How do you find the vertex of #y = x^2-8x+7#?

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Answer 1 · 2016-01-25T17:11:44

#V(4,-9)#

Factor

#y = x^2 -8x +7#.

That's

#y = (x-4)^2 - 9#.

Then, transpose #- 9# to the other side of the equation, that would be

#y+9 = (x-4)^2#.

Remember: #x# is for h and #y# is for k.

Vertex will be #V(4,-9)#.