Question 160400
Find the line of symmetry (sp!), vertex, and the min. or max. of the curve:
{{{f(x) = (1/4)(x+2)^2+2}}}
Your quadratic equation is already written in the "vertex" form of:{{{y = a(x-h)^2+k}}} where the vertex is located at the point (h, k), so you can write the coordinates of the vertex directly from your equation: (h = -2 and k = 2)
Vertex is at (-2, 2)
The line of symmetry is the vertical line that passes through the point x = -2, so the equation of the line of symmetry is: {{{highlight(x = -2)}}}.
The parabola represented by your equation open upwards which you can tell because the coefficient of {{{x^2}}} i.e. ({{{a = 1/4}}}) is positive, so the vertex is a minimum. 
See the graph below:
{{{graph(400,400,-10,5,-2,10,(1/4)(x+2)^2+2)}}}