How do you find the vertex of #y= x^2 - 6x + 14#?

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Answer 1 · 2016-03-19T15:20:01

Vertex#" "->(x,y) ->(3.5)#

The question does not define the method to be used so you have freedom of choice.

Let me show you a 'sort of' cheat method. Actually it forms part completing the square but does not take to the end of the process.

Consider standard form#" " y=ax^2+bx+c#

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

Then you have #x-("vertex")=(-1/2)xxb/a#

For this question:
#a=1#
#b/a =(-6)/1 =-6#

#=>" "color(blue)(x_("vertex")=(-1/2)xx(-6) = +3)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
By substitution

#color(brown)(y=x^2-6x+14 color(blue)(->y_("vertex")=(3)^2-6(3)+14))#

#color(blue)(y_("vertex")=9-18+14=5)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B