How do you find the points where the graph of the function # f(x) = x^3 + 9x^2 + x + 19# has horizontal tangents?

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Answer 1 · 2016-06-21T22:20:50

Horizontal tangents occur when the tangents have a slope of #0#.

Differentiating, by the power rule:

#f'(x) = 3x^2 + 18x + 1#

The slope of the tangent is given by evaluting #f(a)#, where a is the point #x = a# where the tangent intersects the function.

We can therefore say #f(x) =0# and solve the equation.

#0 = 3x^2 + 18x + 1#

#-1 = 3(x^2 + 6x + 9 - 9)#

#-1 = 3(x + 3)^2 -27#

#26/3 = (x + 3)^2#

#+-sqrt(26/3) - 3 = x#

Therefore, the tangent line is horizontal at the points #x = sqrt(26/3) - 3# and #x = -sqrt(26/3) - 3#.

Hopefully this helps!