Resposta :
[tex]\boxed{\begin{array}{l}\sf f(x)=x^2+4x+3\\\sf\Delta=16-12=4\\\sf x_V=-\dfrac{4}{2\cdot1}=-2\\\sf y_V=-\dfrac{4}{4\cdot1}=-1\\\sf V(-2,-1)\end{array}}[/tex]
[tex]\boxed{\begin{array}{l}\sf f(x)=x^2+4x+3\\\sf\Delta=16-12=4\\\sf x_V=-\dfrac{4}{2\cdot1}=-2\\\sf y_V=-\dfrac{4}{4\cdot1}=-1\\\sf V(-2,-1)\end{array}}[/tex]