Question 459952
First we write the equation of the Parabola in standard form:

{{{f(x)=a(x-h)^2+k }}},where (h, k) is the vertex of the Parabola.

In our case we have: {{{f(x)=(x+2)^2+0}}}, and the vertex is (-2, 0).

The value of x that produces the minimum is x=-2 and the value of the minimum is f(-2)=0. See the graph below:

{{{graph(300, 300, -5, 5, -5, 5, x^2+4x+4)}}}.