Bonsoir,
[tex]\textnormal{La forme canonique d'un trinome est sous la forme : $a(x-\alpha )^2 + \beta $}[/tex]
[tex]\alpha = -\frac{b}{2a}[/tex] [tex]\beta = f(\alpha )[/tex]
[tex]1)[/tex]
[tex]\alpha = -\frac{b}{2a} \iff -\frac{6}{2} = -3[/tex]
[tex]\beta \iff f(-3) = (-3)^2 + 6 \times (-3) - 8 = 9 -18 - 8 = -17[/tex]
[tex]a(x-\alpha )^2 + \beta \iff (x-(-3))^2 - 16 \iff \boxed{(x+3)^2 - 17}[/tex]
[tex]2)[/tex]
[tex]\alpha = -\frac{b}{2a} \iff - (\frac{-5}{2}) = \frac{5}{2}[/tex]
[tex]\beta \iff f(\frac{5}{2}) = (\frac{5}{2} )^2 -5 \times \frac{5}{2} + 3 = -\frac{13}{4}[/tex]
[tex]a(x-\alpha )^2 + \beta \iff \boxed{(x-\frac{5}{2} )^2 - \frac{13}{4} }[/tex]
[tex]3)[/tex]
[tex]\alpha = -\frac{b}{2a} \iff -\frac{6}{4} = -\frac{3}{2}[/tex]
[tex]\beta \iff f(-\frac{3}{2}) = 2 \times (-\frac{3}{2})^2 + 6 \times (-\frac{3}{2}) + 4 = -\frac{1}{2}[/tex]
[tex]a(x-\alpha )^2 + \beta \iff 2(x-(-\frac{3}{2}) )^2 - \frac{1}{2} \iff \boxed{2(x+\frac{3}{2} )^2 - \frac{1}{2}}[/tex]