- Minimum at (−3, −4); Maximum at (−1, 0); Point of inflection at (−2, −2).
First, let’s find the y-intercept. We set x = 0 to get: y = −(0)^3 − 6(0)^2 − 9(0) − 4 = −4.
Therefore, the y-intercept is (0, −4). Next, we find any critical points using the first derivative.
The derivative is = −3x^2 − 12x − 9. If we set this equal to zero and solve for x, we get x =
−1 and x = −3. Plug x = −1 and x = −3 into the original equation to find the y-coordinates of the
critical points: When x = −1, y = −(−1)^3 − 6(−1)^2 − 9(−1) − 4 = 0. When x = −3, y = −(−3)^3 −
6(−3)^2 − 9(−3) − 4 = −4. Thus, we have critical points at (−1, 0) and (−3, −4). Next, we take
the second derivative to find any points of inflection. The second derivative is = −6x −
12, which is equal to zero at x = −2. Plug x = −2 into the original equation to find the y-
coordinate: y = −(−2)^3 − 6(−2)^2 − 9(−2) − 4 = −2, so there is a point of inflection at (−2, −2).
Next, we need to determine if each critical point is maximum, minimum, or something else. If
