How do you write #y=x^2-2x+1# in vertex form and identify the vertex, y intercept and x intercept?

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Answer 1 · 2015-06-07T21:30:37

#y= x^2-2x+1 = (x-1)^2 + 0#

This has vertex at #x=1#, #y=0#, that is #(1, 0)#.

The intercept with the #y# axis is where #x=0#, so substituting #x=0# into the formula for #y# we get:

#y = (-1)^2+0 = 1#

So the #y# intercept is at #(0, 1)#

The intercept with the #x# axis is at the vertex #(1, 0)# where the parabola touches it.

graph{x^2-2x+1 [-10, 10, -5, 5]}