Interpreting Quadratic Functions in Vertex Form

We have a quadratic function written in standard form, and we want to rewrite it in vertex form. $$\begin{array}{c}\underline{\text{Standard form}}\\ y=a{x}^2+bx+c\\ \underline{\text{Given equation}}\\ y=\text{-}{x}^2-4x+2\iff y=\text{-}1x^2+(\text{-}4)x+2\end{array}$$ In the given equation, $a=\text{-}1,$ $b=\text{-}4,$ and $c=2.$ Let's now recall the vertex form of a quadratic function. $$\begin{array}{c}\text{Vertex form:}y=a(x-h{)}^2+k\end{array}$$ In this equation, $a$ is the leading coefficient of the quadratic function, and the point $(h,k)$ is the vertex of the parabola. By substituting our given values for $a$ and $b$ into the expression $\text{-}\frac{b}{2a},$ we can find $h.$ So far, we know that the vertex lies at $(\text{-}2,k).$ To find the $y\text{-}$coordinate $k,$ we will substitute $\text{-}2$ for $x$ in the given function. Therefore, the vertex is $(\text{-}2,6).$ Moreover, since we already know that $a=\text{-}1,$ we can rewrite the given function in vertex form. $$\begin{array}{c}y=\text{-}1(x-(\text{-}2){)}^2+6\iff y=\text{-}(x+2{)}^2+6\end{array}$$