What is the equation of the tangent to the pa | vedclass

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(A) Given the equation of the parabola: $y = 2 + 4x - 4x^2$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx}

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$4x + y - 6 = 0$

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(A) Given the equation of the parabola: $y = 2 + 4x - 4x^2$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(2 + 4x - 4x^2) = 4 - 8x$. We are given that the slope of the tangent is $-4$. So,$4 - 8x = -4$. $-8x = -8 \Rightarrow x = 1$. Now,find the corresponding $y$-coordinate by substituting $x = 1$ into the original equation: $y = 2 + 4(1) - 4(1)^2 = 2 + 4 - 4 = 2$. The point of tangency is $(1, 2)$. The equation of the tangent line with slope $m = -4$ passing through $(1, 2)$ is given by: $y - y_1 = m(x - x_1)$ $y - 2 = -4(x - 1)$ $y - 2 = -4x + 4$ $4x + y - 6 = 0$.

What is the equation of the tangent to the parabola $y = 2 + 4x - 4x^2$ having a slope of $-4$?

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