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

Answer 1

Julia Adams

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

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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Answer 2

Nathan Axel

The vertex of the equation ( x = (y - 6)^2 - y + 1 ) is (7, 6).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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