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To match the given function with its graph, we will find its zeros. To do so, we will rewrite it a little bit. $\begin{gathered} y=\dfrac{1}{4}(x+5)(x-3) \ \ \Leftrightarrow \ \ y=\dfrac{1}{4}(x-(\text{-} 5))(x-3) \end{gathered}$ Note that now the quadratic function is written in factored form. \begin{aligned} \textbf{Factored form:}&\ y={\color{#0000FF}{a}}(x-{\color{#009600}{p}})(x-{\color{#FF0000}{q}}) \\ \textbf{Given equation:}&\ y={\color{#0000FF}{\dfrac{1}{4}}}(x-({\color{#009600}{\text{-} 5}}))(x-{\color{#FF0000}{3}}) \end{aligned} In this form, the zeros of the function are ${\color{#009600}{p}}$ and ${\color{#FF0000}{q}}.$ Therefore, the zeros of the given function are ${\color{#009600}{\text{-} 5}}$ and ${\color{#FF0000}{3}}.$ This matches choice V.