Question

Determine  as coordenadas do vértice e trace o grafico da função quadrada y=x²-6x+5

Answer 1

$x^2-6x+5=0\\\Delta=(-6)^2-4.1.5\\\Delta=16$

$Vx= -\frac{b}{2a} ~~ Vx= \frac{6}{2} ~~\boxed{Vx=3}$

$Vy=- \frac{\Delta}{4a} ~~Vy=- \frac{16}{4} ~~ \boxed{Vy=-4}$

$x'~\to~1\\x''~\to~5$