Here's a question to get the discussions started. (This one actually comes from a view-an-example in the HW, but one like this may be on the exam.) The height (in ft.) of a flare shot upward from the ground is given by s = 144t - 16.0t2, where t is the time (in sec.). What is the greatest height to which the flare goes? Notice that you could arrive at the answer using Algebra, not Calculus. Using Algebra, note that this highest point is the Vertex of a Parabola, and the x-coordinate (or, here the t-coordinate) of the Vertex is x = (-b/2a) . .. Then, the height would be the y-coordinate (or here, the s-coordinate) could be found by plugging in the value just found (plugging it into the original equation). This is one way to check your answer if Calculus is used. Now, let's use Calculus to solve: Notice that the quantity to be maximized is the "s," and its equation is s = 144t - 16.0t2 . . . So, maximize this equation. How do you do this? [Answer: By finding the derivative --- with respect to time --- and setting this derivative equal to zero.] So, what is the derivative? [Answer: s' or (ds/dt) = 144 - 32t . .. This is the 1st. derivative.] Now that you've found the derivative, what do you do in order to find the maximum? [Answer: You set this derivative = 0 and solve for the t.] So 0 = 144 - 32t . . . I'll leave for you to solve. Your answer will be the same as finding (-b/2a) from above. Note that this is the TIME it takes to reach its maximum height . . . The problem asks for the GREATEST HEIGHT, which would be found by plugging in the value just found (plugging it into the original equation). Notice that I haven't shown a 2nd. derivative test to prove that this is a maximum, since the equation is that of a Parabola opening downward, as the coefficient of the t-squared term is negative. When you work this one, it only can have a maximum (since, again, it's a Parabola opening down). [Here are the final answers . .. Make sure that you can arrive at them . . . Feel free to ask if there are any questions. t = 4.5 seconds . . . s = 324 feet] I hope this helps. I know, I've given the answer(s), but discuss/show the computations for how these values are found