Find the minimum value of the parabola y=x^2-4x+6 Simplify your answer and write it as a proper fraction, improper fraction, or integer. square

To find the minimum value of the parabola $y=x^{2}-4x+6$, we can use the vertex form of a quadratic equation, which is $y=a(x-h)^{2}+k$, where $(h,k)$ is the vertex of the parabola.<br /><br />First, let's rewrite the given equation in vertex form by completing the square:<br /><br />$y=x^{2}-4x+6$<br />$y=(x^{2}-4x)+6$<br />$y=(x^{2}-4x+4-4)+6$<br />$y=(x-2)^{2}-4+6$<br />$y=(x-2)^{2}+2$<br /><br />Now, we can see that the vertex of the parabola is $(2,2)$. Since the coefficient of $(x-2)^{2}$ is positive, the parabola opens upwards, and the vertex represents the minimum point.<br /><br />Therefore, the minimum value of the parabola is $2$.<br /><br />Answer: $\boxed{2}$