Interpreting Quadratic Functions in Vertex Form

To determine the error, we will find the vertex by ourselves. To do so, we will first express the quadratic function in vertex form, $y=a(x-h{)}^2+k,$ where $a,$ $h,$ and $k$ are either positive or negative constants. $$\begin{array}{c}y=\text{-}(x+5{)}^2\iff y=\text{-}1(x-(\text{-}5){)}^2+0\end{array}$$ Let's compare the general formula for the vertex form with our equation. $$\begin{array}{rl}\text{General formula: }y=& a(x-h{)}^2+k\\ \text{Equation: }y=& \text{-}1(x-(\text{-}5){)}^2+0\end{array}$$ We can see that $a=\text{-}1,$ $h=\text{-}5,$ and $k=0.$ The vertex of a quadratic function written in vertex form is the point $(h,k).$ Therefore, the vertex of this parabola is $(\text{-}5,0).$ Let's now examine the given solution.

We see that the value of $h$ was found correctly. However, the interpretation of the vertex form is not correct. Based on our operations at the beginning, we can correct the solution as follows.