1. **State the problem:** Convert the quadratic equation $y = x^2 - 4x + 6$ into vertex form. 2. **Recall the vertex form:** The vertex form of a quadratic is $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. 3. **Complete the square:** Start with $$y = x^2 - 4x + 6$$ 4. Factor out the coefficient of $x^2$ (which is 1 here, so no change): $$y = x^2 - 4x + 6$$ 5. Take half of the coefficient of $x$, which is $-4$, half is $-2$, and square it: $$(-2)^2 = 4$$ 6. Add and subtract 4 inside the equation to complete the square: $$y = (x^2 - 4x + 4) + 6 - 4$$ 7. Rewrite the trinomial as a perfect square: $$y = (x - 2)^2 + 2$$ 8. **Final vertex form:** $$y = (x - 2)^2 + 2$$ which shows the vertex at $(2, 2)$. This form is useful for graphing and understanding the parabola's vertex and direction.