Begin by grouping the terms involving #x# together.

#y=(-4x^2-x)-3#

Factor out #-4# from the #x# terms.

#y=-4(x^2+1/4x)-3#

Complete the square. Using the formula #(b/2)^2# we get #((-1/4)/2)^2=(-1/8)^2=1/64#.

We now know that to complete the square by adding #1/64# within the parentheses. Because we are adding #1/64#, we must also subtract the amount by which it changed the problem.

#y=-4(x^2+1/4x+1/6464/)-3+1/16#

Since the #1/16# is within the parentheses, it is multiplied by #-4#, meaning overall, it changes the problem by #-1/16#. To undo this change, we add #1/16# outside the parentheses.

Now that we completed the square, the terms involving #x# can be factored like so:

#y=-4(x+1/8)^2-47/16#

The equation is now written in vertex form.