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X^2 - 6X + 12 = 0

X^2 - 6X = -12

halve the linear term (-6), square it and add it to both sides

X^2 - 6X + 9 = -12 + 9

gather terms right side, factor terms left side

(X - 3)^2 = -3

(X - 3)^2 + 3 = 0

(3,3) are the vector coordinates

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Q: Where is the vertex of the parabola y equals x2 - 6x plus 12?
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"y = 2x2 - 12x + 6" is a quadratic equation which describes a parabola whose vertex occurs at the point (3, -12) and which has a range of -12 → ∞. It intercepts the x-axis at the points (3 - √6, 0) and (3 + √6, 0).


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What is the x-coordinate of the of the vertex of y equals -3x2 plus 12x-5?

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What is the vertex of the parabola y equals -2x squared plus 12x -13?

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