1) f:R→R f(x)=x²-2mx+3m, are a=1; b=-2m, c=3m, cum a=1>0 ⇒f are un minim =
-Δ/4a, deci -(b²-4ac)/4a=2, sau -(4m²-4*3m)/4=2, ⇒-4m²+12m-8=0, impartim cu -4, ⇒m²-3m+2=0, ⇒ m∈{1; 2}.
2) f(x)=ax²+bx+c, are forma canonica: f(x)=a[tex][(x+ \frac{b}{2a} )^2- \frac{b^2-4ac}{4a^2} ][/tex]. In cazul de fata f(x)=2x²+4x+6=2[x²+2x+3]=2[(x+1)²+2].