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(D) Given the curve equation is $x^2 = -4y$. Differentiating both sides with respect to $x$,we get $2x = -4 \frac{dy}{dx}$. Thus,$\frac{dy}{dx} = -\fr

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$2x - y + 4 = 0$

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(D) Given the curve equation is $x^2 = -4y$. Differentiating both sides with respect to $x$,we get $2x = -4 \frac{dy}{dx}$. Thus,$\frac{dy}{dx} = -\frac{x}{2}$. At the point $(-4, -4)$,the slope of the tangent is $m = \left( \frac{dy}{dx} \right)_{(-4, -4)} = -\frac{-4}{2} = 2$. The equation of the tangent line passing through $(x_1, y_1) = (-4, -4)$ with slope $m = 2$ is given by $(y - y_1) = m(x - x_1)$. Substituting the values,we get $y - (-4) = 2(x - (-4))$. $y + 4 = 2(x + 4)$. $y + 4 = 2x + 8$. $2x - y + 4 = 0$.

What is the equation of the tangent to the curve $x^2 = -4y$ at the point $(-4, -4)$?

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