# How do you write f(x)=2x^2+12x+12 in vertex form?

Sep 3, 2016

$y = 2 {\left(x + 3\right)}^{2} - 6$

Vertex form is also known as completing the square

#### Explanation:

$\textcolor{b l u e}{\text{Step 1}}$

Write as:

$y = 2 \left({x}^{2} + 6 x\right) + 12 + k$

When we start to change things the equation becomes untrue. So we need to introduce the correction $k$ to compensate for this. The value of $k$ is calculated at the end.

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$\textcolor{b l u e}{\text{Step 2}}$

Move the power of 2 from ${x}^{2}$ to outside the brackets.

$y = 2 {\left(x + 6 x\right)}^{2} + 12 + k$

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$\textcolor{b l u e}{\text{Step 3}}$

Halve the 6 from $6 x$

$y = 2 {\left(x + 3 x\right)}^{2} + 12 + k$

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$\textcolor{b l u e}{\text{Step 3}}$

Remove the $x$ from $3 x$

$y = 2 {\left(x + 3\right)}^{2} + 12 + k$ .....................Equation(1)
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$\textcolor{b l u e}{\text{Step 4}}$

Determine the value of $k$

If you were to square the bracket we would have $\textcolor{m a \ge n t a}{{3}^{2}}$ form $\textcolor{g r e e n}{2} {\left(x + \textcolor{m a \ge n t a}{3}\right)}^{2}$. Also this is multiplied by the $\textcolor{g r e e n}{2}$ from outside the bracket giving:

$\textcolor{g r e e n}{2 \times} \textcolor{m a \ge n t a}{{3}^{2}} \leftarrow \text{ this is the error we introduced}$

So we write:

$\textcolor{g r e e n}{2 \times} \textcolor{m a \ge n t a}{{3}^{2}} + k = 0$

$\implies 18 + k = 0 \to k = - 18$

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$\textcolor{b l u e}{\text{Step 5 - The final equation}}$

Substitute this into equation(1)

$y = 2 {\left(x + 3\right)}^{2} + 12 - 18$

$y = 2 {\left(x + 3\right)}^{2} - 6$

$\textcolor{red}{\text{I have superimposed both graphs so that you can see they are the same.}}$