# How do you write  y=2x^2+4x+4 in vertex form?

May 7, 2016

$y = 2 {\left(x + 1\right)}^{2} + 2$ $\text{ }$The vertex will be at (-1, 2)

#### Explanation:

Use the same process as for "Completing a square",

Treat the right hand side as an expression, rather than as an equation.

Vertex form is also known as the form $a {\left(x + b\right)}^{2} + c$.
From this we can get the vertex (or the turning point) of a parabola as being $\left(- b , c\right)$

$y = 2 {x}^{2} + 4 x + 4$

Step 1. Divide each term by 2 to change $2 {x}^{2}$ to just make ${x}^{2}$

$y = 2 \left({x}^{2} + \textcolor{red}{2} x + 2\right)$

Step 2. Complete the square by adding ${\left(\frac{\textcolor{red}{2}}{2}\right)}^{2}$ = 1
If this was an equation, we would add the same to both sides. but as we are working with an expression, we will use additive inverses.
$+ 1 - 1 = 0$

$y = 2 \left({x}^{2} + 2 x + 1 - 1 + 2\right)$

The first 3 terms inside the bracket now represent the square of a binomial, so we can write them like that, and simplify the last 2 terms.

$y = 2 \left[{\left(x + 1\right)}^{2} + 1\right]$

Step 3. Multiply the 2 into the bracket and the answer is now in vertex form.

$y = 2 {\left(x + 1\right)}^{2} + 2$