[tex]4x^2+12x+13=(2x)^2+2\cdot2x\cdot3+3^2+4=(2x+3)^2+4\geq4[/tex].
Deci valoarea minima este 4 si aceasta se ia pentru x=-3/2
Altfel:
Daca notez [tex]f(x)=4x^2+12x+13\Rightarrow f_{min}=-\dfrac{\Delta}{4a}[/tex]
[tex]unde\ \Delta =b^2-4ac=144-208=-64[/tex]
Deci [tex]f_{min}=\dfrac{64}{16}=4[/tex]