Question

coordenadas do vértice da parábola da função f(x)=x2-4

Answer 1

Resolvendo:

$f(x)=x^{2} -4\\\\a=1\:\:\:b=0\:\:\:c=-4\\\\x_{v} =\frac{-b}{2a} \\\\x_{v} =\frac{0}{1} \\\\x_{v} =0\\\\\\y_{v} =\frac{-(b^{2}-4ac) }{4a} \\\\y_{v}=\frac{-(0-4\:.\:1\:.\:(-4)) }{4}\\\\y_{v}=\frac{-(16) }{4}\\\\y_{v}=-4$

As coordenadas do vértice = (0; -4)

Espero ter ajudado!

Desculpe qualquer erro.