To find the vertex, complete the square:
#f(x)=(x+6/2)^2-6#, which is
#(x+3)^2-6#
This is clearly a minimum value of #-6# when #x=-3#, so the vertex is at #(-3,-6)#, because #(-3)^2+6(-3)+3=-6#.
The intercept with the #y#-axis is when #x=0#, #y# as #0^2+6(0)+3#, giving the intercept as #(0,3)# (i.e., just read off the constant term).
The intercepts, if any, with the #x#-axis have #x#-coordinates which are the roots of the equation #f(x)=0#, which are the roots of #(x+3)^-6=0#. Clearly these are at #(x+3)^2=6#, that is #(x+3)=+-sqrt(6)# so #x=-3+-sqrt(6)#.
Alternatively, use the quadratic formula to get #x=(-(6)+-sqrt((6)^2-4xx(1)xx(3)))/(2xx(1))#
#x=-3+-sqrt(6)# because #sqrt(24)=2xxsqrt(6)#
