Question

determine as coordenadas do vértice da parábola representada pela função 2 grau: f(x)=4x²-4x+1

Answer 1

$f ( x ) =4x^2-4x+1 \\ \\ a=4~~~~b=-4~~~~c=1 \\ \\ \Delta=(-4)^2-4*4*1 \\ \\ \Delta=16-16 \\ \\ \Delta=0~~~(x'=x''~~~Possui~raizes~iguais) \\ \\ Yv= \frac{-\Delta}{4a} = \frac{-0}{16} = 0 \\ \\ Xv= \frac{-b}{2a} = \frac{4}{8} = \frac{1}{2} \\ \\ V= {( \frac{1}{2} ,0)$