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

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Answer 1 · 2017-11-14T11:43:02

Vertex is at #(2.5,0.75) #

#y=x-2+(x-3)^2 or y=x-2+x^2-6x+9# or

# y=x^2-5x+7 or y=(x^2-5x) +7# or

#y={x^2-5x +(5/2)^2}-25/4 +7# or

#y=(x-2.5)^2+3/4 or y={x-2.5)^2+0.75 #

Comparing with vertex form of equation

#y = a(x-h)^2+k ; (h,k)# being vertex we find

here #h=2.5 , k=0.75 :.# Vertex is at #(2.5,0.75) # .

graph{(x-2)+(x-3)^2 [-10, 10, -5, 5]} [Ans]