We can't square each side because that creates a radical function, which we don't want. Instead, let's divide each side by 16:
#y^2/16=(16x)/16#
#x=y^2/16#
Let's rewrite this as #x=1/16 y^2# since it will be easier to help us find what we need.
We don't have much to go off of, but we can find #p#. You might know this as the distance from the vertex of a parabola to both its focus and its directrix. We can find #p# by setting the scale factor equal to #1/(4p)#:
#1/16=1/(4p)#
#4p=16#
#p=4#
Before we go any further, we should know that the graph faces the right. How do we know this? 1) The slope is positive, and 2) #x#, not #y#, is isolated on one side of the equation.
The vertex is #(0,0)# because there are no translations in the function. When a parabola has a vertex at the origin, the focus is #(0,p)#. The focus is always within the parabola, so a right-facing parabola has a focus to the right of the vertex. Our focus, therefore, would be 4 units to the right of the vertex and is #(0,4)#. Finally, the directrix is #x=-4# because directrix is #p# units away from the parabola, but in the opposite direction.
