Question

How do you graph the parabola `y=2x^2` using vertex, intercepts and additional points?

Answer 1

See below.

Explanation:

`y=2x^2` is a quadratic function of the form `y=ax^2+bx+c`
Where: `a=2, b=0 and c=0`

Since `y` is a quadratic, its graph will be a parabola.

Also, since `a>0`, `y` will have a single minimum at it vertex.

The vertex of `y` will be on its axis of symmetry where `x=(-b)/(2a)` which is `x=0` in this case. Hence, the vertex of `y` is `(0,0)`

`:. y_min = 0`

`->` the graph of `y` has no intercepts other than `(0,0)`

We will need additional point to plot the graph.

We will use two points each side of the `y-` axis. (Remember that the graph is symmetric about the `y-`axis.).

`x=+-1 -> y=2`
`x=+-2 -> y=8`

From which we can plot the graph below.

graph{2x^2 [-11.58, 10.93, -1.165, 10.075]}