Juste la 1ere question svp

Bonjour Hayhay13

f est définie par f(x) = ax² + bx + c

[tex]a)\ \left\{\begin{matrix}f(1)=8\\f(-1)=6 \\f(2)=0 \end{matrix}\right.\ \ \ \ \ \left\{\begin{matrix}a\times1^2+b\times1+c=8\\a\times(-1)^2+b\times(-1)+c=6 \\a\times2^2+b\times2+c=0 \end{matrix}\right.\\\\\\\left\{\begin{matrix}a+b+c=8\ \ \ (1)\\a-b+c=6\ \ \ (2) \\4a+2b+c=0 \end{matrix}\right.\ \ \ \ \ \ \left\{\begin{matrix}(1)-(2)\Longrightarrow 2b=2\\a-b+c=6\\4a+2b+c=0 \end{matrix}\right.\\\\\\\left\{\begin{matrix}b=1\\a-1+c=6\\4a+2+c=0 \end{matrix}\right.[/tex]

[tex]\left\{\begin{matrix}b=1\\a+c=7\ \ \ (3)\\4a+c=-2\ \ \ (4) \end{matrix}\right.\ \ \ \ \left\{\begin{matrix}b=1\\(4)-(3)\Longrightarrow 3a=-9\\a+c=7 \end{matrix}\right.\\\\\\\left\{\begin{matrix}b=1\\a=-3\\a+c=7 \end{matrix}\right.\ \ \ \ \left\{\begin{matrix}b=1\\a=-3\\-3+c=7 \end{matrix}\right.\\\\\\\\\boxed{\left\{\begin{matrix}b=1\\a=-3\\c=10 \end{matrix}\right.}[/tex]

Par conséquent, [tex]\\\\\boxed{f(x)=-3x^2+x+10}[/tex]

b) Forme canonique de f(x)

[tex]f(x)=-3x^2+x+10\\\\f(x)=-3(x^2-\dfrac{1}{3}x-\dfrac{10}{3})\\\\f(x)=-3[\bigl( x^2-\dfrac{1}{3}x+(\dfrac{1}{6})^2\end{smallmatrix}\bigr)-(\dfrac{1}{6})^2-\dfrac{10}{3}]\\\\f(x)=-3[ (x-\dfrac{1}{6})^2-\dfrac{1}{36}-\dfrac{10}{3}]\\\\f(x)=-3[ (x-\dfrac{1}{6})^2-\dfrac{1}{36}-\dfrac{120}{36}][/tex]

[tex]f(x)=-3[(x-\dfrac{1}{6})^2-\dfrac{121}{36}]\\\\\boxed{f(x)=-3(x-\dfrac{1}{6})^2+\dfrac{121}{12}}[/tex]

[tex]c)\ f(x)=-3(x-\dfrac{1}{6})^2+\dfrac{121}{12}\\\\f(x)-\dfrac{121}{12}=-3(x-\dfrac{1}{6})^2\\\\Or\ \ -3\ \textless \ 0\ et\ (x-\dfrac{1}{6})^2\ge0\Longrightarrow -3(x-\dfrac{1}{6})^2\le0\\\\D'o\grave{u}\ \ f(x)-\dfrac{121}{12}\le0\\\\\boxed{f(x)\le\dfrac{121}{12}}[/tex]

Par conséquent, f admet un maximum égal à 121/12.

De plus, 
[tex]f(x)=\dfrac{121}{12}\\\\-3(x-\dfrac{1}{6})^2+\dfrac{121}{12}=\dfr\\\\-3(x-\dfrac{1}{6})^2=0\\\\x-\dfrac{1}{6}=0\\\\x=\dfrac{1}{6}[/tex]

Ce maximum de f est atteint pour x = 1/6

d) Tableau de variation de f:

[tex]\begin{array}{|c|ccccc|} x&-\infty&&\dfrac{1}{6}&&+\infty \\\\ f(x)&&\nearrow&\dfrac{121}{12}&\searrow&\\ \end{array}[/tex]

[tex]e)\ f(x)=0\\\\-3(x-\dfrac{1}{6})^2+\dfrac{121}{12}=0\\\\-3[(x-\dfrac{1}{6})^2-\dfrac{121}{36}]=0\\\\(x-\dfrac{1}{6})^2-\dfrac{121}{36}=0\\\\(x-\dfrac{1}{6})^2-(\dfrac{11}{6})^2=0\\\\(x-\dfrac{1}{6}-\dfrac{11}{6})(x-\dfrac{1}{6}+\dfrac{11}{6})=0[/tex]

[tex]\\\\(x-\dfrac{12}{6})(x+\dfrac{10}{6})=0\\\\(x-2)(x+\dfrac{5}{3})=0\\\\x-2=0\ \ ou\ \ x+\dfrac{5}{3}=0\\\\\boxed{x=2\ \ ou\ \ x=-\dfrac{5}{3}}[/tex]