Choice D is correct. It's given that $f(x) = (x - 14)(x + 19)$ , which can be rewritten as $f(x) = x^2 + 5x - 266$. Since the coefficient of the $x^2$-term is positive, the graph of $y = f(x)$ in the $xy$-plane opens upward and reaches its minimum value at its vertex. The $x$-coordinate of the vertex is the value of $x$ such that $f(x)$ reaches its minimum. For an equation in the form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, the $x$-coordinate of the vertex is $-\dfrac{b}{2a}$. For the equation $f(x) = x^2 + 5x - 266$, $a = 1$, $b = 5$, and $c = -266$. It follows that the $x$-coordinate of the vertex is $-\dfrac{5}{2(1)}$, or $-\dfrac{5}{2}$. Therefore, $f(x)$ reaches its minimum when the value of $x$ is $-\dfrac{5}{2}$.

Alternate approach: The value of $x$ for the vertex of a parabola is the $x$-value of the midpoint between the two $x$-intercepts of the parabola. Since it's given that $f(x) = (x - 14)(x + 19)$, it follows that the two $x$-intercepts of the graph of $y = f(x)$ in the $xy$-plane occur when $x = 14$ and $x = -19$, or at the points $\left(14, 0\right)$ and$\left(-19, 0\right)$. The midpoint between two points, $\left(x\_1, y\_1\right)$ and $\left(x\_2, y\_2\right)$  , is $\left(\dfrac{x\_1 + x\_2}{2}, \dfrac{y\_1 + y\_2}{2}\right)$. Therefore, the midpoint between $\left(14, 0\right)$ and $\left(-19, 0\right)$ is $\left(\dfrac{14 + (-19)}{2}, \dfrac{0 + 0}{2}\right)$, or $\left( -\dfrac{5}{2}, 0 \right)$. It follows that $f(x)$ reaches its minimum when the value of $x$ is $-\dfrac{5}{2}$.

Choice A is incorrect. This is the $y$ -coordinate of the $y$-intercept of the graph of $y = f(x)$ in the $xy$-plane.

Choice B is incorrect. This is one of the $x$ -coordinates of the $x$-intercepts of the graph of $y = f(x)$ in the $xy$-plane.

Choice C is incorrect and may result from conceptual or calculation errors.