1. ## Completing the square

Hi!

Apparently this is wrong, how do I get the correct form?

Complete the square for $\displaystyle y = -x^2-2x$

$\displaystyle \frac{-2}{2(-1)} = 1$

$\displaystyle -(x^2-2x) = -(x^2-2x+1)-1$

$\displaystyle -(x+1)^2-1$

2. ## Re: Completing the square

\displaystyle \displaystyle \begin{align*} y &= -x^2 - 2x \\ &= - \left( x^2 + 2x \right) \\ &= - \left[ x^2 + 2x + 1^2 - 1^2 \right] \\ &= - \left[ \left( x + 1 \right) ^2 - 1 \right] \\ &= - \left( x + 1 \right) ^2 + 1 \end{align*}

3. ## Re: Completing the square

Why is it necessary to factor out the negative 1?

$\displaystyle - \left[ x^2 + 2x + 1^2 - 1^2 \right]$
Could this be written as: $\displaystyle - \left[ x^2 + 2x + 1^2\right] - 1^2$ OR $\displaystyle - \left[ x^2 + 2x + 1^2\right] - 1^2 + 0$?

4. ## Re: Completing the square

To complete the square you require that your coefficient of \displaystyle \displaystyle \begin{align*} x^2 \end{align*} is 1, so if your coefficient is something different, you have to take it out as a factor and then complete the square on the leftover stuff.