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

A slight variant on method
##color(blue)(Vertex ->(xy)->(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)##
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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 ->(xy)->(2-1)##