Suppose you want the equation of the line tangent to the curve y = x^3 - 3x^2 + 2x + 5 at x - 4.  For the equation of a line you need the slope and y-intercept. You can get the slope at x=4  by differentiating and substituting 4 for x.

y' = 3x^2 -6x + 2

y' at x = 2 is 3(2)^2 -6(2) + 2 = 2

We know the slope at x= 4 is 2.

plug 4 for x into the original equation to get the y-coordinate

y = 4^3 - 3(4)^2 + 2(4) + 5 = 29

no put 2 for m, 4 for x and 29 for y into y = mx + b and solve for b

29 = 2(4) + b

b = 21

The equation is y = 2x + 21