Question

what is the equation of the line tangent to the graph of f at (3,79), where f is given by f(x)= 5x^3-7x^2+7

Answer 1

$\bf f(x)= 5x^3-7x^2+7\implies \cfrac{dy}{dx}=15x^2-14x \\\\\\ \textit{what's the slope when x = 3?} \\\\\\ \cfrac{dy}{dx}=15(3)^2-14(3)\implies \cfrac{dy}{dx}=93 \\\\\\ \textit{so, now we know that when x =3 } \begin{cases} x=3\\ y=79\\ m=93 \end{cases} \\\\\\ \textit{using the point-slope form then} \\\\\\ y-79=93(x-3)\implies y=93x-200$