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

1 Answer
Jun 26, 2018

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]}