The vertex is #(-1,0)#. The equation is in vertex form, which looks like
#y = a(x-h)^2 + k#
where #h# is the #x#-coordinate of the vertex and #k# is the #y#-coordinate. The equation above has no #k# value, so it's just #0#.
The axis of symmetry is #x = -1#. The axis of symmetry is always the #x# value of the vertex. You could also find it through the standard form of a parabola that looks like
#Ax^2 + Bx + C =0#
by calculating #-B/(2A)#.
The #x#-intercept is #(-1,0)#. You find this by setting #y = 0# and solving for #x#.
The #x#-intercept is the factors of the quadratic equation getting set equal to #0#. For example, the #x#-intercepts for the quadratic
#(x+4)(x-3) = 0#
are #(-4,0)# and #(3,0)#. You get this by doing
#x + 4 = 0" "# and #" " x -3 = 0#
and solving for #x#. For the equation above, there is only one different factor. That would only give one #x#-intercept. The graph would bounce on the point #(-1,0)# due to the even multiplicity.
graph{(x+1)^2 [-13.89, 12.96, -10.4, 48]}
