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To match the function with its graph, we should first identify the vertex and then determine whether the parabola opens upward or downward. Note that the equation is already expressed in vertex form where $a,$ $h,$ and $k$ are either positive or negative numbers. $\begin{gathered} y=\text{-}2\big(x-1\big)^2+1.5 \end{gathered}$ Let's compare the general formula for the vertex form with our equation. \begin{aligned} \textbf{General formula: }y=&\ \phantom{\text{-}}{\color{#FF0000}{a}} \, (x-{\color{#0000FF}{h}} )^2 +\textcolor{magenta}{k} \\ \textbf{Equation: }y=&\ {\color{#FF0000}{\text{-}2}}\big(x-{\color{#0000FF}{1}}\big)^2+\textcolor{magenta}{1.5} \end{aligned} We can see that ${\color{#FF0000}{a}}={\color{#FF0000}{\text{-}2}},$ ${\color{#0000FF}{h}}={\color{#0000FF}{1}},$ and $\textcolor{magenta}{k}=\textcolor{magenta}{1.5}.$ Since the vertex of a quadratic function written in vertex form is the point $({\color{#0000FF}{h}},\textcolor{magenta}{k}),$ the vertex of our function is $({\color{#0000FF}{1}},\textcolor{magenta}{1.5}).$ Let's now determine the direction of the parabola. Recall that if ${\color{#FF0000}{a}}>0,$ the parabola opens upwards. Conversely, if ${\color{#FF0000}{a}}<0,$ the parabola opens downwards.
In the given function, we have ${\color{#FF0000}{a}}={\color{#FF0000}{\text{-}2}},$ which is less than $0.$ Therefore, the parabola opens downward. As a result, the graph of the function is the graph given in choice A.