4) Find the equation of the tangent line to the graph of \( 8 y^{4}+8 x y^{24}+6 x^{3}-64=8 y^{2} \) at \( (2,-1) \).

To find the equation of the tangent line, we first need to compute the derivative of the implicit function defined by the equation \(8 y^{4} + 8 x y^{24} + 6 x^{3} - 64 = 8 y^{2}\). We can use implicit differentiation. Differentiating both sides with respect to \(x\): \[ 32y^{3} \frac{dy}{dx} + 8(y^{24} + 24xy^{23} \frac{dy}{dx}) + 18x^{2} = 16y \frac{dy}{dx} \] Now, substitute \( x = 2 \) and \( y = -1 \): 1. Calculate \( 32(-1)^{3} \frac{dy}{dx} + 8((-1)^{24} + 24(2)(-1)^{23} \frac{dy}{dx}) + 18(2)^{2} - 16(-1) \frac{dy}{dx} = 0 \). 2. This results in \( -32 \frac{dy}{dx} + 8(1 - 48 \frac{dy}{dx}) + 72 + 16 \frac{dy}{dx} = 0 \). 3. Combining similar terms gives us \( -32 \frac{dy}{dx} - 384 \frac{dy}{dx} + 72 + 0 = 0 \). 4. Set \( \frac{dy}{dx}(-(32 + 384)) + 72 = 0 \) and solve for \( \frac{dy}{dx} \). After simplifying, we find \( \frac{dy}{dx} = -\frac{72}{416} = -\frac{9}{52} \). Now use the point-slope form of the equation of the tangent line at the point \( (2, -1) \): \[ y - (-1) = -\frac{9}{52}(x - 2) \] Thus, the equation of the tangent line is: \[ y + 1 = -\frac{9}{52}x + \frac{18}{52} \implies y = -\frac{9}{52}x + \frac{18}{52} - 1 \implies y = -\frac{9}{52}x - \frac{34}{52} \]