f(x)=x²-5
a=1>0 => f admite mimim
[tex]y \: min = - \frac{delta}{4a}[/tex]
Δ=b²-4ac
Δ=0²-4·1·(-5)
Δ=20
[tex]y \: min = - \frac{20}{4 \times 1} = - 5[/tex]
f(x)=-12x²+31x-1
a=-12<0 => f admite maxim
[tex]y \: max = - \frac{delta}{4a} [/tex]
Δ=b²-4ac
Δ=31²-4·(-12)·(-1)
Δ=961-48
Δ=913
[tex]y \: max = - \frac{913}{4 \times ( - 12)} = \frac{913}{48} [/tex]