How do you find the vertex of the parabola by completing the square $x^2-6x+8=y$ ?

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Answer 1 · 2015-06-03T19:58:13

$y=x^2-6x+8=x^2-6x+9-1=(x-3)^2-1$

This has its vertex where $x-3 = 0$ , that is $x=3$ .

When $x=3$ we have $y = 0^2-1 = -1$

So the vertex is at $(3, -1)$

How do you find the vertex of the parabola by completing the square x^2-6x+8=yx^2-6x+8=y?