What is the vertex form of #y= 4x^2 – 36x+ 81 #?

1 Answer
Jul 30, 2017

See a solution process below:

Explanation:

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

First, we must isolate the #x# terms:

#y - color(red)(81) = 4x^2 - 36x + 81 - color(red)(81)#

#y - 81 = 4x^2 - 36x#

We need a leading coefficient of #1# for completing the square, so factor out the current leading coefficient of 2.

#y - 81 = 4(x^2 - 9x)#

Next, we need to add the correct number to both sides of the equation to create a perfect square. However, because the number will be placed inside the parenthesis on the right side we must factor it by #4# on the left side of the equation. This is the coefficient we factored out in the previous step.

#y - 81 + (4 * ?) = 4(x^2 - 9x + ?)#

#y - 81 + (4 * 81/4) = 4(x^2 - 9x + 81/4)#

#y - 81 + 81 = 4(x^2 - 9x + 81/4)#

#y - 0 = 4(x^2 - 9x + 81/4)#

#y = 4(x^2 - 9x + 81/4)#

Then, we need to create the square on the right hand side of the equation:

#y = 4(x - 9/2)^2#

Because the #y# term is already isolated we can write this in precise form as:

#y = 4(x - color(red)(9/2))^2 + color(blue)(0)#