AP Precalculus -1.5 Polynomial Functions and Complex Zeros- FRQ Exam Style Questions - Effective Fall 2023

a. Graphing the function
Using a graphing utility for \(P(x) = -0.0013x^3 + 0.3507x^2 – 0.4591x – 421.888\) on the interval \([0, 200]\) reveals a cubic curve shape.
The graph starts with a slight dip to a local minimum near \(x=0\), then rises steeply to a local maximum near \(x=180\), before falling again.

b. Average rate of change between extrema
First, find the derivative: \(P'(x) = -0.0039x^2 + 0.7014x – 0.4591\).
Set \(P'(x) = 0\) and use the quadratic formula to find the extrema: \(x \approx 0.66\) (local min) and \(x \approx 179.19\) (local max).
Calculate the profit at these points: \(P(0.66) \approx -422.04\) and \(P(179.19) \approx 3276.57\).
The average rate of change is: \(\frac{3276.57 – (-422.04)}{179.19 – 0.66} = \frac{3698.61}{178.53} \approx 20.72\).

c. Equation of the secant line
The slope \(m\) was found in part (b) to be approximately \(20.72\).
Using the point-slope form \(y – y_1 = m(x – x_1)\) with the minimum point \((0.66, -422.04)\):
\(y – (-422.04) = 20.72(x – 0.66)\)
\(y = 20.72x – 13.68 – 422.04\)
The equation is approximately \(y = 20.72x – 435.72\).

d. Inflection point
Find the second derivative: \(P”(x) = -0.0078x + 0.7014\).
Set \(P”(x) = 0\) to find the change in concavity: \(0 = -0.0078x + 0.7014 \Rightarrow x = \frac{0.7014}{0.0078} \approx 89.92\).
Find the corresponding y-value: \(P(89.92) \approx 1427.27\).
The inflection point is approximately \((89.92, 1427.27)\).

e. Variation of rate of change
The rate of change is represented by the derivative, \(P'(x)\).
Before the inflection point (\(x < 89.92\)), the graph is concave up (\(P”(x) > 0\)), so the rate of change is increasing.
After the inflection point (\(x > 89.92\)), the graph is concave down (\(P”(x) < 0\)), so the rate of change is decreasing.