What is the the vertex of #y = (x-1)^2+4x-3#?

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Answer 1 · 2017-05-06T16:31:53

Vertex #(-1, -3)#

First distribute: #" "y = x^2 - 2x + 1 + 4x -3#

Add like terms: #" "y = x^2 +2x -2#

This equation is now in #y = Ax^2 + Bx^+C = 0#

The vertex is found when #x = -B/(2A) = -2/2 = -1#

and #y = (-1)^2 + 2(-1) - 2 = 1 -2 - 2 = -3#

You can also use completing of the square:

#y = (x^2 + 2x) - 2#

Half the x-term and complete the square by subtracting the square of that value:

#y = (x +1)^2 - 2 - (2/2)^2#

#y = (x+1)^2 - 3#

Standard form #y = (x-h)^2 -k#, where the vertex is #(h, k)#

vertex # = (-1, -3)#