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

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Answer 1 · 2018-06-26T07:07:04

See below.

#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]}