Find the equation of the tangent lines to the graph of the curve y=x^3-12x through the point (1,-12) not on the graph.

Answer 1

This problem is a little variation from normal:

Find the point on the curve that has the slope equal to the slope that connects the point to (1, -12)

Slope is dy/dx = 3x² - 12 for all (x,y)

So,

slope from derivative 3x² - 12 = (x³ - 12x + 12)/(x-1) slope to point 1,-12 from (x,y)

multiply both sides by x-1 and distribute, join terms. You get x = 0, 3/2 which both seem to work.