Where is the vertex of the parabola #y = x^2 - 6x + 12#?

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Answer 1 · 2016-03-13T20:49:11

#color(blue)(" "x_("vertex")=+3)#
I will let you solve for #y_("vertex")# by substitution.

Consider the standard form equation of:
#" "y=ax^2+bx+c#

Write this as
#" "y=a(x^2+b/ax)+c#

In your case #a=1# so we have

#b/a =-6#

Apply #(-1/2xxb/a) -> (-1/2)xx(-6/1) = +3#

#color(blue)(" "x_("vertex")=+3)#

By substitution of #x=3# in the original equation you can determine
#y_("vertex")#