What is the vertex form of #x= 4y^2 + 16y + 16#?

1 Answer
Jul 25, 2017

See a solution process below:

Explanation:

To convert a quadratic from #x = ay^2 + by + c# form to vertex form, #x = a(y - color(red)(h))^2+ color(blue)(k)#, you use the process of completing the square.

This equation is already a perfect square. We can factor out a #4# and complete the square:

#x = 4y^2 + 16y + 16 - color(red)(16)#

#x = 4(y^2 + 4y + 4)#

#x = 4(y + 2)^2#

Or, in precise form:

#x = 4(y + (-2))^2 + 0#