How do you find the vertex of the parabola whose equation is #y = x^2 - 4x + 3#?

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Answer 1 · 2016-02-08T07:21:33

Vertex is at #(2, -1)#

the given
#y=x^2-4x+3#
completing the square method

add and subtract 4 to form a "perfect square trinomial"

#y=x^2-4x+4-4+3 #

#y=(x^2-4x+4)-4+3 # group the trinomial

#y=(x-2)^2-4+3 # write the equivalent

#y=(x-2)^2-1# simplify

#y+1=(x-2)^2# transpose the -1 to the left side

#y--1=(x-2)^2# rearrange the signs

following the form #4p(y-k)=(x-h)^2#

it is clearly seen that vertex #(h, k)=(2, -1)#