# How do you write  f(x)= x^2 + 6x + 12 in vertex form?

Apr 14, 2017

$f \left(x\right) = {\left(x + 3\right)}^{2} + 3$

#### Explanation:

The equation of a parabola in $\textcolor{b l u e}{\text{vertex form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where " (h,k)" are the coordinates of the vertex}$ and a is a constant.

$\text{Rearrange " f(x)=x^2+6x+12" into this form}$

$\text{using the method of "color(blue)" completing the square}$

$f \left(x\right) = \left({x}^{2} + 6 x \textcolor{red}{+ 9}\right) \textcolor{red}{- 9} + 12$

$\textcolor{w h i t e}{f \left(x\right)} = {\left(x + 3\right)}^{2} + 3 \leftarrow \textcolor{red}{\text{ in vertex form}}$

Apr 14, 2017

$f \left(x\right) = {\left(x + 3\right)}^{2} + 3$ (with some simplification, see below)

#### Explanation:

The general vertex form is
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given $f \left(x\right) = {x}^{2} + 6 x + 12$

Extracting the "spread" factor $\textcolor{g r e e n}{m}$
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \textcolor{g r e e n}{1} \left({x}^{2} + 6 x\right) + 12$

Completing the square
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \textcolor{g r e e n}{{x}^{2} + 6 x + \textcolor{m a \ge n t a}{{3}^{2}}} + 12 - \textcolor{g r e e n}{1} \cdot \textcolor{m a \ge n t a}{{3}^{2}}$

Re-writing with a squared binomial and some simplification:
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \textcolor{g r e e n}{1} {\left(x + 3\right)}^{2} + 3$

...or in explicit vertex form:
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \textcolor{g r e e n}{1} {\left(x - \textcolor{red}{\left(- 3\right)}\right)}^{2} + \textcolor{b l u e}{3}$
(note that I dropped the $\textcolor{g r e e n}{1}$ and simplified the binomial in my "Answer" even though I consider this to be the more accurate version.

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In order to check that I didn't make any mechanical errors, I've plotted the original equation, and it does appear to have the vertex indicated by the derived equation: