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

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

\begin{document}

\titreCours{6}{Fonctions Affines}{}{}

\begin{frame}{1. Les fonctions linéaires}

\underline{Exemple 1.1} Un litre d'essence coûte 1.50. Quel est le prix de:
\begin{enumerate}
\item \dots  $\,\ro{2}$ litres d'essence? \hunc{2}{$y = 1.5 \cdot \ro{2} = 3$.-}
\item<3-> \dots  $\,\ro{10}$ litres? \hunc{4}{$y =1.5 \cdot \ro{10} = 15$.-}
\item<5-> \dots  $\,\ro{x}$ litres? \hunc{6}{$y =1.5 \cdot \ro{x} \uncover<7->{= 1.5x}$}
\end{enumerate} 

\vfill

\uncover<8->{\underline{Définition 1.1} On dit que deux variables $x$ et $y$ sont \bl{proportionelles} s'il existe $\ve{m}$ tel que $\boxed{y=\ve{m}\cdot x}$. \uncover<9->{$\ve{m}$ est le \ve{facteur de proportion}.}}

\vfill

\uncover<10->{\underline{Exemple 1.1 (suite)} Que vaut le \ve{facteur de proportion} ? 
\hunc{11}{\ve{m=1.5}} }

\vfill 

\uncover<12->{\underline{Définition 1.2} On représente une \ve{relation de proportionalité} par une \bl{fonction linéaire}  \uncover<13->{que l'on note $\boxed{\bl{f(x)= mx}}$.}}

\vfill

\uncover<14->{\underline{Exemple 1.1 (suite)} Donner la fonction représentant cette relation.  
\hunc{15}{On a \bl{f(x)=\ve{1.5}x}} }

\end{frame} 

\begin{frame}

\underline{Exemple 1.1 (suite)} Compléter le tableau de valeurs suivant et placer les points sur la figure. Tracer la courbe correspondant à la fonction $\bl{f(x) = 1.5x}$. \\

\vfill

\uncover<2->{
 \begin{minipage}{0.3\textwidth}
 \begin{tabular}{|c | c|}
 \hline 
	 \ve{Litres} &  \ro{Prix} \\
 \hline
	 \ve{0} & \hunc{3}{\ro{0.00}}\\
 \hline
	 \ve{1} & \hunc{4}{\ro{1.50}}\\
\hline
	 \ve{2} & \hunc{5}{\ro{3.00}}\\
\hline
	 \ve{3} & \hunc{6}{\ro{4.50}}\\
\hline
	 \ve{4} & \hunc{7}{\ro{6.00}}\\
 \hline
 \end{tabular}
\end{minipage}\begin{minipage}{0.7\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[gray, very thin] (-0.9,-0.9) grid (6.9,6.9);
\draw[-latex, dgreen] (-1,0) -- (7,0) node[right]{Litres};
\draw[-latex, dred] (0,-1) -- (0,7) node[above]{Prix};
\draw (1, -0.1) -- (1, 0.1) node[above]{1};
\draw (0.1,1) -- (-0.1,1) node[left]{1};

\hunc{7}{
\clip (-1.9,-1) rectangle (7,7);
\uncover<7->{\fill[dblue] (0,0) circle (2pt) node[above left]{(\ve{0};\ro{0})};}
\uncover<9->{\fill[dgreen] (1,0) circle (2pt) node[below]{1};}
\uncover<10->{\draw[dgreen, dashed, thick] (1,0) -- (1,7);}
\uncover<11->{\fill[dred] (0,1.5) circle (2pt) node[left]{1.50};}
\uncover<12->{\draw[dred, dashed, thick] (0,1.5) -- (7,1.5);}
\uncover<13->{\fill[dblue] (1,1.5) circle (2pt) node[below right]{(\ve{1},\ro{1.50})};}
\uncover<14->{\fill[dgreen] (2,0) circle (2pt) node[below]{2};}
\uncover<15->{\draw[dgreen, dashed, thick] (2,0) -- (2,7);}
\uncover<16->{\fill[dred] (0,3.00) circle (2pt) node[left]{3.00};}
\uncover<17->{\draw[dred, dashed, thick] (0,3.00) -- (7,3.00);}
\uncover<17->{\fill[dblue] (2,3.00) circle (2pt) node[below right]{(\ve{2},\ro{3.00})};}
\uncover<19->{\fill[dgreen] (3,0) circle (2pt) node[below]{3};}
\uncover<20->{\draw[dgreen, dashed, thick] (3,0) -- (3,7);}
\uncover<21->{\fill[dred] (0,4.50) circle (2pt) node[left]{4.50};}
\uncover<22->{\draw[dred, dashed, thick] (0,4.50) -- (7,4.50);}
\uncover<23->{\fill[dblue] (3,4.50) circle (2pt) node[below right]{(\ve{3},\ro{4.50})};}
\uncover<24->{\fill[dgreen] (4,0) circle (2pt) node[below]{4};}
\uncover<25->{\draw[dgreen, dashed, thick] (4,0) -- (4,7);}
\uncover<26->{\fill[dred] (0,6) circle (2pt) node[left]{6.00};}
\uncover<27->{\draw[dred, dashed, thick] (0,6) -- (7,6);}
\uncover<28->{\fill[dblue] (4,6) circle (2pt) node[below right]{(\ve{4},\ro{6.00})};}
\uncover<29->{\draw[dblue, thick, samples=2] plot(\x,{1.5*\x});}
}
\end{tikzpicture}
\end{center}
\end{minipage}}

\end{frame}

\begin{frame}

\underline{Définition 1.3} La courbe représentant une fonction linéaire $\bl{f(x)=mx}$ est une \bl{droite $y=mx$ passant par l'origine $(0,0)$}. \uncover<2->{Le facteur de proportion $\bl{m}$ représente la \bl{pente} de la droite. }

\uncover<3->{
\vfill
On distingue trois cas:\\[0.5cm]

\begin{minipage}{0.32\textwidth}
\begin{center}
\hunc{4}{\bl{$\boxed{m>0}$}}\\[0.2cm]
\begin{tikzpicture}[scale=0.5]
\draw[-latex, dgreen] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred] (0,-3) -- (0,3) node[above]{$y$};
\clip (-3,-3) rectangle (3,3);
\hunc{5}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {1.5*\x});}
\end{tikzpicture}
\hunc{6}{La droite "monte"}
\end{center}
\end{minipage}
\begin{minipage}{0.32\textwidth}
\begin{center}
\hunc{7}{\bl{$\boxed{m=0}$}}\\[0.2cm]
\begin{tikzpicture}[scale=0.5]
\draw[-latex, dgreen] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred] (0,-3) -- (0,3) node[above]{$y$};
\clip (-3,-3) rectangle (3,3);
\hunc{8}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {0});}
\end{tikzpicture}
\hunc{9}{La droite est "plate"}
\end{center}
\end{minipage}
\begin{minipage}{0.32\textwidth}
\begin{center}
\hunc{10}{\bl{$\boxed{m<0}$}}\\[0.2cm]
\begin{tikzpicture}[scale=0.5]
\draw[-latex, dgreen] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred] (0,-3) -- (0,3) node[above]{$y$};
\clip (-3,-3) rectangle (3,3);
\hunc{11}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {-1.5*\x});}
\end{tikzpicture}
\hunc{12}{La droite "descend"}
\end{center}
\end{minipage}
}
\end{frame}

\begin{frame}

\begin{center}
 \begin{tikzpicture}[scale=0.35]
 
 \small
 
 \coordinate (O) at (0,0);
 \def\max{8};
 \def\min{-8};
 
 \def\qmax{7};
 \def\qmin{-7};
 
 % axes
 \draw[->, very thick] (0:\min) -- (0:\max) node[right]{$x$};
 \draw[->, very thick] (90:\min) -- (90:\max) node[above]{$y$};
 
 \clip (\min,\min) rectangle (\max,\max);
 
 % quadrillage
 
 \foreach \x in {\qmin, ..., \qmax}{
 \draw[black!50, very thin] (\x, \min) -- (\x, \max);}
 
 \foreach \y in {\qmin, ..., \qmax}{
 \draw[black!50,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); 
 \fill (0,0) circle (4pt); 
 
 \clip (\min, \min) rectangle (\max, \max);
 
 \hunc{1}{
 \tiny
 \foreach \a/\n in {0.5/1, 1/2, 2/3}{
 \uncover<\n->{\draw[dblue, samples=2, domain=\min:\max, thick] plot(\x,{\a*\x });
 \node[dblue, below right, fill=white] at (4, {\a*4}) {$m=\a$};}}

 
 \foreach \a/\n in {-0.5/4, -1/5, -2/6}{
 \uncover<\n->{\draw[dblue, samples=2, domain=\min:\max, thick] plot(\x,{\a*\x});
 \node[dblue, above right, fill=white] at (4, {\a*4}) {$m=\a$};}}}
 
 \end{tikzpicture}
 \end{center}
 
 \vspace{0.25cm}
 
 \hunc{7}{
 \uncover<7->{La pente $\bl{m}$ influence l'\bl{orientation} de la droite.}\uncover<8->{ Plus la pente est grande en valeur absolue, plus la droite est "raide".} }

\end{frame}

\begin{frame}
\underline{Exercice 1.1} Associer les fonctions linéaires données avec les droites correspondantes:

\vfill

\begin{minipage}{0.65\textwidth}

\begin{tikzpicture}[scale=0.5]

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

\def\max{6};
\def\min{-6};

% axes

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

\clip (-6,-6) rectangle (6,6);
\draw[thick, dblue, samples=2, domain=-6:6] plot(\x, {0}); \node[dblue, above] at (5,0) {d};
\draw[thick, dblue, samples=2, domain=-6:6] plot(\x, {-\x}); \node[dblue, above] at (5,-5) {e};
\draw[thick, dblue, samples=2, domain=-6:6] plot(\x, {0.5*\x}); \node[dblue, above] at (5,2.5) {c};
\draw[thick, dblue, samples=2, domain=-6:6] plot(\x, {2*\x}); \node[dblue, above] at (2.5,5) {a};
\draw[thick, dblue, samples=2, domain=-6:6] plot(\x, {\x}); \node[dblue, above] at (5,5) {b};
\draw[thick, dblue, samples=2, domain=-6:6] plot(\x, {-2*\x}); \node[dblue, above] at (2.5,-5) {f};


\end{tikzpicture}

\end{minipage}\begin{minipage}{0.33\textwidth}

\begin{enumerate}[(1)]
\item $f(x) = 0$ \hunc{2}{\bl{d}}
\item $f(x) = - x$ \hunc{7}{\bl{e}}
\item $f(x) = \frac{1}{2}x$ \hunc{5}{\bl{c}}
\item $f(x) = 2x$ \hunc{3}{\bl{a}}
\item $f(x) = x$ \hunc{4}{\bl{b}}
\item $f(x) = -2x$ \hunc{6}{\bl{f}}
\end{enumerate}

\end{minipage}

\end{frame}

\begin{frame}{2. Fonction affines}

\underline{Exemple 2.1} Un station d'essence propose un système d'abonnement. Pour \ro{50.-} par année, le prix de l'essence passe de 1.50 à \bl{1.40}. Si l'on prend l'abonnement, combien coûtera en tout (abonnement compris) \dots
\begin{enumerate}
\item \dots \ve{0} litre d'essence? \hunc{2}{ \ro{50.-}}  
\item<3-> \dots \ve{10} litres d'essence? \hunc{4}{ $\bl{1.40} \cdot \ve{10} + \ro{50} \hunc{5}{= 14 + 50 = 64.-}$}  
\item<6-> \dots \ve{50} litres d'essence? \hunc{7}{ $\bl{1.40} \cdot \ve{50} + \ro{50} \hunc{8}{= 70 + 50 = 120.-}$}  
\item<9-> \dots \ve{x} litres d'essences? \hunc{10}{ $\bl{1.40} \cdot \ve{x} + \ro{50} \hunc{11}{= \bl{1.4}\ve{x} + \ro{50}}$} 
\end{enumerate}

\vfill 

\uncover<12->{
\underline{Définition 2.1} On appelle \bl{fonctions affines} les relations du type $$\boxed{f(x)=\bl{m}\ve{x} + \ro{h}}.$$ 
\uncover<13->{Ces fonctions sont représentées par des droites $y=\bl{m}\ve{x}+\ro{h}$ de \bl{pente} $\bl{m}$ et passant par le point $(0, \ro{h})$.}\uncover<14->{ $\ro{h}$ est l'\ro{ordonnée à l'origine}.}}\uncover<15->{ Si \ro{$h=0$}, la fonction est linéaire.}

\end{frame}

\begin{frame}

\underline{Exemple 2.1 (suite)} Dans la fonction $f(x)=\bl{1.4}x+ \ro{50}$,
\begin{enumerate}
\item que vaut la pente de la droite correspondante? \hunc{2}{ $\bl{m=1.4}$}
\item<3-> que vaut l'ordonnée à l'origine? \hunc{4}{ $\ro{h=50}$}
\end{enumerate}
\uncover<5->{
Représenter la droite correspondante sur la figure ci-dessous. 
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[gray, very thin] (-0.9,-0.9) grid (8.9,8.9);
\draw[-latex, dgreen, thick] (-1,0) -- (9,0) node[right]{Litres};
\draw[dred, dashed, thick] (0,0) -- (0,1);
\draw[dred, thick] (0, -1) -- (0,0);
\draw[-latex, dred, thick] (0,1) -- (0,9) node[above]{Prix};
\draw (1, -0.1) -- (1, 0.1) node[below]{10};
\draw (0.1,1) -- (-0.1,1) node[left]{50};
\draw (0.1,2) -- (-0.1,2) node[left]{60};

\hunc{6}{
\clip (-1.9,-1) rectangle (9,9);
\uncover<6->{\fill[dblue] (0,1) circle (2pt) node[below right]{(\ve{0};\ro{50})};}
\uncover<7->{\fill[dgreen] (1,0) circle (2pt);}
\uncover<8->{\draw[dgreen, dashed, thick] (1,0) -- (1,9);}
\uncover<9->{\fill[dred] (0,2.4) circle (2pt) node[left]{64};}
\uncover<10->{\draw[dred, dashed, thick] (0,2.4) -- (9,2.4);}
\uncover<11->{\fill[dblue] (1,2.4) circle (2pt) node[below right]{(\ve{10},\ro{64})};}
\uncover<12->{\fill[dgreen] (5,0) circle (2pt) node[below]{50};}
\uncover<13->{\draw[dgreen, dashed, thick] (5,0) -- (5,9);}
\uncover<14->{\fill[dred] (0,8) circle (2pt) node[left]{120};}
\uncover<15->{\draw[dred, dashed, thick] (0,8) -- (9,8);}
\uncover<16->{\fill[dblue] (5,8) circle (2pt) node[below right]{(\ve{50},\ro{120})};}
\uncover<17->{\draw[dblue, thick, samples=2, domain=0:6] plot(\x,{1.4*\x+1});}
}
\end{tikzpicture}
\end{center}}

\end{frame}


\begin{frame}

\underline{Exercice 2.1} Soit $f(x)=2x-1$ une fonction affine. Que vaut:
\begin{enumerate}
\item $f(\ve{2})$? 
\hunc{2}{ $f(\ve{2}) = 2\cdot\ve{2} -1 \uncover<3->{= 4 -1 = \ro{3}}$
\uncover<11->{ $\bl{\rightarrow} (\ve{2},\ro{3})$}} 
\item<4-> $f(\ve{0})$? \hunc{5}{ $f(\ve{0}) = 2\cdot\ve{0} -1 \uncover<6->{= \ro{-1}}$\uncover<12->{ $\bl{\rightarrow } (\ve{0},\ro{-1})$}}  
\item<7-> $f(\ve{-3})$? \hunc{8}{ $f(\ve{-3}) = 2\cdot\ve{(-3)} -1 \uncover<9->{= -6 -1 = \ro{-7}}$ \uncover<13->{$\bl{\rightarrow} (\ve{-3},\ro{-7})$}}  
\end{enumerate}
\uncover<10->{et esquisser la droite associée. }

\uncover<10->{
\begin{center}
\begin{tikzpicture}[scale=0.5]

\draw[gray, very thin] (-7.9,-7.9) grid (6.9,3.9);
\draw[-latex, dgreen, thick] (-8,0) -- (7,0) node[right]{$x$};
\draw[-latex, dred, thick] (0,-8) -- (0,4) node[above]{$y$};
\draw (1, -0.1) -- (1, 0.1) node[below]{1};
\draw (0.1,1) -- (-0.1,1) node[left]{1};

\hunc{14}{
\clip (-8,-8) rectangle (7,4);

\uncover<14->{\fill[dgreen] (2,0) circle (2pt) node[below]{2};}
\uncover<15->{\draw[dgreen, dashed, thick] (2,-8) -- (2,8);}
\uncover<16->{\fill[dred] (0,3) circle (2pt) node[left]{3};}
\uncover<17->{\draw[dred, dashed, thick] (-8,3) -- (8,3);}
\uncover<18->{\fill[dblue] (2,3) circle (2pt) node[below right]{(\ve{2},\ro{3})};}
\uncover<19->{\fill[dgreen] (0,0) circle (2pt) node[below]{0};}
\uncover<20->{\draw[dgreen, dashed, thick] (0,-8) -- (0,8);}
\uncover<21->{\fill[dred] (0,-1) circle (2pt) node[left]{-1};}
\uncover<22->{\draw[dred, dashed, thick] (-8,-1) -- (8,-1);}
\uncover<23->{\fill[dblue] (0,-1) circle (2pt) node[below right]{(\ve{0},\ro{-1})};}
\uncover<24->{\fill[dgreen] (-3,0) circle (2pt) node[below]{-3};}
\uncover<25->{\draw[dgreen, dashed, thick] (-3,-8) -- (-3,8);}
\uncover<26->{\fill[dred] (0,-7) circle (2pt) node[left]{-7};}
\uncover<27->{\draw[dred, dashed, thick] (-8,-7) -- (8,-7);}
\uncover<28->{\fill[dblue] (-3,-7) circle (2pt) node[above left]{(\ve{-3},\ro{-7})};}
\uncover<29->{\draw[dblue, thick, samples=2] plot(\x,{2*\x-1});}
}
\end{tikzpicture}
\end{center}

}

\end{frame}

%%%%%%%%%%%%% AJOUTER PAGE SUR LES ZEROS D'UNE FONCTION...  %%%%%%%%%%%%%%


\begin{frame}{3. Interprétation graphique}

\underline{Exemple 3.1} Soit la fonction $f(x)= \bl{2}x \ro{+ 1}$. Représenter graphiquement la pente et l'ordonnée à l'origine.

\begin{center}
\begin{tikzpicture}[scale=0.4]
\draw[gray, very thin] (-5.9,-5.9) grid (5.9,5.9);
\draw[-latex, dgreen, thick] (-6,0) -- (6,0) node[right]{$x$};
\draw[-latex, dred, thick] (0,-6) -- (0,6) node[above]{$y$};
\draw (1, -0.1) -- (1, 0.1) node[below]{\ve{1}};
\draw (0.1,1) -- (-0.1,1) node[left]{\ro{1}};
\clip (-6,-6) rectangle (12,6);
\draw[dblue, thick, samples=2, domain=-6:6] plot(\x, {2*\x+1});
\node[left] at (0,2) {$\bl{y=2x+1}$};
\hunc{2}{
\uncover<2->{\draw[dblue, thick, fill=blue!30] (1,3) -- node[below]{\ve{1}} (2,3) -- node[right]{\ro{2}} (2,5) --(1,3);}
\uncover<3->{\node [right] at (3,4) {\bl{Pente} $\bl{m=\dfrac{\ro{2}}{\ve{1}}\only<4->{=2}}$};}
\uncover<5->{\draw[dblue, thick, fill=blue!30] (-3, -5) -- node[below]{\ve{2}} (-1, -5) --node[right]{\ro{4}}  (-1,-1) --(-3,-5);}
\uncover<6->{\node [right] at (0,-3) {\bl{Pente} $\bl{m=\dfrac{\ro{4}}{\ve{2}}\only<7->{=2}}$};}
\uncover<8->{\fill[dred, left] (0,1) circle (2pt) node[right] {Ordonnée à l'origine}; }
}
\end{tikzpicture}
\end{center}
\hunc{9}{L'\ro{ordonnée à l'origine} est donnée par $f(0)$, c'est la "hauteur" à laquelle la droite coupe l'axe de $y$. }
\end{frame}

\begin{frame}

\underline{Exemple 3.2} Dessiner la droite associée aux fonctions suivantes

\begin{minipage}{0.38\textwidth}
1. $f(x) = \dfrac{1}{2}x -1$\\[0.2cm]
\hunc{2}{
$\bl{m = \uncover<3->{\dfrac{1}{2}}}$, $\ro{h = \uncover<4->{-1}}$
}
\end{minipage}\begin{minipage}{0.6\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[gray, very thin] (-2.9,-2.9) grid (2.9,2.9);
\draw[-latex, dgreen, thick] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred, thick] (0,-3) -- (0,3) node[above]{$y$};
\draw (1, -0.1) -- (1, 0.1) node[above]{\ve{1}};
\draw (0.1,1) -- (-0.1,1) node[left]{\ro{1}};
\clip (-3,-3) rectangle (3,3);
\hunc{5}{
\uncover<5->{\fill[dred] (0,-1) circle (2pt) node[above left]{$h=-1$};}
\uncover<6->{\draw[dblue, thick] (0,-1) -- node[below]{\ve{2}} (2,-1);}
\uncover<7->{\draw[dblue, thick] (2,-1) -- node[right]{\ro{1}} (2,0);}
\uncover<8->{\fill[dblue] (2,0) circle (2pt);}
\uncover<9->{\draw[dblue, thick] (0,-1) -- (2,0);}
\uncover<10->{\draw[dblue, thick, samples=2, domain=-6:6] plot(\x, {0.5*\x-1});}
}
\end{tikzpicture}
\end{center}
\end{minipage}

\uncover<11->{
\begin{minipage}{0.38\textwidth}
2. $f(x) = 2 \hunc{13}{ = 0\cdot x +2}$\\[0.2cm]
\hunc{12}{
$\bl{m = \uncover<14->{0}}$, $\ro{h = \uncover<15->{2}}$
}
\end{minipage}\begin{minipage}{0.6\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[gray, very thin] (-2.9,-2.9) grid (2.9,2.9);
\draw[-latex, dgreen, thick] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred, thick] (0,-3) -- (0,3) node[above]{$y$};
\draw (1, -0.1) -- (1, 0.1) node[below]{\ve{1}};
\draw (0.1,1) -- (-0.1,1) node[left]{\ro{1}};
\clip (-3,-3) rectangle (3,3);
\hunc{16}{
\uncover<16->{\fill[dred] (0,2) circle (2pt) node[above left]{$h=2$};}
\uncover<17->{\draw[dblue, thick] (0,2) -- node[below]{\ve{1}} (1,2);}
\uncover<18->{\node[dblue, above right] at (1,2) {\ro{0}} ;}
\uncover<19->{\fill[dblue] (1,2) circle (2pt);}
\uncover<20->{\draw[dblue, thick, samples=2, domain=-6:6] plot(\x, {2});}
}
\end{tikzpicture}
\end{center}
\end{minipage}}

\end{frame}

\begin{frame}
\begin{minipage}{0.38\textwidth}
3. $f(x) = -x +1$\\[0.2cm]
\hunc{2}{
$\bl{m = \uncover<3->{-1}}$, $\ro{h = \uncover<4->{1}}$\\
\uncover<5->{La pente est négative, donc la droite descend!} 
}
\end{minipage}\begin{minipage}{0.6\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[gray, very thin] (-2.9,-2.9) grid (2.9,2.9);
\draw[-latex, dgreen, thick] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred, thick] (0,-3) -- (0,3) node[above]{$y$};
\draw (1, -0.1) -- (1, 0.1) node[below]{\ve{1}};
\draw (0.1,1) -- (-0.1,1) node[left]{\ro{1}};
\clip (-3,-3) rectangle (3,3);
\hunc{6}{
\uncover<6->{\fill[dred] (0,1) circle (2pt) node[above right]{$h=1$};}
\uncover<7->{\draw[dblue, thick] (0,1) -- node[below]{\ve{1}} (1,1);}
\uncover<8->{\draw[dblue, thick] (1,1) -- node[right]{\ro{-1}} (1,0);}
\uncover<9->{\fill[dblue] (1,0) circle (2pt);}
\uncover<10->{\draw[dblue, thick] (0,1) -- (1,0);}
\uncover<11->{\draw[dblue, thick, samples=2, domain=-6:6] plot(\x, {-(\x)+1});}
}
\end{tikzpicture}
\end{center}
\end{minipage}

\vfill
\uncover<12->{\underline{Exercice 3.1} Trouver la fonction associée à la droite ci-dessous:
\begin{minipage}{0.6\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[gray, very thin] (-2.9,-2.9) grid (2.9,2.9);
\draw[-latex, dgreen, thick] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred, thick] (0,-3) -- (0,3) node[above]{$y$};
\draw (1, -0.1) -- (1, 0.1) node[below]{\ve{1}};
\draw (0.1,1) -- (-0.1,1) node[left]{\ro{1}};
\clip (-3,-3) rectangle (3,3);
\draw[dblue, thick, samples=2, domain=-6:6] plot(\x, {-2*\x)+1});
\hunc{13}{
\uncover<13->{\fill[dred] (0,1) circle (2pt) node[below left]{\ro{h=1}};}
\uncover<14->{\draw[dblue, thick] (0,1) -- node[below]{\ve{1}} (1,1);}
\uncover<15->{\draw[dblue, thick] (1,1) -- node[right]{\ro{-2}} (1,-1);};
%\uncover<16->{\node[dblue, above left] (0,1) {Pente $m=\only<17->{\dfrac{\ro{-2}}{\ve{1}}}\only<18->{=-2}$};}
}
\end{tikzpicture}
\end{center}
\end{minipage}\begin{minipage}{0.38\textwidth}
\hunc{16}{
Pente $m=\only<17->{\dfrac{\ro{-2}}{\ve{1}}}\only<18->{=-2}$\\[0.2cm]
\uncover<19->{
On a $\bl{m=-2}$ et $\ro{h=1}$, donc 
$$f(x)\uncover<20->{=\bl{-2}x\ro{+1}}$$}}
\end{minipage}}

\end{frame}

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

\underline{Remarque 4.1} Un point ($\ve{x};\ro{y}$) appartient à une droite s'il satisfait son équation $\ro{y}=m\ve{x}+h$. 

\vfill
\uncover<2->{\underline{Exemple 4.1} Le point $(\ve{5}; \ro{6})$ appartient-il à la droite $\ro{y}=2\ve{x}+1$?\\[0.2cm]

\hunc{3}{ $\ro{y} \stackrel{?}{=} 2\ve{x} +1$ 
\uncover<4->{$\Rightarrow \ro{6} \stackrel{?}{=} 2 \cdot \ve{5} + 1$} 
\uncover<5->{$\Leftrightarrow \ro{6} \stackrel{?}{=} 11$} \uncover<6->{$\Rightarrow$ Non! }}}

\vfill
\uncover<7->{\underline{Exercice 4.1} Le point $(\ve{-4}; \ro{2})$ appartient-il à la droite $\ro{y}=-2\ve{x}-6$?\\[0.2cm]

\hunc{8}{ $\ro{y} \stackrel{?}{=} -2\ve{x} -6$ 
\uncover<9->{$\Rightarrow \ro{2} \stackrel{?}{=} (-2) \cdot (\ve{-4}) -6$} 
\uncover<10->{$\Leftrightarrow \ro{2} \stackrel{?}{=} 2$} \uncover<11->{$\Rightarrow$ Oui! }}

}

\end{frame}

\begin{frame} 

\underline{Exemple 4.2} Trouver par calcul l'équation de la droite passant par les points $A(\ve{5},\ro{2})$ et $B(\ve{-1},\ro{4})$. \\[0.5cm]

\begin{minipage}{0.49\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.4]
\draw[gray, very thin] (-5.9,-5.9) grid (5.9,5.9);
\draw[-latex, dgreen, thick] (-6,0) -- (6,0) node[right]{$x$};
\draw[-latex, dred, thick] (0,-6) -- (0,6) node[above]{$y$};
\draw (1, -0.1) -- (1, 0.1) node[below]{\ve{1}};
\draw (0.1,1) -- (-0.1,1) node[left]{\ro{1}};
\clip (-6,-6) rectangle (6,6);
\fill[dblue] (5,2) circle (4pt) node[below left]{A(\ve{5},\ro{2})};
\fill[dblue] (-1,4) circle (4pt) node[below left]{B(\ve{-1},\ro{4})};
\draw[dblue, samples=2, domain=-6:6] plot(\x, {-1/3*(\x)+11/3});
\end{tikzpicture}
\end{center}
\end{minipage}\begin{minipage}{0.49\textwidth}
\hunc{2}{
Soit $\ro{y}=\bl{m}\ve{x}+\ora{h}$ l'équation de la droite. \uncover<3->{Si les deux points appartiennent à la droite, ils satisfont les deux l'équation $\ro{y}=\bl{m}\ve{x}+\ora{h}$.}}
\end{minipage}
\hunc{4}{
\uncover<4->{ On a donc:
$$\begin{cases}
\uncover<5->{\ro{2}} = \bl{m}\cdot \uncover<5->{\ve{5}} + \ora{h} \\
\uncover<6->{\uncover<7->{\ro{4}} = \bl{m}\cdot \uncover<7->{(\ve{-1})} + \ora{h}} \end{cases}$$

\uncover<8->{Il faut résoudre le système pour trouver $\bl{m}$ et $\ora{h}$.} 
}}

\end{frame}

\begin{frame}

\hunc{1}{
On résoud par combinaison linéaire
$$\begin{cases}
\ro{2} = \bl{m}\cdot \ve{5} + \ora{h} \\
\ro{4} = \bl{m}\cdot (\ve{-1}) + \ora{h} \end{cases} 
\uncover<2->{\Leftrightarrow 
\begin{cases}
2 = 5 \bl{m} + \ora{h} \\
4 = -\bl{m} + \ora{h} \uncover<3->{\,\,\boxed{\cdot (-1)}}\end{cases}}
\uncover<4->{\Leftrightarrow 
\begin{cases}
2 = 5 \bl{m} + \ora{h} \\
-4 = \bl{m} - \ora{h} \end{cases}}
$$

\uncover<5->{
On additionne les deux équations
$$\begin{array}{c c c c}
&2 &=& 5 \bl{m} + \ora{h} \\
\boxed{+} &-4 &=& \bl{m} - \ora{h}\\
\hline
\uncover<6->{& -2 &=& 6\bl{m}}
\end{array}
$$}
\uncover<7->{On a donc $\bl{m}=\frac{-2}{6} = \bl{-\frac{1}{3}}$.}\uncover<8->{On remplace dans l'une des équations:
$$2 = 5\bl{m} + \ora{h} 
\uncover<9->{\Rightarrow 2 = 5 \cdot (\bl{-\frac{1}{3}}) + \ora{h}}
\uncover<10->{\Leftrightarrow \ora{h} = 2 -5 \cdot (\bl{-\frac{1}{3}})}
\uncover<11->{\Leftrightarrow \ora{h} = \dfrac{6}{3} + \dfrac{5}{3}}
\uncover<12->{\Leftrightarrow \ora{h = \dfrac{11}{3}}}
$$
\uncover<13->{On a donc $y = \bl{-\frac{1}{3}}x + \ora{\frac{11}{3}}$.}
}}

\end{frame}

\begin{frame}

\underline{Exemple 4.3} Calculer l'intersection $I(\ve{x_I}, \ro{y_I})$ des droites 
$\ro{y}=3\ve{x} + 2$ et $\ro{y}=-2\ve{x} + 4$. \\

\vfill

\hunc{2}{
Le point $I(\ve{x_I}, \ro{y_I})$ appartient aux deux droites, on a donc 
\uncover<3->{$$\begin{cases}
\ro{y_I}=3\ve{x_I} + 2  \\
\ro{y_I}=-2\ve{x_I} + 4 
\end{cases}$$}
\uncover<4->{On a donc, par substitution, 
\begin{equa}
3\ve{x_I} + 2  &=& -2\ve{x_I} + 4  & \uncover<5->{-2 + 2\ve{x_I}}\\
\uncover<6->{\LRq 5\ve{x_I} &=& 2  & \uncover<7->{\div 5}}\\
\uncover<8->{\LRq \ve{x_I} &=& \ve{\dfrac{2}{5}} \uzero }
\end{equa}}
\uncover<9->{On remplace dans l'une des équations: 
$$\ro{y_I} = 3\cdot \ve{\dfrac{2}{5}} + 2 
\uncover<10->{\Leftrightarrow \ro{y_I}= \dfrac{6}{5} + \dfrac{10}{5}}
\uncover<11->{\Leftrightarrow \ro{y_I}= \ro{\dfrac{16}{5}}}
$$}
\uncover<12->{On a donc $I(\ve{\frac{2}{5}}, \ro{\frac{16}{5}})$.}

}


\end{frame}

\begin{frame}

\underline{Remarque 4.2} La pente et l'ordonnée à l'origine de deux droites $y=m_1x+h_1$ et $y=m_2x+h_2$ nous permettent de déduire la \bl{position relative} des droites.

\uncover<2->{
\vfill
On distingue trois cas:\\
\only<beamer>{\small}
\begin{minipage}{0.32\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw[-latex, dgreen] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred] (0,-3) -- (0,3) node[above]{$y$};
\clip (-3,-3) rectangle (3,3);
\hunc{5}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {1.5*\x});}
\hunc{5}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {-2*(\x)+1});}
\hunc{5}{\fill (1/3.5,1.5/3.5) circle (2pt) node[right]{\footnotesize $I(x_I; y_I)$};}
\end{tikzpicture}
\hunc{6}{Droites sécantes\\
$m_1 \neq m_2$\\
\hspace{1cm}}\\[0.1cm]
\hunc{13}{Une intersection \\  $S=\{(x_I,y_I)\}$}
\end{center}
\end{minipage}
\begin{minipage}{0.32\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw[-latex, dgreen] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred] (0,-3) -- (0,3) node[above]{$y$};
\clip (-3,-3) rectangle (3,3);
\hunc{8}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {1.5*\x});}
\hunc{8}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {1.5*\x+1});}
\end{tikzpicture}
\hunc{9}{Droites parallèles\\
$m_1 = m_2$\\
$h_1 \neq h_2$\\[0.1cm]
}
\hunc{14}{Pas d'intersections \\ $S=\emptyset$}
\end{center}
\end{minipage}
\begin{minipage}{0.32\textwidth}
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw[-latex, dgreen] (-3,0) -- (3,0) node[right]{$x$};
\draw[-latex, dred] (0,-3) -- (0,3) node[above]{$y$};
\clip (-3,-3) rectangle (3,3);
\hunc{11}{\draw[samples=2, dblue, thick, domain=-3:3] plot(\x, {-1.5*\x});}
\end{tikzpicture}
\hunc{12}{Droites confondues\\
$m_1 = m_2$\\
$h_1 = h_2$}\\[0.1cm]
\hunc{15}{Infinité d'intersections \\ \footnotesize $S~=~\{(x,y) | y~=~m_1x~+~h_1 \}$ }
\end{center}
\end{minipage}
}
\end{frame}

\end{document}






