Question

Find the two horizontal tangents on the graph of f(x)=2x^3+12x^2-72x+16

Answer 1

find the derivative of this function to be:

f'(x) = 6x^2 + 24x - 72

Horizontal tangents occur when the slope of the tangent line is 0 (horizontal lines have a slope of 0). So, substitute 0 for f'(x) and solve for x.

0 = 6x^2 + 24x - 72
0 = x^2 + 4x - 12
0 = (x + 6)(x - 2)
x = 2, -6

One horizontal tangent line intersects the function at x = 2, and the other at x = -6. Find there corresponding f(x) values to be f(2) = -64 and f(-6) = 448. So, the horizontal tangents are y = -64 and y = 448

Answer 2

o great, thanks!!