#### Math problems and solutions > Derivative > Basic examples > Solved exercises > Intermediate level > Solution 7

Math exercise and derivative exercises online : How calculate the x,y coordinate of a point ? How calculate the slope (angular coefficient) of a straight line and the slope of a tangent line using mathematics and derivatives ?

### Derivative : solution exercise 7 (Intermediate math level exercises)

Find out the coordinates of the points for which the slope of the tangent line to the curve
y = x3 - 12x + 1 is zero.

Given y = f(x) = x3 - 12x + 1
f '(x) = 3x2 - 12

The derivative of f(x) at x = x1 (and y = y1) is equal to the slope of the tangent line to the curve f(x), at (x1 , y1). But that slope must be equal to zero, thus:

f '(x1) = 0 = 3(x1)2 - 12

3(x1)2 = 12

(x1)2 = 12/3 = 4

x1 = +/- 2

y1 = f(x1) = (+/- 2)3 - 12(+/- 2) + 1 = -15 ou 17

The slope of the tangent line to the curve is equal to 0 at the point (2 ; -15) and at the point (-2 ; 17).

Check this on the following graph :

You can clearly see that both tangent lines have a slope equal to zero because the tangent lines are horizontal at x = 2 and at x = -2.