How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y= -2x^2#?

1 Answer
Alan P.
Aug 21, 2015

vertex at #(0,0)#
axis of symmetry: #x=0#
y-intercept: #0#
x-intercept: #0#

Explanation:

General vertex form for a quadratic is
#color(white)("XXXX")y=m(x-a)+b#
#color(white)("XXXXXXXX")#with the vertex at #(a,b)#

#y=-2x^2# can be written in vertex form as
#color(white)("XXXX")y = (-2)(x-0)^2+0#
#color(white)("XXXXXXX")#with vertex at #(0,0)#

The equation is that of a standard parabola (opening downward)
so the axis of symmetry is a vertical line passing through the vertex
i.e. #x=0#

The y-intercept is the value of #y# when #x=0#

The x-intercept(s) is/are the value of #x# when #y=0# (in this case there is only one solution to #0=-2x^2#)

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