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