Question

How do you find the points where the graph of the function `y=x^3+3x^2+x+3` has horizontal tangents?

Answer 1

`x=-1+-1/3sqrt(6)`

Explanation:

The slope of the tangent line to the curve is given by the first derivative of the curve. Incidentally, a horizontal tangent (zero slope) refers to the stationary (or turning) points of the function.

`(dy)/(dx) = 3x^2 + 6x + 1`

Slope is zero so

`3x^2 + 6x + 1 = 0`

Using the quadratic formula

`x = (-6+-sqrt(6^2 - 4(3)(1)))/(2(3)) = (-6+-sqrt(36-12))/6`

`x = (-6+-sqrt(24))/6 = (-6+-2sqrt(6))/6 = -1+-1/3sqrt(6)`