Question

dada a função quadratica f x -2.x ao quadrado + 4.x -9 as coordenadas do vértice do gráfico da parabola definida por f(x) , é

Answer 1

$f(x)=-2x^2+4x-9\\\\\\ -2x^2+4x-9=0\\\\ a=-2~;~b=4~;~c=-9\\\\\\Coeficiente~~angular~~negativo, ~~concavidade~~voltada~~para~~baixo\\\\\\ \Delta=b^2-4ac\to \Delta=4^4-4.(-2).(-9)\to \Delta=16-72 \to \boxed{\Delta=-56}\\\\\\\\x_v= \dfrac{-b}{2a} \to~~x_v= \dfrac{-4}{2.(-2)} \to~~ x_v= \dfrac{-4}{-4} \to~~ \large\boxed{x_v=1} \\\\\\ y_v= \dfrac{-\Delta}{4a} \to~~y_v= \dfrac{-(-56)}{4.(-2)} \to~~ y_v= \dfrac{56}{-8} \to~~ \large\boxed{y_v=-7} \\\\\\\\\\ \large\boxed{\boxed{V=\{1~;~-7\} }}$