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

Jul 30, 2017

See a solution process below:

#### Explanation:

To convert a quadratic from $y = a {x}^{2} + b x + c$ form to vertex form, $y = a {\left(x - \textcolor{red}{h}\right)}^{2} + \textcolor{b l u e}{k}$, you use the process of completing the square.

First, we must isolate the $x$ terms:

$y - \textcolor{red}{81} = 4 {x}^{2} - 36 x + 81 - \textcolor{red}{81}$

$y - 81 = 4 {x}^{2} - 36 x$

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

$y - 81 = 4 \left({x}^{2} - 9 x\right)$

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 + ?)

$y - 81 + \left(4 \cdot \frac{81}{4}\right) = 4 \left({x}^{2} - 9 x + \frac{81}{4}\right)$

$y - 81 + 81 = 4 \left({x}^{2} - 9 x + \frac{81}{4}\right)$

$y - 0 = 4 \left({x}^{2} - 9 x + \frac{81}{4}\right)$

$y = 4 \left({x}^{2} - 9 x + \frac{81}{4}\right)$

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

$y = 4 {\left(x - \frac{9}{2}\right)}^{2}$

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

$y = 4 {\left(x - \textcolor{red}{\frac{9}{2}}\right)}^{2} + \textcolor{b l u e}{0}$