Example Question - slope of a function

Finding Tangent Line Equation Using Calculus

\begin{align*} g(x) &= \frac{16}{x} - 4\sqrt{x}\\ g'(x) &= -\frac{16}{x^2} - \frac{4}{2\sqrt{x}}\\ g'(x) &= -\frac{16}{x^2} - \frac{2}{\sqrt{x}}\\ g'(-4) &= -\frac{16}{(-4)^2} - \frac{2}{\sqrt{-4}}\\ g'(-4) &= -\frac{16}{16} - \frac{2}{2i}\\ g'(-4) &= -1 - \frac{1}{i}\\ g'(-4) &= -1 + i\\ g(-4) &= \frac{16}{-4} - 4\sqrt{-4}\\ g(-4) &= -4 - 4 \cdot 2i\\ g(-4) &= -4 - 8i\\ \text{Point-slope form: } y - y_1 &= m(x - x_1)\\ y - (-4 - 8i) &= (-1 + i)(x - (-4))\\ y + 4 + 8i &= (-1 + i)(x + 4)\\ y &= (-1 + i)x - 4 + 1 - 4i - 8i\\ y &= (-1 + i)x - 3 - 12i\\ \text{Tangent line equation: } y &= (-1 + i)x - 3 - 12i \end{align*}