Cracking The Ap Calculus ab Exam 2018

y   =   (0)^3   −   12(0)   =   0

The curve has a y-intercept at (0, 0).

There are no asymptotes, because there’s no place where the curve is undefined (you won’t have
asymptotes for curves that are polynomials).

Step 2: Take the derivative of the function to find the critical points.

    =   3x^2    −   12

Set the derivative equal to zero, and solve for x.

3 x^2   −   12  =   0

3(x^2   −   4)  =   0

3(x −   2)(x    +   2)  =   0

so x = 2, −2.

Next, plug x = 2, −2 into the original equation to find the y-coordinates of the critical points.

y   =   (2)^3   −   12(2)   =   −16

y   =   (−2)^3  −   12(−2)  =   16

Thus, we have critical points at (2, −16) and (−2, 16).

Step 3: Now, take the second derivative to find any points of inflection.

    =   6x

This equals zero at x = 0. We already know that when x = 0, y = 0, so the curve has a point of inflection at
(0, 0).

Now, plug the critical values into the second derivative to determine whether each is a maximum or a
minimum. f′′(2) = 6(2) = 12. This is positive, so the curve has a minimum at (2, −16), and the curve is
concave up at that point. f′′(−2) = 6(−2) = −12. This value is negative, so the curve has a maximum at (−2,
16) and the curve is concave down there.

Armed with this information, we can now plot the graph.