Question

Find an equation for the tangent to the curve at the given point y=x^3 , (2,8)

Answer 1

$\bf y=x^3\implies \left. \cfrac{dy}{dx}=3x^2 \right|_{x=2}\implies \stackrel{\stackrel{m}{\downarrow }}{12} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{8})~\hspace{10em} slope = m\implies 12 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-8=12(x-2) \\\\\\ y-8=12x-24\implies y=12x-16$