# How do you write in vertex form for y=x^2 + .6x - 11 by completing the square?

Jun 2, 2015

The general squared binomial is ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$

Given the equation $y = {x}^{2} + .6 x - 11$

If the coefficient of $x$ is $\left(0.6\right)$
then
in terms of the general form $a = 0.3$ and ${a}^{2} = 0.09$

Writing as a completed square:
$\textcolor{w h i t e}{\text{XXXXX}}$$y = {\left(x + 0.3\right)}^{2} - 11 - 0.09$

$\textcolor{w h i t e}{\text{XXXXX}}$$y = {\left(x - \left(- 0.3\right)\right)}^{2} + \left(- 11.09\right)$

which is the vertex form
$\textcolor{w h i t e}{\text{XXXXX}}$with the vertex at $\left(- 0.3 , - 11.09\right)$