How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y= -2x^2#?

#y=-2x^2#
and be re-written in explicit vertex form as
#y= (-2)(x-color(red)(0))^2 + color(blue)(0)#
#color(white)("XXX")#with vertex at #(color(red)(0),color(blue)(0))#

Any parabola with the form:
#color(white)("XXX")y=ax^2+bx+c#
has a vertical axis of symmetry (through the vertex)
and opens upward if #a > 0# or downward if #a < 0#
Therefore the axis of symmetry for #y=-2x^2 (+0x+0)#
is #x=0#
(and opens downward).

Note that
#color(white)("XXX")#x-intercept (value of #x# when #y=0#) is #0#
and
#color(white)("XXX")#y-intercept (value of #y# when #x=0#) is #0#
so
#color(white)("XXX")#the x and y intercepts do not give use any additional point to help us plot the graph.
Therefore it will be necessary to pick a few other values for #x# and evaluate their corresponding #y# values in order to plot the graph.

#{: (x,,y), (0,,0), (-1,,-2), (+1,,-2), (-2,,-8), (+2,,-8) :}#

By plotting these point-pairs on the Cartesian plane we should be able to sketch the graph for this function.

(Unfortunately, at this time, there seems to be a bug with Socratic.org that prevents posting a graph. I will try to check again later and add the graph if possible).