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(B) The point at which the tangent line to the curve is horizontal implies that the slope of the tangent is zero.Given the equation of the curve: $y =

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$-2, -12$

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(B) The point at which the tangent line to the curve is horizontal implies that the slope of the tangent is zero.Given the equation of the curve: $y = \frac{16}{x} - x^2$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{16}{x} - x^2) = -\frac{16}{x^2} - 2x$.For the tangent to be horizontal, set $\frac{dy}{dx} = 0$:$-\frac{16}{x^2} - 2x = 0$.$-\frac{16 + 2x^3}{x^2} = 0$.$16 + 2x^3 = 0 \implies 2x^3 = -16 \implies x^3 = -8$.Thus, $x = -2$.Now, substitute $x = -2$ into the original equation to find $y$:$y = \frac{16}{-2} - (-2)^2 = -8 - 4 = -12$.Therefore, the required point is $(-2, -12)$.

The point at which the tangent line to the curve of $y=\frac{16}{x}-x^2$ is horizontal, is

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