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

To find the minimum value of the parabola $y=x^{2}+2$, we need to determine the vertex of the parabola.<br /><br />The vertex form of a parabola is given by $y=a(x-h)^{2}+k$, where $(h,k)$ is the vertex of the parabola.<br /><br />In this case, the equation of the parabola is $y=x^{2}+2$, which can be rewritten as $y=1(x-0)^{2}+2$.<br /><br />Comparing this with the vertex form, we can see that the vertex of the parabola is $(0,2)$.<br /><br />Since the coefficient of $x^{2}$ is positive, the parabola opens upwards, and the vertex represents the minimum point.<br /><br />Therefore, the minimum value of the parabola $y=x^{2}+2$ is $2$.<br /><br />Answer: $\boxed{2}$