Déterminer la forme canonique d'un trinôme Exercice

P\left(x\right)=\left(x+3\right)^{2}-4

P\left(x\right)=\left(x+3\right)^{2}-1

P\left(x\right)=\left(x+3\right)^{2}+14

P\left(x\right)=\left(x+2\right)^{2}+1

P\left(x\right)=-4\left(x-4\right)^2+72

P\left(x\right)=-4\left(x+4\right)^2+24

P\left(x\right)=-4\left(x-4\right)^2-8

P\left(x\right)=-4\left(x+4\right)^2-56

P\left(x\right)=-2\left(x+\dfrac{3}{2}\right)^{2}-\dfrac{1}{2}

P\left(x\right)=-2\left(x-\dfrac{3}{2}\right)^{2}-\dfrac{19}{2}

P\left(x\right)=-2\left(x-\dfrac{3}{2}\right)^{2}-\dfrac{1}{2}

P\left(x\right)=-2\left(x+\dfrac{3}{2}\right)^{2}-\dfrac{29}{4}

P\left(x\right)=-3\left(x-\dfrac{1}{3}\right)^{2}+\dfrac{8}{9}

P\left(x\right)=-3\left(x-\dfrac{1}{3}\right)^{2}+\dfrac{10}{9}

P\left(x\right)=-3\left(x-\dfrac{1}{3}\right)^{2}+\dfrac{4}{3}

P\left(x\right)=-3\left(x-\dfrac{1}{3}\right)^{2}+\dfrac{2}{3}

P\left(x\right)=5\left(x-2\right)^2+14

P\left(x\right)=5\left(x-2\right)^2+6

P\left(x\right)=5\left(x-2\right)^2+30

P\left(x\right)=5\left(x-2\right)^2-10

P\left(x\right)=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{55}{8}

P\left(x\right)=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{73}{8}

P\left(x\right)=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{137}{16}

P\left(x\right)=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{119}{8}