To rewrite the quadratic expression $x^2 - 6x + 12$ in vertex form, we complete the square. The vertex form of a quadratic equation is:

$y = a(x-h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

Step-by-Step Solution:

1. Start with the given equation: $y = x^2 - 6x + 12$

2. Factor out the coefficient of $x^2$ (if necessary, but here $a = 1$, so no factoring is needed): $y = (x^2 - 6x) + 12$

3. Complete the square:

   • Take half the coefficient of $x$, which is $-6$, divide by 2, and square it: $\left(\frac{-6}{2}\right)^2 = 9$
   • Add and subtract 9 inside the parentheses to complete the square: $y = (x^2 - 6x + 9 - 9) + 12$

4. Rewrite the trinomial as a perfect square: $y = ((x - 3)^2 - 9) + 12$

5. Simplify the expression: $y = (x - 3)^2 + 3$

Final Vertex Form:

$y = (x - 3)^2 + 3$

Vertex:

The vertex of the parabola is $(3, 3)$.

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Would you like further clarification or additional examples? Here are some related questions:

1. How do you find the vertex form of a quadratic equation in general?
2. What is the significance of the vertex in a parabola?
3. How can you determine if a parabola opens upward or downward?
4. How do you convert vertex form back to standard form?
5. Can you graph $y = (x - 3)^2 + 3$ and identify key points?

Tip: Completing the square is useful not only for rewriting quadratics but also for solving quadratic equations and analyzing their properties!