Question

(ANGLO) O vértice da parábola Y=2X^- 8X+5 É

Answer 1

$\large\boxed{\begin{array}{l}\rm o\,v\acute ertice\,da\,func_{\!\!,}\tilde ao~y=ax^2+bx+c\\\rm \acute e\,o\,ponto\,V(x_V,y_V)\\\rm tal\,que~x_V=-\dfrac{b}{2a}~e~y_V=-\dfrac{\Delta}{4a}\\\rm onde\,\Delta=b^2-4ac\end{array}}$

$\large\boxed{\begin{array}{l}\rm y=2x^2-8x+5\\\rm\Delta=b^2-4ac\\\rm\Delta=(-8)^2-4\cdot2\cdot5\\\rm\Delta=64-40\\\rm\Delta=24\\\rm x_V=-\dfrac{b}{2a}\\\\\rm x_V=-\dfrac{-8}{2\cdot2}=\dfrac{8}{4}=2\\\\\rm y_V=-\dfrac{\Delta}{4a}\\\\\rm y_V=-\dfrac{24}{4\cdot2}=-\dfrac{24}{8}=-3\\\\\rm V(2,-3)\end{array}}$