Interpreting Quadratic Functions in Vertex Form

We have a quadratic function written in standard form, and we want to rewrite it in vertex form. $$\begin{array}{c}\underline{\text{Standard form}}\\ y=a{x}^2+bx+c\\ \underline{\text{Given equation}}\\ y=3x^2+6x-1\iff y=3x^2+6x+(\text{-}1)\end{array}$$ In the given equation, $a=3,$ $b=6,$ and $c=\text{-}1.$ 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 $(\text{-}1,k).$ To find the $y\text{-}$coordinate $k,$ we will substitute $\text{-}1$ for $x$ in the given function. Therefore, the vertex is $(\text{-}1,\text{-}4).$ Moreover, since we already know that $a=3,$ we can rewrite the given function in vertex form. $$\begin{array}{c}y=3(x-(\text{-}1){)}^2+(\text{-}4)\iff y=3(x+1{)}^2-4\end{array}$$