Hi!
Apparently this is wrong, how do I get the correct form?
Complete the square for $\displaystyle y = -x^2-2x$
$\displaystyle \frac{-2}{2(-1)} = 1$
$\displaystyle -(x^2-2x) = -(x^2-2x+1)-1$
$\displaystyle -(x+1)^2-1$
$\displaystyle \displaystyle \begin{align*} y &= -x^2 - 2x \\ &= - \left( x^2 + 2x \right) \\ &= - \left[ x^2 + 2x + 1^2 - 1^2 \right] \\ &= - \left[ \left( x + 1 \right) ^2 - 1 \right] \\ &= - \left( x + 1 \right) ^2 + 1 \end{align*}$
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 $\displaystyle - \left[ x^2 + 2x + 1^2\right] - 1^2 + 0$?
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.