Question

Quais os vértices dessa função f(x)= -4x^2+4x+5 ?

Answer 1

Resolução pelas coordenadas da função, veja:

$Xv = \dfrac{-b}{2a}\\\\\\ Xv = \dfrac{-4}{2~.~(-4)}}\\\\\\ Xv = \dfrac{-4}{-8}\\\\\\ \large\boxed{\boxed{\boxed{Xv = \dfrac{1}{2}.}}}}}}\\\\\\\\\\\\\\\\ Yv = \dfrac{-\Delta}{4a}\\\\\\ \Delta= {4^{2}-4~.~(-4)~.~5}\\\\\\ \Delta = 16+80\\\\\\\ \Delta= 96\\\\\\\\ Yv = \dfrac{-\Delta}{4a}\\\\\\ Yv = \dfrac{-96}{4~.~(-4)}\\\\\\\\ Yv = \dfrac{-96}{-16}\\\\\\ \large\boxed{\boxed{\boxed{\boxed{Yv=6.}}}}}}}$

Espero que te ajude :).