Graphing
posted by Nada .
What are the steps to graphing this quadratic equation? Y=(2x)^2

It is a parabola U. See what happens with the y's when you use 0, +1, and 1 for the x's.

Y = (2X)^2 = 4X^2.
1. Calculate coordinates of the vertex:
h = Xv = b / 2a = o/8 = 0.
K = Yv = 4(0)^2 = 0. So the vertex is
at the origin. V(h.k) = (0 , 0).
2. Select values of X on both sides
of vertex and cal. the corresponding
value of Y using the given Eq.
(2 , 16) , (1 , 4) , (0 , 0) ,
(1 , 4) , (2 , 16).
The graph should show a Yparabola
that opens upward with vertex at the
origin.