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What is the the vertex of #x = (y - 6)^2 - y+1#?

1 Answer

Shwetank Mauria

Dec 28, 2017

Vertex is #(-5 1/4,-6 1/2)#

Explanation:

We can write #x=(y-6)^2-y+1# as

#x=y^2-12y+36-y+1#

= #y^2-13y+(13/2)^2-169/4+37#

= #(y-13/2)^2-(169-148)/4#

= #(y-13/2)^2-21/4#

Hence vertex is #(-21/4,-13/2)# or #(-5 1/4,-6 1/2)#

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