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# Interpreting Quadratic Functions in Factored Form

## Interpreting Quadratic Functions in Factored Form 1.10 - Solution

We want to write the equation of the quadratic function that passes through the points $(\textcolor{darkorange}{\text{-} 3},0),$ $(1,\text{-} 72),$ and $({\color{#FF0000}{4}},0).$ Note that the $y\text{-}$coordinate of two of the points is $0.$ Therefore, we know that $\textcolor{darkorange}{\text{-} 3}$ and ${\color{#FF0000}{4}}$ are the $x\text{-}$intercepts of the parabola.

$\bm{x}$ $\bm{y}$
$\textcolor{darkorange}{\text{-}3}$ $0$
$1$ $\text{-}72$
${\color{#FF0000}{4}}$ $0$
Consider the factored form of a quadratic function. $\begin{gathered} y=a(x-\textcolor{darkorange}{p})(x-{\color{#FF0000}{q}}) \end{gathered}$ In this form, $\textcolor{darkorange}{p}$ and ${\color{#FF0000}{q}}$ are the $x\text{-}$intercepts. Therefore, we know that, for our equation, it is $\textcolor{darkorange}{p}$ $=$ $\textcolor{darkorange}{\text{-} 3}$ and ${\color{#FF0000}{q}}$ $=$ ${\color{#FF0000}{4}}.$ $\begin{gathered} y=a(x-(\textcolor{#ff8c00}{\text{-}3}))(x-{\color{#FF0000}{4}}) \ \ \ \Leftrightarrow \ \ \ y=a(x+3)(x-4) \end{gathered}$ Since the parabola passes through the point $(1,\text{-}72),$ we can substitute $1$ for $x$ and $\text{-}72$ for $y$ in our partial equation, and solve for $a.$
$y=a(x+3)(x-4)$
${\color{#009600}{\text{-}72}}=a({\color{#0000FF}{1}}+3)({\color{#0000FF}{1}}-4)$
Solve for $a$
$\text{-}72=a(4)(\text{-}3)$
$\text{-}72=a(\text{-}12)$
$\dfrac{\text{-}72}{\text{-}12}=a$
$\dfrac{72}{12}=a$
$6=a$
$a=6$
Knowing that $a=6,$ we can write the full equation of the parabola. $\begin{gathered} y=6(x+3)(x-4) \end{gathered}$