\max, dorange, very thick] plot(\x, {-0.5*((\x)^2) -(\x) + 4});} 
\draw[gray] (\qmin,\qmin) grid[step=1] (\qmax,\qmax);
\end{scope}

\small
\hunc{2}{\fill[dblue] (-1,4.5) circle (3pt) node[above left]{ Sommet $\mathcal{S}\left(-1;\frac{9}{2}\right)$};}
\hunc{3}{\draw[thick, dblue] (-1, \min) -- (-1, \max); \node[right, dblue] at (-1, -3.75) {Axe de symétrie $x=-1$};}
\hunc{4}{\fill[dgreen] (0,4) circle (3pt) node[above right]{Ordonnée à l'origine  $\mathcal{H}\left(0;4\right)$};}
\hunc{5}{\fill[dred] (-4,0) circle (3pt) node[above right]{Zéro  $\mathcal{Z}_1\left(-4;0\right)$};}
\hunc{6}{\fill[dred] (2,0) circle (3pt) node[above right]{Zéro  $\mathcal{Z}_2\left(2;0\right)$};}

\end{tikzpicture}
\end{center}

\end{frame}}

\begin{frame}

\underline{Remarque 3.1} La première coordonnée du sommet $x_{\mathcal{S}}$ d'une parabole est toujours égale à la moyenne des premières coordonnées des zéros $x_{\mathcal{Z}_1}$ et $x_{\mathcal{Z}_2}$
\uncover<2->{$$ x_{\mathcal{S}} =  \frac{x_{\mathcal{Z}_1} + x_{\mathcal{Z}_2}}{2} $$}
\uncover<3->{S'il n'y a qu'un zéro, on a $x_\mathcal{S} = x_{\mathcal{Z}}$.}

\vfill
\uncover<4->{
\underline{Exemple 3.2} Dans l'exemple précédent, on avait 
$$\mathcal{S}\left(-1;\frac{9}{2}\right), \quad  \mathcal{Z}_1\left(-4;0\right) \mbox{ et }\mathcal{Z}_2\left(2;0\right)$$
Vérifier la formule de la remarque précédente. }\\[0.2cm]

\hunc{5}{On a $x_{\mathcal{S}}=\hunc{6}{-1}$, $x_{\mathcal{Z}_1}=\hunc{7}{-4}$ et $x_{\mathcal{Z}_2}=\hunc{8}{2}$. \hunc{9}{Donc
$$ x_{\mathcal{S}} \stackrel{?}{=}  \frac{x_{\mathcal{Z}_1} + x_{\mathcal{Z}_2}}{2}  
\hunc{10}{\Rightarrow -1 \stackrel{?}{=}  \frac{-4 + 2}{2}}  
\hunc{11}{\Leftrightarrow -1 \stackrel{?}{=}  -1 }\hunc{12}{ \qquad \myCheck}$$ }}


\end{frame}


\begin{frame}{4. Calcul avec les coordonnées}

\underline{Rappel 4.1} Un point $(x;y)$ fait partie d'une courbe si ses coordonnées satisfont l'équation de cette courbe. 

\vfill
\uncover<2->{\underline{Exemple 4.1} Le point $(-2;4)$ fait-il partie de la parabole $y= -2x^2 - 5x + 1$?}\\

\hunc{3}{
\hunc{4}{$4 \stackrel{?}{=} - 2\cdot (-2)^2 - 5 \cdot (-2) + 1
\hunc{5}{\Leftrightarrow 4 \stackrel{?}{=} -8 + 10 + 1}
\hunc{6}{\Leftrightarrow 4 \stackrel{?}{=} 3}\hunc{7}{\Rightarrow \mbox{ Non}}$}}

\vfill
\uncover<8->{
\underline{Rappel 4.2} On appelle \bl{zéro} les valeurs telles que $f(x)=0$. }

\vfill
\uncover<9->{\underline{Exemple 4.2} Quels sont les zéros de la fonction $f(x)= 2x^2 - 12x + 18$?}\\

\hunc{10}{On résoud l'équation $f(x)=0$:
\begin{equa}
2x^2 - 12x + 18 &=& 0 & \hunc{11}{MEE}\\
\hunc{12}{\LRq 2(x^2 - 6x + 9) &=& 0 & \hunc{13}{PR}}\\
\hunc{14}{\LRq 2(x - 3)^2 &=& 0 \uzero \hunc{15}{\qqR S=\{3\} }}\\
\end{equa}
\hunc{16}{La fonction n'a qu'un zéro: $x=3$.}
}

\end{frame}

\begin{frame}{Intersection entre une droite et une parabole}

\underline{Exemple 4.3} Calculer les coordonnées des points d'intersection entre la parabole d'équation  $\ve{y_f= x^2 - 2x - 1}$ et la droite d'équation $\bl{y_g = x -1}$.

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-3};

\def\qmax{4};
\def\qmin{-2};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 -2*(\x)-1});
\draw[thick, samples=2, domain=\min:\max, dblue] plot(\x, {\x-1});

\hunc{2}{
\uncover<2->{
\fill[dred] (0,-1) circle (0.1cm) node[left] {$I_1(0;-1)$}; 
\fill[dred]  (3,2) circle (0.1cm) node[right] {$I_2(3;2)$}; 
}}

\end{tikzpicture}
\end{minipage}\begin{minipage}{0.5\textwidth}

\hunc{3}{
On cherche les valeurs de x pour lesquelles $\ve{y_f}=\bl{y_g}$:
\begin{equa}
\uncover<4->{\ve{x^2 -2x -1} &=& \bl{x -1} & \uncover<5->{-x + 1}}\\
\uncover<6->{\Leftrightarrow x^2 -3x &=& 0 &\uncover<5->{\mbox{CL}}}\\
\uncover<7->{\Leftrightarrow x(x -3) &=& 0 \uzero \uncover<8->{\Rightarrow \ro{S=\{0,3\}} }}
\end{equa}}

\end{minipage}
\hunc{9}{
\uncover<9->{On remplace dans $\bl{y_g}$ pour trouver la deuxième coordonnée;}
\begin{eqnarray*}
\uncover<10->{\ro{x=0} &\Rightarrow& \bl{y = x -1} \uncover<11->{= \ro{0} -1 = \ro{-1}}} \uncover<12->{\Rightarrow \ro{I_1(0;-1)}}\\
\uncover<13->{\ro{x=3} &\Rightarrow& \bl{y = x -1}  \uncover<14->{= \ro{3} -1 = \ro{2}}}\uncover<15->{\Rightarrow \ro{I_2(3;2)}}\\
\end{eqnarray*}}

\end{frame}

\begin{frame}

Lorsque que l'on cherche les points d'intersection entre une droite et une parabole, trois cas sont possibles:\\


\begin{minipage}[t]{0.3\textwidth}
\uncover<2->{\begin{tikzpicture}[scale=0.35]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-3};

\def\qmax{4};
\def\qmin{-2};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 -2*(\x)-1});

\draw[thick, samples=2, domain=\min:\max, dblue] plot(\x, {\x-1});

\end{tikzpicture}}

\hunc{3}{Deux intersections}

\end{minipage} \quad \begin{minipage}[t]{0.3\textwidth}
\uncover<4->{\begin{tikzpicture}[scale=0.35]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-3};

\def\qmax{4};
\def\qmin{-2};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 -2*(\x)-1});

\draw[thick, samples=2, domain=\min:\max, violet] plot(\x, {2*\x -5});

\end{tikzpicture}}

\hunc{5}{Une intersection\\
\uncover<6->{La droite et la parabole sont tangentes}} 
\end{minipage} \quad \begin{minipage}[t]{0.3\textwidth}
\uncover<7->{\begin{tikzpicture}[scale=0.35]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-3};

\def\qmax{4};
\def\qmin{-2};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 -2*(\x)-1});

\draw[thick, samples=2, domain=\min:\max, dred] plot(\x, {2/3*\x -3});

\end{tikzpicture}}

\hunc{8}{Aucune intersection}
\end{minipage}

\end{frame}


\begin{frame}{Intersection entre deux paraboles distinctes}

\begin{minipage}{0.38\textwidth}
\begin{tikzpicture}[scale=0.3]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-7};

\def\qmax{4};
\def\qmin{-6};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 +6*(\x)+9});
\draw[thick, samples=\mySample, domain=\min:\max, dblue] plot(\x, {-1*(\x)^2 - 2*\x + 1});

\hunc{2}{
\uncover<2->{
\fill[dred] (-2,1) circle (0.1cm) node[above left] {$I_1(-2;1)$}; }
}

\end{tikzpicture}
\end{minipage}\begin{minipage}{0.6\textwidth}

\underline{Exemple 4.4} Calculer les coordonnées des points d'intersection des paraboles d'équation $\ve{y_f= x^2 + 6x + 9}$ et $\bl{y_g=-x^2-2x+1}$. 

\end{minipage}

\hunc{3}{
On cherche les valeurs de x pour lesquelles $\ve{y_f}=\bl{y_g}$:
\begin{equa}
\uncover<4->{\ve{x^2 + 6x + 9} &=& \bl{-x^2-2x+1} \uncover<5->{& +x^2+2x-1}}\\
\uncover<6->{\Leftrightarrow 2x^2 + 8x + 8 &=& 0 \uncover<7->{& \mbox{CL}}}\\
\uncover<8->{\Leftrightarrow 2(x^2 + 4x + 4) &=& 0 \uncover<9->{& \mbox{CL}}}\\
\uncover<10->{\Leftrightarrow 2(x + 2)^2 &=& 0} \uzero\uncover<11->{\Rightarrow S=\{\ro{-2} \} } \\
\end{equa}

\vspace{0.2cm}
\uncover<12->{On remplace dans $\ve{y_f}$ pour trouver la deuxième coordonnée;}
\begin{eqnarray*}
\uncover<13->{\ro{x=-2} &\Rightarrow& \ve{y = x^2 + 6x + 9} 
\uncover<14->{= (\ro{-2})^2 + 6\cdot(\ro{-2}) + 9 } \\
\uncover<15->{&&= 4 -12 + 9 = \ro{1}}} 
\uncover<16->{\Rightarrow \ro{I_1(-2;1)}}\\
\end{eqnarray*}
}

\end{frame}


\begin{frame}

Lorsque l'on calcule l'intersection entre deux paraboles distinctes, trois cas sont possibles:\\

\begin{minipage}[t]{0.3\textwidth}
\uncover<2->{\begin{tikzpicture}[scale=0.25]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-7};

\def\qmax{4};
\def\qmin{-6};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 +6*(\x)+9});
\draw[thick, samples=\mySample, domain=\min:\max, dblue] plot(\x, {-1*(\x+4)^2 + 4});

\end{tikzpicture}}
\hunc{3}{Deux intersections} 
\end{minipage} \quad \begin{minipage}[t]{0.3\textwidth}
\uncover<4->{\begin{tikzpicture}[scale=0.25]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-7};

\def\qmax{4};
\def\qmin{-6};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 +6*(\x)+9});
\draw[thick, samples=\mySample, domain=\min:\max, violet] plot(\x, {-1*(\x)^2 - 2*\x + 1});

\end{tikzpicture}}
\hunc{5}{Une intersection \\
\hunc{6}{Les paraboles sont tangentes}}
\end{minipage} \quad
\begin{minipage}[t]{0.3\textwidth}
\uncover<7->{\begin{tikzpicture}[scale=0.25]

\scriptsize

\coordinate (O) at (0,0);

\def\max{5};
\def\min{-7};

\def\qmax{4};
\def\qmin{-6};

% quadrillage

\foreach \x in {\qmin, ..., \qmax}{
\draw[black!30, very thin] (\x, \min) -- (\x, \max);}

\foreach \y in {\qmin, ..., \qmax}{
\draw[black!30,very thin]  (\min , \y) -- (\max, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\min) -- (0:\max) node[right]{$x$};
\draw[->] (90:\min) -- (90:\max) node[above]{$y$};

\clip (\min,\min) rectangle (\max,\max);
\draw[thick, samples=\mySample, domain=\min:\max, dgreen] plot(\x, {(\x)^2 +6*(\x)+9});
\draw[thick, samples=\mySample, domain=\min:\max, dred] plot(\x, {-1*(\x-1)^2 - 3});

\end{tikzpicture}}
\hunc{8}{Aucune intersection}
\end{minipage}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}{5. Application pratique}

\underline{Exemple 5.1} Une balle est tirée en l'air à partir du sol. La hauteur $h$ (en mètres) de la balle en fonction du temps $t$ (en secondes) est donnée par $h(t) = \bl{-4}t^2 + \ve{28}t$.

\begin{minipage}{0.35\textwidth}

\uncover<3->{
\begin{tikzpicture}[scale=0.3, y=1cm, x=1cm]

\coordinate (O) at (0,0);
\def\xmax{9};
\def\ymax{9};
\def\xmin{-1};
\def\ymin{-1};
\def\qxmax{8};
\def\qymax{8};
\def\qxmin{0};
\def\qymin{0};
\foreach \x in {\qxmin,..., \qxmax}{
\draw[black!60] (\x, \ymin) -- (\x, \ymax);}
\foreach \y in{\qymin,..., \qymax}{
\draw[black!60]  (\xmin , \y) -- (\xmax, \y );}

\node[below left] at (O) {O};
\node[left] at (0,1) {10};
\node[below] at (1,0) {1};
\draw[very thick] (-0.2, 1) -- (0.2, 1); 
\draw[very thick] (1,-0.2) -- (1, 0.2); 
\fill (0,0) circle (0.5pt); 

\draw[-latex, very thick] (0:\xmin) -- (0:\xmax) node[right]{t [s]};
\draw[-latex, very thick] (90:\ymin) -- (90:\ymax) node[above]{h [m]};

\clip (0,0) rectangle (\xmax,\ymax);
\draw[domain=\xmin:\xmax, samples=\mySample, very thick] plot(\x, {-0.4*(\x)^2+2.8*\x}); 

\hunc{5}{
\uncover<5->{\fill[dorange] (3.5, 4.9) circle (0.2cm) node[above, fill=white, yshift=0.2cm]{\only<13->{S(3.5; 49)}};}}

\end{tikzpicture}}


\end{minipage} \quad \begin{minipage}{0.55\textwidth}

\uncover<2->{a) Calculer la hauteur maximale atteinte par la balle.}

\hunc{4}{La hauteur maximale de la balle correspond au \vi{sommet} de la parabole.\uncover<5->{ On calcule donc les coordonnées du sommet $S\left(-\frac{\ve{b}}{2\bl{a}};-\frac{\Delta}{4\bl{a}}\right)$.} }

\end{minipage}

\hunc{6}{
\uncover<6->{
La balle atteindra le sommet au temps $\ora{t} = -\frac{\ve{b}}{2\bl{a}} \uncover<7->{= -\frac{\ve{28}}{2\cdot\bl{(-4)}} \uncover<8->{= -\frac{28}{-8} = \ora{3.5}$ s.}}} 

\uncover<9->{Pour trouver la hauteur, on peut remplacer $\ora{t=3.5}$ dans l'équation $h(t)$: 
\uncover<10->{$h(\ora{3.5}) = \uncover<11->{\bl{-4}\cdot (\ora{3.5})^2 + \ve{28}\cdot \ora{3.5} \uncover<12->{= \ora{49}$ }}}}

\uncover<14->{La hauteur maximale de la balle (atteinte après 3.5 secondes) sera donc de $\boxed{\ora{49 \mbox{ mètres}}}$.}}

\end{frame}

\begin{frame}
b) Calculer le temps que met la balle pour retomber au sol.\\[0.2cm]

\hunc{2}{
\uncover<2->{Nous devons calculer pour quelles valeurs de $t$ l'on a $h(t)= \bl{-4}t^2 +\ve{28}t = 0$.}\\
\uncover<3->{On calcule $\Delta$:}
\uncover<4->{$$\Delta = \ve{b}^2 - 4 \bl{a}\ro{c} \uncover<5->{= \ve{28}^2 - 4\cdot\bl{(-4)}\cdot\ro{0} }\uncover<6->{=784 >0}$$}

\uncover<7->{ On a donc deux solutions:
$$\ora{x_1} = \frac{-\ve{b} + \sqrt{\Delta}}{2\bl{a}} \uncover<8->{= \frac{-\ve{28} + 28 }{2\cdot (-4)} }\uncover<9->{=\ora{0}}$$
$$\ora{x_2} = \frac{-\ve{b} - \sqrt{\Delta}}{2\bl{a}} \uncover<10->{= \frac{-\ve{28} - 28 }{2\cdot (-4)} }\uncover<11->{=\frac{-56}{-8}} \uncover<12->{= \ora{7}}$$

}

\uncover<13->{La balle met donc $\boxed{\ora{7 \mbox{ secondes}}}$ pour retomber sur le sol. }}

\end{frame}


\begin{frame}{6. Optimisation}

\underline{Exemple 6.1} Robert veut faire un parc rectangulaire pour son chien. Il a 10 mètres de barrière. De quelle taille doivent être la \ve{longueur $L$} et la \bl{largeur $x$} du parc pour maximiser son aire $\ro{\mathcal{A}}$? Exprimer l'aire en fonction de $x$ et tracer le graphe de la fonction.

\hunc{2}{
\vfill
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}
\draw[very thick, dgreen](0,0) -- node[below]{L\only<7->{=2-x}}(2,0);
\draw[very thick, dgreen](0,1.5) -- node[above]{L\only<7->{=2-x}}(2,1.5);
\draw[very thick, dblue](0,0) -- node[left]{x}(0,1.5);
\draw[very thick, dblue](2,0) -- node[right]{x}(2,1.5);
\end{tikzpicture}
\end{minipage}\begin{minipage}{0.58\textwidth}

Soit \bl{x la largeur} du parc et \ve{L sa longueur}. \uncover<2->{Le périmètre du parc vaut \uncover<3->{$$P = 2\bl{x} + 2\ve{L} = 10.$$}}\uncover<4->{On cherche à maximiser l'aire $\mathcal{\ro{A}}=\bl{x}\cdot \ve{L}$.}



\end{minipage}
\vfill 
\hunc{5}{On exprime $\ve{L}$ en fonction de $\bl{x}$:
$2\bl{x} + 2\ve{L} = 10 \uncover<6->{\Rightarrow 2\ve{L} = 10 - 2\bl{x} }\uncover<7->{\Rightarrow \ve{L} = 5 - \bl{x}}$}\\[0.2cm]
\hunc{8}{L'aire du parc sera donc de
$$\mathcal{\ro{A}}=\bl{x}\cdot \ve{L} \uncover<9->{= \bl{x}\ve{(5-x)}}\uncover<10->{ = 5x -x^2}\uncover<11->{ = -x^2 + 5x}$$}}
\end{frame}


\begin{frame}

\begin{minipage}{0.5\textwidth}

\begin{tikzpicture}[scale=0.5]

\coordinate (O) at (0,0);

\def\xmax{7};\def\xmin{-1};\def\ymax{7};\def\ymin{-1};
\def\xqmax{6};\def\xqmin{0};\def\yqmax{6};\def\yqmin{0};

% quadrillage

\foreach \x in {\xqmin, ..., \xqmax}{
\draw[black!30, very thin] (\x, \ymin) -- (\x, \ymax);}

\foreach \y in {\yqmin, ..., \yqmax}{
\draw[black!30,very thin]  (\xmin , \y) -- (\xmax, \y );}

% origine
\node[below left] at (O) {O};
\node[left] at (0,1) {1};
\node[below] at (1,0) {1};
\draw[thick] (-0.1, 1) -- (0.1, 1); 
\draw[thick] (1,-0.1) -- (1, 0.1); 

% axes
\draw[->] (0:\xmin) -- (0:\xmax) node[right]{$x$};
\draw[->] (90:\ymin) -- (90:\ymax) node[above]{$\mathcal{\ro{A}}$};

\clip (\xmin,\ymin) rectangle (\xmax,\ymax);
\hunc{2}{
\uncover<18->{\draw[thick, samples=\mySample, domain=0:5, dred] plot(\x, {-(\x)^2 +5*(\x)});}

\tiny
\uncover<8->{
\fill[dblue] (0,0) circle (2pt) node[above right]{$\mathcal{Z}_1(0,0)$};
\fill[dblue] (5,0) circle (2pt) node[above left]{$\mathcal{Z}_2(5,0)$};}

\uncover<17->{
\fill[dblue] (2.5,6.25) circle (2pt) node[above]{$\mathcal{S}(2.5,6.25)$};}}


\end{tikzpicture}

\end{minipage} \begin{minipage}{0.48\textwidth}

\hunc{2}{
\uncover<2->{Traçons le graphe de la fonction. Pour cela, calculons les coordonnées des zéros et du sommet. }
\uncover<3->{\begin{equa}
-x^2 + 5x &=& 0 &\hunc{4}{MEE}\\
\hunc{5}{\LRq -x(x - 5) &=& 0 \uzero}\\
\end{equa}
\hunc{6}{On a S=\{0;5\} \hunc{7}{ et donc $\mathcal{Z}_1(0,0)$ et $\mathcal{Z}_2(5,0)$.}
}}}

\end{minipage}

\hunc{9}{
\uncover<9->{
On calcule ensuite les coordonnées du sommet S($\frac{-b}{2a}, -\frac{\Delta}{4a}$):
$$ -\frac{b}{2a} \uncover<10->{= - \frac{5}{-2}} \uncover<11->{= 2.5} \qquad
\uncover<12->{-\frac{\Delta}{4a}} \uncover<13->{=-\frac{5^2-4\cdot (-1) \cdot 0}{4\cdot(-1)} \uncover<14->{= \dfrac{15}{4}=\hunc{16}{6.25}}}$$
}

\uncover<18->{
L'aire est maximale quand la largeur vaut \bl{$x=2.5$ m} \hunc{20}{ et la longueur  \ve{$y=5-x = 5-2.5 = 2.5$ m}.}
\uncover<21->{ Le parc doit donc être carré pour que l'aire soit maximale! }
}}
\end{frame}


\end{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%












