Find the horizontal tangent for x^3/3+3x^2-16x+9

The two horiz. tang. lines here are y = -3 and y = 192.

Step-by-step explanation:

Remember that the slope of a tangent line to the graph of a function is given by the derivative of that function.  Thus, we find f '(x):

f '(x) = x^2 + 6x - 16.  This is the formula for the slope.  We set this = to 0 and determine for which x values the tangent line is horizontal:

f '(x) = x^2 + 6x - 16 = 0.  Use the quadratic formula to determine the roots here:  a = 1; b = 6 and c = -16:  the discriminant is b^2-4ac, or 36-4(1)(-16), which has the value 100; thus, the roots are:

      -6 plus or minus √100

x = = 2 and -8.

                         2

Evaluating y = x^3/3+3x^2-16x+9 at x = 2 results in y = -3.  So one point of tangency is (2, -3).  Remembering that the tangent lines in this problem are horizontal, we need only the y-coefficient of (2, -3) to represent this first tangent line:  it is y = -3.

Similarly, find the y-coeff. of the other tangent line, which is tangent to the curve at x = -8.  The value of x^3/3+3x^2-16x+9 at x = -8 is  192, and so the equation of the 2nd tangent line is y=192 (the slope is zero).

step-by-step explanation:

i am not sure sorry : c