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 to the graph of \[ 8 y^{4} + 8 x y^{24} + 6 x^{3} - 64 = 8 y^{2} \] at the point \((2, -1)\), follow these steps: ### 1. Differentiate Implicitly Differentiate both sides of the equation with respect to \(x\): \[ \frac{d}{dx}\left(8 y^{4}\right) + \frac{d}{dx}\left(8 x y^{24}\right) + \frac{d}{dx}\left(6x^{3}\right) - \frac{d}{dx}\left(64\right) = \frac{d}{dx}\left(8 y^{2}\right) \] Compute each derivative term-by-term: - \(\frac{d}{dx}(8 y^{4}) = 32 y^{3} \frac{dy}{dx}\) - \(\frac{d}{dx}(8 x y^{24}) = 8 y^{24} + 192 x y^{23} \frac{dy}{dx}\) (using the product rule) - \(\frac{d}{dx}(6x^{3}) = 18x^{2}\) - \(\frac{d}{dx}(64) = 0\) - \(\frac{d}{dx}(8 y^{2}) = 16 y \frac{dy}{dx}\) Putting it all together: \[ 32 y^{3} \frac{dy}{dx} + 8 y^{24} + 192 x y^{23} \frac{dy}{dx} + 18x^{2} = 16 y \frac{dy}{dx} \] ### 2. Solve for \(\frac{dy}{dx}\) Collect all terms involving \(\frac{dy}{dx}\) on one side: \[ \left(32 y^{3} + 192 x y^{23} - 16 y\right) \frac{dy}{dx} = -8 y^{24} - 18x^{2} \] Thus, \[ \frac{dy}{dx} = \frac{-8 y^{24} - 18x^{2}}{32 y^{3} + 192 x y^{23} - 16 y} \] ### 3. Evaluate at \((2, -1)\) Substitute \(x = 2\) and \(y = -1\): **Numerator:** \[ -8 (-1)^{24} - 18(2)^{2} = -8(1) - 72 = -80 \] **Denominator:** \[ 32 (-1)^{3} + 192(2)(-1)^{23} - 16(-1) = -32 - 384 + 16 = -400 \] So, \[ \frac{dy}{dx}\bigg|_{(2, -1)} = \frac{-80}{-400} = \frac{1}{5} \] ### 4. Write the Tangent Line Equation Using the point-slope form of a line: \[ y - y_1 = m(x - x_1) \] Substitute \(m = \frac{1}{5}\) and \((x_1, y_1) = (2, -1)\): \[ y - (-1) = \frac{1}{5}(x - 2) \\ y + 1 = \frac{1}{5}x - \frac{2}{5} \\ y = \frac{1}{5}x - \frac{7}{5} \] **Final Equation:** \[ y = \frac{1}{5}x - \frac{7}{5} \] **Alternatively**, multiplying through by 5 to eliminate fractions: \[ x - 5y = 7 \] **Answer:** After simplifying, an equation for the tangent is y = (1⁄5)x – 7⁄5