How do you find the vertex for #y = x^2 - 2x#?

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Answer 1 · 2018-04-03T07:47:10

The vertex is at #(1,-1)#

We can quite easily see where the vertex of the quadratic function is if we write it in vertex form:

#a(x-h)^2+k# with vertex at #(h,k)#

To complete the square, we need #h# to be half the #x# coefficient, so in this case we have #-2 \/ 2=-1#:

#(x-1)^2+k=x^2-2x#

#x^2-2x+1+k=x^2-2x#

#k=-1#

This means the vertex form of our quadratic function is:

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

And therefore the vertex is at #(1,-1)#