Find the equation of the tangent line to the graph of f(x)=4x^3-1x^2+1 at x=-2.
find f(-2) to get the y of the point
so now you have a point
f '(x) = 12x^2 - 2x
find f '(-2) to get the slope of the line
now you have the slope and a point on the line,
use the method you learned to find the equation of that line