Question

encontre as coordenadas do vértice f(x)=x2-4x+3

Answer 1

$f(x)=x^2-4x+3\\\\ a=1;b=-4;c=3\\ \\\\ x_v= \frac{-b}{2a} \to x_v= \frac{-(-4)}{2.1} \to x_v= \frac{4}{2} \to \boxed{x_v=2}\\\\\\\\ \Delta=b^2-4ac\to \Delta=(-4)^2-4.1.3\to \Delta=16-12\to \Delta=4\\\\\\ y_v= \frac{-\Delta}{4a} \to y_v= \frac{-4}{4.1} \to y_v= \frac{-4}{4} \to \boxed{y_v=-1}\\\\\\\\ \boxed{\boxed{V=\{2;-1\}}}$