Question

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

Answer 1

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`)