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# Interpreting Quadratic Functions in Standard Form

## Interpreting Quadratic Functions in Standard Form 1.15 - Solution

We have a quadratic function written in standard form. \begin{aligned} h(x)=\textcolor{#ff8c00}{a}x^2+\textcolor{#9400D3}{b}x+\textcolor{magenta}{c} \end{aligned} This kind of equation can give us a lot of information about the parabola by observing the values of $\textcolor{#ff8c00}{a},$ $\textcolor{#9400D3}{b},$ and $\textcolor{magenta}{c}.$ $\begin{gathered} h(x)=\text{-} \dfrac{1}{2}x^2-11x+6 \ \Leftrightarrow \ h(x)=\textcolor{#ff8c00}{\text{-} \dfrac{1}{2}}x^2+(\textcolor{#9400D3}{\text{-}11})x+\textcolor{magenta}{6} \end{gathered}$ We see that for the given equation $\textcolor{#ff8c00}{a}=\textcolor{#ff8c00}{\text{-}\frac{1}{2}},$ $\textcolor{#9400D3}{b}=\textcolor{#9400D3}{\text{-}11},$ and $\textcolor{magenta}{c}=\textcolor{magenta}{6}.$ These values will give us information about the parabola.

### $x\text{-}$value of the Vertex

Consider the point at which the curve of the parabola changes direction.

This point is the vertex of the parabola, and defines the axis of symmetry. If we want to calculate the $x\text{-}$value of this point, we can substitute the given values of $\textcolor{#ff8c00}{a}$ and $\textcolor{#9400D3}{b}$ into the expression $\text{-}\frac{b}{2a}$ and simplify.
$\text{-} \dfrac{b}{2a}$
$\text{-} \dfrac{\textcolor{#9400D3}{\text{-}11}}{2\left(\textcolor{#ff8c00}{\text{-}\dfrac{1}{2}}\right)}$
Simplify
$\text{-} \dfrac{\text{-}11}{\text{-}2\left(\dfrac{1}{2}\right)}$
$\text{-} \dfrac{\text{-}11}{\text{-}2/2}$
$\text{-} \dfrac{\text{-}11}{\text{-}1}$
$\text{-} \dfrac{11}{1}$
$\text{-}11$

### $y\text{-}$value of the Vertex

The point at which the graph of a parabola changes direction also defines the maximum or minimum point of the graph. Whether the parabola has a minimum or maximum is determined by the value of $\textcolor{#ff8c00}{a}.$

Since the given value of $\textcolor{#ff8c00}{a}$ is negative, the parabola has a maximum value at the vertex. To find this value, think of $y$ as a function of $x,$ $y=h(x).$ By substituting the $x\text{-}$value of the vertex into the given equation and simplifying, we will get the $y\text{-}$value of the vertex.
$h(x)=\text{-} \dfrac{1}{2}x^2-11x+6$
$h({\color{#0000FF}{\text{-}11}})=\text{-} \dfrac{1}{2}({\color{#0000FF}{\text{-}11}})^2-11({\color{#0000FF}{\text{-}11}})+6$
Simplify right-hand side
$h(\text{-}11)=\text{-} \dfrac{1}{2}(121)-11(\text{-}11)+6$
$h(\text{-}11)=\text{-} \dfrac{121}{2}-11(\text{-}11)+6$
$h(\text{-}11)=\text{-} \dfrac{121}{2}+121+6$
$h(\text{-}11)=\text{-} \dfrac{121}{2}+\dfrac{242}{2}+\dfrac{12}{2}$
$h(\text{-}11)=\dfrac{133}{2}$
$h(\text{-}11)=66\frac{1}{2}$