Question

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

Answer 1

Vertex: `(0,0)`
Axis of symmetry: `x=0`
(See below for graph)

Explanation:

`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).