How do you find the vertex of #y = x^2 - 6x + 7#?

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Answer 1 · 2018-04-14T22:08:15

#(3, -2)#

The equation is in standard form: #y=ax^2+bx+c#

Finding the vertex of an equation in standard form: #((-b)/(2a), f((-b)/(2a)))#

Here, #a# is #1# and #b# is #-6#

#(-(-6))/(2*1)#

#6/2#

#3 rarr# The x-coordinate of the vertex is #3#.

#f(3)=3^2-6*3+7#

#f(3)=9-6*3+7#

#f(3)=9-18+7#

#f(3)=-9+7#

#f(3)=-2#