AP DSC School Assistant Math Model Test 2

Concept:The slope of a curve at any point is given by the first derivative $\frac{dy}{dx}$. To find the maximum value of the slope, we set the second derivative $\frac{d^2y}{dx^2} = 0$ (critical point) and check that the third derivative is negative (to confirm a maximum).Explanation:We are given the curve $y = -x^{3} + 3x^{2} + 9x - 27$.Step 1: Compute the slope $m = \frac{dy}{dx}$.$m = -3x^{2} + 6x + 9$.Step 2: Differentiate again to find $\frac{d^2y}{dx^2}$.$\frac{d^2y}{dx^2} = -6x + 6 = -6(x - 1)$.Step 3: Set $\frac{d^2y}{dx^2} = 0$ to locate possible extremum of slope.$-6(x - 1) = 0 \implies x = 1$.Step 4: Confirm it is a maximum using the third derivative.$\frac{d^3y}{dx^3} = -6 < 0$ for all $x$, so the slope is maximum at $x = 1$.Step 5: Find the maximum slope value by substituting $x = 1$ into $m$:$m_{max} = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$.Answer:The maximum slope of the curve is 12 (Option B).