Question 625217
Find the vertex, line of symmetry,and the maximum or minimum value of f(x), graph the function. 
f(x)=(1)/(3)(x+1)^(2)+6
Given equation is that of a parabola that opens upwards (has a minimum).
Its form of equation: A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.
For given equation: (1/3)(x+1)^2+6
Vertex: (-1,6)
Line of symmetry: x=-1
minimum value=6
see graph below:
 {{{ graph( 300, 300, -10,10, -10, 10,(1/3)(x+1)^2+6) }}}