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)