How do you find the vertex of the parabola ##y=2x^2-8x+7##?

A slight variant on method

##color(blue)("Vertex" ->(x,y)->(2,-1)##

Given:##" "2x^2-8x+7##

Write as:##" " 2(x^2-color(red)(8/2)x)+7##

Consider the ##color(red)(-8/2)" from "-8/2x##

##color(blue)(x_("vertex")=(-1/2)xx(color(red)(-8/2)) = +8/4 = 2)##

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute ##x=color(magenta)(2)## in the original equation to find ##y_("vertex")##

##y=2x^2-8x+7" "->" "y_("vertex")=2(color(magenta)(2))^2-8(color(magenta)(2))+7##

##color(blue)(y_("vertex")= 8-16+7=-1)##

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ##color(blue)("Vertex" ->(x,y)->(2,-1)##